[removed] Make an array of value pairs form an arr

问题:

Is there an elegant, functional way to turn this array:

[ 1, 5, 9, 21 ]

into this

[ [1, 5], [5, 9], [9, 21] ]

I know I could forEach the array and collect the values to create a new array. Is there an elegant way to do that in _.lodash without using a forEach?

回答1:

You could map a spliced array and check the index. If it is not zero, take the predecessor, otherwise the first element of the original array.

var array = [1, 5, 9, 21],
    result = array.slice(1).map((a, i, aa) => [i ? aa[i - 1] : array[0], a]);

console.log(result);

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An even shorter version, as suggested by Bergi:

var array = [1, 5, 9, 21],
    result = array.slice(1).map((a, i) => [array[i], a]);

console.log(result);

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回答2:

A fast approach using map would be:

const arr = [ 1, 5, 9, 21 ];

const grouped = arr.map((el, i) => [el, arr[i+1]]).slice(0, -1);

console.log(grouped);

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回答3:

This is easily done with array.reduce. What the following does is use an array as aggregator, skips the first item, then for each item after that pushes previous item and the current item as a pair to the array.

const arr = [ 1, 5, 9, 21 ];
const chunked = arr.reduce((p, c, i, a) => i === 0 ? p : (p.push([c, a[i-1]]), p), []);

console.log(chunked);

An expanded version would look like:

const arr = [1, 5, 9, 21];
const chunked = arr.reduce(function(previous, current, index, array) {
  if(index === 0){
    return previous;
  } else {
    previous.push([ current, array[index - 1]]);
    return previous;
  }
}, []);

console.log(chunked);

回答4:

If you're willing to use another functional library 'ramda', aperture is the function you're looking for.

Example usage taken from the ramda docs:

R.aperture(2, [1, 2, 3, 4, 5]); //=> [[1, 2], [2, 3], [3, 4], [4, 5]]
R.aperture(3, [1, 2, 3, 4, 5]); //=> [[1, 2, 3], [2, 3, 4], [3, 4, 5]]
R.aperture(7, [1, 2, 3, 4, 5]); //=> []

回答5:

You may do as follows with just a sinle liner of .reduce() with no initial;

var arr = [ 1, 5, 9, 21 ],
  pairs = arr.reduce((p,c,i) => i == 1 ? [[p,c]] : p.concat([[p[p.length-1][1],c]]));
console.log(pairs);

回答6:

I'm sure there is an elegant way, programmatically, but, mathematically I can't help seeing that each new pair has an index difference of 1 from the original array.

If you (later) have the need to turn your array [ 1, 5, 9, 21, 33 ] into [ [1, 9], [5, 21], [9, 33] ], you can use the fact that the difference between the indices is 2.

If you create code for the index difference of 1, extending this would be easy.

回答7:

Here's slide which has two parameters to control the size of the slice and how many elements are dropped between slices

slide differs from other answers here by giving you these control parameters. other answers here are limited to producing only a slices of 2, or incrementing the slice by 1 each time

// take :: (Int, [a]) -> [a]
const take = (n, xs) =>
  xs.slice(0, n)

// drop :: (Int, [a]) -> [a]
const drop = (n, xs) =>
  xs.slice(n)
  
// slice :: (Int, Int, [a]) -> [[a]]
const slide = (m, n, xs) =>
  xs.length > m
    ? [take(m, xs), ...slide(m, n, drop(n, xs))]
    : [xs]
    
const arr = [0,1,2,3,4,5,6]

// log helper improves readability of output in stack snippet
const log = x => console.log(JSON.stringify(x))

log(slide(1, 1, arr))
// [[0],[1],[2],[3],[4],[5],[6]]

log(slide(1, 2, arr))
// [[0],[2],[4],[6]]

log(slide(2, 1, arr))
// [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]]

log(slide(2, 2, arr))
// [[0,1],[2,3],[4,5],[6]]

log(slide(3, 1, arr))
// [[0,1,2],[1,2,3],[2,3,4],[3,4,5],[4,5,6]]

log(slide(3, 2, arr))
// [[0,1,2],[2,3,4],[4,5,6]]

log(slide(3, 3, arr))
// [[0,1,2],[3,4,5],[6]] 

If for some reason you didn't want slide to include partial slices, (slices smaller than m), we could edit it as such

// slice :: (Int, Int, [a]) -> [[a]]
const slide = (m, n, xs) =>
  xs.length > m
    ? [take(m, xs), ...slide(m, n, drop(n, xs))]
    : [] // <- return [] instead of [xs]

log(slide(2, 2, arr))
// now prints: [[0,1],[2,3],[4,5]]
// instead of: [[0,1],[2,3],[4,5],[6]]

回答8:

I noticed that the current solutions, in a way, all look ahead or behind (arr[i + 1] or arr[i - 1]).

It might be useful to also explore an approach that uses reduce and an additional array, defined within a function's closure, to store a to-be-completed partition.

Notes:

• Not a one liner, but hopefully easy to understand
• part doesn't have to be an array when working with only 2 items, but by using an array, we extend the method to work for n-sized sets of items
• If you're not a fan of shift, you can use a combination of slice and redefine part, but I think it's safe here.
• partitions with a length less than the required number of elements are not returned

const partition = partitionSize => arr => {
  const part = [];
  
  return arr.reduce((parts, x) => {
    part.push(x);
    
    if (part.length === partitionSize) {
      parts.push(part.slice());
      part.shift();
    }
    
    return parts;
  }, []);
};

const makePairs = partition(2);
const makeTrios = partition(3);

const pairs = makePairs([1,2,3,4,5,6]);
const trios = makeTrios([1,2,3,4,5,6]);


console.log("partition(2)", JSON.stringify(pairs));
console.log("partition(3)", JSON.stringify(trios));