We Keep Coding

Serializing Array of Many Duplicates

So I have a series of arrays, each of which are 2500 long, and I need to serialize and store all them in very limited space.
Since I have many duplicates, I wanted to cut them down to something like below.
[0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0]
// to
[0x4,2,7,3x2,0x9]
I wrote a couple one-liners (utilising Lodash' _.repeat) to convert to and from this pattern, however converting to doesn't seem to work in most/all cases.
let serialized = array.toString().replace(/((?:(\d)+,?)((?:\2+,?){2,}))/g, (m, p1, p2) => p2 + 'x' + m.replace(/,/g, '').length);
let parsed = serialized.replace(/(\d+)x(\d+),?/g, (z, p1, p2) => _.repeat(p1 + ',', +p2)).split(',');
I don't know why it doesn't work. It may be due to some of the numbers in the array. Eye-balling, the largest one is 4294967295, however well over 90% is just 0.
What am I missing in my RegEx that's preventing it from working correctly? Is there a simpler way that I'm too blind to see?
I'm fairly confident with converting it back from the serialized state, just need a hand getting it to the state.

Straight forward and simple serialization:
let serialize = arr => {
const elements = [];
const counts = []
let last = undefined;
[0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0].forEach((el,i,arr)=>{
if (el!==last) {
elements.push(el);
counts.push(1);
} else {
counts[counts.length-1]++;
}
last = el;
})
return elements.map((a,i)=>counts[i]>1?`${a}x${counts[i]}`:a).join(",");
};
console.log(serialize([0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0]));
UPDATE
Pure functional serialize one:
let serialize = arr => arr
.reduce((memo, element, i) => {
if (element !== arr[i - 1]) {
memo.push({count: 1, element});
} else {
memo[memo.length - 1].count++;
}
return memo;
},[])
.map(({count, element}) => count > 1 ? `${count}x${element}` : element)
.join(",");
console.log(serialize([0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0]));
Pure functional deserialize:
const deserialize = str => str
.split(",")
.map(c => c.split("x").reverse())
.reduce((memo, [el, count = 1]) => memo.concat(Array(+count).fill(+el)), []);
console.log(deserialize("4x0,2,7,2x3,9x0"))
In order to avoid using .reverse() in this logic, I'd recommend to change serialization from 4x0 to 0x4

Try this
var arr = [0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0];
var finalArray = []; //array into which count of values will go
var currentValue = ""; //current value for comparison
var tmpArr = []; //temporary array to hold values
arr.forEach( function( val, index ){
if ( val != currentValue && currentValue !== "" )
{
finalArray.push( tmpArr.length + "x" + tmpArr[0] );
tmpArr = [];
}
tmpArr.push(val);
currentValue = val;
});
finalArray.push( tmpArr.length + "x" + tmpArr[0] );
console.log(finalArray);
Another version without temporary array
var arr = [0, 0, 0, 0, 2, 7, 3, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0];
var finalArray = []; //array into which count of values will go
var tmpCount = 0; //temporary variable to hold count
arr.forEach(function(val, index) {
if ( (val != arr[ index - 1 ] && index !== 0 ) )
{
finalArray.push(tmpCount + "x" + arr[ index - 1 ] );
tmpCount = 0;
}
tmpCount++;
if ( index == arr.length - 1 )
{
finalArray.push(tmpCount + "x" + arr[ index - 1 ] );
}
});
console.log(finalArray);

Do not use RegEx. Just use regular logic. I recommend array.reduce for this job.
const arr1 = [0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0]
const arr2 = ['0x4','2','7','3x2','0x9'];
const compact = arr => {
const info = arr.reduce((c, v) =>{
if(c.prevValue !== v){
c.order.push(v);
c.count[v] = 1;
c.prevCount = 1;
c.prevValue = v;
} else {
c.prevCount = c.prevCount + 1;
c.count[v] = c.count[v] + 1;
};
return c;
},{
prevValue: null,
prevCount: 0,
count: {},
order: []
});
return info.order.map(v => info.count[v] > 1 ? `${v}x${info.count[v]}` : `${v}`);
}
const expand = arr => {
return arr.reduce((c, v) => {
const split = v.split('x');
const value = +split[0];
const count = +split[1] || 1;
Array.prototype.push.apply(c, Array(count).fill(value));
return c;
}, []);
}
console.log(compact(arr1));
console.log(expand(arr2));

