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I have an array of objects
const objects = [a, b, c, d, e, f, g ... ]
and I want them to turn into
const result = [a, [b, c], d, [e, f], g ... ]
Any ideas?
[Edit] My apologies. This is my first post, didn't know I have to show my attempts. I don't think I deserve the mean comments either, be nice people. I solved it after a head-banging 4 hours. Here is my solution:
const result = []
const method = array => {
for (let i = 0; i < array.length; i += 3) {
const set = new Set([array[i + 1], array[i + 2]])
if (i !== array.length - 1) {
result.push(array[i])
result.push(Array.from(set))
} else {
result.push(array[i])
}
}
}
Thanks for the responses guys! I read every single one of them.
You could take a while loop and push either an item or a pair of items.
var array = ['a', 'b', 'c', 'd', 'e', 'f', 'g'],
grouped = [],
i = 0;
while (i < array.length) {
grouped.push(array[i++]);
if (i >= array.length) break;
grouped.push(array.slice(i, i += 2));
}
console.log(grouped);
You can do this with plain for loop and % modulo operator.
const objects = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
const result = []
for(let i = 0; i < objects.length; i++) {
if(i % 3 === 0) {
const arr = objects.slice(i + 1, i + 3)
result.push(objects[i])
if(arr.length) result.push(arr)
}
}
console.log(result)
this is my solution:
const objects = ["a", "b", "c", "d", "e", "f", "g"];
let result = [];
let toGroup = false;
for(let i = 0; i < objects.length ; i++){
if(toGroup){
result.push([objects[i], objects[++i]]);
}
else result.push(objects[i]);
toGroup = !toGroup;
}
this has a particular case that you have not specified, where it doesn't work, for example if inside objects there are 2 elements, and so i don't know what you would like to do in that case
Say I have an array like so:
const alphabet = ['a', 'b', 'c', 'd'];
This represents 4 political candidates and a rank choice vote, where candidate a is first choice, b is second choice, etc.
I want to shuffle this into a bunch of random orders, but in this case I want a to appear first with probably 60%, b second with probability 20%, and c third with probability 10%, and all the other ordering with probably 10%. Is there some lodash and ramda functionality that can accomplish this or?
This is for testing a rank choice voting algorithm. Shuffling the array randomly yields candidates that all have pretty much identical vote counts which doesn't mirror most reality (although I will test for that too).
I have this pretty horrible routine which will generate one random array:
const getValues = function () {
const results = [];
const remaining = new Set(alphabet);
const probabilities = [0.6, 0.2, 0.1, 0.1];
for(let i = 0; i < alphabet.length; i++){
const r = Math.random();
const letter = alphabet[i];
if(r < probabilities[i] && remaining.has(letter)){
results.push(letter);
remaining.delete(letter);
}
else{
const rand = Math.floor(Math.random()*remaining.size);
const x = Array.from(remaining)[rand];
remaining.delete(x);
results.push(x);
}
}
return results;
};
this "works" but doesn't quite order things according to the specified probabilities, because of conditional probability. Does someone know of a good way to have the order appear with certain probability, as I described above?
Here is some sample output that I am looking for:
[ [ 'd', 'b', 'a', 'c' ],
[ 'a', 'b', 'c', 'd' ],
[ 'a', 'd', 'b', 'c' ],
[ 'd', 'b', 'a', 'c' ],
[ 'b', 'c', 'a', 'd' ],
[ 'a', 'b', 'c', 'd' ],
[ 'd', 'b', 'c', 'a' ],
[ 'c', 'd', 'a', 'b' ],
[ 'd', 'b', 'a', 'c' ],
[ 'a', 'b', 'c', 'd' ] ]
if you generated enough data it wouldn't fit the desired order/distribution.
You could sort them using a shuffle function like this:
const candidates = [
{ name: "a", weight: 6 },
{ name: "b", weight: 2 },
{ name: "c", weight: 1 },
{ name: "d", weight: 1 }
];
const randomShuffleFn = () => Math.random() - .5;
const shuffleFn = (candidateA, candidateB) =>
Math.random() * (candidateB.weight + candidateA.weight) - candidateA.weight;
console.log([...candidates].sort(randomShuffleFn).sort(shuffleFn));
OK, it's not exactly the same, but I think with tweaking the weights you can get the required distribution (as it is, A wins more than 60% of times).
You could take a random part of the array and normalize the remaining possibilities and take another one until all items are take.
As result, you get nearly a wanted result, as you see in the counts of items and their final index.
const
getIndex = (prob) => prob.findIndex((r => p => r < p || (r -= p, false))(Math.random())),
normalized = array => {
var sum = array.reduce((a, b) => a + b, 0);
return array.map(v => v / sum);
};
var items = ['a', 'b', 'c', 'd'],
probabilities = [0.6, 0.2, 0.1, 0.1],
counts = { a: { 0: 0, 1: 0, 2: 0, 3: 0 }, b: { 0: 0, 1: 0, 2: 0, 3: 0 }, c: { 0: 0, 1: 0, 2: 0, 3: 0 }, d: { 0: 0, 1: 0, 2: 0, 3: 0 } },
l = 100,
index,
result = [],
subP,
subI,
temp;
while (l--) {
temp = [];
subP = probabilities.slice();
subI = items.slice();
while (subP.length) {
sum = subP.reduce
index = getIndex(normalized(subP));
temp.push(subI[index]);
subI.splice(index, 1);
subP.splice(index, 1);
}
result.push(temp);
}
console.log(result.map(a => a.join()));
result.forEach(a => a.forEach((v, i) => counts[v][i]++));
console.log(counts);
.as-console-wrapper { max-height: 100% !important; top: 0; }
This might hopefully help you, example incorporated for your situation from https://github.com/substack/node-deck
Example
const normalize = function (weights) {
if (typeof weights !== 'object' || Array.isArray(weights)) {
throw 'Not an object'
}
let keys = Object.keys(weights);
if (keys.length === 0) return undefined;
let total = keys.reduce(function (sum, key) {
let x = weights[key];
if (x < 0) {
throw new Error('Negative weight encountered at key ' + key);
}
else if (typeof x !== 'number') {
throw new TypeError('Number expected, got ' + typeof x);
}
else {
return sum + x;
}
}, 0);
return total === 1
? weights
: keys.reduce(function (acc, key) {
acc[key] = weights[key] / total;
return acc;
}, {})
;
};
const pick = function (xs) {
if (Array.isArray(xs)) {
return xs[Math.floor(Math.random() * xs.length)];
}
else if (typeof xs === 'object') {
// Weighted Sample
let weights = normalize(xs);
if (!weights) return undefined;
var n = Math.random();
var threshold = 0;
var keys = Object.keys(weights);
for (let i = 0; i < keys.length; i++) {
threshold += weights[keys[i]];
if (n < threshold) return keys[i];
}
throw new Error('Exceeded threshold. Something is very wrong.');
}
else {
throw new TypeError('Must be an Array or an object');
}
};
const shuffle = function (xs) {
if (Array.isArray(xs)) {
let res = xs.slice();
for (var i = res.length - 1; i >= 0; i--) {
var n = Math.floor(Math.random() * i);
var t = res[i];
res[i] = res[n];
res[n] = t;
}
return res;
}
else if (typeof xs === 'object') {
// Weighted
let weights = Object.keys(xs).reduce(function (acc, key) {
acc[key] = xs[key];
return acc;
}, {});
let ret = [];
while (Object.keys(weights).length > 0) {
let key = pick(weights);
delete weights[key];
ret.push(key);
}
return ret;
}
else {
throw new TypeError('Must be an Array or an object');
}
};
let results = [];
for (let i = 0; i < 100; i++) {
let weighted = shuffle({
a : 60,
b : 20,
c : 10,
d : 10, // or .1, 100, 1000
});
results.push(weighted);
}
console.log(results);
I think the problem is poorly stated.
