I am trying to get every single combination of elements into an array. Now I can use the method below, and remove the duplicates, but this way is far to slow for my use.
The code below would find every possible combination for 2 digits below 4. Now in the code I actually want to use this for, the least possible code would be 6 for loops (within each other) with the amount being 18 (rememeber this is the minimum).
The code below would execute amount^[amount of for loops], or amount^2 which in this case is 16. That means that in the code I want to use this for, it executes 18^6 times, or 34 million times. And this is the minimum, which would get much higher.
After trying to run my code (with 6 foor loops in which amount = 18), it crashed my browser... My question is: Is there any faster and more efficient (not elegant. I don't care how elegant it is) in which my browser won't crash?
Note: This question is not a duplicate question. All the other questions simply ask for a way to do this, however I already have a way. I am just trying to make it more efficient and faster so that it actually works correctly.
let combinations = [];
let amount = 4;
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
combinations.push(`${a}${b}`);
}
}
console.log(combinations);
Below is a snippet providing a possible example for how my code would work.
let possibilities = [];
let amount = 6; //Amount is set by me, so don't worry about it being incorrect
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
possibilities.push(a + b);
}
}
possibilities = [...new Set(possibilities)]; //Removes duplicates
possibilities.sort((a, b) => b - a); //Sorts in descending order
possibilities = possibilities.slice(0, 3); //Gets top 3 values
console.log(possibilities);
Ok, as discussed in the comments, if you need top 3 values for a particular amount, you could just do something simple like below:
let amount = 6;
let highest = amount - 1,second_highest = amount - 2,third_highest = amount - 3;
let possibilities = [
highest + highest,
highest + second_highest,
highest + third_highest
];
console.log(possibilities);
I don't know the any better solution for this, but yes there are some conditions you need to check first.
If(amount <= 0) return 'Invalid amount, Please enter a valid amount"
So if somebody enters a negative or zero value your loop will goes into infinite loop, and make the situation more worst.
if(amount === 1) return '1 possible combination'
As amount less than 1 is 0 only and combinations for 0 is 1 only, you need not to parse whole loop for 6 digits or n digits for 0 so it will get solve in complexity of 1 instead of N(no. of digits).
And for amount greater then 1 you can create manual loops, like here you created 2 loops for 2 digits, you create 6 loops for 6 digits, better create dynamic logic for this to create number of loops automatically.
You need to consider 1111, 1112 this type of combinations as well right?
Or if only 1234, 2134, 2314 this kind of scenarios are required? This can be done in very less complexity.
For duplication you can store combinations as a key value pair. and then Object.Keys will be your combinations.
I have an array of numbers with 64 indexes (it's canvas image data).
I want to know if my array contains only zero's or anything other than zero.
We can return a boolean upon the first encounter of any number greater than zero (even if the very last index is non-zero and all the others are zero, we should return true).
What is the most efficient way to determine this?
Of course, we could loop over our array (focus on the testImageData function):
// Setup
var imgData = {
data: new Array(64)
};
imgData.data.fill(0);
// Set last pixel to black
imgData.data[imgData.data.length - 1] = 255;
// The part in question...
function testImageData(img_data) {
var retval = false;
for (var i = 0; i < img_data.data.length; i++) {
if (img_data.data[i] > 0) {
retval = true;
break;
}
}
return retval;
}
var result = testImageData(imgData);
...but this could take a while if my array were bigger.
Is there a more efficient way to test if any index in the array is greater than zero?
I am open to answers using lodash, though I am not using lodash in this project. I would rather the answer be native JavaScript, either ES5 or ES6. I'm going to ignore any jQuery answers, just saying...
Update
I setup a test for various ways to check for a non-zero value in an array, and the results were interesting.
Here is the JSPerf Link
Note, the Array.some test was much slower than using for (index) and even for-in. The fastest, of course, was for(index) for(let i = 0; i < arr.length; i++)....
You should note that I also tested a Regex solution, just to see how it compared. If you run the tests, you will find that the Regex solution is much, much slower (not surprising), but still very interesting.
I would like to see if there is a solution that could be accomplished using bitwise operators. If you feel up to it, I would like to see your approach.
Your for loop is the fastest way on Chrome 64 with Windows 10.
I've tested against two other options, here is the link to the test so you can run them on your environment.
