d3.js: extract array in object within array

Question

I would like to obtain array in object within array.

Original structure:
Array [ object, ..., object ]

var dateC = [ 
      {
      key: "2016-01-01",
      values: **[
        {city:"", country:""...},
        {...}
      ]**},
      {
      key: "2016-01-02",
      values: [
        {...},
        {...}
      ]}
]

var dateC2 = dateC.filter(function(d) { return d.key == selected; }).map(function(d) { return d.values })

I extracted the object from dateC that has key: "2016-01-01" using the above code.

Current structure:
Array [ Array[114] ]

var dateC2 = [ 
  {
  key: "2016-01-01",
  values: **[
    {city:"", country:""...},
    {...}
  ]**}
]

Desired structure:
Array [ Object, ..., Object ]

**[{city:"", country:""...}, {...}]**

Array that I want is contained by **

I am not sure how I should use forEach method to get the array from values because I have already use filter and map on dateC to obtain dateC2. Alternatively, is there a quicker way to obtain the desired structure from original structure?

please add some more data, especially how the result should look like. — Nina Scholz
– Nina Scholz, Commented Jul 4, 2016 at 8:09

Nina Scholz · Accepted Answer · 2016-07-04 08:28:24Z

4

You could use a single loop for it with Array#reduce.

var dateC2 = dateC.reduce(function(r, d) {
    return d.key == selected ? 
        r.concat(d.values):
        r;
}, []);

answered Jul 4, 2016 at 8:28

Nina Scholz

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1 Comment

Morteza Tourani Over a year ago

I think forEach would be a better choice. (as you mentioned before, no need to return).

Morteza Tourani · Accepted Answer · 2016-07-04 19:16:18Z

1

I think Array.prototype.forEach would be another choice:

var dateC2 = [];
dateC.forEach(function(d) {
    if(d.key == selected)
         dateC2 = dateC2.concat(d.values);
});

Or

var dateC2 = [];
dateC.forEach(function(d) {
    if(d.key == selected)
        Array.prototype.push.apply(dateC2, d.values);
});

answered Jul 4, 2016 at 19:16

Morteza Tourani

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2 Comments

shawnngtq Over a year ago

Hi, why is it better? Is it because it has a better performance?

Morteza Tourani Over a year ago

@Shawn If this comment is about my comment under Nina's answer, yes. But you should ask it there not here.

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