Related
This is a slightly odd/unique request. I am trying to achieve a result where e.g "yes" becomes, "yyes", "yees", "yess", "yyees", "yyess", "yyeess".
I have looked at this: Find all lowercase and uppercase combinations of a string in Javascript which completes it for capitalisation, however my understanding is prohibiting me from manipulating this into character duplication (if this method is even possible to use in this way).
export function letterDuplication(level: number, input: string){
const houseLength = input.length;
if (level == 1){
var resultsArray: string[] = [];
const letters = input.split("");
const permCount = 1 << input.length;
for (let perm = 0; perm < permCount; perm++) {
// Update the capitalization depending on the current permutation
letters.reduce((perm, letter, i) => {
if (perm & 1) {
letters[i] = (letter.slice(0, perm) + letter.slice(perm-1, perm) + letter.slice(perm));
} else {
letters[i] = (letter.slice(0, perm - 1) + letter.slice(perm, perm) + letter.slice(perm))
}
return perm >> 1;
}, perm);
var result = letters.join("");
console.log(result);
resultsArray[perm] = result;
}
If I haven't explained this particularly well please let me know and I'll clarify. I'm finding it quite the challenge!
General idea
To get the list of all words we can get from ordered array of letters, we need to get all combinations of letters, passed 1 or 2 times into word, like:
word = 'sample'
array = 's'{1,2} + 'a'{1,2} + 'm'{1,2} + 'p'{1,2} + 'l'{1,2} + 'e'{1,2}
Amount of all possible words equal to 2 ^ word.length (8 for "yes"), so we can build binary table with 8 rows that represent all posible combinations just via convering numbers from 0 to 7 (from decimal to binary):
0 -> 000
1 -> 001
2 -> 010
3 -> 011
4 -> 100
5 -> 101
6 -> 110
7 -> 111
Each decimal we can use as pattern for new word, where 0 means letter must be used once, and 1 - letter must be used twice:
0 -> 000 -> yes
1 -> 001 -> yess
2 -> 010 -> yees
3 -> 011 -> yeess
4 -> 100 -> yyes
5 -> 101 -> yyess
6 -> 110 -> yyees
7 -> 111 -> yyeess
Code
So, your code representation may looks like this:
// Initial word
const word = 'yes';
// List of all possible words
const result = [];
// Iterating (2 ^ word.length) times
for (let i = 0; i < Math.pow(2, word.length); i++) {
// Get binary pattern for each word
const bin = decToBin(i, word.length);
// Make a new word from pattern ...
let newWord = '';
for (let i = 0; i < word.length; i++) {
// ... by adding letter 1 or 2 times based on bin char value
newWord += word[i].repeat(+bin[i] ? 2 : 1);
}
result.push(newWord);
}
// Print result (all possible words)
console.log(result);
// Method for decimal to binary convertion with leading zeroes
// (always returns string with length = len)
function decToBin(x, len) {
let rem, bin = 0, i = 1;
while (x != 0) {
rem = x % 2;
x = parseInt(x / 2);
bin += rem * i;
i = i * 10;
}
bin = bin.toString();
return '0'.repeat(len - bin.length) + bin;
}
Maybe this example can help you. It's a bit not neat and not optimal, but it seems to work:
function getCombinations(word = '') {
const allCombination = [];
const generate = (n, arr, i = 0) => {
if (i === n) {
return allCombination.push([...arr]);
} else {
arr[i] = 0;
generate(n, arr, i + 1);
arr[i] = 1;
generate(n, arr, i + 1);
}
}
generate(word.length, Array(word.length).fill(0));
return allCombination.map((el) => {
return el.map((isCopy, i) => isCopy ? word[i].repeat(2) : word[i]).join('')
});
}
console.log(getCombinations('yes'));
console.log(getCombinations('cat'));
How do I create GUIDs (globally-unique identifiers) in JavaScript? The GUID / UUID should be at least 32 characters and should stay in the ASCII range to avoid trouble when passing them around.
I'm not sure what routines are available on all browsers, how "random" and seeded the built-in random number generator is, etc.
[Edited 2021-10-16 to reflect latest best-practices for producing RFC4122-compliant UUIDs]
Most readers here will want to use the uuid module. It is well-tested and supported.
The crypto.randomUUID() function is an emerging standard that is supported in Node.js and an increasing number of browsers. However because new browser APIs are restricted to secure contexts this method is only available to pages served locally (localhost or 127.0.0.1) or over HTTPS. If you're interested in seeing this restriction lifted for crypto.randomUUID() you can follow this GitHub issue.
If neither of those work for you, there is this method (based on the original answer to this question):
function uuidv4() {
return ([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g, c =>
(c ^ crypto.getRandomValues(new Uint8Array(1))[0] & 15 >> c / 4).toString(16)
);
}
console.log(uuidv4());
Note: The use of any UUID generator that relies on Math.random() is strongly discouraged (including snippets featured in previous versions of this answer) for reasons best explained here. TL;DR: solutions based on Math.random() do not provide good uniqueness guarantees.
UUIDs (Universally Unique IDentifier), also known as GUIDs (Globally Unique IDentifier), according to RFC 4122, are identifiers designed to provide certain uniqueness guarantees.
While it is possible to implement RFC-compliant UUIDs in a few lines of JavaScript code (e.g., see #broofa's answer, below) there are several common pitfalls:
Invalid id format (UUIDs must be of the form "xxxxxxxx-xxxx-Mxxx-Nxxx-xxxxxxxxxxxx", where x is one of [0-9, a-f] M is one of [1-5], and N is [8, 9, a, or b]
Use of a low-quality source of randomness (such as Math.random)
Thus, developers writing code for production environments are encouraged to use a rigorous, well-maintained implementation such as the uuid module.
I really like how clean Broofa's answer is, but it's unfortunate that poor implementations of Math.random leave the chance for collision.
Here's a similar RFC4122 version 4 compliant solution that solves that issue by offsetting the first 13 hex numbers by a hex portion of the timestamp, and once depleted offsets by a hex portion of the microseconds since pageload. That way, even if Math.random is on the same seed, both clients would have to generate the UUID the exact same number of microseconds since pageload (if high-perfomance time is supported) AND at the exact same millisecond (or 10,000+ years later) to get the same UUID:
function generateUUID() { // Public Domain/MIT
var d = new Date().getTime();//Timestamp
var d2 = ((typeof performance !== 'undefined') && performance.now && (performance.now()*1000)) || 0;//Time in microseconds since page-load or 0 if unsupported
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
var r = Math.random() * 16;//random number between 0 and 16
if(d > 0){//Use timestamp until depleted
r = (d + r)%16 | 0;
d = Math.floor(d/16);
} else {//Use microseconds since page-load if supported
r = (d2 + r)%16 | 0;
d2 = Math.floor(d2/16);
}
return (c === 'x' ? r : (r & 0x3 | 0x8)).toString(16);
});
}
var onClick = function(){
document.getElementById('uuid').textContent = generateUUID();
}
onClick();
#uuid { font-family: monospace; font-size: 1.5em; }
<p id="uuid"></p>
<button id="generateUUID" onclick="onClick();">Generate UUID</button>
Here's a fiddle to test.
Modernized snippet for ES6
const generateUUID = () => {
let
d = new Date().getTime(),
d2 = ((typeof performance !== 'undefined') && performance.now && (performance.now() * 1000)) || 0;
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, c => {
let r = Math.random() * 16;
if (d > 0) {
r = (d + r) % 16 | 0;
d = Math.floor(d / 16);
} else {
r = (d2 + r) % 16 | 0;
d2 = Math.floor(d2 / 16);
}
return (c == 'x' ? r : (r & 0x7 | 0x8)).toString(16);
});
};
const onClick = (e) => document.getElementById('uuid').textContent = generateUUID();
document.getElementById('generateUUID').addEventListener('click', onClick);
onClick();
#uuid { font-family: monospace; font-size: 1.5em; }
<p id="uuid"></p>
<button id="generateUUID">Generate UUID</button>
broofa's answer is pretty slick, indeed - impressively clever, really... RFC4122 compliant, somewhat readable, and compact. Awesome!
But if you're looking at that regular expression, those many replace() callbacks, toString()'s and Math.random() function calls (where he's only using four bits of the result and wasting the rest), you may start to wonder about performance. Indeed, joelpt even decided to toss out an RFC for generic GUID speed with generateQuickGUID.
But, can we get speed and RFC compliance? I say, YES! Can we maintain readability? Well... Not really, but it's easy if you follow along.
But first, my results, compared to broofa, guid (the accepted answer), and the non-rfc-compliant generateQuickGuid:
Desktop Android
broofa: 1617ms 12869ms
e1: 636ms 5778ms
e2: 606ms 4754ms
e3: 364ms 3003ms
e4: 329ms 2015ms
e5: 147ms 1156ms
e6: 146ms 1035ms
e7: 105ms 726ms
guid: 962ms 10762ms
generateQuickGuid: 292ms 2961ms
- Note: 500k iterations, results will vary by browser/CPU.
So by my 6th iteration of optimizations, I beat the most popular answer by over 12 times, the accepted answer by over 9 times, and the fast-non-compliant answer by 2-3 times. And I'm still RFC 4122 compliant.
Interested in how? I've put the full source on http://jsfiddle.net/jcward/7hyaC/3/ and on https://jsben.ch/xczxS
For an explanation, let's start with broofa's code:
function broofa() {
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
var r = Math.random()*16|0, v = c == 'x' ? r : (r&0x3|0x8);
return v.toString(16);
});
}
console.log(broofa())
So it replaces x with any random hexadecimal digit, y with random data (except forcing the top two bits to 10 per the RFC spec), and the regex doesn't match the - or 4 characters, so he doesn't have to deal with them. Very, very slick.
