I needed 2d arrays, so I made nested array since JavaScript doesn\'t allow them.
They look like that:
var myArray = [
[1, 0],
[1, 1],
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.
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;
}
Short 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
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;
}
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
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.