CODEWITHSUNDEEP
javascriptecmascript-6functional-programming
Pete
Pete

Reputation: 3451

Is there a functional way to init an array in JavaScript ES6?

I finally gave up and wrote a for loop to initialize a simple array of objects where each object has an incremented counter (id) as an attribute of the object. In other words, I just want:

var sampleData = [{id: 1},{id: 2},...];

I was hoping for a compact syntax I could just put on my return statement.

let sampleData = [];
for (var p = 0; p < 25; p++){
    sampleData.push({id: p});
}

return {
    data: sampleData,
    isLoading: true
};

Upvotes: 35

Views: 14627

Answers (6)

jian
jian

Reputation: 1256

What I usually do is this:

const data = Array(10).fill().map((v, i) => ({id: i + 1}));
console.log({data});

fill ensures it can be used with map

Upvotes: 19

Mulan
Mulan

Reputation: 135197

You're looking for an anamorphism, or reverse fold –

// unfold : ((r, state) -> List r, unit -> List r, state) -> List r
const unfold = (f, init) =>
  f ( (x, next) => [ x, ...unfold (f, next) ]
    , () => [] 
    , init
    )
    
// sampleData : List { id: Int }
const sampleData =
  unfold
    ( (next, done, i) =>
        i > 25
          ? done ()
          : next ({ id: i }, i + 1)
    , 0
    )
    
console .log (sampleData)
// [ { id: 0 }, { id : 1 }, ... { id: 25 } ]

You can get an intuition for how unfold works by seeing it used in other common programs –

// unfold : ((r, state) -> List r, unit -> List r, state) -> List r
const unfold = (f, init) =>
  f ( (x, next) => [ x, ...unfold (f, next) ]
    , () => []
    , init
    )
    
// fibseq : Int -> List Int
const fibseq = init =>
  unfold
    ( (next, done, [ n, a, b ]) =>
         n === 0
           ? done ()
           : next (a, [ n - 1, b, a + b ])
    , [ init, 0, 1 ]
    )
    
console .log (fibseq (10))
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ]

The implementation of unfold is just one possibility. Get tinkering and implement it in a way of your choosing –

// type Maybe a = Nothing | Just a    

// Just : a -> Maybe a
const Just = x =>
  ({ match: ({ Just: f }) => f (x) })

// Nothing : unit -> Maybe a
const Nothing = () =>
  ({ match: ({ Nothing: f }) => f () })

// unfold : (state -> Maybe (a, state), state) -> List a  
const unfold = (f, init) =>
  f (init) .match
    ( { Nothing: () => []
      , Just: ([ x, next ]) => [ x, ...unfold (f, next) ]
      }
    )

// fibseq : Int -> List Int
const fibseq = init =>
  unfold
    ( ([ n, a, b ]) =>
        n === 0
          ? Nothing ()
          : Just ([ a, [ n - 1, b, a + b ] ]) // <-- yikes, read more below
    , [ init, 0, 1 ]
    )
    
console .log (fibseq (10))
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ]

I cheated a little above using a [] as a tuple. This kept the program shorter but it's better to explicitly model things and consider their types. You tagged this question with functional-programming so it's worth going the extra inch to remove this kind of implicit handling from our programs. By showing this as a separate step, we isolate a technique that can be applied not just to unfold, but for any program we design –

// type Maybe a = Nothing | Just a
// type Tuple a b = { first: a, second: b }

// Just : a -> Maybe a
const Just = x =>
  ({ match: ({ Just: f }) => f (x) })

// Nothing : unit -> Maybe a
const Nothing = () =>
  ({ match: ({ Nothing: f }) => f () })

// Tuple : (a, b) -> Tuple a b
const Tuple = (first, second) =>
  ({ first, second })

// unfold : (state -> Maybe Tuple (a, state), state) -> List a  
const unfold = (f, init) =>
  f (init) .match
    ( { Nothing: () => []
      , Just: (t) => [ t.first, ...unfold (f, t.second) ] // <-- Tuple
      }
    )

// fibseq : Int -> List Int
const fibseq = init =>
  unfold
    ( ([ n, a, b ]) =>
        n === 0
          ? Nothing ()
          : Just (Tuple (a, [ n - 1, b, a + b ])) // <-- Tuple
    , [ init, 0, 1 ]
    )
    
console .log (fibseq (10))
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ]

Upvotes: 6

Mark
Mark

Reputation: 92440

Array.from() is a nice way to do this. You can pass a {length: somlength} object or some other array-like object and a function that defines each item. The first argument (calling it _ just to indicate it's not used) to that function would be the item from an array we passed in (but we only passed in a length so it doesn't mean much), the second i is the index, which is used for your id:

let sampleData = Array.from({length: 10}, (_, id) => ({id}))

console.log(sampleData)

Upvotes: 59

Aliaksandr Sushkevich
Aliaksandr Sushkevich

Reputation: 12364

You can use a simple recursive process to do that.

const iter = (arr, counter) => {
  if (counter === 25) return arr;
  return iter([...arr, {id:counter}], counter + 1)
}
iter([], 0)

Upvotes: 3

stwilz
stwilz

Reputation: 2397

The .from() example is great but if you really want to get creative check this out.

const newArray = length => [...`${Math.pow(10, length) - 1}`]
newArray(2)
newArray(10)

Massively limited though

newArray(1000)
["I", "n", "f", "i", "n", "i", "t", "y"]

Upvotes: 4

JohanP
JohanP

Reputation: 5472

You can use spread operator with Array and then map each undefined element to the object that you want.

var arr = [...Array(10)].map((_,i)=>({id:i}));
console.log(arr)

Upvotes: 9

Related Questions