I have an array of 3 value arrays, and I need to convert them into a set of arrays where each array is made up of the values which were common between arrays, so my input of
[[2,3,9], [1,4,7], [3,6,9], [1,7,5], [7,5,8], [9,6,10], [3,6,10], [1,8,5]]
becomes
[[2,3,6,9,10],[1,4,5,7,8]]
Order is not important. I've found similar questions such as Group same values in an array within an array in JS but it seems slightly different to my case, but I imagine using reduce is the way to go, but I don't entirely understand how. I have tried creating an object using the following format, but couldn't get from there to an answer:
{
vertex: 3,
triangles: [2,3,9], [3,6,9], [3,6,10]
}
Here is one algorithm. Take first item from array and check first item array has any common array. If they have common item, then merge it and move the merged array to first item of array. If no common item, then add to result array.
const merge = (arr) => {
const result = [];
while (arr.length > 0) {
let first = arr.shift();
const idx = arr.findIndex((items) =>
items.some((item) => first.includes(item))
);
if (idx !== -1) {
first = first.concat(arr[idx]);
arr.splice(idx, 1);
arr.unshift(first);
} else {
result.push(first);
}
}
return result.map(items => [...new Set(items)]);
};
const data = [
[2, 3, 9],
[1, 4, 7],
[3, 6, 9],
[1, 7, 5],
[7, 5, 8],
[9, 6, 10],
[3, 6, 10],
[1, 8, 5],
];
console.log(merge(data));
Related
This question already has answers here:
How to check if an array contains another array?
(6 answers)
javascript search array of arrays
(12 answers)
Closed 3 months ago.
How I can scan my array within array if there is equal array element.
I want to check if its true or false
// the array to be scan
const array = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
]
// the new array
const newArray = [0, 1, 2]
Based on anwser from how-to-compare-arrays-in-javascript,you just need to iterate the first array and then compare them
const array = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
]
// the new array
const newArray1 = [0, 1, 2]
const newArray2 = [0, 1, 3]
const checkArray = (array1,array2) => {
for(let arr of array1){
if(arr.length === array2.length && arr.every((v,i) => v === array2[i])){
return true
}
}
return false
}
console.log(checkArray(array,newArray1))
console.log(checkArray(array,newArray2))
You can use every method to check all the items.
array.filter((arr)=>arr.every((item,index)=>newArray[index]===item))
run a foreach or for loop with a combination of name.find(x => x.ref === value);
or use array.filter((arr)=>arr.every((item,index)=>newArray[index]===item))
const array = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
];
const newArray = [0, 1, 2]
let isPresent = false;
for (let i = 0; i < array.length; i++) {
if (JSON.stringify(array[i]) == JSON.stringify(newArray)) {
isPresent = true
break
}
}
console.log(isPresent)
After stringifying the array you are able to compare directly in js and If it matches I changed the boolean value to true.
In case you have any doubts feel free to comment.
I'm trying to make a tictactoe game using javascript. I have the winning conditions in a nested array
const winConditions = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
[0, 3, 6],
[1, 4, 7],
[2, 5, 8],
[0, 4, 8],
[2, 4, 6]
]
and I have my current player's marked boxes indices in an array
let squaresIndices = [0,4,8]
I'm trying to stop the game when any of the winning conditions occur. I have tried each of the following and I can't figure out why they don't work.
winConditions.forEach((array) => {
if (array === squaresIndices){
alert("Game Over")
}
//////
if (winConditions.includes(squaresIndices)){
alert ("Game Over")
}
//////
winConditions.some((arr) =>{
if (arr.every((square) => {
squaresIndices.includes(square)
})) {
alert("Game Over")
}
})
First of all
Array cannot be compared
Try this example
let a = [0,2,3]
let b = [0,2,3]
alert( a === b )
What you need to understand is that
when you save an array. What you actually doing is created a reference to that array in memory.
Then again try this example
let a = [0,2,3]
let b = a
alert( a === b )
You will get true, Why? Because in first case you are trying to two separate addresses in memory. Meaning that a & b got different addresses but same kind of apartment. But in second case you are comparing the same address.
So the easiest way you can do is convert them into string then try to compare them.
JSON.stringify(a)==JSON.stringify(b)
Then you will get your desire result. While this could work if the order of the properties will always the same for those array instances, this leaves the door open for extremely nasty bugs that could be hard to track down.
Your last example is correct if not for forgetting to return the value and moving the if outside.
const winConditions = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8],
[0, 3, 6],
[1, 4, 7],
[2, 5, 8],
[0, 4, 8],
[2, 4, 6]
];
const squaresIndices = [0, 4, 8];
const isWin = winConditions.some(
(arr) =>
{
// return if this combination is matching or not
return arr.every(
(square) =>
{
// return if the value matches or not
return squaresIndices.includes(square);
}
);
}
);
// Then test the result
if (isWin)
{
alert("Game Over");
}
winConditions.forEach((array) => {
if(JSON.stringify(array)==JSON.stringify(squaresIndices)){
alert("Game Over")
}
})
You cannot compare two arrays directly in javascript. You need to convert it to string for comparison. You can also use toString() instead of JSON.stringify()
Here's a sample of the problem I'm having in JavaScript:
first array [1, 2, 3, 4, 5, 6, 7]
second array [7, 8, 9, 4, 2, 5, 7]
In this case, I need to be able to find and eliminate "4" and "7" from both arrays, eliminating both. This is based on their location and matching value.
I haven't been able to find anything other than eliminating matching values. In this case, however, the values must be in the same place and also be matching.
I've tried this so far:
function findCommonElements3(array1, array2) {
return arr1.some(item => arr2.includes(item))
}
it looks like it only looks for matching elements, whereas I need to find matching corresponding elements and then remove them.