My results are:
// 10776 operations per second (the best)
for (let i = 0; i < arr.length; i++) {
if (arr[i] !== 0) {
break
}
}
// 4131 operations per second
for (const n of arr) {
if (n !== 0) {
break
}
}
// 821 operations per second (the worst)
arr.some(x => x)
There is no faster way than looping through every element in the array. logically in the worst case scenario the last pixel in your array is black, so you have to check all of them. The best algorithm therefore can only have a O(n) runtime. Best thing you can do is write a loop that breaks early upon finding a non-white pixel.
We are trying to create a random number generator to create serial numbers for products on a virtual assembly line.
We got the random numbers to generate, however since they are serial numbers we don't want it to create duplicates.
Is there a way that it can go back and check to see if the number generated has already been generated, and then to tell it that if it is a duplicate to generate a new number, and to repeat this process until it has a "unique" number.
The point of a serial number is that they're NOT random. Serial, by definition, means that something is arranged in a series. Why not just use an incrementing number?
The easiest way to fix this problem is to avoid it. Use something that is monotonically increasing (like time) to form part of your serial number. To that you can prepend some fixed value that identifies the line or something.
So your serial number format could be NNNNYYYYMMDDHHMMSS, where NNNN is a 4-digit line number and YYYY is the 4 digit year, MM is a 2 digit month, ...
If you can produce multiple things per second per line, then add date components until you get to the point where only one per unit time is possible -- or simply add the count of items produced this day to the YYYYMMDD component (e.g., NNNNYYYYMMDDCCCCCC).
With a truly random number you would have to store the entire collection and review it for each number. Obviously this would mean that your generation would become slower and slower the larger the number of keys you generate (since it would have to retry more and more often and compare to a larger dataset).
This is entirely why truly random numbers just are never used for this purpose. For serial numbers the standard is always to just do a sequential number - is there any real real for them to be random?
Unique IDs are NEVER random - GUIDs and the like are based on the system time and (most often) MAC address. They're globally unique because of the algorithm used and the machine specifics - not because of the size of the value or any level of randomness.
Personally I would do everything I could to either use a sequential value (perhaps with a unique prefix if you have multiple channels) or, better, use a real GUID for your purpose.
is this what you are looking for?
var rArray;
function fillArray (range)
{
rArray = new Array ();
for(var x = 0; x < range; x++)
rArray [x] = x;
}
function randomND (range)
{
if (rArray == null || rArray.length < 1)
fillArray (range);
var pos = Math.floor(Math.random()*rArray.length);
var ran = rArray [pos];
for(var x = pos; x < rArray.length; x++)
rArray [x] = rArray [x+1];
var tempArray = new Array (rArray.length-1)
for(var x = 0; x < tempArray.length; x++)
tempArray [x] = rArray [x];
rArray = tempArray;
return ran;
}
I was helping somebody out with his JavaScript code and my eyes were caught by a section that looked like that:
function randOrd(){
return (Math.round(Math.random())-0.5);
}
coords.sort(randOrd);
alert(coords);
My first though was: hey, this can't possibly work! But then I did some experimenting and found that it indeed at least seems to provide nicely randomized results.
Then I did some web search and almost at the top found an article from which this code was most ceartanly copied. Looked like a pretty respectable site and author...
But my gut feeling tells me, that this must be wrong. Especially as the sorting algorithm is not specified by ECMA standard. I think different sorting algoritms will result in different non-uniform shuffles. Some sorting algorithms may probably even loop infinitely...
But what do you think?
And as another question... how would I now go and measure how random the results of this shuffling technique are?
update: I did some measurements and posted the results below as one of the answers.
After Jon has already covered the theory, here's an implementation:
function shuffle(array) {
var tmp, current, top = array.length;
if(top) while(--top) {
current = Math.floor(Math.random() * (top + 1));
tmp = array[current];
array[current] = array[top];
array[top] = tmp;
}
return array;
}
The algorithm is O(n), whereas sorting should be O(n log n). Depending on the overhead of executing JS code compared to the native sort() function, this might lead to a noticable difference in performance which should increase with array sizes.
In the comments to bobobobo's answer, I stated that the algorithm in question might not produce evenly distributed probabilities (depending on the implementation of sort()).
My argument goes along these lines: A sorting algorithm requires a certain number c of comparisons, eg c = n(n-1)/2 for Bubblesort. Our random comparison function makes the outcome of each comparison equally likely, ie there are 2^c equally probable results. Now, each result has to correspond to one of the n! permutations of the array's entries, which makes an even distribution impossible in the general case. (This is a simplification, as the actual number of comparisons neeeded depends on the input array, but the assertion should still hold.)
