Related
I have an array of following strings:
['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
...etc.
I need a solution that will give me following ordered result
['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0'].
I tried to implement a sort so it first sorts by the numbers in the first position, than in case of equality, sort by the numbers in the second position (after the first dot), and so on...
I tried using sort() and localeCompare(), but if I have elements '4.5.0' and '4.11.0', I get them sorted as ['4.11.0','4.5.0'], but I need to get ['4.5.0','4.11.0'].
How can I achieve this?
You could prepend all parts to fixed size strings, then sort that, and finally remove the padding again.
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.split('.').map( n => +n+100000 ).join('.') ).sort()
.map( a => a.split('.').map( n => +n-100000 ).join('.') );
console.log(arr)
Obviously you have to choose the size of the number 100000 wisely: it should have at least one more digit than your largest number part will ever have.
With regular expression
The same manipulation can be achieved without having to split & join, when you use the callback argument to the replace method:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.replace(/\d+/g, n => +n+100000 ) ).sort()
.map( a => a.replace(/\d+/g, n => +n-100000 ) );
console.log(arr)
Defining the padding function once only
As both the padding and its reverse functions are so similar, it seemed a nice exercise to use one function f for both, with an extra argument defining the "direction" (1=padding, -1=unpadding). This resulted in this quite obscure, and extreme code. Consider this just for fun, not for real use:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = (f=>f(f(arr,1).sort(),-1)) ((arr,v)=>arr.map(a=>a.replace(/\d+/g,n=>+n+v*100000)));
console.log(arr);
Use the sort compare callback function
You could use the compare function argument of sort to achieve the same:
arr.sort( (a, b) => a.replace(/\d+/g, n => +n+100000 )
.localeCompare(b.replace(/\d+/g, n => +n+100000 )) );
But for larger arrays this will lead to slower performance. This is because the sorting algorithm will often need to compare a certain value several times, each time with a different value from the array. This means that the padding will have to be executed multiple times for the same number. For this reason, it will be faster for larger arrays to first apply the padding in the whole array, then use the standard sort, and then remove the padding again.
But for shorter arrays, this approach might still be the fastest. In that case, the so-called natural sort option -- that can be achieved with the extra arguments of localeCompare -- will be more efficient than the padding method:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.sort( (a, b) => a.localeCompare(b, undefined, { numeric:true }) );
console.log(arr);
More about the padding and unary plus
To see how the padding works, look at the intermediate result it generates:
[ "100005.100005.100001", "100004.100021.100000", "100004.100022.100000",
"100006.100001.100000", "100005.100001.100000" ]
Concerning the expression +n+100000, note that the first + is the unary plus and is the most efficient way to convert a string-encoded decimal number to its numerical equivalent. The 100000 is added to make the number have a fixed number of digits. Of course, it could just as well be 200000 or 300000. Note that this addition does not change the order the numbers will have when they would be sorted numerically.
The above is just one way to pad a string. See this Q&A for some other alternatives.
If you are looking for a npm package to compare two semver version, https://www.npmjs.com/package/compare-versions is the one.
Then you can sort version like this:
// ES6/TypeScript
import compareVersions from 'compare-versions';
var versions = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var sorted = versions.sort(compareVersions);
You could split the strings and compare the parts.
function customSort(data, order) {
function isNumber(v) {
return (+v).toString() === v;
}
var sort = {
asc: function (a, b) {
var i = 0,
l = Math.min(a.value.length, b.value.length);
while (i < l && a.value[i] === b.value[i]) {
i++;
}
if (i === l) {
return a.value.length - b.value.length;
}
if (isNumber(a.value[i]) && isNumber(b.value[i])) {
return a.value[i] - b.value[i];
}
return a.value[i].localeCompare(b.value[i]);
},
desc: function (a, b) {
return sort.asc(b, a);
}
}
var mapped = data.map(function (el, i) {
return {
index: i,
value: el.split('')
};
});
mapped.sort(sort[order] || sort.asc);
return mapped.map(function (el) {
return data[el.index];
});
}
var array = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0'];
console.log('sorted array asc', customSort(array));
console.log('sorted array desc ', customSort(array, 'desc'));
console.log('original array ', array);
.as-console-wrapper { max-height: 100% !important; top: 0; }
You can check in loop if values are different, return difference, else continue
var a=['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
a.sort(function(a,b){
var a1 = a.split('.');
var b1 = b.split('.');
var len = Math.max(a1.length, b1.length);
for(var i = 0; i< len; i++){
var _a = +a1[i] || 0;
var _b = +b1[i] || 0;
if(_a === _b) continue;
else return _a > _b ? 1 : -1
}
return 0;
})
console.log(a)
Though slightly late this would be my solution;
var arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"],
sorted = arr.sort((a,b) => {var aa = a.split("."),
ba = b.split(".");
return +aa[0] < +ba[0] ? -1
: aa[0] === ba[0] ? +aa[1] < +ba[1] ? -1
: aa[1] === ba[1] ? +aa[2] < +ba[2] ? -1
: 1
: 1
: 1;
});
console.log(sorted);
Here's a solution I developed based on #trincot's that will sort by semver even if the strings aren't exactly "1.2.3" - they could be i.e. "v1.2.3" or "2.4"
function sortSemVer(arr, reverse = false) {
let semVerArr = arr.map(i => i.replace(/(\d+)/g, m => +m + 100000)).sort(); // +m is just a short way of converting the match to int
if (reverse)
semVerArr = semVerArr.reverse();
return semVerArr.map(i => i.replace(/(\d+)/g, m => +m - 100000))
}
console.log(sortSemVer(["1.0.1", "1.0.9", "1.0.10"]))
console.log(sortSemVer(["v2.1", "v2.0.9", "v2.0.12", "v2.2"], true))
This seems to work provided there are only digits between the dots:
var a = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
a = a.map(function (x) {
return x.split('.').map(function (x) {
return parseInt(x)
})
}).sort(function (a, b) {
var i = 0, m = a.length, n = b.length, o, d
o = m < n ? n : m
for (; i < o; ++i) {
d = (a[i] || 0) - (b[i] || 0)
if (d) return d
}
return 0
}).map(function (x) {
return x.join('.')
