Reputation: 8586
I have the following code to combine an array and an array of arrays.
var arr1 = ["A", "B", "C", "D"];
var arr2 = [[1, 2, 3, 4, 3, 2],
[4, 3, 7, 2, 5, 3],
[8, 1, 9, 2, 8, 4],
[2, 5, 9, 8, 7, 6],
]
var result = [], i = -1;
while ( arr1[++i] ) {
result.push( [ arr1[i], arr2[i] ] );
}
Here is how the final array looks
var final =[["A", [1, 2, 3, 4, 3, 2]],
["B", [4, 3, 7, 2, 5, 3]],
["C", [8, 1, 9, 2, 8, 4]],
["D", [2, 5, 9, 8, 7, 6]],
]
I am trying to create a function where I supply the letter and the index of the column # i want to return from the array of arrays.
i.e. If I give it B
and 2
, it would return result[1][1][2]
which is 7
.
If I give it D
and 5
, it would return result[3][1][5]
which is 6
I tried using indexOf
but it is hard to search for since this array has 3 dimensions.
Here is a jsFiddle I made of my sample Any help would be appreciated
Edit: It seems my only problem is returning the column # where the letter occurs. How would I use indexOf to search for the occurence of a letter in first column?
Upvotes: 1
Views: 3391
Reputation: 6467
I'd suggest to change a little bit your logic as follows:
var result = {}; // Make this to be an object
while ( arr1[++i] ) {
result[ arr1[i] ] = arr2[i];
}
This will create an object similar to the following:
result = {
"A": [1,2,3,4,3,2],
"B": [4,3,7,2,5,3],
"C": [8,1,9,2,8,4],
"D": [2,5,9,8,7,6]
}
And the way to access data from it would be:
result['D'][5] // returns 6
Upvotes: 3
Reputation: 10618
How about this function:
function find(text, col) {
for(var i = 0; i < result.length; i++)
if(result[i][0].equals(text)) return result[i][1][col - 1];
return null;
}
console.log(find("B", 3));
var arr1 = ["A", "B", "C", "D"];
var arr2 = [
[1, 2, 3, 4, 3, 2],
[4, 3, 7, 2, 5, 3],
[8, 1, 9, 2, 8, 4],
[2, 5, 9, 8, 7, 6],
]
var result = [],
i = -1;
while (arr1[++i]) {
result.push([arr1[i], arr2[i]]);
}
console.log(result)
//Try to find element where text == "B" and return the 3rd column
console.log(find("B", 3))
function find(text, col) {
for (var i = 0; i < result.length; i++)
if (result[i][0] == text) return result[i][1][col - 1];
return null;
}
Upvotes: 1