A better way to use map to create a new array from array's array

Question

I have a list of projects as followings:

projects = [{name: '1',
             agencies: ['USA', 'China']},
            {name: '2',
             agencies: ['Japan', 'Russia']}]

I want to get the result ['USA', 'China', 'Japan', 'Russia'], currently I use the following code, but I would like to know if there is better solution (one line code) using map function:

const agencies = [];
projects.forEach((project) =>
        project.agencies.forEach((agency) => {
            agencies.push(agency);
        })
);

I'm voting to close this question as off-topic because it is asking for a better way to do already existing code. ask on Code Review — Daniel A. White
– Daniel A. White, Commented Jan 25, 2018 at 20:47
const agencies = [].concat(...projects.map(x => x.agencies)) for a one liner. — CRice
– CRice, Commented Jan 25, 2018 at 20:49
This is definitely what I want. Could you post it as answer? — Thomas
– Thomas, Commented Jan 25, 2018 at 20:51
var agencies = projects.reduce((arr, p) => [...arr, ...p.agencies], []) :D — Fernando Carvajal
– Fernando Carvajal, Commented Jan 25, 2018 at 21:09

CRice · Accepted Answer · 2018-01-25 20:54:26Z

1

Here is a simple one-liner to accomplish this:

const projects = [{
  name: '1',
  agencies: ['USA', 'China']
}, {
  name: '2',
  agencies: ['Japan', 'Russia']
}]

const agencies = [].concat(...projects.map(x => x.agencies));
console.log(agencies)

answered Jan 25, 2018 at 20:54

CRice

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marcelotokarnia · Accepted Answer · 2018-01-25 20:55:24Z

1

returned maps, by definition, will always be of the same length as the array used to generate it.

You should probably use reduce instead, to generate a different length array like you want.

try:

 projects.reduce((acc, project) => [...acc, ...project.agencies], [])

answered Jan 25, 2018 at 20:55

marcelotokarnia

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Jonas Wilms · Accepted Answer · 2018-01-25 21:00:50Z

1

I would use destructuring, reduce and concat:

  agencies.reduce((res, {agencies}) => res.concat(agencies));

answered Jan 25, 2018 at 21:00

Jonas Wilms

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