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Shuffle nested arrays in Javascript

I'm trying to sort multiple arrays within an array (which also has to be shuffled). A simplified example is:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
...
];
const shuffled = shuffle(toShuffle);
// outout would look something like:
// [
// [8, 6, 5, 7, 9],
// [4, 3, 1, 5, 2],
// [19, 26, 10, 67],
// ...
// ]
This needs to be flexible, so any number of arrays with any amount of values should be valid.
Here is what I've tried:
function shuffle(a) {
for (let e in a) {
if (Array.isArray(a[e])) {
a[e] = shuffle(a[e]);
} else {
a.splice(e, 1);
a.splice(Math.floor(Math.random() * a.length), 0, a[e]);
}
}
return a;
}
console.log("Shuffled: " + shuffle([
[1, 2, 3, 4, 5],
[5, 4, 3, 2, 1]
]))
But it's not working as intended. Is their an easier way to do this? Or is my code correct and just buggy.

You can use Array.from() to create a new shallow-copied array and then to shuffle Array.prototype.sort() combined with Math.random()
Code:
const toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
]
const shuffle = a => Array.from(a).sort(() => .5 - Math.random())
const result = toShuffle.map(shuffle)
console.log('Shuffled:', JSON.stringify(result))
console.log('To shuffle:', JSON.stringify(toShuffle))

You almost got it. The problem is that you are removing one item from an array, instead of capturing the removed item and them placing in a random position:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
];
function shuffle(a) {
a = [...a]; //clone array
for (let e in a) {
if (Array.isArray(a[e])) {
a[e] = shuffle(a[e]);
} else {
a.splice(~~(Math.random() * a.length), 0, a.splice(e, 1)[0]);
}
}
return a;
}
console.log(JSON.stringify(shuffle(toShuffle)))
console.log(JSON.stringify(toShuffle))
[EDIT]
The original code did not shuffle the parent array, if you need shuffle everything recursively, you can use this:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
];
function shuffle(a) {
a = a.map(i => Array.isArray(i) ? shuffle(i) : i); //clone array
a.sort(i => ~~(Math.random() * 2) - 1); //shuffle
return a;
}
console.log("shuffled", JSON.stringify(shuffle(toShuffle)))
console.log("original", JSON.stringify(toShuffle))

Delete all elements from an array A which are present in an array B without using a double loop

So, as an input I have two arrays, A and B. Let's suppose that these are the values inside the two:
A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and B = [1, 3, 5, 7, 9]
After the deletion the array A should be [2, 4, 6, 8, 10].
I have written (Javascript) this functioning algorithm to solve this problem:
for (var i=0; i < A.length; i++) {
for (var j=0; j < B.length; j++) {
if(B[j] == A[i])
A.splice(i, 1) // Removes 1 element of the array starting from position i
}
}
I would like to know, is it possible to solve this problem without using a double loop?

What about this:
let A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] ;
const B = [1, 3, 5, 7, 9];
A = A.filter(num => !B.includes(num));

Yes it is. You could use a Set. In terms of Set operations you are computing the difference A \ B.
Using a set which is optimized for lookups in O(1) time will speed up the computing the difference siginificantly from O(n²) when using includes() or double for loop to O(n).
const A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
const B = [1, 3, 5, 7, 9]
const setB = new Set(B);
const difference = A.filter(x => !setB.has(x));
console.log(difference);

Maybe that ?
const
A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
, B = [1, 2, 3, 5, 7, 9] // no gaps in 1,2 and 2,3
;
for (let i =0, j=0 ; i < A.length; i++)
{
if (A[i]===B[j]) { A.splice(i--,1); j++ }
}
document.write( JSON.stringify(A) )
or (faster code)
const
A = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
, B = [ 1, 3, 5, 7, 9 ]
;
for (let i = A.length, j= B.length -1 ; i-- ; )
{
if (A[i]===B[j]) { A.splice(i,1); j-- }
}
document.write( JSON.stringify(A) )

