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How do I find two numbers in an array that are equal to a target number

I'm looking to find two numbers in an array that are equal to a particular target number. I thought this would be a simple task using .filter but for some reason my code only works when I'm looking for a target number of 4 but doesn't work for anything else?
What am I missing here?
var numbers2 = [1,2,3,4];
var target = 3;
var found = numbers2.filter((num) => {
return (num + num) !== target;
});
console returns (4) [1,2,3,4] as opposed to 2[1,2].

var numbers = [1,4,3,2,6,8,12,1,1,1,2,3,4];
var target = 3;
var output = [];
// Use a set to remove duplicate numbers
numbers = [...new Set(numbers)]; // Only do this step if you dont want duplicates ( like 2+2 = 4 so if your target was for 2, would not show up in the list )
// Sort the numbers from lowest to highest
numbers.sort( (a,b) =>a-b);
// Get index of first number that matches the target or is greater than the target
let index;
for( let i =0; i < numbers.length; i++) {
if( numbers[i] >= target ) {
index = i;
break;
}
}
// Remove all numbers from the array starting at the previous index as these are not possible to add up with another number to the target
if( index ) {
numbers.splice(index, numbers.length - index );
}
// Loop through the remianing array to get first number
numbers.forEach( ( num1, index1) => {
// Loop through array again to get second number
numbers.forEach( (num2, index2) => {
// Check if number is same is same index as you dont want to add the same value to itself, then check if the 2 numbers equal the target number
if( index1 !== index2 && num1 + num2 === target ) {
// If number already exists in array dont duplicate otherwise add it to the array
if( output.indexOf( num1 ) == -1 ) {
output.push( num1);
}
// If number already exists in array dont duplicate otherwise add it to the array
if( output.indexOf( num2 ) == -1 ) {
output.push( num2);
}
}
});
});
console.log( output);

You could find the array location of your target number through using a array.forEach, array.indexOf(), array.find(), and array.findIndex():
let numbers2 = [1,2,3,4];
let target = 4;
//Using foreach
numbers2.forEach((item, index)=>{
if (item == target){
console.log("Found the target at array location "+index);
}
});
//Or through using indexOf():
console.log("Found the target at array location "+numbers2.indexOf(target));
//Or through using find():
const found = numbers2.find(element => element == target);
console.log("Found "+target+" in the array.");
//Or through findIndex():
const target1 = (a) => a == target;
console.log("Found the target at array location "+numbers2.findIndex(target1));

Assuming:
you only need one pair
[2,2] does not count when your target is 4 (as '2' only appears once in the array)
One way to go is:
let numbers = [1, 2, 3, 4]
let target = 4;
let output = [];
const N = numbers.length
outer: for (let i = 0; i < N; i++) {
for (let j = i + 1; j < N; j++) {
if (numbers[i] + numbers[j] === target) {
output.push(numbers[i], numbers[j])
break outer;
}
}
}
console.log(output); //[1,3]
Edit: even if you want more than one pair, it's easy to modify to get that effect (now the target is 5):
let numbers = [1, 2, 3, 4]
let target = 5;
let output = [];
const N = numbers.length
for (let i = 0; i < N; i++) {
for (let j = i + 1; j < N; j++) {
if (numbers[i] + numbers[j] === target) {
output.push([numbers[i], numbers[j]])
}
}
}
console.log(output); //[[1,4], [2,3]]

This is an ideal case for the humble for loop. Methods like .forEach() will always try to loop over all the elements in an array, but if we order the data before we start the search we can break early and eliminate a lot of searching.
Ergo...
var numbers = [1,2,3,4];
var target = 5;
var output = [];
// Handling ordered data is much faster than random data so we'll do this first
numbers.sort();
// We want to start the inner search part way up the array, and we also want
// the option to break so use conventional for loops.
for (let i = 0; i<numbers.length; i++) {
for (let j=i+1; j<numbers.length;j++) {
// If the total = target, store the numbers and break the inner loop since later numbers
// will all be too large
if ((numbers[i]+numbers[j])===target) {
output.push([numbers[i], numbers[j]]);
break;
}
}
// Stop searching the first loop once we reach halfway, since any subsequent result
// will already have been found.
if (numbers[i]>(target/2)) {
break;
}
}
console.log( output);

