Javascript - Nice way to create an array from an object array

Question

I have a javascript object array:

array = [ {x:'x1', y:'y1'}, {x:'x2', y:'y2'}, ... {x:'xn', y:'yn'} ]

I want to create a new array of just the x values:

[ 'x1', 'x2', ..., 'xn' ]

I could do this easily in a for loop...:

var newarray = [];
for (var i = 0; i < array.length; i++){
     newarray.push(array[i].x);
}

...but I'm wondering if there's a nice one liner way to do this using jquery or even regular javascript?

The object part of your code is wrong. Your x/y objects don't have any keys. Just thought I'd point that out. — Sasha Chedygov
– Sasha Chedygov, Commented Jun 11, 2009 at 19:25
@musicfreak I think he means that his objects have two properties, x and y: {x: foo, y:bar} — Gabe Moothart
– Gabe Moothart, Commented Jun 11, 2009 at 19:28

Gabe Moothart · Accepted Answer · 2009-06-11 19:17:08Z

11

You can do this with map:

var newarray = jQuery.map(array, function (item) { return item.x; });

answered Jun 11, 2009 at 19:17

Gabe Moothart

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Christoph · Accepted Answer · 2009-06-11 20:46:44Z

3

ECMAScript 5 features a native map() method:

var newArray = array.map(function(value) { return value.x; });

In FF, this should even be faster than looping, but that's not true for all browsers (Opera); others don't even support it (IE).

answered Jun 11, 2009 at 20:46

Christoph

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2 Comments

Prestaul Over a year ago

If he doesn't care about a cross browser solution then in FF (which supports Javascript 1.8) he could do: var arr = array.map(function(v) v.x); or: var arr = [v.x for each (v in array)];

mawaldne Over a year ago

Unfortunately, cross browser compatibility is necessary.

hugoware · Accepted Answer · 2009-06-11 19:21:44Z

0

Granted this is for working with a lot of data, you could use jLinq (disclaimer: my project) to select and query your records and then return what you want.

Again, this is really for when you're working with a lot of objects and making queries, but you might be able to get some ideas looking at the source code.

answered Jun 11, 2009 at 19:21

hugoware

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