picking a unique set from an array of arrays

I attempted to ask a more complicated of this before but I couldn't explain it well so I am trying again with a simplified use case.
I will have an array of arrays like the following
var allData = [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
I need to select 1 element from each array so that I get a unique set like [2,4,1,3,5] easy to do in this case as each array has all values. However this will rarely be the case. Instead I may have
var allData = [[1,2,4],[1,2],[1,2],[2,4,5],[1,2,3,5]]
In this case I couldn't pick 1 or 2 from the first array as that would prevent the 2nd and 3rd from having a unique combination. So something like [4,2,1,5,3] or [4,1,2,5,3] would be the only two possible answers for this combination.
The only way I see to do this is to just go through every combination but these will get fairly large so it doesn't seem reasonable as this happens real time. There are going to be at least 7 arrays, possibly 14 and distantly possible to have 31 so going through every combination would be fairly rough.
The 2nd part is if there is some way to "know" you have the best possible option. Say if there was some way I would know that having a single duplicate is my best case scenario. Even if I have to brute force it if I encounter a 1 duplication solution I would know to stop.
One easy way to get a very simple of this is to just subtract the number of possible choices from the number of elements but this is the correct answer in only the simplest of cases. Is there some type of library or anything to help solve these types of problems? It is a bit beyond my math abilities.
Here is something I have tried but it is too slow for larger sets and can fail. It works sometimes for the 2nd case I presented but only on luck
const allData = [[1,2,4],[1,2],[1,2],[2,4,5],[1,2,3,5]]
var selectedData = []
for (var i in allData){
console.log("length",allData[i].length)
var j = 0
while(j < allData[i].length){
console.log("chekcing",allData[i][j])
if (selectedData.includes(allData[i][j])){
console.log("removing item")
allData[i].splice(j,1)
}
else{j++}
}
var uniqueIds = Object.keys(allData[i])
console.log(uniqueIds)
var randId = Math.floor(Math.random() * uniqueIds.length)
console.log(randId)
selectedData.push(allData[i][randId])
console.log("selectedData",selectedData)
}

You can start with a fairly simple backtracking algorithm:
function pick(bins, n = 0, res = {}) {
if (n === bins.length) {
return res
}
for (let x of bins[n]) {
if (!res[x]) {
res[x] = n + 1
let found = pick(bins, n + 1, res)
if (found)
return found
res[x] = 0
}
}
}
//
let a = [[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 4]]
console.log(pick(a))
This returns a mapping item => bin index + 1, which is easy to convert back to an array if needed.
This should perform relatively well for N < 10, for more/larger bins you can think of some optimizations, for example, avoid the worst case scenario by sorting bins from smallest to longest, or, depending on the nature of elements, represent bins as bitmasks.

You could count all elements and take various comparison with same indices.
function x([...data]) {
while (data.some(Array.isArray)) {
const
counts = data.reduce((r, a, i) => {
if (Array.isArray(a)) a.forEach(v => (r[JSON.stringify(v)] = r[JSON.stringify(v)] || []).push(i));
return r;
}, {}),
entries = Object.entries(counts),
update = ([k, v]) => {
if (v.length === 1) {
data[v[0]] = JSON.parse(k);
return true;
}
};
if (entries.some(update)) continue;
const grouped = entries.reduce((r, [, a]) => {
const key = JSON.stringify(a);
r[key] = (r[key] || 0) + 1;
return r;
}, {});
Object.entries(grouped).forEach(([json, length]) => {
const indices = JSON.parse(json);
if (indices.length === length) {
let j = 0;
indices.forEach(i => data[i] = data[i][j++]);
return;
}
if (length === 1) {
const value = JSON.parse(entries.find(([_, a]) => JSON.stringify(a) === json)[0]);
indices.forEach(i => data[i] = data[i].filter(v => v !== value));
data[indices[0]] = value;
}
});
}
return data;
}
console.log(...x([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]));
console.log(...x([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5]]));
console.log(...x([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]));