The first thing to know is that function calls are expensive, as are regular expressions (though he only uses 1, it has 32 callbacks, one for each match, and in each of the 32 callbacks it calls Math.random() and v.toString(16)).
The first step toward performance is to eliminate the RegEx and its callback functions and use a simple loop instead. This means we have to deal with the - and 4 characters whereas broofa did not. Also, note that we can use String Array indexing to keep his slick String template architecture:
function e1() {
var u='',i=0;
while(i++<36) {
var c='xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'[i-1],r=Math.random()*16|0,v=c=='x'?r:(r&0x3|0x8);
u+=(c=='-'||c=='4')?c:v.toString(16)
}
return u;
}
console.log(e1())
Basically, the same inner logic, except we check for - or 4, and using a while loop (instead of replace() callbacks) gets us an almost 3X improvement!
The next step is a small one on the desktop but makes a decent difference on mobile. Let's make fewer Math.random() calls and utilize all those random bits instead of throwing 87% of them away with a random buffer that gets shifted out each iteration. Let's also move that template definition out of the loop, just in case it helps:
function e2() {
var u='',m='xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx',i=0,rb=Math.random()*0xffffffff|0;
while(i++<36) {
var c=m[i-1],r=rb&0xf,v=c=='x'?r:(r&0x3|0x8);
u+=(c=='-'||c=='4')?c:v.toString(16);rb=i%8==0?Math.random()*0xffffffff|0:rb>>4
}
return u
}
console.log(e2())
This saves us 10-30% depending on platform. Not bad. But the next big step gets rid of the toString function calls altogether with an optimization classic - the look-up table. A simple 16-element lookup table will perform the job of toString(16) in much less time:
function e3() {
var h='0123456789abcdef';
var k='xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx';
/* same as e4() below */
}
function e4() {
var h=['0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'];
var k=['x','x','x','x','x','x','x','x','-','x','x','x','x','-','4','x','x','x','-','y','x','x','x','-','x','x','x','x','x','x','x','x','x','x','x','x'];
var u='',i=0,rb=Math.random()*0xffffffff|0;
while(i++<36) {
var c=k[i-1],r=rb&0xf,v=c=='x'?r:(r&0x3|0x8);
u+=(c=='-'||c=='4')?c:h[v];rb=i%8==0?Math.random()*0xffffffff|0:rb>>4
}
return u
}
console.log(e4())
The next optimization is another classic. Since we're only handling four bits of output in each loop iteration, let's cut the number of loops in half and process eight bits in each iteration. This is tricky since we still have to handle the RFC compliant bit positions, but it's not too hard. We then have to make a larger lookup table (16x16, or 256) to store 0x00 - 0xFF, and we build it only once, outside the e5() function.
var lut = []; for (var i=0; i<256; i++) { lut[i] = (i<16?'0':'')+(i).toString(16); }
function e5() {
var k=['x','x','x','x','-','x','x','-','4','x','-','y','x','-','x','x','x','x','x','x'];
var u='',i=0,rb=Math.random()*0xffffffff|0;
while(i++<20) {
var c=k[i-1],r=rb&0xff,v=c=='x'?r:(c=='y'?(r&0x3f|0x80):(r&0xf|0x40));
u+=(c=='-')?c:lut[v];rb=i%4==0?Math.random()*0xffffffff|0:rb>>8
}
return u
}
console.log(e5())
I tried an e6() that processes 16-bits at a time, still using the 256-element LUT, and it showed the diminishing returns of optimization. Though it had fewer iterations, the inner logic was complicated by the increased processing, and it performed the same on desktop, and only ~10% faster on mobile.
The final optimization technique to apply - unroll the loop. Since we're looping a fixed number of times, we can technically write this all out by hand. I tried this once with a single random variable, r, that I kept reassigning, and performance tanked. But with four variables assigned random data up front, then using the lookup table, and applying the proper RFC bits, this version smokes them all:
var lut = []; for (var i=0; i<256; i++) { lut[i] = (i<16?'0':'')+(i).toString(16); }
function e7()
{
var d0 = Math.random()*0xffffffff|0;
var d1 = Math.random()*0xffffffff|0;
var d2 = Math.random()*0xffffffff|0;
var d3 = Math.random()*0xffffffff|0;
return lut[d0&0xff]+lut[d0>>8&0xff]+lut[d0>>16&0xff]+lut[d0>>24&0xff]+'-'+
lut[d1&0xff]+lut[d1>>8&0xff]+'-'+lut[d1>>16&0x0f|0x40]+lut[d1>>24&0xff]+'-'+
lut[d2&0x3f|0x80]+lut[d2>>8&0xff]+'-'+lut[d2>>16&0xff]+lut[d2>>24&0xff]+
lut[d3&0xff]+lut[d3>>8&0xff]+lut[d3>>16&0xff]+lut[d3>>24&0xff];
}
console.log(e7())
Modualized: http://jcward.com/UUID.js - UUID.generate()
The funny thing is, generating 16 bytes of random data is the easy part. The whole trick is expressing it in string format with RFC compliance, and it's most tightly accomplished with 16 bytes of random data, an unrolled loop and lookup table.
I hope my logic is correct -- it's very easy to make a mistake in this kind of tedious bit work. But the outputs look good to me. I hope you enjoyed this mad ride through code optimization!
Be advised: my primary goal was to show and teach potential optimization strategies. Other answers cover important topics such as collisions and truly random numbers, which are important for generating good UUIDs.
Use:
let uniqueId = Date.now().toString(36) + Math.random().toString(36).substring(2);
document.getElementById("unique").innerHTML =
Math.random().toString(36).substring(2) + (new Date()).getTime().toString(36);
<div id="unique">
</div>
If IDs are generated more than 1 millisecond apart, they are 100% unique.
If two IDs are generated at shorter intervals, and assuming that the random method is truly random, this would generate IDs that are 99.99999999999999% likely to be globally unique (collision in 1 of 10^15).
You can increase this number by adding more digits, but to generate 100% unique IDs you will need to use a global counter.
If you need RFC compatibility, this formatting will pass as a valid version 4 GUID:
let u = Date.now().toString(16) + Math.random().toString(16) + '0'.repeat(16);
let guid = [u.substr(0,8), u.substr(8,4), '4000-8' + u.substr(13,3), u.substr(16,12)].join('-');
let u = Date.now().toString(16)+Math.random().toString(16)+'0'.repeat(16);
let guid = [u.substr(0,8), u.substr(8,4), '4000-8' + u.substr(13,3), u.substr(16,12)].join('-');
document.getElementById("unique").innerHTML = guid;
<div id="unique">
</div>
The above code follow the intention, but not the letter of the RFC. Among other discrepancies it's a few random digits short. (Add more random digits if you need it) The upside is that this is really fast :)
You can test validity of your GUID here
Here's some code based on RFC 4122, section 4.4 (Algorithms for Creating a UUID from Truly Random or Pseudo-Random Number).
function createUUID() {
// http://www.ietf.org/rfc/rfc4122.txt
var s = [];
var hexDigits = "0123456789abcdef";
for (var i = 0; i < 36; i++) {
s[i] = hexDigits.substr(Math.floor(Math.random() * 0x10), 1);
}
s[14] = "4"; // bits 12-15 of the time_hi_and_version field to 0010
s[19] = hexDigits.substr((s[19] & 0x3) | 0x8, 1); // bits 6-7 of the clock_seq_hi_and_reserved to 01
s[8] = s[13] = s[18] = s[23] = "-";
var uuid = s.join("");
return uuid;
}
This is the fastest GUID-like string generator method in the format XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX. It does not generate a standard-compliant GUID.
Ten million executions of this implementation take just 32.5 seconds, which is the fastest I've ever seen in a browser (the only solution without loops/iterations).
The function is as simple as:
/**
* Generates a GUID string.
* #returns {string} The generated GUID.
* #example af8a8416-6e18-a307-bd9c-f2c947bbb3aa
* #author Slavik Meltser.
* #link http://slavik.meltser.info/?p=142
*/
function guid() {
function _p8(s) {
var p = (Math.random().toString(16)+"000000000").substr(2,8);
return s ? "-" + p.substr(0,4) + "-" + p.substr(4,4) : p ;
}
return _p8() + _p8(true) + _p8(true) + _p8();
}
To test the performance, you can run this code:
console.time('t');
for (var i = 0; i < 10000000; i++) {
guid();
};
console.timeEnd('t');
I'm sure most of you will understand what I did there, but maybe there is at least one person that will need an explanation:
The algorithm:
The Math.random() function returns a decimal number between 0 and 1 with 16 digits after the decimal fraction point (for
example 0.4363923368509859).
Then we take this number and convert
it to a string with base 16 (from the example above we'll get
0.6fb7687f).
Math.random().toString(16).
Then we cut off the 0. prefix (0.6fb7687f =>
6fb7687f) and get a string with eight hexadecimal
characters long.
(Math.random().toString(16).substr(2,8).
Sometimes the Math.random() function will return
shorter number (for example 0.4363), due to zeros at the end (from the example above, actually the number is 0.4363000000000000). That's why I'm appending to this string "000000000" (a string with nine zeros) and then cutting it off with substr() function to make it nine characters exactly (filling zeros to the right).