As it is written, A shall be on place 1 with 60% probability, B on place 2 with 20%, C and D on places 3 or 4 with 10% each. There is no distribution which fulfills these probability criteria, so no algorithm can produce it: If in 60% of the cases A is on place 1, either C or D must be on places 3 or 4 in these 60%, so that's way above the required 10% probability.
So, the first task here is to make sense out of what is written in the question (because of course it can make sense, after interpretation).
I guess the 60% for A and 20% for B should not be read as probability but as a kind of popularity. But it cannot be just the quorum for each candidate because in a voting process A will finish on place 1 in 100% of the cases then.
So, let's assume a voting process with some randomness involved which lets A finish on place 1 with 60% probability, B on place 1 (!) with 20% probability, etc. Then we can implement this using a weighted random choice for place 1.
How to continue with the places 2..n? We just keep the weights intact and remove the candidate which already has been chosen. If one of the other candidates has made it to place 1, then this will make A end with a high probability on place 2 which I think makes sense.
I have two arrays:
var array1 = ["A", "B", "C"];
var array2 = ["1", "2", "3"];
How can I set another array to contain every combination of the above, so that:
var combos = ["A1", "A2", "A3", "B1", "B2", "B3", "C1", "C2", "C3"];
Or if you'd like to create combinations with an arbitrary number of arrays of arbitrary sizes...(I'm sure you can do this recursively, but since this isn't a job interview, I'm instead using an iterative "odometer" for this...it increments a "number" with each digit a "base-n" digit based on the length of each array)...for example...
combineArrays([ ["A","B","C"],
["+", "-", "*", "/"],
["1","2"] ] )
...returns...
[
"A+1","A+2","A-1", "A-2",
"A*1", "A*2", "A/1", "A/2",
"B+1","B+2","B-1", "B-2",
"B*1", "B*2", "B/1", "B/2",
"C+1","C+2","C-1", "C-2",
"C*1", "C*2", "C/1", "C/2"
]
...each of these corresponding to an "odometer" value that
picks an index from each array...
[0,0,0], [0,0,1], [0,1,0], [0,1,1]
[0,2,0], [0,2,1], [0,3,0], [0,3,1]
[1,0,0], [1,0,1], [1,1,0], [1,1,1]
[1,2,0], [1,2,1], [1,3,0], [1,3,1]
[2,0,0], [2,0,1], [2,1,0], [2,1,1]
[2,2,0], [2,2,1], [2,3,0], [2,3,1]
The "odometer" method allows you to easily generate
the type of output you want, not just the concatenated strings
like we have here. Besides that, by avoiding recursion
we avoid the possibility of -- dare I say it? -- a stack overflow...
function combineArrays( array_of_arrays ){
// First, handle some degenerate cases...
if( ! array_of_arrays ){
// Or maybe we should toss an exception...?
return [];
}
if( ! Array.isArray( array_of_arrays ) ){
// Or maybe we should toss an exception...?
return [];
}
if( array_of_arrays.length == 0 ){
return [];
}
for( let i = 0 ; i < array_of_arrays.length; i++ ){
if( ! Array.isArray(array_of_arrays[i]) || array_of_arrays[i].length == 0 ){
// If any of the arrays in array_of_arrays are not arrays or zero-length, return an empty array...
return [];
}
}
// Done with degenerate cases...
// Start "odometer" with a 0 for each array in array_of_arrays.
let odometer = new Array( array_of_arrays.length );
odometer.fill( 0 );
let output = [];
let newCombination = formCombination( odometer, array_of_arrays );
output.push( newCombination );
while ( odometer_increment( odometer, array_of_arrays ) ){
newCombination = formCombination( odometer, array_of_arrays );
output.push( newCombination );
}
return output;
}/* combineArrays() */
// Translate "odometer" to combinations from array_of_arrays
function formCombination( odometer, array_of_arrays ){
// In Imperative Programmingese (i.e., English):
// let s_output = "";
// for( let i=0; i < odometer.length; i++ ){
// s_output += "" + array_of_arrays[i][odometer[i]];
// }
// return s_output;
// In Functional Programmingese (Henny Youngman one-liner):
return odometer.reduce(
function(accumulator, odometer_value, odometer_index){
return "" + accumulator + array_of_arrays[odometer_index][odometer_value];
},
""
);
}/* formCombination() */
function odometer_increment( odometer, array_of_arrays ){
// Basically, work you way from the rightmost digit of the "odometer"...
// if you're able to increment without cycling that digit back to zero,
// you're all done, otherwise, cycle that digit to zero and go one digit to the
// left, and begin again until you're able to increment a digit
// without cycling it...simple, huh...?
for( let i_odometer_digit = odometer.length-1; i_odometer_digit >=0; i_odometer_digit-- ){
let maxee = array_of_arrays[i_odometer_digit].length - 1;
if( odometer[i_odometer_digit] + 1 <= maxee ){
// increment, and you're done...
odometer[i_odometer_digit]++;
return true;
}
else{
if( i_odometer_digit - 1 < 0 ){
// No more digits left to increment, end of the line...
return false;
}
else{
// Can't increment this digit, cycle it to zero and continue
// the loop to go over to the next digit...
odometer[i_odometer_digit]=0;
continue;
}
}
}/* for( let odometer_digit = odometer.length-1; odometer_digit >=0; odometer_digit-- ) */
}/* odometer_increment() */
Just in case anyone is looking for Array.map solution
var array1=["A","B","C"];
var array2=["1","2","3","4"];
console.log(array1.flatMap(d => array2.map(v => d + v)))
Seeing a lot of for loops in all of the answers...