This is a typical reducing job. Here is your compress function done in just O(n) time..
var arr = [0,0,0,0,2,7,3,3,0,0,0,0,0,0,0,0,0],
compress = a => a.reduce((r,e,i,a) => e === a[i-1] ? (r[r.length-1][1]++,r) : (r.push([e,1]) ,r),[]);
console.log(JSON.stringify(compress(arr)));

since the motivation here is to reduce the size of the stored arrays, consider using something like gzip-js to compress your data.

Javascript - Generating all combinations of elements in a single array (in pairs)

I've seen several similar questions about how to generate all possible combinations of elements in an array. But I'm having a very hard time figuring out how to write an algorithm that will only output combination pairs. Any suggestions would be super appreciated!
Starting with the following array (with N elements):
var array = ["apple", "banana", "lemon", "mango"];
And getting the following result:
var result = [
"apple banana"
"apple lemon"
"apple mango"
"banana lemon"
"banana mango"
"lemon mango"
];
I was trying out the following approach but this results in all possible combinations, instead only combination pairs.
var letters = splSentences;
var combi = [];
var temp= "";
var letLen = Math.pow(2, letters.length);
for (var i = 0; i < letLen ; i++){
temp= "";
for (var j=0;j<letters.length;j++) {
if ((i & Math.pow(2,j))){
temp += letters[j]+ " "
}
}
if (temp !== "") {
combi.push(temp);
}
}

Here are some functional programming solutions:
Using EcmaScript2019's flatMap:
var array = ["apple", "banana", "lemon", "mango"];
var result = array.flatMap(
(v, i) => array.slice(i+1).map( w => v + ' ' + w )
);
console.log(result);
Before the introduction of flatMap (my answer in 2017), you would go for reduce or [].concat(...) in order to flatten the array:
var array = ["apple", "banana", "lemon", "mango"];
var result = array.reduce( (acc, v, i) =>
acc.concat(array.slice(i+1).map( w => v + ' ' + w )),
[]);
console.log(result);
Or:
var array = ["apple", "banana", "lemon", "mango"];
var result = [].concat(...array.map(
(v, i) => array.slice(i+1).map( w => v + ' ' + w ))
);
console.log(result);

A simple way would be to do a double for loop over the array where you skip the first i elements in the second loop.
let array = ["apple", "banana", "lemon", "mango"];
let results = [];
// Since you only want pairs, there's no reason
// to iterate over the last element directly
for (let i = 0; i < array.length - 1; i++) {
// This is where you'll capture that last value
for (let j = i + 1; j < array.length; j++) {
results.push(`${array[i]} ${array[j]}`);
}
}
console.log(results);
Rewritten with ES5:
var array = ["apple", "banana", "lemon", "mango"];
var results = [];
// Since you only want pairs, there's no reason
// to iterate over the last element directly
for (var i = 0; i < array.length - 1; i++) {
// This is where you'll capture that last value
for (var j = i + 1; j < array.length; j++) {
results.push(array[i] + ' ' + array[j]);
}
}
console.log(results);

In my case, I wanted to get the combinations as follows, based on the size range of the array:
function getCombinations(valuesArray: String[])
{
var combi = [];
var temp = [];
var slent = Math.pow(2, valuesArray.length);
for (var i = 0; i < slent; i++)
{
temp = [];
for (var j = 0; j < valuesArray.length; j++)
{
if ((i & Math.pow(2, j)))
{
temp.push(valuesArray[j]);
}
}
if (temp.length > 0)
{
combi.push(temp);
}
}
combi.sort((a, b) => a.length - b.length);
console.log(combi.join("\n"));
return combi;
}
Example:
// variable "results" stores an array with arrays string type
let results = getCombinations(['apple', 'banana', 'lemon', ',mango']);
Output in console:
The function is based on the logic of the following documentation, more information in the following reference:
https://www.w3resource.com/javascript-exercises/javascript-function-exercise-3.php
if ((i & Math.pow(2, j)))
Each bit of the first value is compared with the second, it is taken as valid if it matches, otherwise it returns zero and the condition is not met.