})
'use strict';
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
Array.prototype.versionSort = function () {
var arr = this;
function isNexVersionBigger (v1, v2) {
var a1 = v1.split('.');
var b2 = v2.split('.');
var len = a1.length > b2.length ? a1.length : b2.length;
for (var k = 0; k < len; k++) {
var a = a1[k] || 0;
var b = b2[k] || 0;
if (a === b) {
continue;
} else
return b < a;
}
}
for (var i = 0; i < arr.length; i++) {
var min_i = i;
for (var j = i + 1; j < arr.length; j++) {
if (isNexVersionBigger(arr[i], arr[j])) {
min_i = j;
}
}
var temp = arr[i];
arr[i] = arr[min_i];
arr[min_i] = temp;
}
return arr;
}
console.log(arr.versionSort());
This solution accounts for version numbers that might not be in the full, 3-part format (for example, if one of the version numbers is just 2 or 2.0 or 0.1, etc).
The custom sort function I wrote is probably mostly what you're looking for, it just needs an array of objects in the format {"major":X, "minor":X, "revision":X}:
var versionArr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var versionObjectArr = [];
var finalVersionArr = [];
/*
split each version number string by the '.' and separate them in an
object by part (major, minor, & revision). If version number is not
already in full, 3-part format, -1 will represent that part of the
version number that didn't exist. Push the object into an array that
can be sorted.
*/
for(var i = 0; i < versionArr.length; i++){
var splitVersionNum = versionArr[i].split('.');
var versionObj = {};
switch(splitVersionNum.length){
case 1:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":-1,
"revision":-1
};
break;
case 2:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":-1
};
break;
case 3:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":parseInt(splitVersionNum[2])
};
}
versionObjectArr.push(versionObj);
}
//sort objects by parts, going from major to minor to revision number.
versionObjectArr.sort(function(a, b){
if(a.major < b.major) return -1;
else if(a.major > b.major) return 1;
else {
if(a.minor < b.minor) return -1;
else if(a.minor > b.minor) return 1;
else {
if(a.revision < b.revision) return -1;
else if(a.revision > b.revision) return 1;
}
}
});
/*
loops through sorted object array to recombine it's version keys to match the original string's value. If any trailing parts of the version
number are less than 0 (i.e. they didn't exist so we replaced them with
-1) then leave that part of the version number string blank.
*/
for(var i = 0; i < versionObjectArr.length; i++){
var versionStr = "";
for(var key in versionObjectArr[i]){
versionStr = versionObjectArr[i].major;
versionStr += (versionObjectArr[i].minor < 0 ? '' : "." + versionObjectArr[i].minor);
versionStr += (versionObjectArr[i].revision < 0 ? '' : "." + versionObjectArr[i].revision);
}
finalVersionArr.push(versionStr);
}
console.log('Original Array: ',versionArr);
console.log('Expected Output: ',['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0']);
console.log('Actual Output: ', finalVersionArr);
Inspired from the accepted answer, but ECMA5-compatible, and with regular string padding (see my comments on the answer):
function sortCallback(a, b) {
function padParts(version) {
return version
.split('.')
.map(function (part) {
return '00000000'.substr(0, 8 - part.length) + part;
})
.join('.');
}
a = padParts(a);
b = padParts(b);
return a.localeCompare(b);
}
Usage:
['1.1', '1.0'].sort(sortCallback);
const arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"];
const sorted = arr.sort((a,b) => {
const ba = b.split('.');
const d = a.split('.').map((a1,i)=>a1-ba[i]);
return d[0] ? d[0] : d[1] ? d[1] : d[2]
});
console.log(sorted);
This can be in an easier way using the sort method without hardcoding any numbers and in a more generic way.
enter code here
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
splitArray = arr.map(elements => elements.split('.'))
//now lets sort based on the elements on the corresponding index of each array
//mapped.sort(function(a, b) {
// if (a.value > b.value) {
// return 1;
// }
// if (a.value < b.value) {
// return -1;
// }
// return 0;
//});
//here we compare the first element with the first element of the next version number and that is [5.1.2,5.7.2] 5,5 and 1,7 and 2,2 are compared to identify the smaller version...In the end use the join() to get back the version numbers in the proper format.
sortedArray = splitArray.sort((a, b) => {
for (i in a) {
if (parseInt(a[i]) < parseInt(b[i])) {
return -1;
break
}
if (parseInt(a[i]) > parseInt(b[i])) {
return +1;
break
} else {
continue
}
}
}).map(p => p.join('.'))
sortedArray = ["5.1.0", "5.1.1", "5.1.1", "5.1.2", "5.7.2.2"]
sort 1.0a notation correct
use native localeCompare to sort 1.090 notation
function log(label,val){
document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}
const sortVersions = (
x,
v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/)
? v(b) < v(a) ? 1 : -1
: a.localeCompare(b, 0, {numeric: true}))
let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1","1.0a"];
log(' input : ',v);
log('sorted: ',sortVersions(v));
log('no dups:',[...new Set(sortVersions(v))]);
In ES6 you can go without regex.
const versions = ["0.4", "0.11", "0.4.1", "0.4", "0.4.2", "2.0.1","2", "0.0.1", "0.2.3"];
const splitted = versions.map(version =>
version
.split('.')