The reason for adding exactly nine zeros is because of the worse case scenario, which is when the Math.random() function will return exactly 0 or 1 (probability of 1/10^16 for each one of them). That's why we needed to add nine zeros to it ("0"+"000000000" or "1"+"000000000"), and then cutting it off from the second index (third character) with a length of eight characters. For the rest of the cases, the addition of zeros will not harm the result because it is cutting it off anyway.
Math.random().toString(16)+"000000000").substr(2,8).
The assembly:
The GUID is in the following format XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX.
I divided the GUID into four pieces, each piece divided into two types (or formats): XXXXXXXX and -XXXX-XXXX.
Now I'm building the GUID using these two types to assemble the GUID with call four pieces, as follows: XXXXXXXX -XXXX-XXXX -XXXX-XXXX XXXXXXXX.
To differ between these two types, I added a flag parameter to a pair creator function _p8(s), the s parameter tells the function whether to add dashes or not.
Eventually we build the GUID with the following chaining: _p8() + _p8(true) + _p8(true) + _p8(), and return it.
Link to this post on my blog
Enjoy! :-)
Here is a totally non-compliant but very performant implementation to generate an ASCII-safe GUID-like unique identifier.
function generateQuickGuid() {
return Math.random().toString(36).substring(2, 15) +
Math.random().toString(36).substring(2, 15);
}
Generates 26 [a-z0-9] characters, yielding a UID that is both shorter and more unique than RFC compliant GUIDs. Dashes can be trivially added if human-readability matters.
Here are usage examples and timings for this function and several of this question's other answers. The timing was performed under Chrome m25, 10 million iterations each.
>>> generateQuickGuid()
"nvcjf1hs7tf8yyk4lmlijqkuo9"
"yq6gipxqta4kui8z05tgh9qeel"
"36dh5sec7zdj90sk2rx7pjswi2"
runtime: 32.5s
>>> GUID() // John Millikin
"7a342ca2-e79f-528e-6302-8f901b0b6888"
runtime: 57.8s
>>> regexGuid() // broofa
"396e0c46-09e4-4b19-97db-bd423774a4b3"
runtime: 91.2s
>>> createUUID() // Kevin Hakanson
"403aa1ab-9f70-44ec-bc08-5d5ac56bd8a5"
runtime: 65.9s
>>> UUIDv4() // Jed Schmidt
"f4d7d31f-fa83-431a-b30c-3e6cc37cc6ee"
runtime: 282.4s
>>> Math.uuid() // broofa
"5BD52F55-E68F-40FC-93C2-90EE069CE545"
runtime: 225.8s
>>> Math.uuidFast() // broofa
"6CB97A68-23A2-473E-B75B-11263781BBE6"
runtime: 92.0s
>>> Math.uuidCompact() // broofa
"3d7b7a06-0a67-4b67-825c-e5c43ff8c1e8"
runtime: 229.0s
>>> bitwiseGUID() // jablko
"baeaa2f-7587-4ff1-af23-eeab3e92"
runtime: 79.6s
>>>> betterWayGUID() // Andrea Turri
"383585b0-9753-498d-99c3-416582e9662c"
runtime: 60.0s
>>>> UUID() // John Fowler
"855f997b-4369-4cdb-b7c9-7142ceaf39e8"
runtime: 62.2s
Here is the timing code.
var r;
console.time('t');
for (var i = 0; i < 10000000; i++) {
r = FuncToTest();
};
console.timeEnd('t');
From sagi shkedy's technical blog:
function generateGuid() {
var result, i, j;
result = '';
for(j=0; j<32; j++) {
if( j == 8 || j == 12 || j == 16 || j == 20)
result = result + '-';
i = Math.floor(Math.random()*16).toString(16).toUpperCase();
result = result + i;
}
return result;
}
There are other methods that involve using an ActiveX control, but stay away from these!
I thought it was worth pointing out that no GUID generator can guarantee unique keys (check the Wikipedia article). There is always a chance of collisions. A GUID simply offers a large enough universe of keys to reduce the change of collisions to almost nil.
Here is a combination of the top voted answer, with a workaround for Chrome's collisions:
generateGUID = (typeof(window.crypto) != 'undefined' &&
typeof(window.crypto.getRandomValues) != 'undefined') ?
function() {
// If we have a cryptographically secure PRNG, use that
// https://stackoverflow.com/questions/6906916/collisions-when-generating-uuids-in-javascript
var buf = new Uint16Array(8);
window.crypto.getRandomValues(buf);
var S4 = function(num) {
var ret = num.toString(16);
while(ret.length < 4){
ret = "0"+ret;
}
return ret;
};
return (S4(buf[0])+S4(buf[1])+"-"+S4(buf[2])+"-"+S4(buf[3])+"-"+S4(buf[4])+"-"+S4(buf[5])+S4(buf[6])+S4(buf[7]));
}
:
function() {
// Otherwise, just use Math.random
// https://stackoverflow.com/questions/105034/how-to-create-a-guid-uuid-in-javascript/2117523#2117523
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
var r = Math.random()*16|0, v = c == 'x' ? r : (r&0x3|0x8);
return v.toString(16);
});
};
It is on jsbin if you want to test it.
Here's a solution dated Oct. 9, 2011 from a comment by user jed at https://gist.github.com/982883:
UUIDv4 = function b(a){return a?(a^Math.random()*16>>a/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,b)}
This accomplishes the same goal as the current highest-rated answer, but in 50+ fewer bytes by exploiting coercion, recursion, and exponential notation. For those curious how it works, here's the annotated form of an older version of the function:
UUIDv4 =
function b(
a // placeholder
){
return a // if the placeholder was passed, return
? ( // a random number from 0 to 15
a ^ // unless b is 8,
Math.random() // in which case
* 16 // a random number from
>> a/4 // 8 to 11
).toString(16) // in hexadecimal
: ( // or otherwise a concatenated string:
[1e7] + // 10000000 +
-1e3 + // -1000 +
-4e3 + // -4000 +
-8e3 + // -80000000 +
-1e11 // -100000000000,
).replace( // replacing
/[018]/g, // zeroes, ones, and eights with
b // random hex digits
)
}
You can use node-uuid. It provides simple, fast generation of RFC4122 UUIDS.
Features:
Generate RFC4122 version 1 or version 4 UUIDs
Runs in Node.js and browsers.
Cryptographically strong random # generation on supporting platforms.
Small footprint (Want something smaller? Check this out!)
Install Using NPM:
npm install uuid
Or using uuid via a browser:
Download Raw File (uuid v1): https://raw.githubusercontent.com/kelektiv/node-uuid/master/v1.js
Download Raw File (uuid v4): https://raw.githubusercontent.com/kelektiv/node-uuid/master/v4.js
Want even smaller? Check this out: https://gist.github.com/jed/982883
Usage:
// Generate a v1 UUID (time-based)
const uuidV1 = require('uuid/v1');
uuidV1(); // -> '6c84fb90-12c4-11e1-840d-7b25c5ee775a'
// Generate a v4 UUID (random)
const uuidV4 = require('uuid/v4');
uuidV4(); // -> '110ec58a-a0f2-4ac4-8393-c866d813b8d1'
// Generate a v5 UUID (namespace)
const uuidV5 = require('uuid/v5');
// ... using predefined DNS namespace (for domain names)
uuidV5('hello.example.com', v5.DNS)); // -> 'fdda765f-fc57-5604-a269-52a7df8164ec'
// ... using predefined URL namespace (for, well, URLs)
uuidV5('http://example.com/hello', v5.URL); // -> '3bbcee75-cecc-5b56-8031-b6641c1ed1f1'
// ... using a custom namespace
const MY_NAMESPACE = '(previously generated unique uuid string)';
uuidV5('hello', MY_NAMESPACE); // -> '90123e1c-7512-523e-bb28-76fab9f2f73d'
ECMAScript 2015 (ES6):
import uuid from 'uuid/v4';
const id = uuid();
var uuid = function() {
var buf = new Uint32Array(4);
window.crypto.getRandomValues(buf);
var idx = -1;
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
idx++;
var r = (buf[idx>>3] >> ((idx%8)*4))&15;
var v = c == 'x' ? r : (r&0x3|0x8);
return v.toString(16);
});
};
This version is based on Briguy37's answer and some bitwise operators to extract nibble sized windows from the buffer.
It should adhere to the RFC Type 4 (random) schema, since I had problems last time parsing non-compliant UUIDs with Java's UUID.
This creates a version 4 UUID (created from pseudo random numbers):
function uuid()
{
var chars = '0123456789abcdef'.split('');
var uuid = [], rnd = Math.random, r;
uuid[8] = uuid[13] = uuid[18] = uuid[23] = '-';
uuid[14] = '4'; // version 4
for (var i = 0; i < 36; i++)
{
if (!uuid[i])
{
r = 0 | rnd()*16;
uuid[i] = chars[(i == 19) ? (r & 0x3) | 0x8 : r & 0xf];
}
}
return uuid.join('');
}
Here is a sample of the UUIDs generated:
682db637-0f31-4847-9cdf-25ba9613a75c
97d19478-3ab2-4aa1-b8cc-a1c3540f54aa
2eed04c9-2692-456d-a0fd-51012f947136
One line solution using Blobs.
window.URL.createObjectURL(new Blob([])).substring(31);
The value at the end (31) depends on the length of the URL.
