Merge property values of 2 objects into an array using underscore chain

Question

I have the following:

a = {x:1, y:3, w:4}
b = {c:2, d:3}

And I want to obtain all the values of these objects iterating only once.

result = [1, 3, 4, 2, 3]

I have the following solution but it has multiple iterations.

result = _.chain(a).values().union(_.values(b)).value();

I would like to avoid the "_.values(b)" and do this using the same chain from a.

I also tried this, but it is not working properly:

result = _.chain({}).extend(a,b).values().value();

You could merge the objects, then get the values from the merged object. — user663031
– user663031, Commented Mar 6, 2015 at 5:52
@torazaburo That will not work if the objects have keys in common, right — thefourtheye
– thefourtheye, Commented Mar 6, 2015 at 6:02
@luis No, it won't, but you yourself showed a similar approach in your last code fragment (which works for me...why do you say "it is not working properly"?). — user663031
– user663031, Commented Mar 6, 2015 at 6:38
The problem comes when you have common keys between the 2 objects — luis
– luis, Commented Mar 6, 2015 at 19:05

score 1 · Accepted Answer · 2015-03-06 06:51:27Z

1

If you're intent on chaining, then

_.chain([a, b])   .         // [ { x: 1, y: 3, w: 4 }, { c: 2, d: 3 } ]
    map(_.values) .         // [ [    1,    3,    4 ], [    2,    3 ] ]
    flatten()     .         // [      1,    3,    4,        2,    3   ]
    uniq()        .         // [      1,    3,    4,        2         ]
    value()

edited Mar 6, 2015 at 6:51

answered Mar 6, 2015 at 6:23

user663031

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Comments

Gus Ortiz · Accepted Answer · 2015-03-06 05:51:53Z

0

How about.

var a = {x:1, y:3, w:4},
    b = {c:2, d:3};

result = _.values(_.extend(a,b));

answered Mar 6, 2015 at 5:51

Gus Ortiz

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4 Comments

Gus Ortiz Over a year ago

Weird, I always get the same result. jsfiddle.net/jrmh35js Are you changing the input?

thefourtheye Over a year ago

@GusOrtiz What if both the objects have same keys?

luis Over a year ago

right that's the issue.. it's fine though, i can make my keys unique for my solution. thanks! is having _.chain({}).extend(a,b).values() any different than your solution? I thought the chain one was faster

Gus Ortiz Over a year ago

Well I updated the fiddle adding the chain one, and timed them both. They usually end in the same time, but we´d need a lot of data to make a better benchmark. I don´t know the internals of underscore, so I wouldn't know which one is faster. jsfiddle.net/jrmh35js/1 Theres the updated fiddle.

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