It makes very little sense to get an array of single numbers, because you'll get all the numbers except for the last one unless the array starts at zero or there are numbers skipped. So I've written a function that'll return an array of single numbers or an array of expressions (strings).
First, make a copy of the array:
const array = [1, 2, 3, 4]
const copy = array.slice(0);
Next, use .flatMap() for the first set of iterations:
array.flatMap(num => { // This is the outer loop of numbers
If the third parameter expression is undefined it will default to false. Then .filter() the copy array, the criteria being that the number from the outer loop plus the current number of the inner loop equals the target number AND the numbers cannot be identical.
copy.filter(n => n !== num && target === n + num);
/*
Iterations on the first iteration of outer loop
1 + 1, 1 + 2, 1 + 3,...
*/
If expression is true, then use .flatMap() to return an expression (string) of whatever equals the target number or an empty array (which returns as nothing since .flatMap() flattens it's returns by a level). If both numbers are identical an empty array will be returned.
copy.flatMap(n => n === num ? [] :
target === n + num ? `${n} + ${num}` :
[]
);
If expression is true half of the array is returned so that there isn't any reversed dupes (ex. 6+2 and 2+6)
let half = result.length / 2;
result = result.slice(0, half);
const log = data => console.log(JSON.stringify(data));
// [1, 2, 3,...10]
const array10 = [...new Array(10)].map((_, i) => i + 1);
// [0, 2, 4, 6,...18]
const arrayEven = [...new Array(10)].map((_, i) => i * 2);
function operands(array, target, expression = false) {
const copy = array.slice(0);
let result = array.flatMap(num => {
if (expression) {
return copy.flatMap((n, i) =>
num === n ? [] :
target === n + num ? `${n} + ${num}` :
[]
);
}
return copy.filter(n => n !== num && target === n + num);
});
if (expression) {
let half = result.length / 2;
result = result.slice(0, half);
}
return result;
}
// Return as an array of single numbers
log(array10);
log('3: '+operands(array10, 3));
log('8: '+operands(array10, 8));
log('5: '+operands(array10, 5));
log(arrayEven);
log('2: '+operands(arrayEven, 2));
log('8: '+operands(arrayEven, 8));
log('15: '+operands(arrayEven, 15));
log('=======================');
// Return as an array of expressions (string)
log(array10);
log('3: '+operands(array10, 3, true));
log('8: '+operands(array10, 8, true));
log('5: '+operands(array10, 5, true));
log(arrayEven);
log('2: '+operands(arrayEven, 2, true));
log('8: '+operands(arrayEven, 8, true));
log('15: '+operands(arrayEven, 15, true));

How to sort version IDs in a drop down list which are in "A.B.C" pattern [duplicate]

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!!!

Javascript merge 2 arrays and sum same key values

I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?

You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})

This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.

a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.

Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.

The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4

JS calculate mean of same elements in 2d array

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;
}

Javascript natural sort array/object and maintain index association

I have an array of items as follows in Javascript:
var users = Array();
users[562] = 'testuser3';
users[16] = 'testuser6';
users[834] = 'testuser1';
users[823] = 'testuser4';
users[23] = 'testuser2';
users[917] = 'testuser5';
I need to sort that array to get the following output:
users[834] = 'testuser1';
users[23] = 'testuser2';
users[562] = 'testuser3';
users[823] = 'testuser4';
users[917] = 'testuser5';
users[16] = 'testuser6';
Notice how it is sorted by the value of the array and the value-to-index association is maintained after the array is sorted (that is critical). I have looked for a solution to this, tried making it, but have hit a wall.
By the way, I am aware that this is technically not an array since that would mean the indices are always iterating 0 through n where n+1 is the counting number proceeding n. However you define it, the requirement for the project is still the same. Also, if it makes a difference, I am NOT using jquery.

