unique javascript array based on one property

Question

I have a multidimensional array in javascript. I want to make it so it shows the first record for each unique "name". So for example this:

[{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},

Would turn into this:

[{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},

Originally It was not multivarite and i could use a basic function, but that method wont work for this one. Any ideas? Pretty desperate! Thanks.

It's actually an array of objects as opposed to a multi-dimensional array... — War10ck
– War10ck, Commented Apr 30, 2014 at 19:40

adeneo · Accepted Answer · 2014-04-30 19:43:48Z

3

Something like this would create a new array with the first unique records only

var obj     = {},
    new_arr = [];

arr.forEach(function(itm) {
    if ( ! (itm.name in obj) ) obj[itm.name] = itm;
});

for (var key in obj) {
    new_arr.push(obj[key]);
}

FIDDLE

answered Apr 30, 2014 at 19:43

adeneo

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2 Comments

Rahil Wazir Over a year ago

Why don't you just do new_arr.push(obj)? Instead of for in loop.

adeneo Over a year ago

@RahilWazir - because I'm not pushing that object, but the objects contained in that object etc. Just using the names as keys to make sure they are unique.

Dean Barnes · Accepted Answer · 2014-04-30 20:22:16Z

0

This should do the trick:

var foos = [{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"}];

var new_foos = [];

var previous_foo = null;

for (var i = 0; i < foos.length; i++) {
    if (previous_foo != foos[i].name)
    {
        new_foos.push(foos[i]);
    }

    previous_foo = foos[i].name;
}

As mentioned in the comments, this requires that foo is sorted. This works if foo is unsorted:

var foos = [{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},

{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Ashmore , Abigail","col1":"Vice President","col2":"test","col3":"test","col4":"99.80%"},
{"name":"Baker , Ian Howard","col1":"Director","col2":"test","col3":"test","col4":"67.20%"},
{"name":"Aellen , Nick","col1":"Director","col2":"test","col3":"test","col4":"96.88%"},
{"name":"Armstrong , Peter","col1":"Director","col2":"test","col3":"test","col4":"95.15%"},{"name":"Adelman , Erica","col1":"Vice President","col2":"test","col3":"test","col4":"73.77%"},
{"name":"Avent , Christopher","col1":"Vice President","col2":"test","col3":"test","col4":"90.11%"}];

var new_foos = [];

for (var i = 0; i < foos.length; i++) {
    var new_foo = true;
    for (var j = 0; j < new_foos.length; j++) {
        if (foos[i].name == new_foos[j].name) {
            new_foo = false;
            break;
        }
    }

    if (new_foo)
    {
        new_foos.push(foos[i]);
    }
}

edited Apr 30, 2014 at 20:22

answered Apr 30, 2014 at 19:37

Dean Barnes

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10 Comments

Rahil Wazir Over a year ago

Why the solution is in Python? :D

Dean Barnes Over a year ago

Apparently it has been a loooong day D:

Katana314 Over a year ago

@War10ck True, I suppose, but it's still usable pseudocode I'd say. I would hope for people to understand the code they're copying and not just copy-paste, so it shouldn't be a big effort to rewrite in JS if you have some level of understanding.

War10ck Over a year ago

@Katana314 While this may be true, I don't think it's entirely clear if you're not familiar with python. I've written both so I understand what you mean. For someone who hasn't though, I think it could be a little difficult to follow. It's for the reason you said though, that I did not downvote the post and merely commented on it.

Jason P Over a year ago

I would note that this algorithm only works if the array is sorted by name.

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