Return second value from multiple arrays Javascript

Question

How can I return the first values from a set of arrays?

var arr = [
    ['2009', 1000, 400, 800],
    ['2010', 1170, 460, 1200],
    ['2011', 660, 1120, 500],
    ['2012', 1030, 540, 800]
], res, i, a, b, c, d, e;

!function doMerge() {
    for(i = 0; i < arr.length; i++) {
        res = arr[i];       
        console.log(res.splice(1));
    }    
}(arr)

// get rid of years
// prints    
[1000, 400, 800] 
[1170, 460, 1200] 
[660, 1120, 500] 
[1030, 540, 800]

I'm trying to build an array of the first of these values [1000, 1170, 660, 1030]

Pascal Belloncle · Accepted Answer · 2013-02-14 02:03:44Z

1

res = [];

function doMerge() {
    for(i = 0; i < arr.length; i++) {
        res.push(arr[i][1]);       
    }    
}

doMerge();

console.log(res);

outputs [ 1000, 1170, 660, 1030 ]

edited Feb 14, 2013 at 2:03

answered Feb 14, 2013 at 1:52

Pascal Belloncle

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4 Comments

Paul S. Over a year ago

Where is [1000, 1170, 660, 1030] produced by this?

Pascal Belloncle Over a year ago

it is in the res variable; this was the minimal change from original code. I would have returned the new array from the function.

Paul S. Over a year ago

You have a syntax error on line 3 (an unmatched {) and even after fixing, I don't think res becomes what you think it becomes (try it and see)

Pascal Belloncle Over a year ago

remove console.log with splice, I missed that.

marteljn · Accepted Answer · 2013-02-14 01:54:04Z

1

var arr = [
    ['2009', 1000, 400, 800],
    ['2010', 1170, 460, 1200],
    ['2011', 660, 1120, 500],
    ['2012', 1030, 540, 800]
];

var newArr = [];

for (var x = 0; x < arr.length; x++){
  newArr.push(arr[x][1]);
}

console.log(newArr);

Here is a solution.

http://jsfiddle.net/wcVZH/

answered Feb 14, 2013 at 1:54

marteljn

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Comments

Paul S. · Accepted Answer · 2013-02-14 01:56:24Z

0

If you don't want a for or while loop you can one line this with Array.prototype.map

var myArr = arr.map(function (a) {return a[1];});

answered Feb 14, 2013 at 1:56

Paul S.

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