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I needed 2d arrays, so I made a nested array since JavaScript doesn't allow them.
They look like this:
var myArray = [
[1, 0],
[1, 1],
[1, 3],
[2, 4]
]
How can I check if this array includes a specific element (i.e. one of these [0,1]
arrays) in vanilla JS?
Here is what I tried, with no success (everything returns false
) (EDIT: I included the answers in the snippet):
var myArray = [
[1, 0],
[1, 1],
[1, 3],
[2, 4]
]
var itemTrue = [2, 4];
var itemFalse = [4, 4];
function contains(a, obj) {
var i = a.length;
while (i--) {
if (a[i] === obj) {
return true;
}
}
return false;
}
// EDIT: first answer's solution
function isArrayInArray(x, check) {
for (var i = 0, len = x.length; i < len; i++) {
if (x[i][0] === check[0] && x[i][1] === check[1]) {
return true;
}
}
return false;
}
// EDIT: accepted answer's solution
function isArrayInArray2(x, check) {
var result = x.find(function(ele) {
return (JSON.stringify(ele) === JSON.stringify(check));
})
return result !=null
}
console.log("true :" + myArray.includes(itemTrue));
console.log("false :" + myArray.includes(itemFalse));
console.log("true :" + (myArray.indexOf(itemTrue) != -1));
console.log("false :" + (myArray.indexOf(itemFalse) != -1));
console.log("true :" + contains(myArray, itemTrue));
console.log("false :" + contains(myArray, itemFalse));
// EDIT: first answer's solution
console.log("true :" + isArrayInArray(myArray, itemTrue));
console.log("false :" + isArrayInArray(myArray, itemFalse));
// EDIT: accepted answer's solution
console.log("true :" + isArrayInArray2(myArray, itemTrue));
console.log("false :" + isArrayInArray2(myArray, itemFalse));
I needed 2d arrays, so I made a nested array since JavaScript doesn't allow them.
They look like this:
var myArray = [
[1, 0],
[1, 1],
[1, 3],
[2, 4]
]
How can I check if this array includes a specific element (i.e. one of these [0,1]
arrays) in vanilla JS?
Here is what I tried, with no success (everything returns false
) (EDIT: I included the answers in the snippet):
var myArray = [
[1, 0],
[1, 1],
[1, 3],
[2, 4]
]
var itemTrue = [2, 4];
var itemFalse = [4, 4];
function contains(a, obj) {
var i = a.length;
while (i--) {
if (a[i] === obj) {
return true;
}
}
return false;
}
// EDIT: first answer's solution
function isArrayInArray(x, check) {
for (var i = 0, len = x.length; i < len; i++) {
if (x[i][0] === check[0] && x[i][1] === check[1]) {
return true;
}
}
return false;
}
// EDIT: accepted answer's solution
function isArrayInArray2(x, check) {
var result = x.find(function(ele) {
return (JSON.stringify(ele) === JSON.stringify(check));
})
return result !=null
}
console.log("true :" + myArray.includes(itemTrue));
console.log("false :" + myArray.includes(itemFalse));
console.log("true :" + (myArray.indexOf(itemTrue) != -1));
console.log("false :" + (myArray.indexOf(itemFalse) != -1));
console.log("true :" + contains(myArray, itemTrue));
console.log("false :" + contains(myArray, itemFalse));
// EDIT: first answer's solution
console.log("true :" + isArrayInArray(myArray, itemTrue));
console.log("false :" + isArrayInArray(myArray, itemFalse));
// EDIT: accepted answer's solution
console.log("true :" + isArrayInArray2(myArray, itemTrue));
console.log("false :" + isArrayInArray2(myArray, itemFalse));
It could look like duplicate but I couldn't find a similar question. If it is, feel free to tag it as such.
Share Improve this question edited Sep 6, 2021 at 20:13 Audwin Oyong 2,5563 gold badges19 silver badges36 bronze badges asked Jan 15, 2017 at 12:50 BillybobbonnetBillybobbonnet 3,2264 gold badges26 silver badges51 bronze badges 5 |6 Answers
Reset to default 13Short and easy, stringify the array and compare as strings
function isArrayInArray(arr, item){
var item_as_string = JSON.stringify(item);
var contains = arr.some(function(ele){
return JSON.stringify(ele) === item_as_string;
});
return contains;
}
var myArray = [
[1, 0],
[1, 1],
[1, 3],
[2, 4]
]
var item = [1, 0]
console.log(isArrayInArray(myArray, item)); // Print true if found
check some documentation here
A nested array is essentially a 2D array, var x = [[1,2],[3,4]]
would be a 2D array since I reference it with 2 index's, eg x[0][1]
would be 2
.
Onto your question you could use a plain loop to tell if they're included since this isn't supported for complex arrays:
var x = [[1,2],[3,4]];
var check = [1,2];
function isArrayInArray(source, search) {
for (var i = 0, len = source.length; i < len; i++) {
if (source[i][0] === search[0] && source[i][1] === search[1]) {
return true;
}
}
return false;
}
console.log(isArrayInArray(x, check)); // prints true
Update that accounts for any length array
function isArrayInArray(source, search) {
var searchLen = search.length;
for (var i = 0, len = source.length; i < len; i++) {
// skip not same length
if (source[i].length != searchLen) continue;
// compare each element
for (var j = 0; j < searchLen; j++) {
// if a pair doesn't match skip forwards
if (source[i][j] !== search[j]) {
break;
}
return true;
}
}
return false;
}
console.log(isArrayInArray([[1,2,3],[3,4,5]], [1,2,3])); // true
Here is an ES6 solution:
myArray.some(
r => r.length == itemTrue.length &&
r.every((value, index) => itemTrue[index] == value)
);
Check the JSFiddle.
Take a look at arrow functions and the methods some and every of the Array object.
You can't do like that .instance you have to do some thing by your own .. first you have to do a foreach from your array that you want to search and run 'compareArray' function for each item of your array .
function compareArray( arrA, arrB ){
//check if lengths are different
if(arrA.length !== arrB.length) return false;
for(var i=0;i<arrA.length;i++){
if(arrA[i]!==arrB[i]) return false;
}
return true;
}
The code provided by D. Young's comment that checks for any length array is faulty. It only checks if the first element is the same.
A corrected version of D. Young's comment:
function isArrayInArray(source, search) {
var searchLen = search.length;
for (var i = 0, len = source.length; i < len; i++) {
// skip not same length
if (source[i].length != searchLen) continue;
// compare each element
for (var j = 0; j < searchLen; j++) {
// if a pair doesn't match skip forwards
if (source[i][j] !== search[j]) {
break;
} else if (j == searchLen - 1) {return true}
}
}
return false;
}
For those who are interested in finding an array inside another and get back an index number, here's a modified version of mohamed-ibrahim's answer:
function findArrayInArray(innerArray, outerArray) {
const innerArrayString = JSON.stringify(innerArray);
let index = 0;
const inArray = outerArray.some(function (element) {
index ++;
return JSON.stringify(element) === innerArrayString;
});
if (inArray) {
return index - 1;
} else {
return -1;
}
}
findArrayInArray([1, 2, 3], [[3, .3], [1, 2, 3], [2]]); // 1
findArrayInArray([1, 2, 3], [[[1], 2, 3], [2]]) // -1
This function returns the index of the array you are searching inside the outer array and -1 if not found.
Checkout this CodePen.
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if (paramArray[i] == searchElement)
, which fails, because you can't compare two arrays using==
, so you can't useincludes
to do this. – pushkin Commented Jan 15, 2017 at 13:02