Although solutions have been found, I post here an algorithm for general case to find all combinations size n of m (m>n) elements. In your case, we have n=2 and m=4.
const result = [];
result.length = 2; //n=2
function combine(input, len, start) {
if(len === 0) {
console.log( result.join(" ") ); //process here the result
return;
}
for (let i = start; i <= input.length - len; i++) {
result[result.length - len] = input[i];
combine(input, len-1, i+1 );
}
}
const array = ["apple", "banana", "lemon", "mango"];
combine( array, result.length, 0);

I ended up writing a general solution to this problem, which is functionally equivalent to nhnghia's answer, but I'm sharing it here as I think it's easier to read/follow and is also full of comments describing the algorithm.
/**
* Generate all combinations of an array.
* #param {Array} sourceArray - Array of input elements.
* #param {number} comboLength - Desired length of combinations.
* #return {Array} Array of combination arrays.
*/
function generateCombinations(sourceArray, comboLength) {
const sourceLength = sourceArray.length;
if (comboLength > sourceLength) return [];
const combos = []; // Stores valid combinations as they are generated.
// Accepts a partial combination, an index into sourceArray,
// and the number of elements required to be added to create a full-length combination.
// Called recursively to build combinations, adding subsequent elements at each call depth.
const makeNextCombos = (workingCombo, currentIndex, remainingCount) => {
const oneAwayFromComboLength = remainingCount == 1;
// For each element that remaines to be added to the working combination.
for (let sourceIndex = currentIndex; sourceIndex < sourceLength; sourceIndex++) {
// Get next (possibly partial) combination.
const next = [ ...workingCombo, sourceArray[sourceIndex] ];
if (oneAwayFromComboLength) {
// Combo of right length found, save it.
combos.push(next);
}
else {
// Otherwise go deeper to add more elements to the current partial combination.
makeNextCombos(next, sourceIndex + 1, remainingCount - 1);
}
}
}
makeNextCombos([], 0, comboLength);
return combos;
}

The best solutions I have found - https://lowrey.me/es6-javascript-combination-generator/
Uses ES6 generator functions, I adapted to TS. Most often you don't need all of the combinations at the same time. And I was getting annoyed by writing loops like for (let i=0; ... for let (j=i+1; ... for (let k=j+1... just to get combos one by one to test if I need to terminate the loops..
export function* combinations<T>(array: T[], length: number): IterableIterator<T[]> {
for (let i = 0; i < array.length; i++) {
if (length === 1) {
yield [array[i]];
} else {
const remaining = combinations(array.slice(i + 1, array.length), length - 1);
for (let next of remaining) {
yield [array[i], ...next];
}
}
}
}
usage:
for (const combo of combinations([1,2,3], 2)) {
console.log(combo)
}
output:
> (2) [1, 2]
> (2) [1, 3]
> (2) [2, 3]

Just to give an option for next who'll search it
const arr = ['a', 'b', 'c']
const combinations = ([head, ...tail]) => tail.length > 0 ? [...tail.map(tailValue => [head, tailValue]), ...combinations(tail)] : []
console.log(combinations(arr)) //[ [ 'a', 'b' ], [ 'a', 'c' ], [ 'b', 'c' ] ]

There are also this answer:
https://stackoverflow.com/a/64414875/19518308
The alghorithm is this answer generates all the possible sets of combination(or choose(n, k)) of n items within k spaces.
The algorhitm:
function choose(arr, k, prefix=[]) {
if (k == 0) return [prefix];
return arr.flatMap((v, i) =>
choose(arr.slice(i+1), k-1, [...prefix, v])
);
}
console.log(choose([0,1,2,3,4], 3));
I had a similar problem and this algorhitm is working very well for me.

