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JavaScript - transform this data structure

When I have this array with this data structure:
[[3, 16, 1]
[3, 17, 1]
[3, 18, 1]
[3, 19, 1]
[3, 20, 1]
[3, 19, 1]
[3, 21, 1]
[3, 23, 1]
[3, 16, 1]
[3, 17, 1]
[2, 24, 1]
[2, 25, 1]
[2, 26, 1]
[2, 24, 1]]
and I want to count array elements and multiply this to the last number of the array, like this:
[[3, 16, 2]
[3, 17, 2]
[3, 18, 1]
[3, 19, 2]
[3, 20, 1]
[3, 21, 1]
[3, 23, 1]
[2, 24, 2]
[2, 25, 1]
[2, 26, 1]]
I tried to do it with:
array.forEach(function(x) {
counts[x] = (counts[x] || 0)+1;
});
but this changes the data structure, is there another form to do it?

I rewrite your function and console.log the output you want.
Whenever I check a new array, I store it in checkedArr and will check against it to prevent duplicate records.
const arr = [
[3, 16, 1],
[3, 17, 1],
[3, 18, 1],
[3, 19, 1],
[3, 20, 1],
[3, 19, 1],
[3, 21, 1],
[3, 23, 1],
[3, 16, 1],
[3, 17, 1],
[2, 24, 1],
[2, 25, 1],
[2, 26, 1],
[2, 24, 1],
];
const checkedArr = [];
const output = [];
for (let i = 0; i < arr.length; i++) {
let count = 0;
if (!isArrayChecked(arr[i])) {
checkedArr.push(arr[i]);
for (let j = i; j < arr.length; j++) {
if (isArraySame(arr[i], arr[j])) {
count++;
}
}
output.push([arr[i][0], arr[i][1], count]);
}
}
function isArrayChecked(array) {
let exist = false;
for (let k = 0; k < checkedArr.length; k++) {
if (isArraySame(array, checkedArr[k])) {
exist = true;
break;
}
}
return exist;
}
function isArraySame(arr1, arr2) {
if (arr1[0] !== arr2[0] || arr1[1] !== arr2[1] || arr1[2] !== arr2[2])
return false;
return true;
}
console.log(output);

The problem with your code is in counts[x]. You can't use an array to index an object.
You can solve the issue by stringifying the array. That is, use:
counts[String(x)] = (counts[String(x)] || 0)+1;

let arr=[
[3, 16, 1],
[3, 17, 1],
[3, 18, 1],
[3,16,1],
[3, 19, 1],
[3, 20, 1],
[3, 19, 1],
[3, 21, 1],
[3, 23, 1],
[3, 16, 1],
[3, 17, 1],
[2, 24, 1],
[2, 25, 1],
[2, 26, 1],
[2, 24, 1]]
let solution = (arr) => {
let obj = {};
let result = [];
for (let i of arr) {
let keyOfObj = JSON.stringify(i);
obj[keyOfObj] ? obj[keyOfObj]++ : obj[keyOfObj] = 1;
}
for (let [key, value] of Object.entries(obj)) {
let objKeyArr = JSON.parse(key);
if (value >= 2) {
let lastElemt = objKeyArr[objKeyArr.length - 1];
objKeyArr.splice(objKeyArr.length - 1, 1, lastElemt * value);
result.push(objKeyArr);
} else {
result.push(objKeyArr);
}
}
return result;
};
solution(arr)

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

Why my variable is not iterable when i loop

I try to loop the 2d arrays, but the I variable is undefined or not iterable, why?
can anyone tell me ??
function sum (arr) {
var total = 0
for(let [a1,a2,a3] of arr){
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
if(typeof a2 == "undefined" && typeof a3 == "undefined"){
a2 = [0]
a3 = [0]
}
}
};
console.log(sum([
[
[10, 10],
[15],
[1, 1]
],
[
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[4],
[9, 11]
],
[
[3, 5, 1],
[1, 5, 3],
[1]
],
[
[90]
]
]));
but when i sum another 2D array, it works, like this :
function sum (arr) {
var total = 0
for(let [a1,a2,a3] of arr){
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
}
return total
}
console.log(sum([
[
[4, 5, 6],
[9, 1, 2, 10],
[9, 4, 3]
],
[
[4, 14, 31],
[9, 10, 18, 12, 20],
[1, 4, 90]
],
[
[2, 5, 10],
[3, 4, 5],
[2, 4, 5, 10]
]
]));
i try to loop 3 times for this 2d arrays, the first top code is each lengths are diffreen in array
and the last code is same,

Cause
let [a1,a2,a3] of [ [90] ])
will result in a2 and a3 being undefined, therefore in the following line it is:
for(const i of [90, undefined, undefined])
And at the second index it does:
for(let j of undefined)
which doesnt work.

You just need to move your if statement that checks if the value is undefined and assigns it to zero if it is ahead of the part of code that iterates over those values. You were getting this error because there wasn't anything there.
function sumTwo(arr) {
var total = 0
for(let [a1,a2,a3] of arr){
if(typeof a2 == "undefined" && typeof a3 == "undefined"){
a2 = [0]
a3 = [0]
}
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
}
return total
};
console.log(sumTwo([
[
[10, 10],
[15],
[1, 1]
],
[
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[4],
[9, 11]
],
[
[3, 5, 1],
[1, 5, 3],
[1]
],
[
[90]
]
])); //prints 237

When you say
let [a1,a2,a3] of [ [90] ])
there is no a2 or a3 there...
My suggestion would be using the code before you get into the first for loop:
if(arr.length < 3){
for(let y = arr.length, y > 3, y++ ){
arr.push(0)
}
}
Cheers!