The order of the elements of an array is defined by the index. So even if you specify the values in a different order, the values will always be stored in the order of their indices and undefined indices are undefined:
> var arr = [];
> arr[2] = 2;
> arr[0] = 0;
> arr
[0, undefined, 2]
Now if you want to store the pair of index and value, you will need a different data structure, maybe an array of array like this:
var arr = [
[562, 'testuser3'],
[16, 'testuser6'],
[834, 'testuser1'],
[823, 'testuser4'],
[23, 'testuser2'],
[917, 'testuser5']
];
This can be sorted with this comparison function:
function cmp(a, b) {
return a[1].localeCompare(b[1]);
}
arr.sort(cmp);
The result is this array:
[
[834, 'testuser1'],
[23, 'testuser2'],
[562, 'testuser3'],
[823, 'testuser4'],
[917, 'testuser5'],
[16, 'testuser6']
]

If I understand the question correctly, you're using arrays in a way they are not intended to be used. In fact, the initialization style
// Don't do this!
var array = new Array();
array[0] = 'value';
array[1] = 'value';
array[2] = 'value';
teaches wrong things about the nature and purpose of arrays. An array is an ordered list of items, indexed from zero up. The right way to create an array is with an array literal:
var array = [
'value',
'value',
'value'
]
The indexes are implied based on the order the items are specified. Creating an array and setting users[562] = 'testuser3' implies that there are at least 562 other users in the list, and that you have a reason for only knowing the 563rd at this time.
In your case, the index is data, and is does not represent the order of the items in the set. What you're looking for is a map or dictionary, represented in JavaScript by a plain object:
var users = {
562: 'testuser3',
16: 'testuser6',
834: 'testuser1',
823: 'testuser4',
23: 'testuser2',
917: 'testuser5'
}
Now your set does not have an order, but does have meaningful keys. From here, you can follow galambalazs's advice to create an array of the object's keys:
var userOrder;
if (typeof Object.keys === 'function') {
userOrder = Object.keys(users);
} else {
for (var key in users) {
userOrder.push(key);
}
}
…then sort it:
userOrder.sort(function(a, b){
return users[a].localeCompare(users[b]);
});
Here's a demo

You can't order arrays like this in Javascript. Your best bet is to make a map for order.
order = new Array();
order[0] = 562;
order[1] = 16;
order[2] = 834;
order[3] = 823;
order[4] = 23;
order[5] = 917;
In this way, you can have any order you want independently of the keys in the original array.
To sort your array use a custom sorting function.
order.sort( function(a, b) {
if ( users[a] < users[b] ) return -1;
else if ( users[a] > users[b] ) return 1;
else return 0;
});
for ( var i = 0; i < order.length; i++ ) {
// users[ order[i] ]
}
[Demo]