Here's a recursive solution I came up with that will find all combinations of N number of arrays by taking 1 element from each array:
const array1=["A","B","C"]
const array2=["1","2","3"]
const array3=["red","blue","green"]
const combine = ([head, ...[headTail, ...tailTail]]) => {
if (!headTail) return head
const combined = headTail.reduce((acc, x) => {
return acc.concat(head.map(h => `${h}${x}`))
}, [])
return combine([combined, ...tailTail])
}
console.log('With your example arrays:', combine([array1, array2]))
console.log('With N arrays:', combine([array1, array2, array3]))
//-----------UPDATE BELOW FOR COMMENT---------
// With objects
const array4=[{letter: "A"}, {letter: "B"}, {letter: "C"}]
const array5=[{number: 1}, {number: 2}, {number: 3}]
const array6=[{color: "RED"}, {color: "BLUE"}, {color: "GREEN"}]
const combineObjects = ([head, ...[headTail, ...tailTail]]) => {
if (!headTail) return head
const combined = headTail.reduce((acc, x) => {
return acc.concat(head.map(h => ({...h, ...x})))
}, [])
return combineObjects([combined, ...tailTail])
}
console.log('With arrays of objects:', combineObjects([array4, array5, array6]))
A loop of this form
combos = [] //or combos = new Array(2);
for(var i = 0; i < array1.length; i++)
{
for(var j = 0; j < array2.length; j++)
{
//you would access the element of the array as array1[i] and array2[j]
//create and array with as many elements as the number of arrays you are to combine
//add them in
//you could have as many dimensions as you need
combos.push(array1[i] + array2[j])
}
}
Assuming you're using a recent web browser with support for Array.forEach:
var combos = [];
array1.forEach(function(a1){
array2.forEach(function(a2){
combos.push(a1 + a2);
});
});
If you don't have forEach, it is an easy enough exercise to rewrite this without it. As others have proven before, there's also some performance advantages to doing without... (Though I contend that not long from now, the common JavaScript runtimes will optimize away any current advantages to doing this otherwise.)
Solution enhancement for #Nitish Narang's answer.
Use reduce in combo with flatMap to support N arrays combination.
const combo = [
["A", "B", "C"],
["1", "2", "3", "4"]
];
console.log(combo.reduce((a, b) => a.flatMap(x => b.map(y => x + y)), ['']))
Here is functional programming ES6 solution:
var array1=["A","B","C"];
var array2=["1","2","3"];
var result = array1.reduce( (a, v) =>
[...a, ...array2.map(x=>v+x)],
[]);
/*---------OR--------------*/
var result1 = array1.reduce( (a, v, i) =>
a.concat(array2.map( w => v + w )),
[]);
/*-------------OR(without arrow function)---------------*/
var result2 = array1.reduce(function(a, v, i) {
a = a.concat(array2.map(function(w){
return v + w
}));
return a;
},[]
);
console.log(result);
console.log(result1);
console.log(result2)
Part II: After my complicated iterative "odometer" solution of July 2018, here's a simpler recursive version of combineArraysRecursively()...
function combineArraysRecursively( array_of_arrays ){
// First, handle some degenerate cases...
if( ! array_of_arrays ){
// Or maybe we should toss an exception...?
return [];
}
if( ! Array.isArray( array_of_arrays ) ){
// Or maybe we should toss an exception...?
return [];
}
if( array_of_arrays.length == 0 ){
return [];
}
for( let i = 0 ; i < array_of_arrays.length; i++ ){
if( ! Array.isArray(array_of_arrays[i]) || array_of_arrays[i].length == 0 ){
// If any of the arrays in array_of_arrays are not arrays or are zero-length array, return an empty array...
return [];
}
}
// Done with degenerate cases...
let outputs = [];
function permute(arrayOfArrays, whichArray=0, output=""){
arrayOfArrays[whichArray].forEach((array_element)=>{
if( whichArray == array_of_arrays.length - 1 ){
// Base case...
outputs.push( output + array_element );
}
else{
// Recursive case...
permute(arrayOfArrays, whichArray+1, output + array_element );
}
});/* forEach() */
}
permute(array_of_arrays);
return outputs;
}/* function combineArraysRecursively() */
const array1 = ["A","B","C"];
const array2 = ["+", "-", "*", "/"];
const array3 = ["1","2"];
console.log("combineArraysRecursively(array1, array2, array3) = ", combineArraysRecursively([array1, array2, array3]) );
Here is another take. Just one function and no recursion.
function allCombinations(arrays) {
const numberOfCombinations = arrays.reduce(
(res, array) => res * array.length,
1
)
const result = Array(numberOfCombinations)
.fill(0)
.map(() => [])
let repeatEachElement
for (let i = 0; i < arrays.length; i++) {
const array = arrays[i]
repeatEachElement = repeatEachElement ?
repeatEachElement / array.length :
numberOfCombinations / array.length
const everyElementRepeatedLength = repeatEachElement * array.length
for (let j = 0; j < numberOfCombinations; j++) {
const index = Math.floor(
(j % everyElementRepeatedLength) / repeatEachElement
)
result[j][i] = array[index]
}
}
return result
}
const result = allCombinations([
['a', 'b', 'c', 'd'],
[1, 2, 3],
[true, false],
])
console.log(result.join('\n'))
Arbitrary number of arrays, arbitrary number of elements.
Sort of using number base theory I guess - the j-th array changes to the next element every time the number of combinations of the j-1 arrays has been exhausted. Calling these arrays 'vectors' here.
let vectorsInstance = [
[1, 2],
[6, 7, 9],
[10, 11],
[1, 5, 8, 17]]
function getCombos(vectors) {
function countComb(vectors) {
let numComb = 1
for (vector of vectors) {
numComb *= vector.length
}
return numComb
}
let allComb = countComb(vectors)
let combos = []
for (let i = 0; i < allComb; i++) {
let thisCombo = []
for (j = 0; j < vectors.length; j++) {
let vector = vectors[j]
let prevComb = countComb(vectors.slice(0, j))
thisCombo.push(vector[Math.floor(i / prevComb) % vector.length])
}
combos.push(thisCombo)
}
return combos
}
console.log(getCombos(vectorsInstance))
While there's already plenty of good answers to get every combination, which is of course the original question, I'd just like to add a solution for pagination. Whenever there's permutations involved, there's the risk of extremely large numbers. Let's say, for whatever reason, we wanted to build an interface where a user could still browse through pages of practically unlimited permutations, e.g. show permutations 750-760 out of one gazillion.