It's probably better to recursively reduce the array using concat until you have a flat array and then reduce that to the sum of it's numbers:
const arr = [
[[10, 10], [15], [1, 1]],
[[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [4], [9, 11]],
[[3, 5, 1], [1, 5, 3], [1]],
[[90]],
];
const flatten = (arr) => {
const recur = (result, item) =>
!Array.isArray(item)
? result.concat(item)
: result.concat(item.reduce(recur, []));
return arr.reduce(recur, []);
};
console.log(
flatten(arr).reduce((result, item) => result + item, 0),
);

How to make _lodash.zip with less code

Definition：Creates an array of grouped elements, the first of which contains the first elements of the given arrays, the second of which contains the second elements of the given arrays, and so on.
Current Solution：
const zip = (...arr) => {
let maxLength = 0
let res = []
for (let el of arr) {
maxLength = Math.max(maxLength, el.length)
}
for (let j = 0; j < maxLength; j++) {
const foo = []
for (let n of arr) {
foo.push(n[j])
}
res.push(foo)
}
return res
}
Test Case:
test(('zip', () => {
expect(zip([1, 2], [4, 5], [9, 1])).toEqual([[1, 4, 9], [2, 5, 1]])
}
test('zip', () => {
expect(zip([1, 2, 3], [4, 5, 6])).toEqual([[1, 4], [2, 5], [3, 6]])
})
test('zip', () => {
expect(zip([1, 2], [], [3, 4, 5])).toEqual([
[1, undefined, 3],
[2, undefined, 4],
[undefined, undefined, 5],
])
})
I want to get a better way to achieve zip, current solution is ugly

See Destructuring Assignment and Array.prototype.map for more info.
// Proof.
const zip = (...args) => [...new Array(Math.max(...args.map(arr => arr.length)))].map((x, i) => args.map((y) => y[i]))
// Proof.
console.log(zip([1, 2], [4, 5], [9, 1])) // [[1, 4, 9], [2, 5, 1]]
console.log(zip([1, 2, 3], [4, 5, 6])) // [[1, 4], [2, 5], [3, 6]]
console.log(zip([1, 2], [], [3, 4, 5])) // [[1, undefined, 3], [2, undefined, 4], [undefined, undefined, 5]]

Type Error: Undefined

I'm working on a tic tac toe game, and though the game is functional, when one player wins, the debugger pops up if they win using certain spaces. I get a type error saying that 'O' can't be set of undefined. Also in my console a negative number will print out for the elementIndex when a player wins by clicking only on certain spaces.
It appears to be coming from this line in my game.win function:
game.winCombos[elementIndex][unitIndex] = game.currentPlayerTurn;
var gameboard = {
initialize: function() {
for (var x = 0; x < 3; x++) {
for (var y = 0; y < 3; y++) {
var unit = $("<div class='unit'></div>");
unit.appendTo('#gameboard');
}
}
console.log(100);
gameboard.addId();
},
addId: function() {
var id = 1
$('.unit').each(function() {
$(this).attr('id', id);
id++;
});
}
};
var players = {
firstPlayer: {
token: 'X'
},
secondPlayer: {
token: 'O'
}
};
var game = {
newGame: function() {
game.winCombos = game.clone();
game.currentPlayerTurn = players.firstPlayer.token;
gameboard.initialize();
game.displayToken();
},
currentPlayerTurn: players.firstPlayer.token,
displayToken: function() {
$('.unit').click(function() {
if (game.currentPlayerTurn === 'X' && !$(this).hasClass('selected')) {
$(this).addClass('selected').removeClass('unit').text("X");
game.currentPlayerTurn = players.secondPlayer.token;
} else if (game.currentPlayerTurn === 'O' && !$(this).hasClass('selected')) {
$(this).addClass('selected').removeClass('unit').text("O");
game.currentPlayerTurn = players.firstPlayer.token;
}
game.win($(this));
})
},
win: function(div) {
game.winCombos.forEach(function(element) {
element.forEach(function(unitIndex) {
if (unitIndex.toString() === div.attr('id').toString()) {
var elementIndex = game.winCombos.indexOf(element);
var unitIndex = element.indexOf(unitIndex);
game.winCombos[elementIndex][unitIndex] = game.currentPlayerTurn;
}
})
var counter = 0
for (var i = 0; i < element.length; i++) {
if (element[i] === game.currentPlayerTurn) {
counter++;
}
if (counter === 3) {
counter = 0;
game.gameOver(true);
}
}
})
},
gameOver: function(bool) {
if (bool === undefined) {
bool = false;
}
$('.unit').remove();
$('.selected').remove();
game.newGame();
},
winCombos: [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
[1, 4, 7],
[2, 5, 8],
[3, 6, 9],
[1, 5, 9],
[3, 5, 7]
],
clone: function() {
return [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
[1, 4, 7],
[2, 5, 8],
[3, 6, 9],
[1, 5, 9],
[3, 5, 7]
]
}