EDIT:
A more compact and universal solution, as suggested by rinogo:
URL.createObjectURL(new Blob([])).substr(-36);
Simple JavaScript module as a combination of best answers in this question.
var crypto = window.crypto || window.msCrypto || null; // IE11 fix
var Guid = Guid || (function() {
var EMPTY = '00000000-0000-0000-0000-000000000000';
var _padLeft = function(paddingString, width, replacementChar) {
return paddingString.length >= width ? paddingString : _padLeft(replacementChar + paddingString, width, replacementChar || ' ');
};
var _s4 = function(number) {
var hexadecimalResult = number.toString(16);
return _padLeft(hexadecimalResult, 4, '0');
};
var _cryptoGuid = function() {
var buffer = new window.Uint16Array(8);
crypto.getRandomValues(buffer);
return [_s4(buffer[0]) + _s4(buffer[1]), _s4(buffer[2]), _s4(buffer[3]), _s4(buffer[4]), _s4(buffer[5]) + _s4(buffer[6]) + _s4(buffer[7])].join('-');
};
var _guid = function() {
var currentDateMilliseconds = new Date().getTime();
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(currentChar) {
var randomChar = (currentDateMilliseconds + Math.random() * 16) % 16 | 0;
currentDateMilliseconds = Math.floor(currentDateMilliseconds / 16);
return (currentChar === 'x' ? randomChar : (randomChar & 0x7 | 0x8)).toString(16);
});
};
var create = function() {
var hasCrypto = crypto != 'undefined' && crypto !== null,
hasRandomValues = typeof(window.crypto.getRandomValues) != 'undefined';
return (hasCrypto && hasRandomValues) ? _cryptoGuid() : _guid();
};
return {
newGuid: create,
empty: EMPTY
};
})();
// DEMO: Create and show GUID
console.log('1. New Guid: ' + Guid.newGuid());
// DEMO: Show empty GUID
console.log('2. Empty Guid: ' + Guid.empty);
Usage:
Guid.newGuid()
"c6c2d12f-d76b-5739-e551-07e6de5b0807"
Guid.empty
"00000000-0000-0000-0000-000000000000"
The version below is an adaptation of broofa's answer, but updated to include a "true" random function that uses crypto libraries where available, and the Alea() function as a fallback.
Math.log2 = Math.log2 || function(n){ return Math.log(n) / Math.log(2); }
Math.trueRandom = (function() {
var crypt = window.crypto || window.msCrypto;
if (crypt && crypt.getRandomValues) {
// If we have a crypto library, use it
var random = function(min, max) {
var rval = 0;
var range = max - min;
if (range < 2) {
return min;
}
var bits_needed = Math.ceil(Math.log2(range));
if (bits_needed > 53) {
throw new Exception("We cannot generate numbers larger than 53 bits.");
}
var bytes_needed = Math.ceil(bits_needed / 8);
var mask = Math.pow(2, bits_needed) - 1;
// 7776 -> (2^13 = 8192) -1 == 8191 or 0x00001111 11111111
// Create byte array and fill with N random numbers
var byteArray = new Uint8Array(bytes_needed);
crypt.getRandomValues(byteArray);
var p = (bytes_needed - 1) * 8;
for(var i = 0; i < bytes_needed; i++ ) {
rval += byteArray[i] * Math.pow(2, p);
p -= 8;
}
// Use & to apply the mask and reduce the number of recursive lookups
rval = rval & mask;
if (rval >= range) {
// Integer out of acceptable range
return random(min, max);
}
// Return an integer that falls within the range
return min + rval;
}
return function() {
var r = random(0, 1000000000) / 1000000000;
return r;
};
} else {
// From https://web.archive.org/web/20120502223108/http://baagoe.com/en/RandomMusings/javascript/
// Johannes Baagøe <baagoe#baagoe.com>, 2010
function Mash() {
var n = 0xefc8249d;
var mash = function(data) {
data = data.toString();
for (var i = 0; i < data.length; i++) {
n += data.charCodeAt(i);
var h = 0.02519603282416938 * n;
n = h >>> 0;
h -= n;
h *= n;
n = h >>> 0;
h -= n;
n += h * 0x100000000; // 2^32
}
return (n >>> 0) * 2.3283064365386963e-10; // 2^-32
};
mash.version = 'Mash 0.9';
return mash;
}
// From http://baagoe.com/en/RandomMusings/javascript/
function Alea() {
return (function(args) {
// Johannes Baagøe <baagoe#baagoe.com>, 2010
var s0 = 0;
var s1 = 0;
var s2 = 0;
var c = 1;
if (args.length == 0) {
args = [+new Date()];
}
var mash = Mash();
s0 = mash(' ');
s1 = mash(' ');
s2 = mash(' ');
for (var i = 0; i < args.length; i++) {
s0 -= mash(args[i]);
if (s0 < 0) {
s0 += 1;
}
s1 -= mash(args[i]);
if (s1 < 0) {
s1 += 1;
}
s2 -= mash(args[i]);
if (s2 < 0) {
s2 += 1;
}
}
mash = null;
var random = function() {
var t = 2091639 * s0 + c * 2.3283064365386963e-10; // 2^-32
s0 = s1;
s1 = s2;
return s2 = t - (c = t | 0);
};
random.uint32 = function() {
return random() * 0x100000000; // 2^32
};
random.fract53 = function() {
return random() +
(random() * 0x200000 | 0) * 1.1102230246251565e-16; // 2^-53
};
random.version = 'Alea 0.9';
random.args = args;
return random;
}(Array.prototype.slice.call(arguments)));
};
return Alea();
}
}());
Math.guid = function() {
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
var r = Math.trueRandom() * 16 | 0,
v = c == 'x' ? r : (r & 0x3 | 0x8);
return v.toString(16);
});
};
JavaScript project on GitHub - https://github.com/LiosK/UUID.js
UUID.js The RFC-compliant UUID generator for JavaScript.
See RFC 4122 http://www.ietf.org/rfc/rfc4122.txt.
Features Generates RFC 4122 compliant UUIDs.
Version 4 UUIDs (UUIDs from random numbers) and version 1 UUIDs
(time-based UUIDs) are available.
UUID object allows a variety of access to the UUID including access to
the UUID fields.
Low timestamp resolution of JavaScript is compensated by random
numbers.
// RFC 4122
//
// A UUID is 128 bits long
//
// String representation is five fields of 4, 2, 2, 2, and 6 bytes.
// Fields represented as lowercase, zero-filled, hexadecimal strings, and
// are separated by dash characters
//
// A version 4 UUID is generated by setting all but six bits to randomly
// chosen values
var uuid = [
Math.random().toString(16).slice(2, 10),
Math.random().toString(16).slice(2, 6),
// Set the four most significant bits (bits 12 through 15) of the
// time_hi_and_version field to the 4-bit version number from Section
// 4.1.3
(Math.random() * .0625 /* 0x.1 */ + .25 /* 0x.4 */).toString(16).slice(2, 6),
// Set the two most significant bits (bits 6 and 7) of the
// clock_seq_hi_and_reserved to zero and one, respectively
(Math.random() * .25 /* 0x.4 */ + .5 /* 0x.8 */).toString(16).slice(2, 6),
Math.random().toString(16).slice(2, 14)].join('-');
Added in: v15.6.0, v14.17.0 there is a built-in crypto.randomUUID() function.
import * as crypto from "crypto";
const uuid = crypto.randomUUID();
In the browser, crypto.randomUUID() is currently supported in Chromium 92+ and Firefox 95+.
For those wanting an RFC 4122 version 4 compliant solution with speed considerations (few calls to Math.random()):
var rand = Math.random;
function UUID() {
var nbr, randStr = "";
do {
randStr += (nbr = rand()).toString(16).substr(3, 6);
} while (randStr.length < 30);
return (
randStr.substr(0, 8) + "-" +
randStr.substr(8, 4) + "-4" +
randStr.substr(12, 3) + "-" +
((nbr*4|0)+8).toString(16) + // [89ab]
randStr.substr(15, 3) + "-" +
randStr.substr(18, 12)
);
}
console.log( UUID() );
The above function should have a decent balance between speed and randomness.
I wanted to understand broofa's answer, so I expanded it and added comments:
var uuid = function () {
return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(
/[xy]/g,
function (match) {
/*
* Create a random nibble. The two clever bits of this code:
*
* - Bitwise operations will truncate floating point numbers
* - For a bitwise OR of any x, x | 0 = x
*
* So:
*
* Math.random * 16
*
* creates a random floating point number
* between 0 (inclusive) and 16 (exclusive) and
*
* | 0
*
* truncates the floating point number into an integer.
*/
var randomNibble = Math.random() * 16 | 0;
/*
* Resolves the variant field. If the variant field (delineated
* as y in the initial string) is matched, the nibble must
* match the mask (where x is a do-not-care bit):
*
* 10xx
*
* This is achieved by performing the following operations in
* sequence (where x is an intermediate result):
*
* - x & 0x3, which is equivalent to x % 3
* - x | 0x8, which is equivalent to x + 8
*
* This results in a nibble between 8 inclusive and 11 exclusive,
* (or 1000 and 1011 in binary), all of which satisfy the variant
* field mask above.
*/
var nibble = (match == 'y') ?
(randomNibble & 0x3 | 0x8) :
randomNibble;
/*
* Ensure the nibble integer is encoded as base 16 (hexadecimal).
*/
return nibble.toString(16);
}
);
};
ES6 sample
const guid=()=> {
const s4=()=> Math.floor((1 + Math.random()) * 0x10000).toString(16).substring(1);
return `${s4() + s4()}-${s4()}-${s4()}-${s4()}-${s4() + s4() + s4()}`;
}
I adjusted my own UUID/GUID generator with some extras here.
I'm using the following Kybos random number generator to be a bit more cryptographically sound.
Below is my script with the Mash and Kybos methods from baagoe.com excluded.