As mentioned in the comments, you may use the splice method to remove one or more elements of an array in JavaScript.
First of all I would store the indexes of the elements I should remove looping the array as so:
const array1 = [1, 2, 3, 4, 5, 6, 7];
const array2 = [7, 8, 9, 4, 2, 5, 7];
//Indexes of same elements
var sameIndexes = [];
function findSameIndexes(element, index) {
if (array1[index] == array2[index]) {
sameIndexes.push(index);
}
}
array1.forEach(findSameIndexes);
Calling console.log(sameIndexes) should give this result:
Array [3, 6]
The problem is that if you loop again the array and remove the elements in that order, the indexes would not correspond to the elements anymore.
For example if you remove the 3rd element, the number 7 wouldn't be at index 6 anymore, to solve this issue I'd use the reverse method so you won't lose track of the indexes
// A simple function to remove the elements in both arrays
function removeElements(index) {
array1.splice(index,1);
array2.splice(index,1);
}
sameIndexes.reverse().forEach(removeElements);
And the final results would be
Array [1, 2, 3, 5, 6]
Array [7, 8, 9, 2, 5]
Which hopefully is what you were looking for, of course there are better ways to write it down, but maybe this will help you find a solution.
You could just use a for loop and use index. something like this
const firstarray = [1, 2, 3, 4, 5, 6, 7]
const secondarray = [7, 8, 9, 4, 2, 5, 7]
for (let i = 0; i <= firstarray.length - 1; i++) {
if (firstarray[i] === secondarray[i]) {
console.log(`found ${firstarray[i]} at index ${i}`);
firstarray.splice(i, 1);
secondarray.splice(i, 1);
}
}
console.log(firstarray, secondarray);
const excludeCommon = (ar1, ar2) => {
const both = [...ar1, ...ar2].filter((v, i, ar) => v !== ar[i + (2 * (i < ar1.length) - 1) * ar1.length]);
return [both.slice(0, both.length / 2), both.slice(both.length / 2)];
}
console.log(excludeCommon([1, 2, 3, 4, 5, 6, 7], [7, 8, 9, 4, 2, 5, 7]));
I need to create a new array made up of unique elements from two separate arrays.
I have converted both arrays into a single array and then converted this into an object to check the frequency of the elements. If the value of an object property is 1 (making it a unique property), I want to return it to an array (minus the value). Is there a straightforward way to achieve this?
Edits: Moved result outside for loop. Expected output should be [4]
function diffArray(arr1, arr2) {
var finalArr = [];
var countObj = {};
var newArr = [...arr1, ...arr2];
for (var i = 0; i < newArr.length; i++) {
if (!countObj[newArr[i]]) countObj[newArr[i]] = 0;
++countObj[newArr[i]];
}
for (var key in countObj) {
if (countObj[key] === 1) {
finalArr.push(key);
}
} return finalArr;
}
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
If I understand correctly, you're wanting to find the difference between arr1 and arr2, and returns that difference (if any) as a new array of items (that are distinct in either array).
There are a number of ways this can be achieved. One approach is as follows:
function diffArray(arr1, arr2) {
const result = [];
const combination = [...arr1, ...arr2];
/* Obtain set of unique values from each array */
const set1 = new Set(arr1);
const set2 = new Set(arr2);
for(const item of combination) {
/* Iterate combined array, adding values to result that aren't
present in both arrays (ie exist in one or the other, "difference") */
if(!(set1.has(item) && set2.has(item))) {
result.push(item);
}
}
return result;
}
console.log(diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]), " should be [4]");
console.log(diffArray([1, 2, 3, 5, 8], [1, 2, 3, 5]), " should be [8]");
console.log(diffArray([1, 2, 3, 5, 8], [1, 2, 3, 5, 9]), " should be [8, 9]");
console.log(diffArray([1, 2], [1, 2]), " should be []");
I want to get values from an array of arrays, and I'm having difficulties doing it.
I have the following:
var id = 1; //value I want to use for the search
var _restrictions = [[1, 2], [2, 4], [5, 1], [1, 6]]; //arrays that I want to check
var arrVal = [];
By using the id, I want to retrieve all of the values, inside the arrays, where the id exits and store them in the array "arrVal".
For example:
_restrictions = [[1, 2], [2, 4], [5, 1], [1, 6]];
//arrVal will be: [2, 5, 6], because the id existing inside the arrays [1,2],
//[5,1] and [1,6]
The "_restrictions" array is a array of arrays that contain restrictions. They are independent values (the first one isn't the index or id).
How can I do that?
Thanks in advance!
Here's a version that will work for any size of nested array. It returns an flattened array of all values not including the id.
var id = 1;
var _restrictions = [[1, 2, 9], [2, 4], [5, 1], [1, 6]];
var arrVal = _restrictions.reduce((acc, c) => {
// Find the index of the id in each nested array
const i = c.findIndex(arr => arr === id);
// If it exists, splice it out and add it
// to the accumulator array
if (i > -1) {
c.splice(i, 1);
acc.push(...c);
}
return acc;
}, []);
console.log(arrVal);
EDIT: Updated code after the question is edited.
The question lacks a bit of clarity. I am assuming you want to filter the sub-arrays which have id in them i.e. contains the value 1.
let id = 1; //value I want to use for the search
let _restrictions = [[1, 2], [2, 4], [5, 1], [1, 6]];
let arrVal = _restrictions.filter((item) => {
return item.includes(id);
});
let new_array = arrVal.concat.apply([], arrVal).filter(x => x !== id);
console.log(new_array);
// [2, 5, 6]