How to return conditional arrays from an array of objects?

Question

I'm wondering if it's possible to achieve the below in a cleaner/more efficient way?

const data = [
    { active: false },
    { active: false },
    { active: true },
    { active: false },
    { active: true },
]

const split = (data) => {
    const activeData = data.filter(({active}) => active).map(element => 1)
    const inactiveData = data.filter(({active}) => !active).map(element => 0)

    return [
        activeData,
        inactiveData,
    ]
}

console.log(split(data))

The problem here, I think, is that this approach is not efficient, because it's filtering and mapping through the same array twice. Is there a way to do it just once?

do you want only ones and zeros? please add the wanted result. — Nina Scholz
– Nina Scholz, Commented Aug 4, 2020 at 14:02
Define data just with array: [false, false, true, false, true] — Bhojendra Rauniyar
– Bhojendra Rauniyar, Commented Aug 4, 2020 at 14:03
The result is not particularly important, it could be mapped to anything. I'm just asking if it's possible to avoid cycling through the same array twice, as in the example. The only requirement here, is that data is an array of objects. — Mike K
– Mike K, Commented Aug 4, 2020 at 14:04

Pointy · Accepted Answer · 2020-08-04 14:56:43Z

1

Unless your array is immense, it's not going to make any difference. However you could indeed do this in one pass through the array:

    const split = (data) =>
      data.reduce((rv, entry) => 
        (rv[+!entry.active].push(+entry.active), rv)
      , [[], []]);
      
    console.log(split([{active: true}, {active: false}, {active: false}]))

The + operators convert the boolean values to numbers (0 or 1). The "rv" parameter to the .reduce() callback is the "return value", or the accumulator into which the .reduce() callback builds up the result. In this code, the callback uses the "active" flag (negated) to pick either the first array (index 0) or the second array (index 1) in the accumulator "array of arrays". The value pushed into the chosen array is then either 0 or 1, again obtained from the "active" value by converting it to a number.

edited Aug 4, 2020 at 14:56

answered Aug 4, 2020 at 14:10

Pointy

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2 Comments

Mike K Over a year ago

This an interesting approach and seems to be what I'm looking for. Can you please elaborate what's going on here? Does rv here refer to both of the initial arrays of the reducer? Also, it doesn't seem to work without the cast number (doing (rv[+!entry.active].push(+entry.active), rv) seems to crash it), does this mean it'll only work with numbers and not boolean values, for example?

Pointy Over a year ago

@MikeK I'll add a little explanation to the answer; it's basically what .reduce() is all about.

Nina Scholz · Accepted Answer · 2020-08-04 14:09:26Z

0

You could take a single loop and reduce the array using an object as target.

const
    data = [{ active: false }, { active: false }, { active: true }, { active: false }, { active: true }],
    { true: activeData, false: inactiveData } = data.reduce((r, { active }) =>
        (r[active].push(+active), r),
        { true: [], false: [] }
    );

console.log(activeData);
console.log(inactiveData);

edited Aug 4, 2020 at 14:09

answered Aug 4, 2020 at 14:04

Nina Scholz

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1 Comment

Mike K Over a year ago

This only returns one array, is it possible to return two, as in the OP approach?

Get Off My Lawn · Accepted Answer · 2020-08-04 14:59:53Z

0

You pretty much have a clean way to do it, however using multiple array functions requires more loops over the data.

Reduce will allow us to do this just once.

const data = [
    { active: false },
    { active: false },
    { active: true },
    { active: false },
    { active: true },
]

const split = (data) =>
  data.reduce((acc, {active}) => acc[+!active].push(+active) && acc, [[],[]])

console.log(split(data))

edited Aug 4, 2020 at 14:59

answered Aug 4, 2020 at 14:35

Get Off My Lawn

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2 Comments

Mike K Over a year ago

Is acc here always both of the arrays? If so, how do you determine which one to push to, if you're always just pushing to the acc?

Get Off My Lawn Over a year ago

This is just another variation of what @Pointy did.

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