//UUID/Guid Generator
// use: UUID.create() or UUID.createSequential()
// convenience: UUID.empty, UUID.tryParse(string)
(function(w){
// From http://baagoe.com/en/RandomMusings/javascript/
// Johannes Baagøe <baagoe#baagoe.com>, 2010
//function Mash() {...};
// From http://baagoe.com/en/RandomMusings/javascript/
//function Kybos() {...};
var rnd = Kybos();
//UUID/GUID Implementation from http://frugalcoder.us/post/2012/01/13/javascript-guid-uuid-generator.aspx
var UUID = {
"empty": "00000000-0000-0000-0000-000000000000"
,"parse": function(input) {
var ret = input.toString().trim().toLowerCase().replace(/^[\s\r\n]+|[\{\}]|[\s\r\n]+$/g, "");
if ((/[a-f0-9]{8}\-[a-f0-9]{4}\-[a-f0-9]{4}\-[a-f0-9]{4}\-[a-f0-9]{12}/).test(ret))
return ret;
else
throw new Error("Unable to parse UUID");
}
,"createSequential": function() {
var ret = new Date().valueOf().toString(16).replace("-","")
for (;ret.length < 12; ret = "0" + ret);
ret = ret.substr(ret.length-12,12); //only least significant part
for (;ret.length < 32;ret += Math.floor(rnd() * 0xffffffff).toString(16));
return [ret.substr(0,8), ret.substr(8,4), "4" + ret.substr(12,3), "89AB"[Math.floor(Math.random()*4)] + ret.substr(16,3), ret.substr(20,12)].join("-");
}
,"create": function() {
var ret = "";
for (;ret.length < 32;ret += Math.floor(rnd() * 0xffffffff).toString(16));
return [ret.substr(0,8), ret.substr(8,4), "4" + ret.substr(12,3), "89AB"[Math.floor(Math.random()*4)] + ret.substr(16,3), ret.substr(20,12)].join("-");
}
,"random": function() {
return rnd();
}
,"tryParse": function(input) {
try {
return UUID.parse(input);
} catch(ex) {
return UUID.empty;
}
}
};
UUID["new"] = UUID.create;
w.UUID = w.Guid = UUID;
}(window || this));
The native URL.createObjectURL is generating an UUID. You can take advantage of this.
function uuid() {
const url = URL.createObjectURL(new Blob())
const [id] = url.toString().split('/').reverse()
URL.revokeObjectURL(url)
return id
}
The better way:
function(
a, b // Placeholders
){
for( // Loop :)
b = a = ''; // b - result , a - numeric variable
a++ < 36; //
b += a*51&52 // If "a" is not 9 or 14 or 19 or 24
? // return a random number or 4
(
a^15 // If "a" is not 15,
? // generate a random number from 0 to 15
8^Math.random() *
(a^20 ? 16 : 4) // unless "a" is 20, in which case a random number from 8 to 11,
:
4 // otherwise 4
).toString(16)
:
'-' // In other cases, (if "a" is 9,14,19,24) insert "-"
);
return b
}
Minimized:
function(a,b){for(b=a='';a++<36;b+=a*51&52?(a^15?8^Math.random()*(a^20?16:4):4).toString(16):'-');return b}
The following is simple code that uses crypto.getRandomValues(a) on supported browsers (Internet Explorer 11+, iOS 7+, Firefox 21+, Chrome, and Android Chrome).
It avoids using Math.random(), because that can cause collisions (for example 20 collisions for 4000 generated UUIDs in a real situation by Muxa).
function uuid() {
function randomDigit() {
if (crypto && crypto.getRandomValues) {
var rands = new Uint8Array(1);
crypto.getRandomValues(rands);
return (rands[0] % 16).toString(16);
} else {
return ((Math.random() * 16) | 0).toString(16);
}
}
var crypto = window.crypto || window.msCrypto;
return 'xxxxxxxx-xxxx-4xxx-8xxx-xxxxxxxxxxxx'.replace(/x/g, randomDigit);
}
Notes:
Optimised for code readability, not speed, so it is suitable for, say, a few hundred UUIDs per second. It generates about 10000 uuid() per second in Chromium on my laptop using http://jsbin.com/fuwigo/1 to measure performance.
It only uses 8 for "y" because that simplifies code readability (y is allowed to be 8, 9, A, or B).
If you just need a random 128 bit string in no particular format, you can use:
function uuid() {
return crypto.getRandomValues(new Uint32Array(4)).join('-');
}
Which will return something like 2350143528-4164020887-938913176-2513998651.
I couldn't find any answer that uses a single 16-octet TypedArray and a DataView, so I think the following solution for generating a version 4 UUID per the RFC will stand on its own here:
const uuid4 = () => {
const ho = (n, p) => n.toString(16).padStart(p, 0); /// Return the hexadecimal text representation of number `n`, padded with zeroes to be of length `p`
const data = crypto.getRandomValues(new Uint8Array(16)); /// Fill the buffer with random data
data[6] = (data[6] & 0xf) | 0x40; /// Patch the 6th byte to reflect a version 4 UUID
data[8] = (data[8] & 0x3f) | 0x80; /// Patch the 8th byte to reflect a variant 1 UUID (version 4 UUIDs are)
const view = new DataView(data.buffer); /// Create a view backed by a 16-byte buffer
return `${ho(view.getUint32(0), 8)}-${ho(view.getUint16(4), 4)}-${ho(view.getUint16(6), 4)}-${ho(view.getUint16(8), 4)}-${ho(view.getUint32(10), 8)}${ho(view.getUint16(14), 4)}`; /// Compile the canonical textual form from the array data
};
I prefer it because:
it only relies on functions available to the standard ECMAScript platform, where possible -- which is all but one procedure
it only uses a single buffer, minimizing copying of data, which should in theory yield performance advantages
At the time of writing this, getRandomValues is not something implemented for the crypto object in Node.js. However, it has the equivalent randomBytes function which may be used instead.
Just another more readable variant with just two mutations.
function uuid4()
{
function hex (s, b)
{
return s +
(b >>> 4 ).toString (16) + // high nibble
(b & 0b1111).toString (16); // low nibble
}
let r = crypto.getRandomValues (new Uint8Array (16));
r[6] = r[6] >>> 4 | 0b01000000; // Set type 4: 0100
r[8] = r[8] >>> 3 | 0b10000000; // Set variant: 100
return r.slice ( 0, 4).reduce (hex, '' ) +
r.slice ( 4, 6).reduce (hex, '-') +
r.slice ( 6, 8).reduce (hex, '-') +
r.slice ( 8, 10).reduce (hex, '-') +
r.slice (10, 16).reduce (hex, '-');
}
Requirement: Algorithm to generate all possible combinations of a set , without duplicates , or recursively calling function to return results.
The majority , if not all of the Answers provided at Permutations in JavaScript? recursively call a function from within a loop or other function to return results.
Example of recursive function call within loop
function p(a, b, res) {
var b = b || [], res = res || [], len = a.length;
if (!len)
res.push(b)
else
for (var i = 0; i < len
// recursive call to `p` here
; p(a.slice(0, i).concat(a.slice(i + 1, len)), b.concat(a[i]), res)
, i++
);
return res
}
p(["a", "b", "c"]);
The current Question attempts to create the given permutation in a linear process , relying on the previous permutation.
For example , given an array
var arr = ["a", "b", "c"];
to determine the total number of possible permutations
for (var len = 1, i = k = arr.length; len < i ; k *= len++);
k should return 6 , or total number of possible permutations of arr ["a", "b", "c"]
With the total number of individual permutations determined for a set , the resulting array which would contain all six permutations could be created and filled using Array.prototype.slice() , Array.prototype.concat() and Array.prototype.reverse()
var res = new Array(new Array(k));
res[0] = arr;
res[1] = res[0].slice(0,1).concat(res[0].slice(-2).reverse());
res[2] = res[1].slice(-1).concat(res[1].slice(0,2));
res[3] = res[2].slice(0,1).concat(res[2].slice(-2).reverse());
res[4] = res[3].slice(-2).concat(res[3].slice(0,1));
res[5] = res[4].slice(0,1).concat(res[4].slice(-2).reverse());
Attempted to reproduce results based on the pattern displayed at the graph for An Ordered Lexicographic Permutation Algorithm based on one published in Practical Algorithms in C++ at Calculating Permutations and Job Interview Questions .
There appears to be a pattern that could be extended if the input set was , for example
["a", "b", "c", "d", "e"]
where 120 permutations would be expected.
An example of an attempt at filling array relying only on previous permutation
// returns duplicate entries at `j`
var arr = ["a", "b", "c", "d", "e"], j = [];
var i = k = arr.length;
arr.forEach(function(a, b, array) {
if (b > 1) {
k *= b;
if (b === i -1) {
for (var q = 0;j.length < k;q++) {
if (q === 0) {
j[q] = array;
} else {
j[q] = !(q % i)
? array.slice(q % i).reverse().concat(array.slice(0, q % i))
: array.slice(q % i).concat(array.slice(0, q % i));
}
}
}
}
})
however have not yet been able to make the necessary adjustments at parameters for .slice() , .concat() , .reverse() at above js to step from one permutation to the next ; while only using the previous array entry within res to determine current permutation , without using recursive.
Noticed even , odd balance of calls and tried to use modulus % operator and input array .length to either call .reverse() or not at ["a", "b", "c", "d", "e"] array , though did not produce results without duplicate entries.
