如何在JavaScript中压缩两个数组？

Question

我有2个数组:

var a = [1, 2, 3]
var b = [a, b, c]

我想得到的结果是:

[[1, a], [2, b], [3, c]]

这看起来很简单,但我无法弄清楚.

我希望结果是一个数组,两个数组中的每个元素都压缩在一起.

Answer 1

使用map方法:

var a = [1, 2, 3]
var b = ['a', 'b', 'c']

var c = a.map(function(e, i) {
  return [e, b[i]];
});

console.log(c)

DEMO

Answer 2

Array.prototype.map() 旧版浏览器不支持

const a = [1, 2, 3],
      b = ['a', 'b', 'c'];

const zip = (arr1, arr2) => arr1.map((k, i) => [k, arr2[i]]);

console.log(zip(a, b)) // [[1, "a"], [2, "b"], [3, "c"]]

这是一个适用于所有地方的替代方案

var a = [1, 2, 3],
    b = ['a', 'b', 'c'];

var zip = [];
for (var i = 0; i < a.length; i++){
   zip.push([a[i], b[i]]);
}

console.log(zip); // [[1, "a"], [2, "b"], [3, "c"]]

Answer 3

利用生成器函数进行压缩

您还可以使用生成器函数来zip().

const a = [1, 2, 3]
const b = ['a', 'b', 'c']

/**
 * Zips any number of arrays. It will always zip() the largest array returning undefined for shorter arrays.
 * @param  {...Array<any>} arrays 
 */
function* zip(...arrays){
  const maxLength = arrays.reduce((max, curIterable) => curIterable.length > max ? curIterable.length: max, 0);
  for (let i = 0; i < maxLength; i++) {
    yield arrays.map(array => array[i]);
  }
}

// put zipped result in an array
const result = [...zip(a, b)]

// or lazy generate the values
for (const [valA, valB] of zip(a, b)) {
  console.log(`${valA}: ${valB}`);  
}

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上面的方法适用于任意数量的数组，并且zip()最长的数组undefined将作为较短数组的值返回。

全部压缩Iterables

这里的函数可用于所有Iterables（例如Maps，Sets或您的自定义Iterable），而不仅仅是数组。

const a = [1, 2, 3];
const b = ["a", "b", "c"];

/**
 * Zips any number of iterables. It will always zip() the largest Iterable returning undefined for shorter arrays.
 * @param  {...Iterable<any>} iterables
 */
function* zip(...iterables) {
  // get the iterator of for each iterables
  const iters = [...iterables].map((iterable) => iterable[Symbol.iterator]());
  let next = iters.map((iter) => iter.next().value);
  // as long as any of the iterables returns something, yield a value (zip longest)
  while(anyOf(next)) {
    yield next;
    next = iters.map((iter) => iter.next().value);
  }

  function anyOf(arr){
    return arr.some(v => v !== undefined);
  }
}

// put zipped result in aa array
const result = [...zip(a, new Set(b))];

// or lazy generate the values
for (const [valA, valB] of zip(a, new Set(b))) {
  console.log(`${valA}: ${valB}`);
}

显然，也可以仅使用[...Iterable]将 any 转换Iterable为数组，然后使用第一个函数。