We could do so using an odometer similar to the one in John's solution. Instead of only incrementing our way through the odometer, we also calculate its initial value, similar to how you'd convert for example seconds into a hh:mm:ss clock.
function getPermutations(arrays, startIndex = 0, endIndex) {
if (
!Array.isArray(arrays) ||
arrays.length === 0 ||
arrays.some(array => !Array.isArray(array))
) {
return { start: 0, end: 0, total: 0, permutations: [] };
}
const permutations = [];
const arrayCount = arrays.length;
const arrayLengths = arrays.map(a => a.length);
const maxIndex = arrayLengths.reduce(
(product, arrayLength) => product * arrayLength,
1,
);
if (typeof endIndex !== 'number' || endIndex > maxIndex) {
endIndex = maxIndex;
}
const odometer = Array.from({ length: arrayCount }).fill(0);
for (let i = startIndex; i < endIndex; i++) {
let _i = i; // _i is modified and assigned to odometer indexes
for (let odometerIndex = arrayCount - 1; odometerIndex >= 0; odometerIndex--) {
odometer[odometerIndex] = _i % arrayLengths[odometerIndex];
if (odometer[odometerIndex] > 0 && i > startIndex) {
// Higher order values in existing odometer are still valid
// if we're not hitting 0, since there's been no overflow.
// However, startIndex always needs to follow through the loop
// to assign initial odometer.
break;
}
// Prepare _i for next odometer index by truncating rightmost digit
_i = Math.floor(_i / arrayLengths[odometerIndex]);
}
permutations.push(
odometer.map(
(odometerValue, odometerIndex) => arrays[odometerIndex][odometerValue],
),
);
}
return {
start: startIndex,
end: endIndex,
total: maxIndex,
permutations,
};
}
So for the original question, we'd do
getPermutations([['A', 'B', 'C'], ['1', '2', '3']]);
-->
{
"start": 0,
"end": 9,
"total": 9,
"permutations": [
["A", "1"],
["A", "2"],
["A", "3"],
["B", "1"],
["B", "2"],
["B", "3"],
["C", "1"],
["C", "2"],
["C", "3"]
]
}
but we could also do
getPermutations([['A', 'B', 'C'], ['1', '2', '3']], 2, 5);
-->
{
"start": 2,
"end": 5,
"total": 9,
"permutations": [
["A", "3"],
["B", "1"],
["B", "2"]
]
}
And more importantly, we could do
getPermutations(
[
new Array(1000).fill(0),
new Array(1000).fill(1),
new Array(1000).fill(2),
new Array(1000).fill(3),
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'],
['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'],
['X', 'Y', 'Z'],
['1', '2', '3', '4', '5', '6']
],
750,
760
);
-->
{
"start": 750,
"end": 760,
"total": 1800000000000000,
"permutations": [
[0, 1, 2, 3, "e", "B", "Z", "1"],
[0, 1, 2, 3, "e", "B", "Z", "2"],
[0, 1, 2, 3, "e", "B", "Z", "3"],
[0, 1, 2, 3, "e", "B", "Z", "4"],
[0, 1, 2, 3, "e", "B", "Z", "5"],
[0, 1, 2, 3, "e", "B", "Z", "6"],
[0, 1, 2, 3, "e", "C", "X", "1"],
[0, 1, 2, 3, "e", "C", "X", "2"],
[0, 1, 2, 3, "e", "C", "X", "3"],
[0, 1, 2, 3, "e", "C", "X", "4"]
]
}
without the computer hanging.
Here's a short recursive one that takes N arrays.
function permuteArrays(first, next, ...rest) {
if (rest.length) next = permuteArrays(next, ...rest);
return first.flatMap(a => next.map(b => [a, b].flat()));
}
Or with reduce (slight enhancement of Penny Liu's):
function multiply(a, b) {
return a.flatMap(c => b.map(d => [c, d].flat()));
}
[['a', 'b', 'c'], ['+', '-'], [1, 2, 3]].reduce(multiply);
Runnable example:
function permuteArrays(first, next, ...rest) {
if (rest.length) next = permuteArrays(next, ...rest);
return first.flatMap(a => next.map(b => [a, b].flat()));
}
const squish = arr => arr.join('');
console.log(
permuteArrays(['A', 'B', 'C'], ['+', '-', '×', '÷'], [1, 2]).map(squish),
permuteArrays(['a', 'b', 'c'], [1, 2, 3]).map(squish),
permuteArrays([['a', 'foo'], 'b'], [1, 2]).map(squish),
permuteArrays(['a', 'b', 'c'], [1, 2, 3], ['foo', 'bar', 'baz']).map(squish),
)
I had a similar requirement, but I needed get all combinations of the keys of an object so that I could split it into multiple objects. For example, I needed to convert the following;
{ key1: [value1, value2], key2: [value3, value4] }
into the following 4 objects
{ key1: value1, key2: value3 }
{ key1: value1, key2: value4 }
{ key1: value2, key2: value3 }
{ key1: value2, key2: value4 }
I solved this with an entry function splitToMultipleKeys and a recursive function spreadKeys;
function spreadKeys(master, objects) {
const masterKeys = Object.keys(master);
const nextKey = masterKeys.pop();
const nextValue = master[nextKey];
const newObjects = [];
for (const value of nextValue) {
for (const ob of objects) {
const newObject = Object.assign({ [nextKey]: value }, ob);
newObjects.push(newObject);
}
}
if (masterKeys.length === 0) {
return newObjects;
}
const masterClone = Object.assign({}, master);
delete masterClone[nextKey];
return spreadKeys(masterClone, newObjects);
}
export function splitToMultipleKeys(key) {
const objects = [{}];
return spreadKeys(key, objects);
}
one more:
const buildCombinations = (allGroups: string[][]) => {
const indexInArray = new Array(allGroups.length);
indexInArray.fill(0);
let arrayIndex = 0;
const resultArray: string[] = [];
while (allGroups[arrayIndex]) {
let str = "";
allGroups.forEach((g, index) => {
str += g[indexInArray[index]];
});
resultArray.push(str);
// if not last item in array already, switch index to next item in array
if (indexInArray[arrayIndex] < allGroups[arrayIndex].length - 1) {
indexInArray[arrayIndex] += 1;
} else {
// set item index for the next array
indexInArray[arrayIndex] = 0;
arrayIndex += 1;
// exclude arrays with 1 element
while (allGroups[arrayIndex] && allGroups[arrayIndex].length === 1) {
arrayIndex += 1;
}
indexInArray[arrayIndex] = 1;
}
}
return resultArray;
};
One example:
const testArrays = [["a","b"],["c"],["d","e","f"]]
const result = buildCombinations(testArrays)
// -> ["acd","bcd","ace","acf"]
My version of the solution by John D. Aynedjian, which I rewrote for my own understanding.