Using map and flatMap the following can be done (flatMap is only supported on chrome and firefox)
var array = ["apple", "banana", "lemon", "mango"]
array.flatMap(x => array.map(y => x !== y ? x + ' ' + y : null)).filter(x => x)

I think it is an answer to all such questions.
/**
*
* Generates all combination of given Array or number
*
* #param {Array | number} item - Item accepts array or number. If it is array exports all combination of items. If it is a number export all combination of the number
* #param {number} n - pow of the item, if given value is `n` it will be export max `n` item combination
* #param {boolean} filter - if it is true it will just export items which have got n items length. Otherwise export all posible length.
* #return {Array} Array of combination arrays.
*
* Usage Example:
*
* console.log(combination(['A', 'B', 'C', 'D'], 2, true)); // [[ 'A','A' ], [ 'A', 'B' ]...] (16 items)
* console.log(combination(['A', 'B', 'C', 'D'])); // [['A', 'A', 'A', 'B' ],.....,['A'],] (340 items)
* console.log(comination(4, 2)); // all posible values [[ 0 ], [ 1 ], [ 2 ], [ 3 ], [ 0, 0 ], [ 0, 1 ], [ 0, 2 ]...] (20 items)
*/
function combination(item, n) {
const filter = typeof n !=='undefined';
n = n ? n : item.length;
const result = [];
const isArray = item.constructor.name === 'Array';
const count = isArray ? item.length : item;
const pow = (x, n, m = []) => {
if (n > 0) {
for (var i = 0; i < count; i++) {
const value = pow(x, n - 1, [...m, isArray ? item[i] : i]);
result.push(value);
}
}
return m;
}
pow(isArray ? item.length : item, n);
return filter ? result.filter(item => item.length == n) : result;
}
console.log("#####first sample: ", combination(['A', 'B', 'C', 'D'], 2)); // with filter
console.log("#####second sample: ", combination(['A', 'B', 'C', 'D'])); // without filter
console.log("#####third sample: ", combination(4, 2)); // gives array with index number

Generating combinations of elements in an array is a lot like counting in a numeral system,
where the base is the number of elements in your array (if you account for the leading zeros that will be missing).
This gives you all the indices to your array (concatenated):
arr = ["apple", "banana", "lemon", "mango"]
base = arr.length
idx = [...Array(Math.pow(base, base)).keys()].map(x => x.toString(base))
You are only interested in pairs of two, so restrict the range accordingly:
range = (from, to) = [...Array(to).keys()].map(el => el + from)
indices = range => range.map(x => x.toString(base).padStart(2,"0"))
indices( range( 0, Math.pow(base, 2))) // range starts at 0, single digits are zero-padded.
Now what's left to do is map indices to values.
As you don't want elements paired with themselves and order doesn't matter,
those need to be removed, before mapping to the final result.
const range = (from, to) => [...Array(to).keys()].map(el => el + from)
const combinations = arr => {
const base = arr.length
return range(0, Math.pow(base, 2))
.map(x => x.toString(base).padStart(2, "0"))
.filter(i => !i.match(/(\d)\1/) && i === i.split('').sort().join(''))
.map(i => arr[i[0]] + " " + arr[i[1]])
}
console.log(combinations(["apple", "banana", "lemon", "mango"]))
With more than ten elements, toString() will return letters for indices; also, this will only work with up to 36 Elements.

Generating combinations is a classic problem. Here's my interpretation of that solution:
const combinations = (elements) => {
if (elements.length == 1) {
return [elements];
} else {
const tail = combinations(elements.slice(1));
return tail.reduce(
(combos, combo) => { combos.push([elements[0], ...combo]); return combos; },
[[elements[0]], ...tail]
);
}
};
const array = ["apple", "banana", "lemon", "mango"];
console.log(combinations(array));