Using the ideas from the comments, I came up with the following solution. The naturalSort function is something I found on google and I modified it to sort a multidimensional array. Basically, I made the users array a multidimensional array with the first index being the user id and the second index being the user name. So:
users[0][0] = 72;
users[0][1] = 'testuser4';
users[1][0] = 91;
users[1][1] = 'testuser2';
users[2][0] = 12;
users[2][1] = 'testuser8';
users[3][0] = 3;
users[3][1] = 'testuser1';
users[4][0] = 18;
users[4][1] = 'testuser7';
users[5][0] = 47;
users[5][1] = 'testuser3';
users[6][0] = 16;
users[6][1] = 'testuser6';
users[7][0] = 20;
users[7][1] = 'testuser5';
I then sorted the array to get the following output:
users_sorted[0][0] = 3;
users_sorted[0][1] = 'testuser1';
users_sorted[1][0] = 91;
users_sorted[1][1] = 'testuser2';
users_sorted[2][0] = 47;
users_sorted[2][1] = 'testuser3';
users_sorted[3][0] = 72;
users_sorted[3][1] = 'testuser4';
users_sorted[4][0] = 20;
users_sorted[4][1] = 'testuser5';
users_sorted[5][0] = 16;
users_sorted[5][1] = 'testuser6';
users_sorted[6][0] = 18;
users_sorted[6][1] = 'testuser7';
users_sorted[7][0] = 12;
users_sorted[7][1] = 'testuser8';
The code to do this is below:
function naturalSort(a, b) // Function to natural-case insensitive sort multidimensional arrays by second index
{
// setup temp-scope variables for comparison evauluation
var re = /(-?[0-9\.]+)/g,
x = a[1].toString().toLowerCase() || '',
y = b[1].toString().toLowerCase() || '',
nC = String.fromCharCode(0),
xN = x.replace( re, nC + '$1' + nC ).split(nC),
yN = y.replace( re, nC + '$1' + nC ).split(nC),
xD = (new Date(x)).getTime(),
yD = xD ? (new Date(y)).getTime() : null;
// natural sorting of dates
if ( yD )
if ( xD < yD ) return -1;
else if ( xD > yD ) return 1;
// natural sorting through split numeric strings and default strings
for( var cLoc = 0, numS = Math.max(xN.length, yN.length); cLoc < numS; cLoc++ ) {
oFxNcL = parseFloat(xN[cLoc]) || xN[cLoc];
oFyNcL = parseFloat(yN[cLoc]) || yN[cLoc];
if (oFxNcL < oFyNcL) return -1;
else if (oFxNcL > oFyNcL) return 1;
}
return 0;
}
// Set values for index
var users = Array();
var temp = Array();
users.push(Array('72', 'testuser4'));
users.push(Array('91', 'testuser2'));
users.push(Array('12', 'testuser8'));
users.push(Array('3', 'testuser1'));
users.push(Array('18', 'testuser7'));
users.push(Array('47', 'testuser3'));
users.push(Array('16', 'testuser6'));
users.push(Array('20', 'testuser5'));
// Sort the array
var users_sorted = Array();
users_sorted = users.sort(naturalSort);

I'd use map once to make a new array of users,
then a second time to return the string you want from the new array.
var users= [];
users[562]= 'testuser3';
users[16]= 'testuser6';
users[834]= 'testuser1';
users[823]= 'testuser4';
users[23]= 'testuser2';
users[917]= 'testuser5';
var u2= [];
users.map(function(itm, i){
if(itm){
var n= parseInt(itm.substring(8), 10);
u2[n]= i;
}
});
u2.map(function(itm, i){
return 'users['+itm+']= testuser'+i;
}).join('\n');
/*returned value: (String)
users[834]= testuser1
users[23]= testuser2
users[562]= testuser3
users[823]= testuser4
users[917]= testuser5
users[16]= testuser6
*/
If you want to avoid any gaps. use a simple filter on the output-
u2.map(function(itm, i){
return 'users['+itm+']= testuser'+i;
}).filter(function(itm){return itm}).join('\n');

Sparse arrays usually spell trouble. You're better off saving key-value pairs in an array as objects (this technique is also valid JSON):
users = [{
"562": "testuser3"
},{
"16": "testuser6"
}, {
"834": "testuser1"
}, {
"823": "testuser4"
}, {
"23": "testuser2"
}, {
"917": "testuser5"
}];
As suggested, you can use a for loop to map the sorting function onto the array.

Array.prototype.sort() takes an optional custom comparison function -- so if you dump all of your users into an array in this manner [ [562, "testuser3"], [16, "testuser6"] ... etc.]
Then sort this array with the following function:
function(comparatorA, comparatorB) {
var userA = comparatorA[1], userB = comparatorB[1]
if (userA > userB) return 1;
if (userA < userB) return -1;
if (userA === userB) return 0;
}
Then rebuild your users object. (Which will loose you your sorting.) Or, keep the data in the newly sorted array of arrays, if that will work for your application.

A oneliner with array of array as a result:
For sorting by Key.
let usersMap = users.map((item, i) => [i, item]).sort((a, b) => a[0] - b[0]);
For sorting by Value. (works with primitive types)
let usersMap = users.map((item, i) => [i, item]).sort((a, b) => a[1] - b[1]);