console.log(getPermutations([["A","B","C"],["1","2","3"]]));
function getPermutations(arrayOfArrays)
{
let permutations=[];
let remainder,permutation;
let permutationCount=1;
let placeValue=1;
let placeValues=new Array(arrayOfArrays.length);
for(let i=arrayOfArrays.length-1;i>=0;i--)
{
placeValues[i]=placeValue;
placeValue*=arrayOfArrays[i].length;
}
permutationCount=placeValue;
for(let i=0;i<permutationCount;i++)
{
remainder=i;
permutation=[];
for(let j=0;j<arrayOfArrays.length;j++)
{
permutation[j]=arrayOfArrays[j][Math.floor(remainder/placeValues[j])];
remainder=remainder%placeValues[j];
}
permutations.push(permutation.reduce((prev,curr)=>prev+curr,"")); }
return permutations;
}
First express arrays as array of arrays:
arrayOfArrays=[["A","B","C"],["a","b","c","d"],["1","2"]];
Next work out the number of permuations in the solution by multiplying the number of elements in each array by each other:
//["A","B","C"].length*["a","b","c","d"].length*["1","2"].length //24 permuations
Then give each array a place value, starting with the last:
//["1","2"] place value 1
//["a","b","c","d"] place value 2 (each one of these letters has 2 possibilities to the right i.e. 1 and 2)
//["A","B","C"] place value 8 (each one of these letters has 8 possibilities to the right i.e. a1,a2,b1,b2,c1,c2,d1,d2
placeValues=[8,2,1]
This allows each element to be represented by a single digit:
arrayOfArrays[0][2]+arrayOfArrays[1][3]+arrayOfArrays[2][0] //"Cc1"
...would be:
2*placeValues[2]+3*placesValues[1]+0*placeValues[2] //2*8+3*2+0*1=22
We actually need to do the reverse of this so convert numbers 0 to the number of permutations to an index of each array using quotients and remainders of the permutation number.
Like so:
//0 = [0,0,0], 1 = [0,0,1], 2 = [0,1,0], 3 = [0,1,1]
for(let i=0;i<permutationCount;i++)
{
remainder=i;
permutation=[];
for(let j=0;j<arrayOfArrays.length;j++)
{
permutation[j]=arrayOfArrays[j][Math.floor(remainder/placeValues[j])];
remainder=remainder%placeValues[j];
}
permutations.push(permutation.join(""));
}
The last bit turns the permutation into a string, as requested.
Make a loop like this
->
let numbers = [1,2,3,4,5];
let letters = ["A","B","C","D","E"];
let combos = [];
for(let i = 0; i < numbers.length; i++) {
combos.push(letters[i] + numbers[i]);
};
But you should make the array of “numbers” and “letters” at the same length thats it!
Data with repeated 'i's followed by 'i's and/or 't's.
data = ['i','t','t','i','i','t','t','t']
Trying to retrieve the index of the last 't' in the pattern ['i','t','t']:
[2,6] // ['i','t','t','i','i','t','t','t'] # position of the returned 't's
// _______ ^ _______ ^
I'm looking for a non-recursive solution using (pure) functions only, using ramdajs for example.
Tried to use reduce and transduce, but unsuccessful sofar.
One approach would be to use R.aperture to iterate over a 3-element sliding window of the data list, then tracking the position of any sub-list that equals the pattern ['i', 't', 't'].
const data = ['i','t','t','i','i','t','t','t']
const isPattern = R.equals(['i', 't', 't'])
const reduceWithIdx = R.addIndex(R.reduce)
const positions = reduceWithIdx((idxs, next, idx) =>
isPattern(next) ? R.append(idx + 2, idxs) : idxs
, [], R.aperture(3, data))
console.log(positions)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.24.1/ramda.min.js"></script>
A point-free version of this approach could look something like the following, though whether this is preferable comes down to a preference of style/readability.
const data = ['i','t','t','i','i','t','t','t']
const isPattern = R.equals(['i', 't', 't'])
const run = R.pipe(
// create sliding window of 3 elements
R.aperture(3),
// zip sliding window with index
R.chain(R.zip, R.compose(R.range(0), R.length)),
// filter matching pattern
R.filter(R.compose(isPattern, R.nth(1))),
// extract index
R.map(R.compose(R.add(2), R.head))
)
console.log(run(data))
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.24.1/ramda.min.js"></script>
You could use a nested approach with a temporary array for checking the same pattern for different starting points. This proposal works with an arbitrary length of pattern and returns the index of the predefined pattern.
This solution features obviously plain Javascript.
index i t i t t i i t t t temp result comment
----- ------------------------------ ------ -------- ------------
0 <i> [0] [] match
1 i <t> [0] [] match
<-> [0] [] no match
2 i t <-> [] [] no match
<i> [2] [] match
3 i <t> [2] [] match
<-> [2] [] no match
4 i t <t> [] [4] pattern found
<-> [] [4] no match
5 <i> [5] [4] match
6 i <-> [] [4] no match
<i> [6] [4] match
7 i <t> [6] [4] match
<-> [6] [4] no match
8 i t <t> [] [4, 8] pattern found
<-> [] [4, 8] no match
9 <-> [] [4, 8] no match
<t> matches 't' at position
<-> does not match at position
function getPatternPos(array, pattern) {
var result = [];
array.reduce(function (r, a, i) {
return r.concat(i).filter(function (j) {
if (i - j === pattern.length - 1 && a === pattern[i - j]) {
result.push(i);
return false;
}
return a === pattern[i - j];
});
}, []);
return result;
}
console.log(getPatternPos(['i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [2, 6]
console.log(getPatternPos(['i','t','i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [4, 8]
console.log(getPatternPos(['a', 'b', 'a', 'b', 'b', 'a', 'b', 'c', 'd'], ['a', 'b', 'c']));
// [7]
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You can do it using Array.prototype.reduce() with a simple condition.
data = ['i','t','t','i','i','t','t','t']
var newData = data.reduce(function (acc, item, index) {
// Check if current element is `t` and the item before it is `i`, `t`
if (item === 't' && data[index - 1] === 'i' && data[index - 2] === 't') {
acc.push(item)
}
return acc;
}, []);
console.log(newData); // ['t', 't']
You can do simply by for loop and check last values of array:
var data = ['i','t','t','i','i','t','t','t'];
var positions = new Array();
for(var i=2; i< data.length; i++){
if(data[i-2] === 'i' && data[i-1] === 't' && data[i] === 't') {
positions.push(i)
}
}
console.log(positions)
data.filter((c, i, d) => c === 't' && d[i - 1] === 't' && d[i - 2] === 'I')
**No negative indexes: **
const matchMaker = () => {
let memo = [‘a’, ‘b’];
return (c, i, d) => {
memo.unshift(c);
return memo[1] + memo[2] + c === 'itt';
}
};
data.filter(matchMaker());
function getPattern(arr, p) {
var r = [],
dir = [];
for (let [i, v] of arr.entries()) {
dir = dir.concat(i).filter(function(x) {
if (v === p[i - x] && i - x === p.length - 1) {
r.push(i);
return false;
}
return v === p[i - x];
})
};
return r;
}
console.log(getPattern(['i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
console.log(getPattern(['i', 't', 'i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
console.log(getPattern(['a', 'b', 'a', 'b', 'b', 'a', 'b', 'c', 'd'], ['a', 'b', 'c']));
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In case anyone wants to see an example without using a library that does not use look ahead or behinds ("i + 1" or "i - 2'", etc.).
