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javascripthaskellrecursion

takeWhile implementation in JavaScript - Looking for better ideas


Haskell has a takeWhile function:

Prelude> takeWhile odd [1,3,5,7,8,9]
[1,3,5,7]

It “takes” elements from a list as long as applying the predicate function results in True. At the point it becomes False, it stops.

How can we implement it?

Here is a Haskell recursive approach I came up with:

takewhile::(a->Bool)->[a]->[a]
takewhile _ [] = []
takewhile f (x:xs) | f x == True = x : takewhile f xs
                   | otherwise = []

It keeps on calling itself as long as predicate f x is True, otherwise it returns an empty list without calling itself.

I could come up with the following implementation for JavaScript. It is a bit verbose and invokes defining another function to pass the intermediate result around:

function takeWhile(f, xs) {
 return take(f, xs, [])
}

function take(f, xs, arr) {
 if(!xs || xs.length === 0) {
 return arr
 }
 x = xs.shift()
 if(f(x)) {
   arr.push(x)
   return take(f, xs, arr)
 } else {
   return arr
 }
}

takeWhile((x)=>{
 return x % 2 !== 0
},[1,3,5,7,9,11])

Are there better ideas for implementing it in JavaScript?


Solution

  • If you want your takeWhile to perform like in HS, i.e. lazily, you need generators in JS:

    function* takeWhile(fn, xs) {
        for (let x of xs)
            if (fn(x))
                yield x;
            else
                break;
    }
    
    function* naturalNumbers() {
        let n = 0;
        while (true)
            yield n++;
    }
    
    result = takeWhile(x => x < 10, naturalNumbers())
    console.log([...result])

    A straight port of the HS code is also possible, but it only works with materialized arrays (that is, eagerly):

    // would be nice, but JS sucks ;(
    // let takeWhile = (f, [x, ...xs]) => f(x) ? [x, ...takeWhile(f, xs)] : [];
    
    let takeWhile = (f, xs) => xs.length ? takeWhileNotEmpty(f, xs) : [];
    let takeWhileNotEmpty = (f, [x, ...xs]) =>  f(x) ? [x, ...takeWhile(f, xs)] : [];
    
    
    let odd = x => x % 2
    a = [1,3,5,7,8,9]
    r = takeWhile(odd, a)
    console.log(r)

    Actually, as @naomik shows here there's a better way to deal with empty lists:

    let nil = {};
    let takeWhile = (f, [x = nil, ...xs]) => (x === nil || !f(x)) 
        ? [] : [x, ...takeWhile(f, xs)];
    
    console.log(takeWhile(x => x % 2, [1, 3, 5, 7, 8, 9]));

    Finally, your initial attempt does have a point, because, unlike the above, it's tail-recursive, which is a Good Thing. It can be written more concisely as

    let takeWhile = (f, xs) => take1(f, xs, []);
    let take1 = (f, xs, acc) => xs.length ? take2(f, xs, acc) : acc;
    let take2 = (f, [x, ...xs], acc) => f(x) ? take1(f, xs, acc.concat(x)) : acc;
    

    A combination of both approaches (that is, a recursive generator) left as an exercise...