Here is an non-mutating ES6 approach combining things (TS):
function combine (tail: any[], length: number, head: any[][] = [[]]): any[][] {
return tail.reduce((acc, tailElement) => {
const tailHeadVariants = head.reduce((acc, headElement: any[]) => {
const combination = [...headElement, tailElement]
return [...acc, combination]
}, [])
if (length === 1) return [...acc, tailHeadVariants]
const subCombinations = combine(tail.filter(t => t !== tailElement), length - 1, tailHeadVariants)
return [...acc, ...subCombinations]
}, [])
}

As this post is well indexed on Google under the keywords "generate all combinations", lots of people coming here simply need to generate all the unique combinations, regardless of the size of the output (not only pairs).
This post answers this need.
All unique combinations, without recursion:
const getCombos = async (a) => {
const separator = '';
const o = Object();
for (let i = 0; i < a.length; ++i) {
for (let j = i + 1; j <= a.length; ++j) {
const left = a.slice(i, j);
const right = a.slice(j, a.length);
o[left.join(separator)] = 1;
for (let k = 0; k < right.length; ++k) {
o[[...left, right[k]].join(separator)] = 1;
}
}
}
return Object.keys(o);
}
const a = ['a', 'b', 'c', 'd'];
const b = await getCombos(a);
console.log(b);
// (14) ['a', 'ab', 'ac', 'ad', 'abc', 'abd', 'abcd',
// 'b', 'bc', 'bd', 'bcd', 'c', 'cd', 'd']
This code splits the array into 2 sub arrays, left / right, then iterate over the right array to combine it with the left array. The left becomes bigger overtime, while the right becomes smaller. The result has only unique values.

Beating a dead horse a bit, but with smaller sets where recursion limit and performance is not a problem, the general combination generation can be done recursively with "recurse combinations containing the first element in given array" plus "recurse combinations not containing the first element". It gives quite compact implementation as a generator:
// Generator yielding k-item combinations of array a
function* choose(a, k) {
if(a.length == k) yield a;
else if(k == 0) yield [];
else {
for(let rest of choose(a.slice(1), k-1)) yield [a[0], ...rest];
for(let rest of choose(a.slice(1), k)) yield rest;
}
}
And even slightly shorter (and twice faster, 1 M calls of 7 choose 5 took 3.9 seconds with my MacBook) with function returning and array of combinations:
// Return an array of combinations
function comb(a, k) {
if(a.length === k) return [a];
else if(k === 0) return [[]];
else return [...comb(a.slice(1), k-1).map(c => [a[0], ...c]),
...comb(a.slice(1), k)];
}

Resort array of objects on another key without disturbing previous sorting (part-sorting) in JavaScript