I think it works similarly to what the Ramda approach does, but I chose to combine the partitioning and equality check in the same loop:
For every step in reduce
Take a section of the array matching the pattern length
Check if it is equal to the pattern
If it is, add the index of the last element in the section to the result of reduce
The code, in which pattern and data are both arrays of strings:
const findPattern = (pattern, data) => data.reduce(
(results, _, i, all) =>
// Check if a slice from this index equals the pattern
arrEqual(all.slice(i, i + pattern.length), pattern)
// Add the last index of the pattern to our results
? (results.push(i + pattern.length - 1), results)
// or, return what we had
: results,
[]);
// Utility method to check array equality
const arrEqual = (arr1, arr2) =>
arr1.length === arr2.length &&
arr1.every((x, i) => x === arr2[i]);
I tested on several data sets and think it meets all requirements:
const findPattern = (pattern, data) => data.reduce(
(results, _, i, all) =>
arrEqual(all.slice(i, i + pattern.length), pattern)
? push(results, i + pattern.length - 1)
: results,
[]);
const arrEqual = (arr1, arr2) =>
arr1.length === arr2.length &&
arr1.every((x, i) => x === arr2[i]);
const push = (xs, x) => (xs.push(x), xs);
// For just string patterns we could also do:
// const arrEqual = (arr1, arr2) => arr1.join("") === arr2.join("");
// Test cases
const dataSets = [
// (i) marks a matching index
// [i] marks a last matching index that should be returned
// | marks a new start
{ pattern: ["i","t","t"], input: ['i','t','t','i','i','t','t','t'], output: [2, 6] },
// |(0) (1) [2]| 3 -(4) (5) [6]| 7
{ pattern: ["i","t"], input: ['i','t','i','t','t','i','i','t','t','t'], output: [1, 3, 7] },
// |(0) [1]|(2) [3]| 4 | 5 |(6) [7]| 8 | 9
{ pattern: ["i","t","t"], input: ['i','t','i','t','t','i','i','t','t','t'], output: [4, 8] },
// |(0) (1)|(2) (3) [4]| 5 |(6) (7) [8]| 9
{ pattern: ["i","t","i"], input: ['i','t','i','t','i','t','i','t','i','t'], output: [2, 4, 6, 8] }
// |(0) (1) [2]| |(6) (7) [8]| 9
// |(2) (3) [4]
// |(4) (5) [6]
];
dataSets.forEach(({pattern, input, output}) =>
console.log(
"| input:", input.join(" "),
"| control:", output.join(", "),
"| answer:", findPattern(pattern, input).join(", ")
)
)
Two years later, lost traveler stumbles upon this question and notices that for variable sized (and especially large pattern with even larger input array or multiple input arrays), classical KMP algorithm would be great.
I think it is worth studing this algorithm.
We will start with simple imperative implementation. Then switch to (at least for me) more intuitive (but probably slightly less optimal, and definitely less optimal when it comes to memory) version with finite automaton. At the end, we'll see something that looks like functional but it is not 100% pure. I wasn't in a mood to torture my self with pure functional implementation of KMP in JS :).
Prefix function KMP, imperative implementation:
function getPatternPos(array, pattern) {
const result = [];
// trying to explain this is a waste of time :)
function createPrefix(pattern) {
// initialize array with zeros
const prefix = Array.apply(null, Array(pattern.length)).map(Number.prototype.valueOf, 0);
let s = 0;
prefix[0] = 0;
for (let i = 1; i < pattern.length; ++i) {
while (s > 0 && pattern[s] !== pattern[i]) {
s = prefix[s - 1];
}
if (pattern[i] === pattern[s]) {
++s;
}
prefix[i] = s;
}
return prefix;
}
const prefix = createPrefix(pattern);
let s = 0;
for (let i = 0; i < array.length; ++i) {
while (s > 0 && pattern[s] !== array[i]) {
s = prefix[s - 1];
}
if (array[i] === pattern[s]) {
++s;
}
if (s === pattern.length) {
result.push(i);
s = 0;
}
}
return result;
}
console.log(getPatternPos(['i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [2, 6]
console.log(getPatternPos(['i','t','i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [4, 8]
console.log(getPatternPos(['a', 'b', 'a', 'b', 'b', 'a', 'b', 'c', 'd'], ['a', 'b', 'c']));
// [7]
console.log(getPatternPos("ababxabababcxxababc".split(""), "ababc".split("")));
// [11, 18]
console.log(getPatternPos("abababcx".split(""), "ababc".split("")));
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Finate automaton KMP implementation:
function getPatternPos(array, pattern) {
const result = [];
function patternCode(i) {
return pattern[i].charCodeAt(0);
}
function createStateMachine(pattern) {
// return single dimensional array as matrix instead of array of arrays,
// for better perfomanse (locality - cache optimizations) and memory usage.
// zero initialize matrix
const sm = Array.apply(null, Array(256 * pattern.length)).map(Number.prototype.valueOf, 0);
let s = 0;
sm[patternCode(0) * pattern.length + 0] = 1;
for (let i = 1; i < pattern.length; ++i) {
// go to same states as if we would go after backing up, so copy all
for (let code = 0; code < 256; ++code)
sm[code * pattern.length + i] = sm[code * pattern.length + s];
// only in case of current symbol go to different/next state
sm[patternCode(i) * pattern.length + i] = i + 1;
// update the state that fallows backup path
s = sm[patternCode(i) * pattern.length + s];
}
return sm;
}
const sm = createStateMachine(pattern);
numStates = pattern.length;
let s = 0;
// now simply fallow state machine
for (let i = 0; i < array.length; ++i) {
s = sm[array[i].charCodeAt(0) * numStates + s];
if (s === pattern.length) {
result.push(i);
s = 0;
}
}
return result;
}
console.log(getPatternPos(['i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [2, 6]
console.log(getPatternPos(['i','t','i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [4, 8]
console.log(getPatternPos(['a', 'b', 'a', 'b', 'b', 'a', 'b', 'c', 'd'], ['a', 'b', 'c']));
// [7]
console.log(getPatternPos("ababxabababcxxababc".split(""), "ababc".split("")));
// [11, 18]
console.log(getPatternPos("abababcx".split(""), "ababc".split("")));
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Funcational-ish KMP implementation:
function getPatternPos(array, pattern) {
// pure function that creates state machine,
// but it's implementation is not complitely pure internally.
