We all know that array are zero-based.
In your first Example to access 'g'
You need to access:
- the second position (index
[1]
) which is ['e','f',['g','h']]
- then access the third position (index
[2]
) which is ['g','h']
- finally the first position (index
[0]
) and you get 'g'
.
var nestedArrays = [['a','b','c','d'],['e','f',['g','h']]];
let x = nestedArrays[1][2][0];
console.log(x); // g
So from that we can say that to access 'g'
We need to follow this road of indices [1,2,0]
var nestedArrays = [['a','b','c','d'],['e','f',['g','h']]];
let roadToG = [1,2,0];
for(let i = 0; i < roadToG.length; i++){
nestedArrays = nestedArrays[roadToG[i]];
}
console.log(nestedArrays); // g
BUT if "g" is nested in 100 of arrays i have to write a lot of square brackets as it var g = nestedArrays[2][3][1][?][?].....[?];
In case we want to access 'g'
dynamically without hard code the road of indices; We must loop recursively in array till we find 'g'
, in looping we keep track of road
function loopRecursively( arr, x ,roadToX = []){
if(JSON.stringify(arr).indexOf(x) == -1) return null; //in edge case where there is no x in array; we just return null!
for ( let i = 0; i < arr.length; i++ ) {
if( arr[i] == x ){
roadToX.push(i);
} else if( Array.isArray( arr[i] ) && JSON.stringify(arr[i]).indexOf(x) != -1 ){
roadToX.push(i);
loopRecursively( arr[i] , x, roadToX);
}
}
return roadToX;
}
let nestedArrays = [['a','b','c','d'],['e','f',['g','h']]];
console.log( loopRecursively( nestedArrays, 'g' ) ); // [1, 2, 0]