.map(i => +i))
.map(i => {
let items;
if (i.length === 1) {
items = [0, 0]
i.push(...items)
}
if (i.length === 2) {
items = [0]
i.push(...items)
}
return i
})
.sort((a, b) => {
for(i in a) {
if (a[i] < b[i]) {
return -1;
}
if (a[i] > b[i]) {
return +1;
}
}
})
.map(item => item.join('.'))
const sorted = [...new Set(splitted)]
If ES6 I do this:
versions.sort((v1, v2) => {
let [, major1, minor1, revision1 = 0] = v1.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
let [, major2, minor2, revision2 = 0] = v2.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
if (major1 != major2) return parseInt(major1) - parseInt(major2);
if (minor1 != minor2) return parseInt(minor1) - parseInt(major2);
return parseInt(revision1) - parseInt(revision2);
});
**Sorted Array Object by dotted version value**
var sampleData = [
{ name: 'Edward', value: '2.1.2' },
{ name: 'Sharpe', value: '2.1.3' },
{ name: 'And', value: '2.2.1' },
{ name: 'The', value: '2.1' },
{ name: 'Magnetic', value: '2.2' },
{ name: 'Zeros', value: '0' },
{ name: 'Zeros', value: '1' }
];
arr = sampleData.map( a => a.value).sort();
var requireData = [];
arr.forEach(function(record, index){
var findRecord = sampleData.find(arr => arr.value === record);
if(findRecord){
requireData.push(findRecord);
}
});
console.log(requireData);
[check on jsfiddle.net][1]
[1]: https://jsfiddle.net/jx3buswq/2/
It is corrected now!!!
Running a test on my code. It should return 0 if the property passed into the function does not exist. But its not returning anything. Did I make a typo?
var obj = {
key: [1, 2, 3]
};
function getAverageOfElementsAtProperty(obj, key) {
if (obj[key].length === 0 || Array.isArray(obj[key]) === false || obj.hasOwnProperty(key) === false) {
return 0;
}
var average = 0;
for (var i = 0; i < obj[key].length; i++) {
average += obj[key][i];
}
average /= obj[key].length;
return average;
}
console.log(getAverageOfElementsAtProperty(obj, 'notKey'));
If you don't pass an array to your function, it fails when it tries to get the length of the object that was passed, so that shouldn't be how you test initially. You just need to see if the property exists and that is done by simply attempting to access the property in an if condition.
Now, if you are going to add tests to see if the property does exist and that it is an array, then you have to add more tests to check the items in the array to see if they are numbers, otherwise trying to get a numerical average will fail.
Since there are really two things to do (check if the property is there and get math average), I would break this into two functions:
var obj1 = {
key: [1, 2, 3]
};
var obj2 = {
key: [1, "test", 3]
};
function getAverageOfElementsAtProperty(obj, key) {
if (obj[key] && Array.isArray(obj[key])) {
// Ok, we have the right property and there is an array there,
// now we have to test that each item in the array is a number
var numbers = obj[key].every(function (currentValue) {
return typeof currentValue === "number";
});
// Return the average if we have all numbers or 0 if not
return numbers ? getAverage(obj[key]) : 0;
} else {
return 0;
}
}
function getAverage(arr){
var average = 0;
// Array.forEach() is much simpler than counting loops
arr.forEach(function(item) {
average += item;
});
return average / arr.length;
}
console.log(getAverageOfElementsAtProperty(obj1, 'notkey')); // 0
console.log(getAverageOfElementsAtProperty(obj1, 'key')); // 2
console.log(getAverageOfElementsAtProperty(obj2, 'key')); // 0 - obj2 does not contain all numbers
Since obj['notKey'] does not exist, it does not return an array. Therefore you cannot do .length of undefined. I would change it to typeof to see if its defined or not.
var obj = {
key: [1, 2, 3]
};
function getAverageOfElementsAtProperty(obj, key) {
if (typeof obj[key] == 'undefined' || Array.isArray(obj[key]) === false || obj.hasOwnProperty(key) === false) {
return 0;
}
var average = 0;
for (var i = 0; i < obj[key].length; i++) {
average += obj[key][i];
}
average /= obj[key].length;
return average;
}
console.log(getAverageOfElementsAtProperty(obj, 'notKey'));
I have an array of arrays as follows:
[[3, 4], [1, 2], [3, 4]]
I wish to create a new array of arrays that has no duplicates, and has a count of the number of occurrences of each element in the first array:
[[3,4,2], [1,2,1]]
here is what I have so far:
var alreadyAdded = 0;
dataset.forEach(function(data) {
From = data[0];
To = data[1];
index = 0;
newDataSet.forEach(function(newdata) {
newFrom = newData[0];
newTo = newData[1];
// check if the point we are looking for is already added to the new array
if ((From == newFrom) && (To == newTo)) {
// if it is, increment the count for that pair
var count = newData[2];
var newCount = count + 1;
newDataSet[index] = [newFrom, newTo, newCount];
test = "reached here";
alreadyAdded = 1;
}
index++;
});
// the pair was not already added to the new dataset, add it
if (alreadyAdded == 0) {
newDataSet.push([From, To, 1]);
}
// reset alreadyAdded variable
alreadyAdded = 0;
});
I am very new to Javascript, can someone help explain to me what I'm doing wrong? I'm sure there is a more concise way of doing this, however I wasn't able to find an example in javascript that dealt with duplicate array of arrays.