I have an array of objects like this:
var booksArray = [
{Id:1,Rate:5,Price:200,Name:"History"},
{Id:2,Rate:5,Price:150,Name:"Geographic"},
{Id:3,Rate:1,Price:75,Name:"Maths"},
{Id:4,Rate:2,Price:50,Name:"Statistics"},
{Id:5,Rate:3,Price:120,Name:"Drawing"},
{Id:6,Rate:2,Price:100,Name:"Arts"},
{Id:7,Rate:3,Price:110,Name:"Chemistry"},
{Id:8,Rate:4,Price:160,Name:"Biology"},
{Id:9,Rate:4,Price:90,Name:"Software Engineering"},
{Id:10,Rate:4,Price:80,Name:"Algorithm"},
{Id:11,Rate:1,Price:85,Name:"Urdu"},
{Id:12,Rate:1,Price:110,Name:"International Relations"},
{Id:13,Rate:1,Price:120,Name:"Physics"},
{Id:14,Rate:3,Price:230,Name:"Electronics"},
{Id:15,Rate:3,Price:250,Name:"Jihad"},
{Id:16,Rate:3,Price:200,Name:"Functional Programming"},
{Id:17,Rate:2,Price:190,Name:"Computer Science"},
{Id:18,Rate:2,Price:50,Name:"Problem solving Techniques"},
{Id:19,Rate:6,Price:150,Name:"C#"},
{Id:20,Rate:7,Price:250,Name:"ASP.Net"}
]
I am sorting it with this function:
function sortBy(key, reverse) {
var moveSmaller = reverse ? 1 : -1;
var moveLarger = reverse ? -1 : 1;
return function(a, b) {
if (a[key] < b[key]) {
return moveSmaller;
}
if (a[key] > b[key]) {
return moveLarger;
}
return 0;
};
}
booksArray.sort(sortBy('Rate', false))
console.log(JSON.stringify(booksArray))
And this is producing this result:
[
{Id:3,Rate:1,Price:75,Name:"Maths"},
{Id:11,Rate:1,Price:85,Name:"Urdu"},
{Id:12,Rate:1,Price:110,Name:"International Relations"},
{Id:13,Rate:1,Price:120,Name:"Physics"},
{Id:4,Rate:2,Price:50,Name:"Statistics"},
{Id:6,Rate:2,Price:100,Name:"Arts"},
{Id:17,Rate:2,Price:190,Name:"Computer Science"},
{Id:18,Rate:2,Price:50,Name:"Problem solving Techniques"},
{Id:5,Rate:3,Price:120,Name:"Drawing"},
{Id:7,Rate:3,Price:110,Name:"Chemistry"},
{Id:14,Rate:3,Price:230,Name:"Electronics"},
{Id:15,Rate:3,Price:250,Name:"Jihad"},
{Id:16,Rate:3,Price:200,Name:"Functional Programming"},
{Id:8,Rate:4,Price:160,Name:"Biology"},
{Id:9,Rate:4,Price:90,Name:"Software Engineering"},
{Id:10,Rate:4,Price:80,Name:"Algorithm"},
{Id:1,Rate:5,Price:200,Name:"History"},
{Id:2,Rate:5,Price:150,Name:"Geographic"},
{Id:19,Rate:6,Price:150,Name:"C#"},
{Id:20,Rate:7,Price:250,Name:"ASP.Net"}
]
You can see it is sorting on the Rate field which is fine. Now I want to resort again the parts of array without disturbing Rate sorting. I need output like this:
[
{Id:12,Rate:1,Price:110,Name:"International Relations"},
{Id:3,Rate:1,Price:75,Name:"Maths"},
{Id:13,Rate:1,Price:120,Name:"Physics"},
{Id:11,Rate:1,Price:85,Name:"Urdu"},
{Id:6,Rate:2,Price:100,Name:"Arts"},
{Id:17,Rate:2,Price:190,Name:"Computer Science"},
{Id:18,Rate:2,Price:50,Name:"Problem solving Techniques"},
{Id:4,Rate:2,Price:50,Name:"Statistics"},
{Id:7,Rate:3,Price:110,Name:"Chemistry"},
{Id:5,Rate:3,Price:120,Name:"Drawing"},
{Id:14,Rate:3,Price:230,Name:"Electronics"},
{Id:16,Rate:3,Price:200,Name:"Functional Programming"},
{Id:15,Rate:3,Price:250,Name:"Jihad"},
{Id:10,Rate:4,Price:80,Name:"Algorithm"},
{Id:8,Rate:4,Price:160,Name:"Biology"},
{Id:9,Rate:4,Price:90,Name:"Software Engineering"},
{Id:2,Rate:5,Price:150,Name:"Geographic"},
{Id:1,Rate:5,Price:200,Name:"History"},
{Id:19,Rate:6,Price:150,Name:"C#"},
{Id:20,Rate:7,Price:250,Name:"ASP.Net"}
]
Here you can see it is sorted on Rate as well as Name.
Why am I not doing sorting with multiple keys once and for all? Because I am creating a function which sorts the array of objects and can be called multiple times.
Example:
myarray.order('Rate') // single sort
myarray.order('Rate').order('Name') // multiple sort
I can use some more parameters in my function to track if array has already been sorted.
Sample Fiddle