As Jon pointed out, this alone is no reason to prefer Fisher-Yates over using sort(), as the random number generator will also map a finite number of pseudo-random values to the n! permutations. But the results of Fisher-Yates should still be better:
Math.random() produces a pseudo-random number in the range [0;1[. As JS uses double-precision floating point values, this corresponds to 2^x possible values where 52 ≤ x ≤ 63 (I'm too lazy to find the actual number). A probability distribution generated using Math.random() will stop behaving well if the number of atomic events is of the same order of magnitude.
When using Fisher-Yates, the relevant parameter is the size of the array, which should never approach 2^52 due to practical limitations.
When sorting with a random comparision function, the function basically only cares if the return value is positive or negative, so this will never be a problem. But there is a similar one: Because the comparison function is well-behaved, the 2^c possible results are, as stated, equally probable. If c ~ n log n then 2^c ~ n^(a·n) where a = const, which makes it at least possible that 2^c is of same magnitude as (or even less than) n! and thus leading to an uneven distribution, even if the sorting algorithm where to map onto the permutaions evenly. If this has any practical impact is beyond me.
The real problem is that the sorting algorithms are not guaranteed to map onto the permutations evenly. It's easy to see that Mergesort does as it's symmetric, but reasoning about something like Bubblesort or, more importantly, Quicksort or Heapsort, is not.
The bottom line: As long as sort() uses Mergesort, you should be reasonably safe except in corner cases (at least I'm hoping that 2^c ≤ n! is a corner case), if not, all bets are off.
It's never been my favourite way of shuffling, partly because it is implementation-specific as you say. In particular, I seem to remember that the standard library sorting from either Java or .NET (not sure which) can often detect if you end up with an inconsistent comparison between some elements (e.g. you first claim A < B and B < C, but then C < A).
It also ends up as a more complex (in terms of execution time) shuffle than you really need.
I prefer the shuffle algorithm which effectively partitions the collection into "shuffled" (at the start of the collection, initially empty) and "unshuffled" (the rest of the collection). At each step of the algorithm, pick a random unshuffled element (which could be the first one) and swap it with the first unshuffled element - then treat it as shuffled (i.e. mentally move the partition to include it).
This is O(n) and only requires n-1 calls to the random number generator, which is nice. It also produces a genuine shuffle - any element has a 1/n chance of ending up in each space, regardless of its original position (assuming a reasonable RNG). The sorted version approximates to an even distribution (assuming that the random number generator doesn't pick the same value twice, which is highly unlikely if it's returning random doubles) but I find it easier to reason about the shuffle version :)
This approach is called a Fisher-Yates shuffle.
I would regard it as a best practice to code up this shuffle once and reuse it everywhere you need to shuffle items. Then you don't need to worry about sort implementations in terms of reliability or complexity. It's only a few lines of code (which I won't attempt in JavaScript!)
The Wikipedia article on shuffling (and in particular the shuffle algorithms section) talks about sorting a random projection - it's worth reading the section on poor implementations of shuffling in general, so you know what to avoid.
I did some measurements of how random the results of this random sort are...
My technique was to take a small array [1,2,3,4] and create all (4! = 24) permutations of it. Then I would apply the shuffling function to the array a large number of times and count how many times each permutation is generated. A good shuffling algoritm would distribute the results quite evenly over all the permutations, while a bad one would not create that uniform result.
Using the code below I tested in Firefox, Opera, Chrome, IE6/7/8.
Surprisingly for me, the random sort and the real shuffle both created equally uniform distributions. So it seems that (as many have suggested) the main browsers are using merge sort. This of course doesn't mean, that there can't be a browser out there, that does differently, but I would say it means, that this random-sort-method is reliable enough to use in practice.
EDIT: This test didn't really measured correctly the randomness or lack thereof. See the other answer I posted.
But on the performance side the shuffle function given by Cristoph was a clear winner. Even for small four-element arrays the real shuffle performed about twice as fast as random-sort!