Depending on how large the dataset is that you're iterating over I'd be cautious of looping over it so many times. You can avoid having to do that by creating an 'index' for each element in the original dataset and then using it to reference the elements in your grouping. This is the approach that I took when I solved the problem. You can see it here on jsfiddle. I used Array.prototype.reduce to create an object literal which contained the grouping of elements from the original dataset. Then I iterated over it's keys to create the final grouping.
var dataSet = [[3,4], [1,2], [3,4]],
grouping = [],
counts,
keys,
current;
counts = dataSet.reduce(function(acc, elem) {
var key = elem[0] + ':' + elem[1];
if (!acc.hasOwnProperty(key)) {
acc[key] = {elem: elem, count: 0}
}
acc[key].count += 1;
return acc;
}, {});
keys = Object.keys(counts);
for (var i = 0, l = keys.length; i < l; i++) {
current = counts[keys[i]];
current.elem.push(current.count);
grouping.push(current.elem);
}
console.log(grouping);
Assuming order of sub array items matters, assuming that your sub arrays could be of variable length and could contain items other than numbers, here is a fairly generic way to approach the problem. Requires ECMA5 compatibility as it stands, but would not be hard to make it work on ECMA3.
Javascript
// Create shortcuts for prototype methods
var toClass = Object.prototype.toString.call.bind(Object.prototype.toString),
aSlice = Array.prototype.slice.call.bind(Array.prototype.slice);
// A generic deepEqual defined by commonjs
// http://wiki.commonjs.org/wiki/Unit_Testing/1.0
function deepEqual(a, b) {
if (a === b) {
return true;
}
if (toClass(a) === '[object Date]' && toClass(b) === '[object Date]') {
return a.getTime() === b.getTime();
}
if (toClass(a) === '[object RegExp]' && toClass(b) === '[object RegExp]') {
return a.toString() === b.toString();
}
if (a && typeof a !== 'object' && b && typeof b !== 'object') {
return a == b;
}
if (a.prototype !== b.prototype) {
return false;
}
if (toClass(a) === '[object Arguments]') {
if (toClass(b) !== '[object Arguments]') {
return false;
}
return deepEqual(aSlice(a), aSlice(b));
}
var ka,
kb,
length,
index,
it;
try {
ka = Object.keys(a);
kb = Object.keys(b);
} catch (eDE) {
return false;
}
length = ka.length;
if (length !== kb.length) {
if (Array.isArray(a) && Array.isArray(b)) {
if (a.length !== b.length) {
return false;
}
} else {
return false;
}
} else {
ka.sort();
kb.sort();
for (index = 0; index < length; index += 1) {
if (ka[index] !== kb[index]) {
return false;
}
}
}
for (index = 0; index < length; index += 1) {
it = ka[index];
if (!deepEqual(a[it], b[it])) {
return false;
}
}
return true;
};
// Recursive function for counting arrays as specified
// a must be an array of arrays
// dupsArray is used to keep count when recursing
function countDups(a, dupsArray) {
dupsArray = Array.isArray(dupsArray) ? dupsArray : [];
var copy,
current,
count;
if (a.length) {
copy = a.slice();
current = copy.pop();
count = 1;
copy = copy.filter(function (item) {
var isEqual = deepEqual(current, item);
if (isEqual) {
count += 1;
}
return !isEqual;
});
current.push(count);
dupsArray.push(current);
if (copy.length) {
countDups(copy, dupsArray);
}
}
return dupsArray;
}
var x = [
[3, 4],
[1, 2],
[3, 4]
];
console.log(JSON.stringify(countDups(x)));
Output
[[3,4,2],[1,2,1]]
on jsFiddle
After fixing a typo I tried your solution in the debugger; it works!
Fixed the inner forEach-loop variable name to match case. Also some var-keywords added.
var alreadyAdded = 0;
dataset.forEach(function (data) {
var From = data[0];
var To = data[1];
var index = 0;
newDataSet.forEach(function (newData) {
var newFrom = newData[0];
var newTo = newData[1];
// check if the point we are looking for is already added to the new array
if ((From == newFrom) && (To == newTo)) {
// if it is, increment the count for that pair
var count = newData[2];
var newCount = count + 1;
newDataSet[index] = [newFrom, newTo, newCount];
test = "reached here";
alreadyAdded = 1;
}
index++;
});
// the pair was not already added to the new dataset, add it
if (alreadyAdded == 0) {
newDataSet.push([From, To, 1]);
}
// reset alreadyAdded variable
alreadyAdded = 0;
});
const x = [[3, 4], [1, 2], [3, 4]];
const with_duplicate_count = [
...x
.map(JSON.stringify)
.reduce( (acc, v) => acc.set(v, (acc.get(v) || 0) + 1), new Map() )
.entries()
].map(([k, v]) => JSON.parse(k).concat(v));
console.log(with_duplicate_count);
I'm looking for an efficient way to find out whether two arrays contain same amounts of equal elements (in the == sense), in any order:
foo = {/*some object*/}
bar = {/*some other object*/}
a = [1,2,foo,2,bar,2]
b = [bar,2,2,2,foo,1]
sameElements(a, b) --> true
PS. Note that pretty much every solution in the thread uses === and not == for comparison. This is fine for my needs though.
Update 5
I posted a new answer with a different approach.
