Zip arrays in JavaScript? - Stack Overflow

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I have 2 arrays:

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

What I want to get as a result is:

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

It seems simple but I just can't figure out.

I want the result to be one array with each of the elements from the two arrays zipped together.

I have 2 arrays:

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

What I want to get as a result is:

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

It seems simple but I just can't figure out.

I want the result to be one array with each of the elements from the two arrays zipped together.

• javascript
• arrays

Share Improve this question Follow edited Feb 12, 2023 at 12:54 Roko C. Buljan 206k4141 gold badges325325 silver badges335335 bronze badges asked Feb 25, 2014 at 13:30 userMod2userMod2 8,9401717 gold badges7272 silver badges133133 bronze badges 3

• 3 Note that array map() is not supported in IE8, if that is a problem. Polyfill here developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… – TimHayes Commented Feb 25, 2014 at 13:37
• 20 This question is NOT a duplicate as it only asks for 2 arrays to be zipped instead of N arrays. Thus it is a special case for which there are specialized, more performant solutions. – le_m Commented Jun 4, 2016 at 0:52
• Just to say, the zip analogy comes from python – Matt Commented Sep 15, 2024 at 8:26

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6 Answers 6

Sorted by: Reset to default 388

Use the map method:

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

Zip Arrays of same length:

Using Array.prototype.map()

const zip = (a, b) => a.map((k, i) => [k, b[i]]);

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

Zip Arrays of different length:

Using Array.from()

const zip = (a, b) => Array.from(Array(Math.max(b.length, a.length)), (_, i) => [a[i], b[i]]);

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

Using Array.prototype.fill() and Array.prototype.map()

const zip = (a, b) => Array(Math.max(b.length, a.length)).fill().map((_,i) => [a[i], b[i]]);

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

Zip Multiple (n) Arrays:

const zip = (...arr) => Array(Math.max(...arr.map(a => a.length))).fill().map((_,i) => arr.map(a => a[i]));  
console.log(zip([1,2], [3,4], [5,6])); // [[1,3,5], [2,4,6]]

Zipping by leveraging generator functions

You can also use a generator function to 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}`);  
}

.as-console-wrapper { max-height: 100% !important; top: 0; }

The above works for any number of arrays and will zip() the longest array so undefined is returned as a value for shorter arrays.

Zipping of all Iterables

Here a function which can be used for all Iterables (e.g. Maps, Sets or your custom Iterable), not just arrays.

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}`);
}

Obviously it would also be possible to just use [...Iterable] to transform any Iterable to an array and then use the first function.

Using the reduce method:

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

var c = a.reduce((acc, curr, ind) => {
  acc.push([curr, b[ind]]);
  return acc;
}, []);
console.log(c)

With forEach method:

const a = [1, 2, 3]
const b = ['a', 'b', 'c']
const c = [];
a.forEach((el, ind) => {
    c.push([el, b[ind]])
});
console.log(c)

Providing a solution with imperative programming by a simple for loop.

This performs better when doing the zip operation on huge data sets compared to the convenient array functions like map() and forEach().

Example:

const a = [1, 2, 3];
const b = ['a', 'b', 'c'];
const result = [];
for (let i = 0; i < a.length; i++) {
  result.push([a[i], b[i]]);
}
console.log(result);

And if you want a 1 line simpler solution then you can use a library like ramda which has a zip function.

Example:

const a = [1, 2, 3];
const b = ['a', 'b', 'c'];
const result = R.zip(a, b);
console.log(result);

Zip function for 2D arrays

Here is a function that zips multiple arrays. However, It only works with 2D arrays only.

function zip(listOfLists) {
  return listOfLists[0].map((value, index) => {
      let result = [];

      for(let i=0; i <= listOfLists.length-1 ; i++){
          result[i] = listOfLists[i][index]
      }
      return result;
  });
}

This is how you run it.

console.log(
  zip(
    [[1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]]
  )
);

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