Here is an implementation based around counting occurrences across the arrays.
It first creates a map indexed by value counting the number of inner arrays each value occurs in. It then sorts by inner array length to prioritize shorter arrays, and then iterates over each inner array, sorting by occurrence and selecting the first non-duplicate with the lowest count, or, if there are no unique values, the element with the lowest count.
const
occurrencesAcrossArrays = (arr) =>
arr
.reduce((a, _arr) => {
[...new Set(_arr)].forEach(n => {
a[n] = a[n] || 0;
a[n] += 1;
});
return a;
}, {}),
generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr)
return arr
.sort((a, b) => a.length - b.length)
.reduce((a, _arr) => {
_arr.sort((a, b) => dist[a] - dist[b]);
let m = _arr.find(n => !a.includes(n));
if (m !== undefined) {
a.push(m);
} else {
a.push(_arr[0]);
}
return a;
}, []);
};
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
Edit
In response to your comment – The situation seems to be emerging because the values all have the same occurrence count and are sequential.
This can be solved by keeping a running count of each value in the result array, and sorting each inner array by both by this running occurrence count as well as the original distribution count.This adds complexity to the sort, but allows you to simply access the first element in the array (the element with the lowest rate of occurrence in the result with the lowest occurrence count across all arrays).
const
occurrencesAcrossArrays = (arr) =>
arr
.reduce((a, _arr) => {
[...new Set(_arr)].forEach(n => {
a[n] = a[n] || 0;
a[n] += 1;
});
return a;
}, {}),
generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr)
return arr
.sort((a, b) => a.length - b.length)
.reduce((acc, _arr) => {
_arr.sort((a, b) => (acc.occurrences[a] || 0) - (acc.occurrences[b] || 0) || dist[a] - dist[b]);
let m = _arr[0]
acc.occurrences[m] = acc.occurrences[m] || 0;
acc.occurrences[m] += 1;
acc.result.push(m);
return acc;
}, { result: [], occurrences: {} })
.result; // return the .result property of the accumulator
};
console.log(generateCombination([[2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6]]).toString());
// 2,3,4,5,6,2,3
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
// 1,2,3,4,5
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
// 1,2,4,5,3
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
//1,2,4,5,3,9,6,10,7,8
console.log(generateCombination([[1], [2, 3,], [3, 4, 5], [3, 4, 5, 6], [2, 3, 4, 5, 6, 7]]).toString());
// 1,2,4,6,7
A note on .reduce()
If you're having trouble getting your head around .reduce() you can rewrite all the instances of it in this example using .forEach() and declaring accumulator variables outside of the loop. (This will not always be the case, depending on how you manipulate the accumulator value within a reduce() call).
Example below:
const occurrencesAcrossArrays = (arr) => {
const occurrences = {};
arr.forEach(_arr => {
[...new Set(_arr)].forEach(n => {
occurrences[n] = occurrences[n] || 0;
occurrences[n] += 1;
});
});
return occurrences;
};
const generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr);
const result = [];
const occurrences = {};
arr.sort((a, b) => a.length - b.length);
arr.forEach(_arr => {
_arr.sort((a, b) => (occurrences[a] || 0) - (occurrences[b] || 0) || dist[a] - dist[b]);
let m = _arr[0]
occurrences[m] = occurrences[m] || 0;
occurrences[m] += 1;
result.push(m);
});
return result;
};
console.log(generateCombination([[2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6]]).toString());
// 2,3,4,5,6,2,3
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
// 1,2,3,4,5
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
// 1,2,4,5,3
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
//1,2,4,5,3,9,6,10,7,8
console.log(generateCombination([[1], [2, 3,], [3, 4, 5], [3, 4, 5, 6], [2, 3, 4, 5, 6, 7]]).toString());
// 1,2,4,6,7

You could solve this problem using a MILP-model. Here is one implementation in MiniZinc (data has been extended to seven days):
int: Days = 7;
int: Items = 5;
set of int: DAY = 1..Days;
set of int: ITEM = 1..Items;
array[DAY, ITEM] of 0..1: A = % whether item k is allowed on day i
[| 1, 1, 0, 1, 0
| 1, 1, 0, 0, 0
| 1, 1, 0, 0, 0
| 0, 1, 0, 1, 1
| 1, 1, 0, 0, 0
| 0, 1, 0, 1, 1
| 1, 1, 1, 0, 1 |];
array[DAY, ITEM] of var 0..1: x; % 1 if item selected k on day i, otherwise 0
array[DAY, DAY, ITEM] of var 0..1: w; % 1 if item k selected on both day i and day j, otherwise 0
% exactly one item per day
constraint forall(i in DAY)
(sum(k in ITEM)(x[i, k]) = 1);
% linking variables x and w
constraint forall(i, j in DAY, k in ITEM where i < j)
(w[i, j, k] <= x[i, k] /\ w[i, j, k] <= x[j, k] /\ w[i, j, k] >= x[i, k] + x[j, k] - 1);
% try to minimize duplicates and if there are duplicates put them as far apart as possible
var int: obj = sum(i, j in DAY, k in ITEM where i < j)(((Days - (j - i))^2)*w[i, j, k]);
solve minimize obj;
output
["obj="] ++ [show(obj)] ++
["\nitem="] ++ [show([sum(k in ITEM)(k*x[i, k]) | i in DAY])];
Running gives:
obj=8
item=[2, 1, 5, 4, 3, 2, 1]
The following package looks promising for a JavaScript implementation: https://www.npmjs.com/package/javascript-lp-solver

Reorder with recursive function

How to use recursive function to reordered array with even indices first and then odd indices?
For example:
input: [5, 2, 4, 9]
output: [5, 4, 2, 9]
I want to improve this code.
let arrayB = [],
arrayOdd = [],
arrayEven = [];
let i = 0;
const reorder = (arrayA) => {
if (arrayA.length >= 2) {
if (i < arrayA.length) {
i === 0 || i % 2 === 0 ? arrayEven.push(arrayA[i]) : arrayOdd.push(arrayA[i]);
i++;
arrayB = [...arrayEven, ...arrayOdd];
reorder(arrayA);
} else {
arrayOdd = [];
arrayEven = [];
i = 0;
}
return arrayB;
}
}
console.log(reorder([4, 8, 12, 16]));
console.log(reorder([1, 2, 3, 4, 5, 6, 7, 8, 9]));