Update
I extended the code to have the possibility of either checking by reference or equality
just pass true as second parameter to do a reference check.
Also I added the example to Brunos JSPerf
It runs at about 11 ops/s doing a reference check
I will comment the code as soon(!) as I get some spare time to explain it a bit more, but at the moment don't have the time for that, sry. Done
Update 2.
Like Bruno pointed out in the comments sameElements([NaN],[NaN]) yields false
In my opinion this is the correct behaviour as NaN is ambigious and should always lead to a false result,at least when comparing NaN.equals(NaN). But he had quite a good point.
Whether
[1,2,foo,bar,NaN,3] should be equal to [1,3,foo,NaN,bar,2] or not.
Ok.. honestly I'm a bit torn whether it should or not, so i added two flags.
Number.prototype.equal.NaN
If true
NaN.equals(NaN) //true
Array.prototype.equal.NaN
If true
[NaN].equals([NaN],true) //true
note this is only for reference checks. As a deep check would invoke Number.prototype.equals anyway
Update 3:
Dang i totally missed 2 lines in the sort function.
Added
r[0] = a._srt; //DANG i totally missed this line
r[1] = b._srt; //And this.
Line 105 in the Fiddle
Which is kind of important as it determines the consistent order of the Elements.
Update 4
I tried to optimize the sort function a bit, and managed to get it up to about 20 ops/s.
Below is the updated code, as well as the updated fiddle =)
Also i chose to mark the objects outside the sort function, it doesn't seem to make a performance difference anymore, and its more readable
Here is an approach using Object.defineProperty to add equals functions to
Array,Object,Number,String,Boolean's prototype to avoid typechecking in one function for
performance reasons. As we can recursively call .equals on any element.
But of course checking Objects for equality may cause performance issues in big Objects.
So if anyone feels unpleasant manipulating native prototypes, just do a type check and put it into one function
Object.defineProperty(Boolean.prototype, "equals", {
enumerable: false,
configurable: true,
value: function (c) {
return this == c; //For booleans simply return the equality
}
});
Object.defineProperty(Number.prototype, "equals", {
enumerable: false,
configurable: true,
value: function (c) {
if (Number.prototype.equals.NaN == true && isNaN(this) && c != c) return true; //let NaN equals NaN if flag set
return this == c; // else do a normal compare
}
});
Number.prototype.equals.NaN = false; //Set to true to return true for NaN == NaN
Object.defineProperty(String.prototype, "equals", {
enumerable: false,
configurable: true,
value: Boolean.prototype.equals //the same (now we covered the primitives)
});
Object.defineProperty(Object.prototype, "equals", {
enumerable: false,
configurable: true,
value: function (c, reference) {
if (true === reference) //If its a check by reference
return this === c; //return the result of comparing the reference
if (typeof this != typeof c) {
return false; //if the types don't match (Object equals primitive) immediately return
}
var d = [Object.keys(this), Object.keys(c)],//create an array with the keys of the objects, which get compared
f = d[0].length; //store length of keys of the first obj (we need it later)
if (f !== d[1].length) {//If the Objects differ in the length of their keys
return false; //immediately return
}
for (var e = 0; e < f; e++) { //iterate over the keys of the first object
if (d[0][e] != d[1][e] || !this[d[0][e]].equals(c[d[1][e]])) {
return false; //if either the key name does not match or the value does not match, return false. a call of .equal on 2 primitives simply compares them as e.g Number.prototype.equal gets called
}
}
return true; //everything is equal, return true
}
});
Object.defineProperty(Array.prototype, "equals", {
enumerable: false,
configurable: true,
value: function (c,reference) {
var d = this.length;
if (d != c.length) {
return false;
}
var f = Array.prototype.equals.sort(this.concat());
c = Array.prototype.equals.sort(c.concat(),f)
if (reference){
for (var e = 0; e < d; e++) {
if (f[e] != c[e] && !(Array.prototype.equals.NaN && f[e] != f[e] && c[e] != c[e])) {
return false;
}
}
} else {
for (var e = 0; e < d; e++) {
if (!f[e].equals(c[e])) {
return false;
}
}
}
return true;
}
});
Array.prototype.equals.NaN = false; //Set to true to allow [NaN].equals([NaN]) //true
Object.defineProperty(Array.prototype.equals,"sort",{
enumerable:false,
value:function sort (curr,prev) {
var weight = {
"[object Undefined]":6,
"[object Object]":5,
"[object Null]":4,
"[object String]":3,
"[object Number]":2,
"[object Boolean]":1
}
if (prev) { //mark the objects
for (var i = prev.length,j,t;i>0;i--) {
t = typeof (j = prev[i]);
if (j != null && t === "object") {
j._pos = i;
} else if (t !== "object" && t != "undefined" ) break;
}
}
curr.sort (sorter);
if (prev) {
for (var k = prev.length,l,t;k>0;k--) {
t = typeof (l = prev[k]);
if (t === "object" && l != null) {
delete l._pos;
} else if (t !== "object" && t != "undefined" ) break;
}
}
return curr;
function sorter (a,b) {
var tStr = Object.prototype.toString
var types = [tStr.call(a),tStr.call(b)]
var ret = [0,0];
if (types[0] === types[1] && types[0] === "[object Object]") {
if (prev) return a._pos - b._pos
else {
return a === b ? 0 : 1;
}
} else if (types [0] !== types [1]){
return weight[types[0]] - weight[types[1]]
}
return a>b?1:a<b?-1:0;
}
}
});
With this we can reduce the sameElements function to
function sameElements(c, d,referenceCheck) {
return c.equals(d,referenceCheck); //call .equals of Array.prototype.