// The shuffle function posted by Cristoph.
var shuffle = function(array) {
var tmp, current, top = array.length;
if(top) while(--top) {
current = Math.floor(Math.random() * (top + 1));
tmp = array[current];
array[current] = array[top];
array[top] = tmp;
}
return array;
};
// the random sort function
var rnd = function() {
return Math.round(Math.random())-0.5;
};
var randSort = function(A) {
return A.sort(rnd);
};
var permutations = function(A) {
if (A.length == 1) {
return [A];
}
else {
var perms = [];
for (var i=0; i<A.length; i++) {
var x = A.slice(i, i+1);
var xs = A.slice(0, i).concat(A.slice(i+1));
var subperms = permutations(xs);
for (var j=0; j<subperms.length; j++) {
perms.push(x.concat(subperms[j]));
}
}
return perms;
}
};
var test = function(A, iterations, func) {
// init permutations
var stats = {};
var perms = permutations(A);
for (var i in perms){
stats[""+perms[i]] = 0;
}
// shuffle many times and gather stats
var start=new Date();
for (var i=0; i<iterations; i++) {
var shuffled = func(A);
stats[""+shuffled]++;
}
var end=new Date();
// format result
var arr=[];
for (var i in stats) {
arr.push(i+" "+stats[i]);
}
return arr.join("\n")+"\n\nTime taken: " + ((end - start)/1000) + " seconds.";
};
alert("random sort: " + test([1,2,3,4], 100000, randSort));
alert("shuffle: " + test([1,2,3,4], 100000, shuffle));
Interestingly, Microsoft used the same technique in their pick-random-browser-page.
They used a slightly different comparison function:
function RandomSort(a,b) {
return (0.5 - Math.random());
}
Looks almost the same to me, but it turned out to be not so random...
So I made some testruns again with the same methodology used in the linked article, and indeed - turned out that the random-sorting-method produced flawed results. New test code here:
function shuffle(arr) {
arr.sort(function(a,b) {
return (0.5 - Math.random());
});
}
function shuffle2(arr) {
arr.sort(function(a,b) {
return (Math.round(Math.random())-0.5);
});
}
function shuffle3(array) {
var tmp, current, top = array.length;
if(top) while(--top) {
current = Math.floor(Math.random() * (top + 1));
tmp = array[current];
array[current] = array[top];
array[top] = tmp;
}
return array;
}
var counts = [
[0,0,0,0,0],
[0,0,0,0,0],
[0,0,0,0,0],
[0,0,0,0,0],
[0,0,0,0,0]
];
var arr;
for (var i=0; i<100000; i++) {
arr = [0,1,2,3,4];
shuffle3(arr);
arr.forEach(function(x, i){ counts[x][i]++;});
}
alert(counts.map(function(a){return a.join(", ");}).join("\n"));
I have placed a simple test page on my website showing the bias of your current browser versus other popular browsers using different methods to shuffle. It shows the terrible bias of just using Math.random()-0.5, another 'random' shuffle that isn't biased, and the Fisher-Yates method mentioned above.
You can see that on some browsers there is as high as a 50% chance that certain elements will not change place at all during the 'shuffle'!
Note: you can make the implementation of the Fisher-Yates shuffle by #Christoph slightly faster for Safari by changing the code to:
function shuffle(array) {
for (var tmp, cur, top=array.length; top--;){
cur = (Math.random() * (top + 1)) << 0;
tmp = array[cur]; array[cur] = array[top]; array[top] = tmp;
}
return array;
}
Test results: http://jsperf.com/optimized-fisher-yates
I think it's fine for cases where you're not picky about distribution and you want the source code to be small.
In JavaScript (where the source is transmitted constantly), small makes a difference in bandwidth costs.
It's been four years, but I'd like to point out that the random comparator method won't be correctly distributed, no matter what sorting algorithm you use.
Proof:
For an array of n elements, there are exactly n! permutations (i.e. possible shuffles).
Every comparison during a shuffle is a choice between two sets of permutations. For a random comparator, there is a 1/2 chance of choosing each set.
Thus, for each permutation p, the chance of ending up with permutation p is a fraction with denominator 2^k (for some k), because it is a sum of such fractions (e.g. 1/8 + 1/16 = 3/16).
For n = 3, there are six equally-likely permutations. The chance of each permutation, then, is 1/6. 1/6 can't be expressed as a fraction with a power of 2 as its denominator.
Therefore, the coin flip sort will never result in a fair distribution of shuffles.
The only sizes that could possibly be correctly distributed are n=0,1,2.
As an exercise, try drawing out the decision tree of different sort algorithms for n=3.