No idea why you'd need a recursive function. This should do it:
const reorder = (a) => [...a.filter((_, i) => !(i % 2)), ...a.filter((_, i) => i % 2)];
console.log(reorder([4, 8, 12, 16]));
console.log(reorder([1, 2, 3, 4, 5, 6, 7, 8, 9]));
Because you stipulate your in comments that you have instructions to use a recursive function with minimum space requirements, here's a solution that uses Array.prototype.splice() to modify the array in place:
const reorder = (a, offset = 2) => {
if (offset < a.length) {
a.splice(offset / 2, 0, ...a.splice(offset, 1));
return reorder(a, offset + 2);
}
return a;
};
console.log(reorder([4, 8, 12, 16]));
console.log(reorder([1, 2, 3, 4, 5, 6, 7, 8, 9]));
Note that, like the Array.prototype.sort() function, the above function only returns the array for convenience. No new array is created.

I think you can do this by using sort method of array. but for that you might need to use map to get index and then converting back to values.
const isOdd = n => n % 2 === 1;
const reorder = input => input.map((value, index) => ({index, value})).sort((a,b) =>
isOdd(a.index) ? 1 : isOdd(b.index) ? -1 : 0).map(({value}) => value);
console.log(reorder([4, 8, 12, 16]));
console.log(reorder([1, 2, 3, 4, 5, 6, 7, 8, 9]));

Avoid using global variables.
const reorder = (arrayA, arrayOdd = [], arrayEven = []) => {
if (arrayA.length >= 1)
arrayEven.push(arrayA.shift());
if (arrayA.length >= 1) {
arrayOdd.push(arrayA.shift());
reorder(arrayA, arrayOdd, arrayEven);
}
return [...arrayEven, ...arrayOdd];
}
console.log(reorder([4, 8, 12, 16]));
console.log(reorder([1, 2, 3, 4, 5, 6, 7, 8, 9]));

I don't know why you would choose to do this recursively, but this seems to do it:
const evenThenOdd = ([e = undefined, o = undefined, ...xs], evens = [], odds = []) =>
e == undefined
? [...evens, ...odds]
: o == undefined
? evenThenOdd ( xs, [...evens, e] [odds] )
: evenThenOdd ( xs, [...evens, e], [...odds, o] )
console .log (
evenThenOdd ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
//~> [0, 2, 4, 6, 8, 1, 3, 5, 7, 9]
)
We keep evens and odds arrays of values. The base case, when there is nothing left in the array, is to just concatenate those two arrays. The other two cases are when there is one item left and where there are two or more items left. If the undefined values are legitimate elements of your array you want to handle, then add const None = Symbol(), and replace each undefined with None.
A cleaner, non-recursive solution might look like this:
const evenThenOdd = (xs) => [
...xs.filter((_, i) => i % 2 == 0),
...xs.filter((_, i) => i % 2 == 1),
]

Merge elements of array with comma with max counter

So I have an array of ids something like this:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
I need a function that will be called like mergeArray(arr, 3), and it should return comma separated values with maximum of 3 elements like this:
const newArr = ['1,2,3', '4,5,6', '7,8,9', '10,11'];
How can I do this? If possible with ES6 functions for simpler code.

slice your array into 3 lengths arrays and directly join them
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
const mergeArray = (arr, size) => {
let res = [];
for (i = 0; i < arr.length; i += size) {
res.push(arr.slice(i, i + size).join(','));
}
return res;
}
console.log(mergeArray(arr, 3));

You can split() the array into the specific size and join() them before pushing into the resulting array:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
var i, j, newArr=[], size = 3;
for (i=0,j=arr.length; i<j; i+=size) {
newArr.push(arr.slice(i, i+size).join());
}
console.log(newArr);

One of the ways to do it is with Array.prototype.reduce and Array.prototype.map:
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
function mergeArray(arr, n) {
return arr
.reduce((all, el, i) => {
const ind = Math.floor(i/n);
all[ind] = [...all[ind] || [], el]
return all;
},[])
.map(a => a.join(','))
}
console.log(mergeArray(arr, 3));

You could join the array and match the wanted parts with a regular expression.
var data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
result = data.join(',').match(/\d+(,\d+(,\d+)?)?/g)
console.log(result);