}
Note. of course you could put all equal functions into the sameElements function, for the cost of the typechecking.
Now here are 3 examples: 1 with deep checking, 2 with reference checking.
var foo = {
a: 1,
obj: {
number: 2,
bool: true,
string: "asd"
},
arr: [1, 2, 3]
};
var bar = {
a: 1,
obj: {
number: 2,
bool: true,
string: "asd"
},
arr: [1, 2, 3]
};
var foobar = {
a: 1,
obj: {
number: 2,
bool: true,
string: "asd"
},
arr: [1, 2, 3, 4]
};
var a = [1, 2, foo, 2, bar, 2];
var b = [foo, 2, 2, 2, bar, 1];
var c = [bar, 2, 2, 2, bar, 1];
So these are the Arrays we compare. And the output is
Check a and b with references only.
console.log (sameElements ( a,b,true)) //true As they contain the same elements
Check b and c with references only
console.log (sameElements (b,c,true)) //false as c contains bar twice.
Check b and c deeply
console.log (sameElements (b,c,false)) //true as bar and foo are equal but not the same
Check for 2 Arrays containing NaN
Array.prototype.equals.NaN = true;
console.log(sameElements([NaN],[NaN],true)); //true.
Array.prototype.equals.NaN = false;
Demo on JSFiddle
You can implement the following algorithm:
If a and b do not have the same length:
Return false.
Otherwise:
Clone b,
For each item in a:
If the item exists in our clone of b:
Remove the item from our clone of b,
Otherwise:
Return false.
Return true.
With Javascript 1.6, you can use every() and indexOf() to write:
function sameElements(a, b)
{
if (a.length != b.length) {
return false;
}
var ourB = b.concat();
return a.every(function(item) {
var index = ourB.indexOf(item);
if (index < 0) {
return false;
} else {
ourB.splice(index, 1);
return true;
}
});
}
Note this implementation does not completely fulfill your requirements because indexOf() uses strict equality (===) internally. If you really want non-strict equality (==), you will have to write an inner loop instead.
Like this perhaps?
var foo = {}; var bar=[];
var a = [3,2,1,foo]; var b = [foo,1,2,3];
function comp(a,b)
{
// immediately discard if they are of different sizes
if (a.length != b.length) return false;
b = b.slice(0); // clone to keep original values after the function
a.forEach(function(e) {
var i;
if ((i = b.indexOf(e)) != -1)
b.splice(i, 1);
});
return !b.length;
}
comp(a,b);
UPDATE
As #Bergi and #thg435 point out my previous implementation was flawed so here is another implementation:
function sameElements(a, b) {
var objs = [];
// if length is not the same then must not be equal
if (a.length != b.length) return false;
// do an initial sort which will group types
a.sort();
b.sort();
for ( var i = 0; i < a.length; i++ ) {
var aIsPrimitive = isPrimitive(a[i]);
var bIsPrimitive = isPrimitive(b[i]);
// NaN will not equal itself
if( a[i] !== a[i] ) {
if( b[i] === b[i] ) {
return false;
}
}
else if (aIsPrimitive && bIsPrimitive) {
if( a[i] != b[i] ) return false;
}
// if not primitive increment the __count property
else if (!aIsPrimitive && !bIsPrimitive) {
incrementCountA(a[i]);
incrementCountB(b[i]);
// keep track on non-primitive objects
objs.push(i);
}
// if both types are not the same then this array
// contains different number of primitives
else {
return false;
}
}
var result = true;
for (var i = 0; i < objs.length; i++) {
var ind = objs[i];
// if __aCount and __bCount match then object exists same
// number of times in both arrays
if( a[ind].__aCount !== a[ind].__bCount ) result = false;
if( b[ind].__aCount !== b[ind].__bCount ) result = false;
// revert object to what it was
// before entering this function
delete a[ind].__aCount;
delete a[ind].__bCount;
delete b[ind].__aCount;
delete b[ind].__bCount;
}
return result;
}
// inspired by #Bergi's code
function isPrimitive(arg) {
return Object(arg) !== arg;
}
function incrementCountA(arg) {
if (arg.hasOwnProperty("__aCount")) {
arg.__aCount = arg.__aCount + 1;
} else {
Object.defineProperty(arg, "__aCount", {
enumerable: false,
value: 1,
writable: true,
configurable: true
});
}
}
function incrementCountB(arg) {
if (arg.hasOwnProperty("__bCount")) {
arg.__bCount = arg.__bCount + 1;
} else {
Object.defineProperty(arg, "__bCount", {
enumerable: false,
value: 1,
writable: true,
configurable: true
});
}
}
Then just call the function
sameElements( ["NaN"], [NaN] ); // false
// As "1" == 1 returns true
sameElements( [1],["1"] ); // true
sameElements( [1,2], [1,2,3] ); //false
The above implement actually defines a new property called "__count" that is used to keep track of non-primitive elements in both arrays. These are deleted before the function returns so as to leave the array elements as before.
Fiddle here
jsperf here.
The reason I changed the jsperf test case was that as #Bergi states the test arrays, especially the fact there were only 2 unique objects in the whole array is not representative of what we are testing for.
One other advantage of this implementation is that if you need to make it compatible with pre IE9 browsers instead of using the defineProperty to create a non-enumerable property you could just use a normal property.