There is a gap in the proof: If a sort algorithm depends on the consistency of the comparator, and has unbounded runtime with an inconsistent comparator, it can have an infinite sum of probabilities, which is allowed to add up to 1/6 even if every denominator in the sum is a power of 2. Try to find one.
Also, if a comparator has a fixed chance of giving either answer (e.g. (Math.random() < P)*2 - 1, for constant P), the above proof holds. If the comparator instead changes its odds based on previous answers, it may be possible to generate fair results. Finding such a comparator for a given sorting algorithm could be a research paper.
It is a hack, certainly. In practice, an infinitely looping algorithm is not likely.
If you're sorting objects, you could loop through the coords array and do something like:
for (var i = 0; i < coords.length; i++)
coords[i].sortValue = Math.random();
coords.sort(useSortValue)
function useSortValue(a, b)
{
return a.sortValue - b.sortValue;
}
(and then loop through them again to remove the sortValue)
Still a hack though. If you want to do it nicely, you have to do it the hard way :)
If you're using D3 there is a built-in shuffle function (using Fisher-Yates):
var days = ['Lundi','Mardi','Mercredi','Jeudi','Vendredi','Samedi','Dimanche'];
d3.shuffle(days);
And here is Mike going into details about it:
http://bost.ocks.org/mike/shuffle/
No, it is not correct. As other answers have noted, it will lead to a non-uniform shuffle and the quality of the shuffle will also depend on which sorting algorithm the browser uses.
Now, that might not sound too bad to you, because even if theoretically the distribution is not uniform, in practice it's probably nearly uniform, right? Well, no, not even close. The following charts show heat-maps of which indices each element gets shuffled to, in Chrome and Firefox respectively: if the pixel (i, j) is green, it means the element at index i gets shuffled to index j too often, and if it's red then it gets shuffled there too rarely.
These screenshots are taken from Mike Bostock's page on this subject.
As you can see, shuffling using a random comparator is severely biased in Chrome and even more so in Firefox. In particular, both have a lot of green along the diagonal, meaning that too many elements get "shuffled" somewhere very close to where they were in the original sequence. In comparison, a similar chart for an unbiased shuffle (e.g. using the Fisher-Yates algorithm) would be all pale yellow with just a small amount of random noise.
Here's an approach that uses a single array:
The basic logic is:
Starting with an array of n elements
Remove a random element from the array and push it onto the array
Remove a random element from the first n - 1 elements of the array and push it onto the array
Remove a random element from the first n - 2 elements of the array and push it onto the array
...
Remove the first element of the array and push it onto the array
Code:
for(i=a.length;i--;) a.push(a.splice(Math.floor(Math.random() * (i + 1)),1)[0]);
Can you use the Array.sort() function to shuffle an array – Yes.
Are the results random enough – No.
Consider the following code snippet:
/*
* The following code sample shuffles an array using Math.random() trick
* After shuffling, the new position of each item is recorded
* The process is repeated 100 times
* The result is printed out, listing each item and the number of times
* it appeared on a given position after shuffling
*/
var array = ["a", "b", "c", "d", "e"];
var stats = {};
array.forEach(function(v) {
stats[v] = Array(array.length).fill(0);
});
var i, clone;
for (i = 0; i < 100; i++) {
clone = array.slice();
clone.sort(function() {
return Math.random() - 0.5;
});
clone.forEach(function(v, i) {
stats[v][i]++;
});
}
Object.keys(stats).forEach(function(v, i) {
console.log(v + ": [" + stats[v].join(", ") + "]");
});
Sample output:
a: [29, 38, 20, 6, 7]
b: [29, 33, 22, 11, 5]
c: [17, 14, 32, 17, 20]
d: [16, 9, 17, 35, 23]
e: [ 9, 6, 9, 31, 45]
Ideally, the counts should be evenly distributed (for the above example, all counts should be around 20). But they are not. Apparently, the distribution depends on what sorting algorithm is implemented by the browser and how it iterates the array items for sorting.
There is nothing wrong with it.
The function you pass to .sort() usually looks something like
function sortingFunc( first, second )
{
// example:
return first - second ;
}
Your job in sortingFunc is to return:
a negative number if first goes before second
a positive number if first should go after second
and 0 if they are completely equal
The above sorting function puts things in order.
If you return -'s and +'s randomly as what you have, you get a random ordering.
Like in MySQL:
SELECT * from table ORDER BY rand()