function createStateMachine(pattern) {
const initState = Object.create(null);
initState[pattern[0]] = Object.create(initState);
const {currState: finalState} = pattern.slice(1).reduce(function(acc, cval, cidx) {
const newFallbackState = acc.fallbackState[cval] || initState;
// WARNING: non-functional/immutable part,
// to make it complitely pure we would probably need to
// complicate our lives with better data structures or
// lazy evalutaion.
acc.currState[cval] = Object.create(newFallbackState);
return {currState: acc.currState[cval], fallbackState: newFallbackState};
}, {currState: initState[pattern[0]], fallbackState: initState});
return {initState: initState, finalState: finalState};
}
const {initState, finalState} = createStateMachine(pattern);
return array.reduce(function (acc, cval, cidx, array) {
const newState = acc.currState[cval];
if (typeof newState === 'undefined') {
return {currState: initState, result: acc.result};
}
if (newState === finalState) {
// WARNING: not purly functional/immutable,
// still implemenations of JS pure functional/immutable libraries
// probaly use mutation under the hood, and just make it look pure,
// this is what happens here also :)
acc.result.push(cidx);
return {currState: initState, result: acc.result};
}
return {currState: newState, result: acc.result};
}, {currState: initState, result: []}).result;
}
console.log(getPatternPos(['i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [2, 6]
console.log(getPatternPos(['i','t','i', 't', 't', 'i', 'i', 't', 't', 't'], ['i', 't', 't']));
// [4, 8]
console.log(getPatternPos(['a', 'b', 'a', 'b', 'b', 'a', 'b', 'c', 'd'], ['a', 'b', 'c']));
// [7]
console.log(getPatternPos("ababxabababcxxababc".split(""), "ababc".split("")));
// [11, 18]
console.log(getPatternPos("abababcx".split(""), "ababc".split("")));
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How can I produce all of the combinations of the values in N number of JavaScript arrays of variable lengths?
Let's say I have N number of JavaScript arrays, e.g.
var first = ['a', 'b', 'c', 'd'];
var second = ['e'];
var third = ['f', 'g', 'h', 'i', 'j'];
(Three arrays in this example, but its N number of arrays for the problem.)
And I want to output all the combinations of their values, to produce
aef
aeg
aeh
aei
aej
bef
beg
....
dej
EDIT: Here's the version I got working, using ffriend's accepted answer as the basis.
var allArrays = [['a', 'b'], ['c', 'z'], ['d', 'e', 'f']];
function allPossibleCases(arr) {
if (arr.length === 0) {
return [];
}
else if (arr.length ===1){
return arr[0];
}
else {
var result = [];
var allCasesOfRest = allPossibleCases(arr.slice(1)); // recur with the rest of array
for (var c in allCasesOfRest) {
for (var i = 0; i < arr[0].length; i++) {
result.push(arr[0][i] + allCasesOfRest[c]);
}
}
return result;
}
}
var results = allPossibleCases(allArrays);
//outputs ["acd", "bcd", "azd", "bzd", "ace", "bce", "aze", "bze", "acf", "bcf", "azf", "bzf"]
This is not permutations, see permutations definitions from Wikipedia.
But you can achieve this with recursion:
var allArrays = [
['a', 'b'],
['c'],
['d', 'e', 'f']
]
function allPossibleCases(arr) {
if (arr.length == 1) {
return arr[0];
} else {
var result = [];
var allCasesOfRest = allPossibleCases(arr.slice(1)); // recur with the rest of array
for (var i = 0; i < allCasesOfRest.length; i++) {
for (var j = 0; j < arr[0].length; j++) {
result.push(arr[0][j] + allCasesOfRest[i]);
}
}
return result;
}
}
console.log(allPossibleCases(allArrays))
You can also make it with loops, but it will be a bit tricky and will require implementing your own analogue of stack.
I suggest a simple recursive generator function as follows:
// Generate cartesian product of given iterables:
function* cartesian(head, ...tail) {
let remainder = tail.length ? cartesian(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// Example:
const first = ['a', 'b', 'c', 'd'];
const second = ['e'];
const third = ['f', 'g', 'h', 'i', 'j'];
console.log(...cartesian(first, second, third));
You don't need recursion, or heavily nested loops, or even to generate/store the whole array of permutations in memory.
Since the number of permutations is the product of the lengths of each of the arrays (call this numPerms), you can create a function getPermutation(n) that returns a unique permutation between index 0 and numPerms - 1 by calculating the indices it needs to retrieve its characters from, based on n.
How is this done? If you think of creating permutations on arrays each containing: [0, 1, 2, ... 9] it's very simple... the 245th permutation (n=245) is "245", rather intuitively, or:
arrayHundreds[Math.floor(n / 100) % 10]
+ arrayTens[Math.floor(n / 10) % 10]
+ arrayOnes[Math.floor(n / 1) % 10]
The complication in your problem is that array sizes differ. We can work around this by replacing the n/100, n/10, etc... with other divisors. We can easily pre-calculate an array of divisors for this purpose. In the above example, the divisor of 100 was equal to arrayTens.length * arrayOnes.length. Therefore we can calculate the divisor for a given array to be the product of the lengths of the remaining arrays. The very last array always has a divisor of 1. Also, instead of modding by 10, we mod by the length of the current array.
Example code is below:
var allArrays = [first, second, third, ...];
// Pre-calculate divisors
var divisors = [];
for (var i = allArrays.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * allArrays[i + 1].length : 1;
}
function getPermutation(n) {
var result = "", curArray;
for (var i = 0; i < allArrays.length; i++) {
curArray = allArrays[i];
result += curArray[Math.floor(n / divisors[i]) % curArray.length];
}
return result;
}
Provided answers looks too difficult for me. So my solution is:
var allArrays = new Array(['a', 'b'], ['c', 'z'], ['d', 'e', 'f']);
function getPermutation(array, prefix) {
prefix = prefix || '';
if (!array.length) {
return prefix;
}
var result = array[0].reduce(function(result, value) {
return result.concat(getPermutation(array.slice(1), prefix + value));
}, []);
return result;
}
console.log(getPermutation(allArrays));
You could take a single line approach by generating a cartesian product.
result = items.reduce(
(a, b) => a.reduce(
(r, v) => r.concat(b.map(w => [].concat(v, w))),
[]
)
);
var items = [['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']],
result = items.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(' ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
Copy of le_m's Answer to take Array of Arrays directly:
function *combinations(arrOfArr) {
let [head, ...tail] = arrOfArr
let remainder = tail.length ? combinations(tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
Hope it saves someone's time.