Thanks everyone for sharing ideas! I've came up with the following
function sameElements(a, b) {
var hash = function(x) {
return typeof x + (typeof x == "object" ? a.indexOf(x) : x);
}
return a.map(hash).sort().join() == b.map(hash).sort().join();
}
This isn't the fastest solution, but IMO, most readable one so far.
i wasn't sure if "===" is ok, the question is a bit vauge...
if so, this is quite a bit faster and simpler than some other possible ways of doing it:
function isSame(a,b){
return a.length==b.length &&
a.filter(function(a){ return b.indexOf(a)!==-1 }).length == b.length;
}
Edit 2
1) Thanks to user2357112 for pointing out the Object.prototype.toString.call issue
this also showed, the reason it was that fast, that it didn't consider Arrays ...
I fixed the code,it should be working now :), unfortunately its now at about 59ops/s on chrome and 45ops/s on ff.
Fiddle and JSPerf is updated.
Edit
1)
I fixed the code, it supports mutliple variables referencing the same Object now.
A little bit slower than before, but still over 100ops/s on chrome.
2)
I tried using a bitmask instead of an array to keep multiple positions of the objects, but its nearly 15ops/s slow
3) As pointed ot in the comments i forgot to reset tmp after [[get]] is called fixed the code, the fiddle, and the perf.
So thanks to user2357112 with his Answer heres another approach using counting
var sameElements = (function () {
var f, of, objectFlagName;
of = objectFlagName = "__pos";
var tstr = function (o) {
var t = typeof o;
if (o === null)
t = "null";
return t
};
var types = {};
(function () {
var tmp = {};
Object.defineProperty(types, tstr(1), {
set: function (v) {
if (f)
tmp[v] = -~tmp[v];
else
tmp[v] = ~-tmp[v];
},
get: function () {
var ret = 1;
for (var k in tmp) {
ret &= !tmp[k];
}
tmp = {};
return ret;
}
});
})();
(function () {
var tmp = {};
Object.defineProperty(types, tstr(""), {
set: function (v) {
if (f) {
tmp[v] = -~tmp[v];
} else {
tmp[v] = ~-tmp[v];
}
},
get: function () {
var ret = 1;
for (var k in tmp) {
ret &= !tmp[k];
}
tmp = {};
return ret;
}
});
})();
(function () {
var tmp = [];
function add (v) {
tmp.push(v);
if (v[of]===undefined) {
v[of] = [tmp.length -1];
} else {
v[of].push(tmp.length -1)
}
}
Object.defineProperty(types, tstr({}), {
get: function () {
var ret = true;
for (var i = tmp.length - 1; i >= 0; i--) {
var c = tmp[i]
if (typeof c !== "undefined") {
ret = false
delete c[of]
}
}
tmp = [];
return ret;
},
set: function (v) {
var pos;
if (f) {
add (v);
} else if (!f && (pos = v[of]) !== void 0) {
tmp[pos.pop()] = undefined;
if (pos.length === 0)
delete v[of];
} else {
add (v);
}
}
});
}());
(function () {
var tmp = 0;
Object.defineProperty(types, tstr(undefined), {
get: function () {
var ret = !tmp;
tmp = 0;
return ret;
},
set: function () {
tmp += f ? 1 : -1;
}
});
})();
(function () {
var tmp = 0;
Object.defineProperty(types, tstr(null), {
get: function () {
var ret = !tmp;
tmp = 0;
return ret;
},
set: function () {
tmp += f ? 1 : -1;
}
});
})();
var tIt = [tstr(1), tstr(""), tstr({}), tstr(undefined), tstr(null)];
return function eq(a, b) {
f = true;
for (var i = a.length - 1; i >= 0; i--) {
var v = a[i];
types[tstr(v)] = v;
}
f = false;
for (var k = b.length - 1; k >= 0; k--) {
var w = b[k];
types[tstr(w)] = w;
}
var r = 1;
for (var l = 0, j; j = tIt[l]; l++) {
r &= types [j]
}
return !!r;
}
})()
Here is a JSFiddle and a JSPerf (it uses the same Arrays a and b as in the previous answers perf) with this code vs the Closure compiled
Heres the output. note: it doesn't support a deep comparison anymore, as is
var foo = {a:2}
var bar = {a:1};
var a = [1, 2, foo, 2, bar, 2];
var b = [foo, 2, 2, 2, bar, 1];
var c = [bar, 2, 2, 2, bar, 1];
console.log(sameElements([NaN],[NaN])); //true
console.log (sameElements ( a,b)) //true
console.log (sameElements (b,c)) //false
Using efficient lookup tables for the counts of the elements:
function sameElements(a) { // can compare any number of arrays
var map, maps = [], // counting booleans, numbers and strings
nulls = [], // counting undefined and null
nans = [], // counting nans
objs, counts, objects = [],
al = arguments.length;
// quick escapes:
if (al < 2)
return true;
var l0 = a.length;
if ([].slice.call(arguments).some(function(s) { return s.length != l0; }))
return false;
for (var i=0; i<al; i++) {
var multiset = arguments[i];
maps.push(map = {}); // better: Object.create(null);
objects.push({vals: objs=[], count: counts=[]});
nulls[i] = 0;
nans[i] = 0;
for (var j=0; j<l0; j++) {
var val = multiset[j];
if (val !== val)
nans[i]++;
else if (val === null)
nulls[i]++;
else if (Object(val) === val) { // non-primitive
var ind = objs.indexOf(val);
if (ind > -1)
counts[ind]++;
else
objs.push(val), counts.push(1);
} else { // booleans, strings and numbers do compare together
if (typeof val == "boolean")
val = +val;
if (val in map)
map[val]++;
else
map[val] = 1;
}
}
}
// testing if nulls and nans are the same everywhere
for (var i=1; i<al; i++)
if (nulls[i] != nulls[0] || nans[i] != nans[0])
return false;
// testing if primitives were the same everywhere
var map0 = maps[0];
for (var el in map0)
for (var i=1; i<al; i++) {
if (map0[el] !== maps[i][el])
return false;
delete maps[i][el];
}
for (var i=1; i<al; i++)
for (var el in maps[i])
return false;
// testing if objects were the same everywhere
var objs0 = objects[0].vals,
ol = objs0.length;
counts0 = objects[0].count;
for (var i=1; i<al; i++)
if (objects[i].count.length != ol)
return false;
for (var i=0; i<ol; i++)
for (var j=1; j<al; j++)
if (objects[j].count[ objects[j].vals.indexOf(objs0[i]) ] != counts0[i])
return false;
// else, the multisets are equal:
return true;
}
It still uses indexOf search amongst all objects, so if you have multisets with many different objects you might want to optimize that part as well. Have a look at Unique ID or object signature (and it's duplicate questions) for how to get lookup table keys for them. And if you don't have many primitive values in the multisets, you might just store them in arrays and sort those before comparing each item-by-item (like #Bruno did).