Assuming that you already have bookArray sorted by Rate. It can be sorted on second level using following. Can be fine tuned better for easy usability.
var arr = bookArray;
var firstSortKey = 'Rate';
var secondSortKey = 'Name';
var count = 0 ;
var firstSortValue = arr[count][firstSortKey];
for(var i=count+1; i<arr.length; i++){
if(arr[i][firstSortKey]!==firstSortValue){
var data = arr.slice(count, i);
data = data.sort(function(a, b){
return a[secondSortKey]>b[secondSortKey];
});
var argsArray = [count, i-count];
argsArray.push.apply(argsArray, data);
Array.prototype.splice.apply(arr, argsArray);
count = i;
firstSortValue = arr[i][firstSortKey];
}
}
Fiddle Demo

Change your sortBy function to accept an array of keys. Something like:
function sortBy(keys, reverse) {
var moveSmaller = reverse ? 1 : -1;
var moveLarger = reverse ? -1 : 1;
var key = keys.shift();
return function(a, b) {
if (a[key] < b[key]) {
return moveSmaller;
}
if (a[key] > b[key]) {
return moveLarger;
}
return keys.length ? (sortBy(keys, reverse))(a, b) : 0;
};
}
booksArray.sort(sortBy(['Rate', 'Name'], false))
(Untested, jsfiddle is down for me)

Just for fun: here's a generic sorter for multiple levels of field values within an Array of Objects.
Syntax
XSort().create([arrOfObjects])
.orderBy(
{key: [keyname], [descending=0/1] },
{key: [keyname], [descending=0/1]},
... );
See jsFiddle for an example using booksArray.
function XSort() {
const multiSorter = sortKeys => {
if (!sortKeys || sortKeys[0].constructor !== Object) {
throw new TypeError("Provide at least one {key: [keyname]} to sort on");
}
return function (val0, val1) {
for (let sortKey of sortKeys) {
const v0 = sortKey.key instanceof Function
? sortKey.key(val0) : val0[sortKey.key];
const v1 = sortKey.key instanceof Function
? sortKey.key(val1) : val1[sortKey.key];
const isString = v0.constructor === String || v1.constructor === String;
const compare = sortKey.descending ?
isString ? v1.toLowerCase().localeCompare(v0.toLowerCase()) : v1 - v0 :
isString ? v0.toLowerCase().localeCompare(v1.toLowerCase()) : v0 - v1;
if ( compare !== 0 ) { return compare; }
}
};
}
const Sorter = function (array) {
this.array = array;
};
Sorter.prototype = {
orderBy: function(...sortOns) {
return this.array.slice().sort(multiSorter(sortOns));
},
};
return {
create: array => new Sorter(array)
};
}

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;
}

Finding longest string in array

Is there a short way to find the longest string in a string array?
Something like arr.Max(x => x.Length);?

Available since Javascript 1.8/ECMAScript 5 and available in most older browsers:
var longest = arr.reduce(
function (a, b) {
return a.length > b.length ? a : b;
}
);
Otherwise, a safe alternative:
var longest = arr.sort(
function (a, b) {
return b.length - a.length;
}
)[0];

A new answer to an old question: in ES6 you can do shorter:
Math.max(...(x.map(el => el.length)));

I would do something like this
var arr = [
'first item',
'second item is longer than the third one',
'third longish item'
];
var lgth = 0;
var longest;
for (var i = 0; i < arr.length; i++) {
if (arr[i].length > lgth) {
var lgth = arr[i].length;
longest = arr[i];
}
}
console.log(longest);

Maybe not the fastest, but certainly pretty readable:
function findLongestWord(array) {
var longestWord = "";
array.forEach(function(word) {
if(word.length > longestWord.length) {
longestWord = word;
}
});
return longestWord;
}
var word = findLongestWord(["The","quick","brown", "fox", "jumped", "over", "the", "lazy", "dog"]);
console.log(word); // result is "jumped"
The array function forEach has been supported since IE9+.