You can use a typical backtracking:
function cartesianProductConcatenate(arr) {
var data = new Array(arr.length);
return (function* recursive(pos) {
if(pos === arr.length) yield data.join('');
else for(var i=0; i<arr[pos].length; ++i) {
data[pos] = arr[pos][i];
yield* recursive(pos+1);
}
})(0);
}
I used generator functions to avoid allocating all the results simultaneously, but if you want you can
[...cartesianProductConcatenate([['a', 'b'], ['c', 'z'], ['d', 'e', 'f']])];
// ["acd","ace","acf","azd","aze","azf","bcd","bce","bcf","bzd","bze","bzf"]
Easiest way to find the Combinations
const arr1= [ 'a', 'b', 'c', 'd' ];
const arr2= [ '1', '2', '3' ];
const arr3= [ 'x', 'y', ];
const all = [arr1, arr2, arr3];
const output = all.reduce((acc, cu) => {
let ret = [];
acc.map(obj => {
cu.map(obj_1 => {
ret.push(obj + '-' + obj_1)
});
});
return ret;
})
console.log(output);
If you're looking for a flow-compatible function that can handle two dimensional arrays with any item type, you can use the function below.
const getUniqueCombinations = <T>(items : Array<Array<T>>, prepend : Array<T> = []) : Array<Array<T>> => {
if(!items || items.length === 0) return [prepend];
let out = [];
for(let i = 0; i < items[0].length; i++){
out = [...out, ...getUniqueCombinations(items.slice(1), [...prepend, items[0][i]])];
}
return out;
}
A visualisation of the operation:
in:
[
[Obj1, Obj2, Obj3],
[Obj4, Obj5],
[Obj6, Obj7]
]
out:
[
[Obj1, Obj4, Obj6 ],
[Obj1, Obj4, Obj7 ],
[Obj1, Obj5, Obj6 ],
[Obj1, Obj5, Obj7 ],
[Obj2, Obj4, Obj6 ],
[Obj2, Obj4, Obj7 ],
[Obj2, Obj5, Obj6 ],
[Obj2, Obj5, Obj7 ],
[Obj3, Obj4, Obj6 ],
[Obj3, Obj4, Obj7 ],
[Obj3, Obj5, Obj6 ],
[Obj3, Obj5, Obj7 ]
]
You could create a 2D array and reduce it. Then use flatMap to create combinations of strings in the accumulator array and the current array being iterated and concatenate them.
const data = [ ['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j'] ]
const output = data.reduce((acc, cur) => acc.flatMap(c => cur.map(n => c + n)) )
console.log(output)
2021 version of David Tang's great answer
Also inspired with Neil Mountford's answer
const getAllCombinations = (arraysToCombine) => {
const divisors = [];
let permsCount = 1;
for (let i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
permsCount *= (arraysToCombine[i].length || 1);
}
const getCombination = (n, arrays, divisors) => arrays.reduce((acc, arr, i) => {
acc.push(arr[Math.floor(n / divisors[i]) % arr.length]);
return acc;
}, []);
const combinations = [];
for (let i = 0; i < permsCount; i++) {
combinations.push(getCombination(i, arraysToCombine, divisors));
}
return combinations;
};
console.log(getAllCombinations([['a', 'b'], ['c', 'z'], ['d', 'e', 'f']]));
Benchmarks: https://jsbench.me/gdkmxhm36d/1
Here's a version adapted from the above couple of answers, that produces the results in the order specified in the OP, and returns strings instead of arrays:
function *cartesianProduct(...arrays) {
if (!arrays.length) yield [];
else {
const [tail, ...head] = arrays.reverse();
const beginning = cartesianProduct(...head.reverse());
for (let b of beginning) for (let t of tail) yield b + t;
}
}
const first = ['a', 'b', 'c', 'd'];
const second = ['e'];
const third = ['f', 'g', 'h', 'i', 'j'];
console.log([...cartesianProduct(first, second, third)])
You could use this function too:
const result = (arrayOfArrays) => arrayOfArrays.reduce((t, i) => { let ac = []; for (const ti of t) { for (const ii of i) { ac.push(ti + '/' + ii) } } return ac })
result([['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']])
// which will output [ 'a/e/f', 'a/e/g', 'a/e/h','a/e/i','a/e/j','b/e/f','b/e/g','b/e/h','b/e/i','b/e/j','c/e/f','c/e/g','c/e/h','c/e/i','c/e/j','d/e/f','d/e/g','d/e/h','d/e/i','d/e/j']
Of course you can remove the + '/' in ac.push(ti + '/' + ii) to eliminate the slash from the final result. And you can replace those for (... of ...) with forEach functions (plus respective semicolon before return ac), whatever of those you are more comfortable with.
An array approach without recursion:
const combinations = [['1', '2', '3'], ['4', '5', '6'], ['7', '8']];
let outputCombinations = combinations[0]
combinations.slice(1).forEach(row => {
outputCombinations = outputCombinations.reduce((acc, existing) =>
acc.concat(row.map(item => existing + item))
, []);
});
console.log(outputCombinations);
let arr1 = [`a`, `b`, `c`];
let arr2 = [`p`, `q`, `r`];
let arr3 = [`x`, `y`, `z`];
let result = [];
arr1.forEach(e1 => {
arr2.forEach(e2 => {
arr3.forEach(e3 => {
result[result.length] = e1 + e2 + e3;
});
});
});
console.log(result);
/*
output:
[
'apx', 'apy', 'apz', 'aqx',
'aqy', 'aqz', 'arx', 'ary',
'arz', 'bpx', 'bpy', 'bpz',
'bqx', 'bqy', 'bqz', 'brx',
'bry', 'brz', 'cpx', 'cpy',
'cpz', 'cqx', 'cqy', 'cqz',
'crx', 'cry', 'crz'
]
*/
A solution without recursion, which also includes a function to retrieve a single combination by its id:
function getCombination(data, i) {
return data.map(group => {
let choice = group[i % group.length]
i = (i / group.length) | 0;
return choice;
});
}
function* combinations(data) {
let count = data.reduce((sum, {length}) => sum * length, 1);
for (let i = 0; i < count; i++) {
yield getCombination(data, i);
}
}
let data = [['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']];
for (let combination of combinations(data)) {
console.log(...combination);
}