The expected result is that the above pattern could be reduced to two lines called in succession for the duration of the process until all permutations completed, res filled ; one each for call to .reverse() , call without .reverse() ; e.g., after res[0] filled
// odd , how to adjust `.slice()` , `.concat()` parameters
// for array of unknown `n` `.length` ?
res[i] = res[i - 1].slice(0,1).concat(res[i - 1].slice(-2).reverse());
// even
res[i] = res[1 - 1].slice(-1).concat(res[i - 1].slice(0,2));
Question: What adjustments to above pattern are necessary , in particular parameters , or index , passed .slice() , .concat() to produce all possible permutations of a given set without using a recursive call to the currently processing function ?
var arr = ["a", "b", "c"];
for (var len = 1, i = k = arr.length; len < i; k *= len++);
var res = new Array(new Array(k));
res[0] = arr;
res[1] = res[0].slice(0, 1).concat(res[0].slice(-2).reverse());
res[2] = res[1].slice(-1).concat(res[1].slice(0, 2));
res[3] = res[2].slice(0, 1).concat(res[2].slice(-2).reverse());
res[4] = res[3].slice(-2).concat(res[3].slice(0, 1));
res[5] = res[4].slice(0, 1).concat(res[4].slice(-2).reverse());
console.log(res);
Edit, Update
Have found a process to utilize pattern described above to return permutations in lexicographic order for an input up to .length 4 , using a single for loop. Expected results are not returned for array with .length of 5.
The pattern is based on the second chart at "Calculating Permutations and Job Interview Questions"[0].
Would prefer not to use .splice() or .sort() to return results, though used here while attempting to adhere to last "rotate" requirement at each column. The variable r should reference the index of the first element of the next permutation, which it does.
The use of .splice() , .sort() could be included if their usage followed the pattern at the chart ; though at js below, they actually do not.
Not entirely certain that the issue with js below is only the statement following if (i % (total / len) === reset) , though that portion required the most investment of time; yet still does not return expected results.
Specifically, now referring to the chart, at rotating , for example 2 to index 0, 1 to index 2. Attempted to achieve this by using r , which is a negative index, to traverses from right to left to retrieve next item that should be positioned at index 0 of adjacent "column".
At next column, 2 would be placed at index 2 , 3 would be placed at index 0. This is portion, as far as have been able to grasp or debug, so far, is the area where error is occurring.
Again, returns expected results for [1,2,3,4], though not for [1,2,3,4,5]
var arr = [1, 2, 3, 4];
for (var l = 1, j = total = arr.length; l < j ; total *= l++);
for (var i = 1
, reset = 0
, idx = 0
, r = 0
, len = arr.length
, res = [arr]
; i < total; i++) {
// previous permutation
var prev = res[i - 1];
// if we are at permutation `6` here, or, completion of all
// permutations beginning with `1`;
// setting next "column", place `2` at `index` 0;
// following all permutations beginning with `2`, place `3` at
// `index` `0`; with same process for `3` to `4`
if (i % (total / len) === reset) {
r = --r % -(len);
var next = prev.slice(r);
if (r === -1) {
// first implementation used for setting item at index `-1`
// to `index` 0
// would prefer to use single process for all "rotations",
// instead of splitting into `if` , `else`, though not there, yet
res[i] = [next[0]].concat(prev.slice(0, 1), prev.slice(1, len - 1)
.reverse());
} else {
// workaround for "rotation" at from `index` `r` to `index` `0`
// the chart does not actually use the previous permutation here,
// but rather, the first permutation of that particular "column";
// here, using `r` `,i`, `len`, would be
// `res[i - (i - 1) % (total / len)]`
var curr = prev.slice();
// this may be useful, to retrieve `r`,
// `prev` without item at `r` `index`
curr.splice(prev.indexOf(next[0]), 1);
// this is not optiomal
curr.sort(function(a, b) {
return arr.indexOf(a) > arr.indexOf(b)
});
// place `next[0]` at `index` `0`
// place remainder of sorted array at `index` `1` - n
curr.splice(0, 0, next[0])
res[i] = curr
}
idx = reset;
} else {
if (i % 2) {
// odd
res[i] = prev.slice(0, len - 2).concat(prev.slice(-2)
.reverse())
} else {
// even
--idx
res[i] = prev.slice(0, len - (len - 1))
.concat(prev.slice(idx), prev.slice(1, len + (idx)))
}
}
}
// try with `arr` : `[1,2,3,4,5]` to return `res` that is not correct;
// how can above `js` be adjusted to return correct results for `[1,2,3,4,5]` ?
console.log(res, res.length)
Resources:
Generating Permutation with Javascript
(Countdown) QuickPerm Head Lexicography:
(Formally Example_03 ~ Palindromes)
Generating all Permutations [non-recursive]
(Attempt to port to from C++ to javascript jsfiddle http://jsfiddle.net/tvvvjf3p/)
Calculating Permutation without Recursion - Part 2
permutations of a string using iteration
iterative-permutation
Permutations by swapping
Evaluation of permutation algorithms
Permutation algorithm without recursion? Java
Non-recursive algorithm for full permutation with repetitive elements?
String permutations in Java (non-recursive)
Generating permutations lazily
How to generate all permutations of a list in Python
Can all permutations of a set or string be generated in O(n log n) time?
Finding the nth lexicographic permutation of ‘0123456789’
Combinations and Permutations
Here is a simple solution to compute the nth permutation of a string:
function string_nth_permutation(str, n) {
var len = str.length, i, f, res;
for (f = i = 1; i <= len; i++)
f *= i;
if (n >= 0 && n < f) {
for (res = ""; len > 0; len--) {
f /= len;
i = Math.floor(n / f);
n %= f;
res += str.charAt(i);
str = str.substring(0, i) + str.substring(i + 1);
}
}
return res;
}
The algorithm follows these simple steps:
first compute f = len!, there are factorial(len) total permutations of a set of len different elements.
as the first element, divide the permutation number by (len-1)! and chose the element at the resulting offset. There are (len-1)! different permutations that have any given element as their first element.
remove the chosen element from the set and use the remainder of the division as the permutation number to keep going.
perform these steps with the rest of the set, whose length is reduced by one.
This algorithm is very simple and has interesting properties:
It computes the n-th permutation directly.
If the set is ordered, the permutations are generated in lexicographical order.
It works even if set elements cannot be compared to one another, such as objects, arrays, functions...
Permutation number 0 is the set in the order given.
Permutation number factorial(a.length)-1 is the last one: the set a in reverse order.
Permutations outside this range are returned as undefined.
It can easily be converted to handle a set stored as an array:
function array_nth_permutation(a, n) {
var b = a.slice(); // copy of the set
var len = a.length; // length of the set
var res; // return value, undefined
var i, f;
// compute f = factorial(len)
for (f = i = 1; i <= len; i++)
f *= i;
// if the permutation number is within range
if (n >= 0 && n < f) {
// start with the empty set, loop for len elements
for (res = []; len > 0; len--) {
// determine the next element:
// there are f/len subsets for each possible element,
f /= len;
// a simple division gives the leading element index
i = Math.floor(n / f);
// alternately: i = (n - n % f) / f;
res.push(b.splice(i, 1)[0]);
// reduce n for the remaining subset:
// compute the remainder of the above division
n %= f;
// extract the i-th element from b and push it at the end of res
}
}
// return the permutated set or undefined if n is out of range
return res;
}
clarification:
f is first computed as factorial(len).
For each step, f is divided by len, giving exacty the previous factorial.
n divided by this new value of f gives the slot number among the len slots that have the same initial element. Javascript does not have integral division, we could use (n / f) ... 0) to convert the result of the division to its integral part but it introduces a limitation to sets of 12 elements. Math.floor(n / f) allows for sets of up to 18 elements. We could also use (n - n % f) / f, probably more efficient too.
n must be reduced to the permutation number within this slot, that is the remainder of the division n / f.
We could use i differently in the second loop, storing the division remainder, avoiding Math.floor() and the extra % operator. Here is an alternative for this loop that may be even less readable:
// start with the empty set, loop for len elements
for (res = []; len > 0; len--) {
i = n % (f /= len);
res.push(b.splice((n - i) / f, 1)[0]);
n = i;
}
I think this post should help you. The algorithm should be easily translatable to JavaScript (I think it is more than 70% already JavaScript-compatible).
slice and reverse are bad calls to use if you are after efficiency. The algorithm described in the post is following the most efficient implementation of the next_permutation function, that is even integrated in some programming languages (like C++ e.g.)
EDIT
As I iterated over the algorithm once again I think you can just remove the types of the variables and you should be good to go in JavaScript.
EDIT
JavaScript version:
function nextPermutation(array) {
// Find non-increasing suffix
var i = array.length - 1;
while (i > 0 && array[i - 1] >= array[i])
i--;
if (i <= 0)
return false;
// Find successor to pivot
var j = array.length - 1;
while (array[j] <= array[i - 1])
j--;
var temp = array[i - 1];
array[i - 1] = array[j];
array[j] = temp;
// Reverse suffix
j = array.length - 1;
while (i < j) {
temp = array[i];
array[i] = array[j];
array[j] = temp;
i++;
j--;
}
return true;
}
One method to create permutations is by adding each element in all of the spaces between elements in all of the results so far. This can be done without recursion using loops and a queue.