Disclaimer: This solution doesn't try to get the [[PrimitiveValue]] of objects, they will never be counted as equal to primitives (while == would do).
Here is the update on #Bruno's jsperf test of the answers, yet I guess only two objects (each of them present 500 times in the 10k array) and no duplicate primitive values are not representative.
I've got a 'table' of two columns represented as an array. The first column are numbers from 1 to 20 and they are labels, the second column are the corresponding values (seconds):
my_array = [ [ 3,4,5,3,4,5,2 ],[ 12,14,16,11,12,10,20 ] ];
I need the mean (average) for each label:
my_mean_array = [ [ 2,3,4,5 ],[ 20/1, (12+11)/2, (14+12)/2, (16+10)/2 ] ];
// edit: The mean should be a float - the notion above is just for clarification.
// Also the number 'labels' should remain as numbers/integers.
My try:
var a = my_array[0];
var b = my_array[1];
m = [];
n = [];
for( var i = 0; a.length; i++){
m[ a[i] ] += b[i]; // accumulate the values in the corresponding place
n[ a[i] ] += 1; // count the occurences
}
var o = [];
var p = [];
o = m / n;
p.push(n);
p.push(o);
How about this (native JS, will not break on older browsers):
function arrayMean(ary) {
var index = {}, i, label, value, result = [[],[]];
for (i = 0; i < ary[0].length; i++) {
label = ary[0][i];
value = ary[1][i];
if (!(label in index)) {
index[label] = {sum: 0, occur: 0};
}
index[label].sum += value;
index[label].occur++;
}
for (i in index) {
if (index.hasOwnProperty(i)) {
result[0].push(parseInt(i, 10));
result[1].push(index[i].occur > 0 ? index[i].sum / index[i].occur : 0);
}
}
return result;
}
FWIW, if you want fancy I've created a few other ways to do it. They depend on external libraries and are very probably an order of magnitude slower than a native solution. But they are nicer to look at.
It could look like this, with underscore.js:
function arrayMeanUnderscore(ary) {
return _.chain(ary[0])
.zip(ary[1])
.groupBy(function (item) { return item[0]; })
.reduce(function(memo, items) {
var values = _.pluck(items, 1),
toSum = function (a, b) { return a + b; };
memo[0].push(items[0][0]);
memo[1].push(_(values).reduce(toSum) / values.length);
return memo;
}, [[], []])
.value();
}
// --------------------------------------------
arrayMeanUnderscore([[3,4,5,3,4,5,2], [12,14,16,11,12,10,20]]);
// -> [[2,3,4,5], [20,11.5,13,13]]
or like this, with the truly great linq.js (I've used v2.2):
function arrayMeanLinq(ary) {
return Enumerable.From(ary[0])
.Zip(ary[1], "[$, $$]")
.GroupBy("$[0]")
.Aggregate([[],[]], function (result, item) {
result[0].push(item.Key());
result[1].push(item.Average("$[1]"));
return result;
});
}
// --------------------------------------------
arrayMeanLinq([[3,4,5,3,4,5,2], [12,14,16,11,12,10,20]]);
// -> [[3,4,5,2], [11.5,13,13,20]]
As suspected, the "fancy" implementations are an order of magnitude slower than a native implementation: jsperf comparison.
var temp = {};
my_array[0].map(function(label, i) {
if (! temp[label])
{
temp[label] = [];
}
temp[label].push(my_array[1][i]);
});
var result = [ [], [] ];
for (var label in temp) {
result[0].push(label);
result[1].push(
temp[label].reduce(function(p, v) { return p + v }) / temp[label].length
);
}
This function do not sort the resulted array like in your result example. If you need sorting, just say me and i will add it.
function getMeanArray(my_array)
{
m = {}; //id={count,value}
for( var i = 0; i<my_array[0].length; i++){
if (m[my_array[0][i]]===undefined)
{
m[my_array[0][i]]={count:0, value:0};
}
m[ my_array[0][i] ].value += my_array[1][i]; // accumulate the values in the corresponding place
m[ my_array[0][i] ].count++; // count the occurences
}
var my_mean_array=[[],[]];
for (var id in m)
{
my_mean_array[0].push(id);
my_mean_array[1].push(m[id].count!=0?m[id].value/m[id].count:0);
}
return my_mean_array;
}