打印稿版本。演示了将101个随机uid分成10个组

const idArrayLengthLimit = 10;
const randomOneHundredOneIdArray = Array
    .from(Array(101).keys())
    .map(() => generateUid(5));

function generateUid(length: number) {
  const uidString: string[] = [];
  const uidChars = 'abcdefghijklmnopqrstuvwxyz0123456789';
  for (let i = 0; i < length; i++) {
    uidString
      .push(uidChars.charAt(Math.floor(Math.random() * uidChars.length)));
  }
  return uidString.join('');
}

for (let i = 0; i < randomOneHundredOneIdArray.length; i++) {
 if(i % idArrayLengthLimit === 0){
     const result = randomOneHundredOneIdArray
       .filter((_,id) => id >= i && id < i + idArrayLengthLimit);
    // Observe result
    console.log(result);
 }
}

好吧，让我们从一个相当严格的开始:

function chunk(arr, n) {
    return arr.slice(0,(arr.length+n-1)/n|0).
           map(function(c,i) { return arr.slice(n*i,n*i+n); });
}

它是这样使用的:

chunk([1,2,3,4,5,6,7], 2);

然后我们就有了这个紧密的减速器函数:

function chunker(p, c, i) {
    (p[i/this|0] = p[i/this|0] || []).push(c);
    return p;
}

它是这样使用的:

[1,2,3,4,5,6,7].reduce(chunker.bind(3),[]);

因为当我们将它绑定到一个数字时，小猫就死了，我们可以像这样手动curry:

// Fluent alternative API without prototype hacks.
function chunker(n) {
   return function(p, c, i) {
       (p[i/n|0] = p[i/n|0] || []).push(c);
       return p;
   };
}

它是这样使用的:

[1,2,3,4,5,6,7].reduce(chunker(3),[]);

然后是仍然非常紧凑的函数，它可以一次性完成所有操作:

function chunk(arr, n) {
    return arr.reduce(function(p, cur, i) {
        (p[i/n|0] = p[i/n|0] || []).push(cur);
        return p;
    },[]);
}

chunk([1,2,3,4,5,6,7], 3);

这里是整洁和优化的实现chunk()函数。假设默认块大小为10。

var chunk = function(list, chunkSize) {
  if (!list.length) {
    return [];
  }
  if (typeof chunkSize === undefined) {
    chunkSize = 10;
  }

  var i, j, t, chunks = [];
  for (i = 0, j = list.length; i < j; i += chunkSize) {
    t = list.slice(i, i + chunkSize);
    chunks.push(t);
  }

  return chunks;
};

//calling function
var list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12];
var chunks = chunk(list);

嗨，试试这个——

 function split(arr, howMany) {
        var newArr = []; start = 0; end = howMany;
        for(var i=1; i<= Math.ceil(arr.length / howMany); i++) {
            newArr.push(arr.slice(start, end));
            start = start + howMany;
            end = end + howMany
        }
        console.log(newArr)
    }
    split([1,2,3,4,55,6,7,8,8,9],3)

这里是一个仅使用递归和slice()的非突变解决方案。

const splitToChunks = (arr, chunkSize, acc = []) => (
    arr.length > chunkSize ?
        splitToChunks(
            arr.slice(chunkSize),
            chunkSize,
            [...acc, arr.slice(0, chunkSize)]
        ) :
        [...acc, arr]
);

然后简单地像splitToChunks([1,2,3,4,5]， 3)一样使用它来获得[[1,2,3]，[4,5]]。

这里有一个小提琴供你尝试:https://jsfiddle.net/6wtrbx6k/2/

这是我能想到的最有效、最直接的解决方案:

function chunk(array, chunkSize) {
    let chunkCount = Math.ceil(array.length / chunkSize);
    let chunks = new Array(chunkCount);
    for(let i = 0, j = 0, k = chunkSize; i < chunkCount; ++i) {
        chunks[i] = array.slice(j, k);
        j = k;
        k += chunkSize;
    }
    return chunks;
}