JavaScript code:
function ps(a){
var res = [[]];
for (var i=0; i<a.length; i++){
while(res[res.length-1].length == i){
var l = res.pop();
for (var j=0; j<=l.length; j++){
var copy = l.slice();
copy.splice(j,0,a[i]);
res.unshift(copy);
}
}
}
return res;
}
console.log(JSON.stringify(ps(['a','b','c','d'])));
Here could be another solution, inspired from the Steinhaus-Johnson-Trotter algorithm:
function p(input) {
var i, j, k, temp, base, current, outputs = [[input[0]]];
for (i = 1; i < input.length; i++) {
current = [];
for (j = 0; j < outputs.length; j++) {
base = outputs[j];
for (k = 0; k <= base.length; k++) {
temp = base.slice();
temp.splice(k, 0, input[i]);
current.push(temp);
}
}
outputs = current;
}
return outputs;
}
// call
var outputs = p(["a", "b", "c", "d"]);
for (var i = 0; i < outputs.length; i++) {
document.write(JSON.stringify(outputs[i]) + "<br />");
}
Here's a snippet for an approach that I came up with on my own, but naturally was also able to find it described elsewhere:
generatePermutations = function(arr) {
if (arr.length < 2) {
return arr.slice();
}
var factorial = [1];
for (var i = 1; i <= arr.length; i++) {
factorial.push(factorial[factorial.length - 1] * i);
}
var allPerms = [];
for (var permNumber = 0; permNumber < factorial[factorial.length - 1]; permNumber++) {
var unused = arr.slice();
var nextPerm = [];
while (unused.length) {
var nextIndex = Math.floor((permNumber % factorial[unused.length]) / factorial[unused.length - 1]);
nextPerm.push(unused[nextIndex]);
unused.splice(nextIndex, 1);
}
allPerms.push(nextPerm);
}
return allPerms;
};
Enter comma-separated string (e.g. a,b,c):
<br/>
<input id="arrInput" type="text" />
<br/>
<button onclick="perms.innerHTML = generatePermutations(arrInput.value.split(',')).join('<br/>')">
Generate permutations
</button>
<br/>
<div id="perms"></div>
Explanation
Since there are factorial(arr.length) permutations for a given array arr, each number between 0 and factorial(arr.length)-1 encodes a particular permutation. To unencode a permutation number, repeatedly remove elements from arr until there are no elements left. The exact index of which element to remove is given by the formula (permNumber % factorial(arr.length)) / factorial(arr.length-1). Other formulas could be used to determine the index to remove, as long as it preserves the one-to-one mapping between number and permutation.
Example
The following is how all permutations would be generated for the array (a,b,c,d):
# Perm 1st El 2nd El 3rd El 4th El
0 abcd (a,b,c,d)[0] (b,c,d)[0] (c,d)[0] (d)[0]
1 abdc (a,b,c,d)[0] (b,c,d)[0] (c,d)[1] (c)[0]
2 acbd (a,b,c,d)[0] (b,c,d)[1] (b,d)[0] (d)[0]
3 acdb (a,b,c,d)[0] (b,c,d)[1] (b,d)[1] (b)[0]
4 adbc (a,b,c,d)[0] (b,c,d)[2] (b,c)[0] (c)[0]
5 adcb (a,b,c,d)[0] (b,c,d)[2] (b,c)[1] (b)[0]
6 bacd (a,b,c,d)[1] (a,c,d)[0] (c,d)[0] (d)[0]
7 badc (a,b,c,d)[1] (a,c,d)[0] (c,d)[1] (c)[0]
8 bcad (a,b,c,d)[1] (a,c,d)[1] (a,d)[0] (d)[0]
9 bcda (a,b,c,d)[1] (a,c,d)[1] (a,d)[1] (a)[0]
10 bdac (a,b,c,d)[1] (a,c,d)[2] (a,c)[0] (c)[0]
11 bdca (a,b,c,d)[1] (a,c,d)[2] (a,c)[1] (a)[0]
12 cabd (a,b,c,d)[2] (a,b,d)[0] (b,d)[0] (d)[0]
13 cadb (a,b,c,d)[2] (a,b,d)[0] (b,d)[1] (b)[0]
14 cbad (a,b,c,d)[2] (a,b,d)[1] (a,d)[0] (d)[0]
15 cbda (a,b,c,d)[2] (a,b,d)[1] (a,d)[1] (a)[0]
16 cdab (a,b,c,d)[2] (a,b,d)[2] (a,b)[0] (b)[0]
17 cdba (a,b,c,d)[2] (a,b,d)[2] (a,b)[1] (a)[0]
18 dabc (a,b,c,d)[3] (a,b,c)[0] (b,c)[0] (c)[0]
19 dacb (a,b,c,d)[3] (a,b,c)[0] (b,c)[1] (b)[0]
20 dbac (a,b,c,d)[3] (a,b,c)[1] (a,c)[0] (c)[0]
21 dbca (a,b,c,d)[3] (a,b,c)[1] (a,c)[1] (a)[0]
22 dcab (a,b,c,d)[3] (a,b,c)[2] (a,b)[0] (b)[0]
23 dcba (a,b,c,d)[3] (a,b,c)[2] (a,b)[1] (a)[0]
Note that each permutation # is of the form:
(firstElIndex * 3!) + (secondElIndex * 2!) + (thirdElIndex * 1!) + (fourthElIndex * 0!)
which is basically the reverse process of the formula given in the explanation.
I dare to add another answer, aiming at answering you question regarding slice, concat, reverse.
The answer is it is possible (almost), but it would not be quite effective. What you are doing in your algorithm is the following:
Find the first inversion in the permutation array, right-to-left (inversion in this case defined as i and j where i < j and perm[i] > perm[j], indices given left-to-right)
place the bigger number of the inversion
concatenate the processed numbers in reversed order, which will be the same as sorted order, as no inversions were observed.
concatenate the second number of the inversion (still sorted in accordsnce with the previos number, as no inversions were observed)
This is mainly, what my first answer does, but in a bit more optimal manner.
Example
Consider the permutation 9,10, 11, 8, 7, 6, 5, 4 ,3,2,1
The first inversion right-to-left is 10, 11.
And really the next permutation is:
9,11,1,2,3,4,5,6,7,8,9,10=9concat(11)concat(rev(8,7,6,5,4,3,2,1))concat(10)
Source code
Here I include the source code as I envision it:
var nextPermutation = function(arr) {
for (var i = arr.length - 2; i >= 0; i--) {
if (arr[i] < arr[i + 1]) {
return arr.slice(0, i).concat([arr[i + 1]]).concat(arr.slice(i + 2).reverse()).concat([arr[i]]);
}
}
// return again the first permutation if calling next permutation on last.
return arr.reverse();
}
console.log(nextPermutation([9, 10, 11, 8, 7, 6, 5, 4, 3, 2, 1]));
console.log(nextPermutation([6, 5, 4, 3, 2, 1]));
console.log(nextPermutation([1, 2, 3, 4, 5, 6]));
The code is avaiable for jsfiddle here.
A fairly simple C++ code without recursion.
#include <vector>
#include <algorithm>
#include <iterator>
#include <iostream>
#include <string>
// Integer data
void print_all_permutations(std::vector<int> &data) {
std::stable_sort(std::begin(data), std::end(data));
do {
std::copy(data.begin(), data.end(), std::ostream_iterator<int>(std::cout, " ")), std::cout << '\n';
} while (std::next_permutation(std::begin(data), std::end(data)));
}
// Character data (string)
void print_all_permutations(std::string &data) {
std::stable_sort(std::begin(data), std::end(data));
do {
std::copy(data.begin(), data.end(), std::ostream_iterator<char>(std::cout, " ")), std::cout << '\n';
} while (std::next_permutation(std::begin(data), std::end(data)));
}
int main()
{
std::vector<int> v({1,2,3,4});
print_all_permutations(v);
std::string s("abcd");
print_all_permutations(s);
return 0;
}
We can find next permutation of a sequence in linear time.
Here is an answer from #le_m. It might be of help.
The following very efficient algorithm uses Heap's method to generate all permutations of N elements with runtime complexity in O(N!):
function permute(permutation) {
var length = permutation.length,
result = [permutation.slice()],
c = new Array(length).fill(0),
i = 1, k, p;
while (i < length) {
if (c[i] < i) {
k = i % 2 && c[i];
p = permutation[i];
permutation[i] = permutation[k];
permutation[k] = p;
++c[i];
i = 1;
result.push(permutation.slice());
} else {
c[i] = 0;
++i;
}
}
return result;
}
console.log(JSON.stringify(permute([1, 2, 3, 4])));
You can use a stack to go through permutations.
This approach is ideal when dealing with trees or other problems while not leaning on recursion.
You will need to make adjustments to not have any duplicate values.
type permutation = [string, string[]]
function p(str: string): string[]{
const res: string[] = []
const stack: permutation[] = [["", str.split('')]]
while(stack.length){
const [head, tail] = stack.pop()
if(!tail.length){
res.push(head)
continue
}
for(let i = 0; i < tail.length; i++){
let newTail = tail.slice()
newTail.splice(i, 1)
stack.push([head + tail[i], newTail])
}
}
return res
}
In the past I have made a function that generates an unique id (number) from a string. Today I discover that it is not as unique as should be. Never saw a problem before with it. Today two different inputs generates the same id (number).
I use the same technique in Delphi, C++, PHP and Javascript to generate the same id's so there is no difference when different languages are involved to a project. For example this can be handy to communicate, for HTML id's, tempfiles etc.
In general, what I do is calculate a CRC16 of a string, add the sum and return it.
For example, these two strings generate the same id (number):
o.uniqueId( 'M:/Mijn Muziek/Various Artists/Revs & ElBee - Tell It To My Heart.mp3' );
o.uniqueId( 'M:/Mijn Muziek/Various Artists/Dwight Yoakam - The Back Of Your Hand.Mp3');
They both generates an id of 224904.