In ES6 this could be accomplished with a reduce() call in O(n) complexity as opposed to solutions using sort() which is O(nlogn):
const getLongestText = (arr) => arr.reduce(
(savedText, text) => (text.length > savedText.length ? text : savedText),
'',
);
console.log(getLongestText(['word', 'even-longer-word', 'long-word']))

var arr = [ 'fdgdfgdfg', 'gdfgf', 'gdfgdfhawsdgd', 'gdf', 'gdfhdfhjurvweadsd' ];
arr.sort(function (a, b) { return b.length - a.length })[0];

I was inspired of Jason's function and made a little improvements to it and got as a result rather fast finder:
function timo_longest(a) {
var c = 0, d = 0, l = 0, i = a.length;
if (i) while (i--) {
d = a[i].length;
if (d > c) {
l = i; c = d;
}
}
return a[l];
}
arr=["First", "Second", "Third"];
var longest = timo_longest(arr);
Speed results: http://jsperf.com/longest-string-in-array/7

I provide a functional+recursive approach. See comments to understand how it works:
const input1 = ['a', 'aa', 'aaa']
const input2 = ['asdf', 'qwer', 'zxcv']
const input3 = ['asdfasdf fdasdf a sd f', ' asdfsdf', 'asdfasdfds', 'asdfsdf', 'asdfsdaf']
const input4 = ['ddd', 'dddddddd', 'dddd', 'ddddd', 'ddd', 'dd', 'd', 'd', 'dddddddddddd']
// Outputs which words has the greater length
// greatestWord :: String -> String -> String
const greatestWord = x => y =>
x.length > y.length ? x : y
// Recursively outputs the first longest word in a series
// longestRec :: String -> [String] -> String
const longestRec = longestWord => ([ nextWord, ...words ]) =>
// ^^^^^^^^^^^^
// Destructuring lets us get the next word, and remaining ones!
nextWord // <-- If next word is undefined, then it won't recurse.
? longestRec (greatestWord (nextWord) (longestWord)) (words)
: longestWord
// Outputs the first longest word in a series
// longest :: [String] -> String
const longest = longestRec ('')
const output1 = longest (input1)
const output2 = longest (input2)
const output3 = longest (input3)
const output4 = longest (input4)
console.log ('output1: ', output1)
console.log ('output2: ', output2)
console.log ('output3: ', output3)
console.log ('output4: ', output4)

function max( input ) {
return input.reduce((a, b) => a.length <= b.length ? b : a)
}

I see the shortest solution
function findLong(s){
return Math.max.apply(null, s.split(' ').map(w => w.length));
}

If your string is already split into an array, you'll not need the split part.
function findLongestWord(str) {
str = str.split(' ');
var longest = 0;
for(var i = 0; i < str.length; i++) {
if(str[i].length >= longest) {
longest = str[i].length;
}
}
return longest;
}
findLongestWord("The quick brown fox jumped over the lazy dog");

In case you expect more than one maximum this will work:
_.maxBy(Object.entries(_.groupBy(x, y => y.length)), y => parseInt(y[0]))[1]
It uses lodash and returns an array.

With ES6 and it support a duplicate string
var allLongestStrings = arrayOfStrings => {
let maxLng = Math.max(...arrayOfStrings.map( elem => elem.length))
return arrayOfStrings.filter(elem => elem.length === maxLng)
}
let arrayOfStrings = ["aba", "aa", "ad", "vcd","aba"]
console.log(allLongestStrings(arrayOfStrings))

I would do something like this:
function findLongestWord(str) {
var array = str.split(" ");
var maxLength=array[0].length;
for(var i=0; i < array.length; i++ ) {
if(array[i].length > maxLength) maxLength = array[i].length
}
return maxLength;
}
findLongestWord("What if we try a super-long word such as otorhinolaryngology");

Modern browsers support a for...of loop. The fastest and shortest way to solve this problem in Chrome, Safari, Edge, and Firefox is also the clearest:
let largest = '';
for (let item of arr) {
if (item.length > largest.length) largest = item
}
In IE, you can use Array.forEach; that's still faster and clearer than sorting or reducing the array.
var largest = '';
arr.forEach(function(item) {
if (item.length > largest.length) largest = item
});

If you want to know the INDEX of the longest item:
var longest = arr.reduce(
(a, b, i) => arr[a].length < b.length ? i : a,
0
);
(which can be a one-liner for those that love that stuff.... but it's split up here for readabilty)