The following example is a javascript example. My question is, how can i avoid (with a little change) that it generates a duplicate? (In case you might wonder what 'o.' means, it is the object where these functions belongs to):
o.getCrc16 = function(s, bSumPos) {
if(typeof s !== 'string' || s.length === 0) {
return 0;
}
var crc = 0xFFFF,
L = s.length,
sum = 0,
x = 0,
j = 0;
for(var i = 0; i < L; i++) {
j = s.charCodeAt(i);
sum += ((i + 1) * j);
x = ((crc >> 8) ^ j) & 0xFF;
x ^= x >> 4;
crc = ((crc << 8) ^ (x << 12) ^ (x << 5) ^ x) & 0xFFFF;
}
return crc + ((bSumPos ? 1 : 0) * sum);
}
o.uniqueId = function(s, bres) {
if(s == undefined || typeof s != 'string') {
if(!o.___uqidc) {
o.___uqidc = 0;
} else {
++o.___uqidc;
}
var od = new Date(),
i = s = od.getTime() + '' + o.___uqidc;
} else {
var i = o.getCrc16(s, true);
}
return((bres) ? 'res:' : '') + (i + (i ? s.length : 0));
};
How can I avoid duplicates with use of a little change to the code?
All right, did allot of testing and come to this. A relative short unique id generated by the following:
o.lz = function(i,c)
{
if( typeof c != 'number' || c <= 0 || (typeof i != 'number' && typeof i != 'string') )
{ return i; }
i+='';
while( i.length < c )
{ i='0'+i; }
return i;
}
o.getHashCode = function(s)
{
var hash=0,c=(typeof s == 'string')?s.length:0,i=0;
while(i<c)
{
hash = ((hash<<5)-hash)+s.charCodeAt(i++);
//hash = hash & hash; // Convert to 32bit integer
}
return ( hash < 0 )?((hash*-1)+0xFFFFFFFF):hash; // convert to unsigned
};
o.uniqueId = function( s, bres )
{
if( s == undefined || typeof s != 'string' )
{
if( !o.___uqidc )
{ o.___uqidc=0; }
else { ++o.___uqidc; }
var od = new Date(),
i = s = od.getTime()+''+o.___uqidc;
}
else { var i = o.getHashCode( s ); }
return ((bres)?'res:':'')+i.toString(32)+'-'+o.lz((s.length*4).toString(16),3);
};
Examples:
o.uniqueId( 'M:/Mijn Muziek/Various Artists/Revs & ElBee - Tell It To My Heart.mp3' );
o.uniqueId( 'M:/Mijn Muziek/Various Artists/Dwight Yoakam - The Back Of Your Hand.Mp3');
Will produce the following id's:
dh8qi9t-114
je38ugg-120
For my purpose it seems to be unique enough, also the extra length adds some more uniqueness. Test it on filesystem with approx 40.000 mp3 files and did not found any collision.
If you think this is not the way to go, please let me know.
You should increase the number of bits created by your hash function. Assuming that your hash function approximately uniform over the space, you can mathematically derive the probability of observing a collision.
This is strongly related to the birthday paradox. In the case of CRC16, where the hash value is 17 bits (though your implementation may have a mistake; I don't see how you obtained 224094 as that is greater than 2^17), you will have a collision probability above 50% when you store more than approximately 2^8 items. In addition, CRC is not really a great hashing function because it's meant for error detection, not uniform hashing.
This table shows mathematical probabilities of collision based on hash length. For example, if you have a 128-bit hash key, you can store up to 10^31 elements before the collision probability increases beyond 10^-15. As a comparison, this probability is lower than that of your hard drive failing, or maybe your computer being zapped by lightning, so a safe number to use.
Just increase your hash length based on the number of strings you are planning to identify, and a pick a collision probability that is acceptable to you.
If I have an Uint8Array array in JavaScript, how would I get the last four bytes and then convert that to an int? Using C# I would do something like this:
int count = BitConverter.ToInt32(array, array.Length - 4);
Is there an inequivalent way to do this using JavaScript?
Access the underlying ArrayBuffer and create a new TypedArray with a slice of its bytes:
var u8 = new Uint8Array([1,2,3,4,5,6]); // original array
var u32bytes = u8.buffer.slice(-4); // last four bytes as a new `ArrayBuffer`
var uint = new Uint32Array(u32bytes)[0];
If the TypedArray does not cover the entire buffer, you need to be a little trickier, but not much:
var startbyte = u8.byteOffset + u8.byteLength - Uint32Array.BYTES_PER_ELEMENT;
var u32bytes = u8.buffer.slice(startbyte, startbyte + Uint32Array.BYTES_PER_ELEMENT);
This works in both cases.
If the bytes you want fit in the alignment boundary of your underlying buffer for the datatype (e.g., you want the 32-bit value of bytes 4-8 of the underlying buffer), you can avoid copying the bytes with slice() and just supply a byteoffset to the view constructor, as in #Bergi's answer.
Below is a very-lightly-tested function that should get the scalar value of any offset you want. It will avoid copying if possible.
function InvalidArgument(msg) {
this.message = msg | null;
}
function scalarValue(buf_or_view, byteOffset, type) {
var buffer, bufslice, view, sliceLength = type.BYTES_PER_ELEMENT;
if (buf_or_view instanceof ArrayBuffer) {
buffer = buf_or_view;
if (byteOffset < 0) {
byteOffset = buffer.byteLength - byteOffset;
}
} else if (buf_or_view.buffer instanceof ArrayBuffer) {
view = buf_or_view;
buffer = view.buffer;
if (byteOffset < 0) {
byteOffset = view.byteOffset + view.byteLength + byteOffset;
} else {
byteOffset = view.byteOffset + byteOffset;
}
return scalarValue(buffer, view.byteOffset + byteOffset, type);
} else {
throw new InvalidArgument('buf_or_view must be ArrayBuffer or have a .buffer property');
}
// assert buffer instanceof ArrayBuffer
// assert byteOffset > 0
// assert byteOffset relative to entire buffer
try {
// try in-place first
// only works if byteOffset % slicelength === 0
return (new type(buffer, byteOffset, 1))[0]
} catch (e) {
// if this doesn't work, we need to copy the bytes (slice them out)
bufslice = buffer.slice(byteOffset, byteOffset + sliceLength);
return (new type(bufslice, 0, 1))[0]
}
}
You would use it like this:
// positive or negative byte offset
// relative to beginning or end *of a view*
100992003 === scalarValueAs(u8, -4, Uint32Array)
// positive or negative byte offset
// relative to the beginning or end *of a buffer*
100992003 === scalarValue(u8.buffer, -4, Uint32Array)
Do you have an example? I think this would do it:
var result = ((array[array.length - 1]) |
(array[array.length - 2] << 8) |
(array[array.length - 3] << 16) |
(array[array.length - 4] << 24));
Nowadays if you can live with IE 11+ / Chrome 49+ / Firefox 50+, then you can use DataView to make your life almost as easy as in C#:
var u8array = new Uint8Array([0xFF, 0xFF, 0xFF, 0xFF]); // -1
var view = new DataView(u8array.buffer)
console.log("result:" + view.getInt32());
Test it here: https://jsfiddle.net/3udtek18/1/
A little inelegant, but if you can do it manually based on the endianess.
Little endian:
var count = 0;
// assuming the array has at least four elements
for(var i = array.length - 1; i >= array.length - 4; i--)
{
count = count << 8 + array[i];
}
Big endian:
var count = 0;
// assuming the array has at least four elements
for(var i = array.length - 4; i <= array.length - 1 ; i++)
{
count = count << 8 + array[i];
}
This can be extended to other data lengths
Edit: Thanks to David for pointing out my typos
I'm surprised that the other answers don't use the native Buffer object, which provides a lot of this tooling in a simple, native library. It's possible that this library isn't widely used for bitpacking/unpacking simply because people don't think to check here and it took me a while to find it, too, but it's the right tool for bitpacking/unpacking in nodejs/javascript/typescript.
You can use it like so:
// Create a simple array with 5 elements. We'll pop the last 4 and you should expect the end value to be 1 because this is a little-endian array with all zeros other than the 1 in the littlest(?)-endian
const array = [0, 1, 0, 0, 0]
// get the last 4 elements of your array and convert it to a Buffer
const buffer = Buffer.from(array.slice(-4));
// Use the native Buffer type to read the object as an (U) unsigned (LE) little-endian 32 (32 bits) integer
const value = Buffer.readUInt32LE();
Or, more concisely:
const value = Buffer.from(array.slice(-4)).readUInt32LE();
It should be more efficient to just create an Uint32Array view on the same ArrayBuffer and accessing the 32-bit number directly:
var uint8array = new Uint8Array([1,2,3,4,5,6,7,8]);
var uint32array = new Uint32Array(
uint8array.buffer,
uint8array.byteOffset + uint8array.byteLength - 4,
1 // 4Bytes long
);
return uint32array[0];
var a = Uint8Array(6)
a.set([1,2,8,0,0,1])
i1 = a[a.length-4];
i2 = a[a.length-3];
i3 = a[a.length-2];
i4 = a[a.length-1];
console.log(i1<<24 | i2<<16 | i3<<8 | i4);
It's a shame there are not build in ways to do this.
I needed to read variables of variable sizes so based on Imortenson answer I've wrote this little function where p is read position and s is number of bytes to read:
function readUInt(arr, p, s) {
var r = 0;
for (var i = s-1; i >= 0; i--) {
r |= arr[p + i] << (i * 8);
} return r >>> 0;
}
var iable = readUint(arr, arr.length - 4, 4);