使用来自lodash的chunk

lodash.chunk(arr,<size>).forEach(chunk=>{
  console.log(chunk);
})

js

函数splitToBulks(arr, bulkSize = 20) { Const bulks = []; 对于(设I = 0;i < Math.ceil(arr。长度/ bulkSize);我+ +){ bulks.push(加勒比海盗。(i * bulkSize， (i + 1) * bulkSize)); ｝ 返回散货; ｝ console.log(splitToBulks([1,2,3,4,5,6,7]， 3));

打印稿

function splitToBulks<T>(arr: T[], bulkSize: number = 20): T[][] {
    const bulks: T[][] = [];
    for (let i = 0; i < Math.ceil(arr.length / bulkSize); i++) {
        bulks.push(arr.slice(i * bulkSize, (i + 1) * bulkSize));
    }
    return bulks;
}

如果你使用EcmaScript >= 5.1版本，你可以使用array.reduce()实现一个函数版本的chunk()，复杂度为O(N):

function chunk(chunkSize, array) { return array.reduce(function(previous, current) { var chunk; if (previous.length === 0 || previous[previous.length -1].length === chunkSize) { chunk = []; // 1 previous.push(chunk); // 2 } else { chunk = previous[previous.length -1]; // 3 } chunk.push(current); // 4 return previous; // 5 }, []); // 6 } console.log(chunk(2, ['a', 'b', 'c', 'd', 'e'])); // prints [ [ 'a', 'b' ], [ 'c', 'd' ], [ 'e' ] ]

以上每个// nbr的解释:

如果之前的值，即之前返回的块数组是空的，或者如果之前的最后一个块有chunkSize项，则创建一个新的块 将新数据块添加到现有数据块数组中 否则，当前块是块数组中的最后一个块 将当前值添加到块中 返回修改后的块数组 通过传递一个空数组初始化还原

基于chunkSize的curry:

var chunk3 = function(array) {
    return chunk(3, array);
};

console.log(chunk3(['a', 'b', 'c', 'd', 'e']));
// prints [ [ 'a', 'b', 'c' ], [ 'd', 'e' ] ]

你可以将chunk()函数添加到全局Array对象:

Object.defineProperty(Array.prototype, 'chunk', { value: function(chunkSize) { return this.reduce(function(previous, current) { var chunk; if (previous.length === 0 || previous[previous.length -1].length === chunkSize) { chunk = []; previous.push(chunk); } else { chunk = previous[previous.length -1]; } chunk.push(current); return previous; }, []); } }); console.log(['a', 'b', 'c', 'd', 'e'].chunk(4)); // prints [ [ 'a', 'b', 'c' 'd' ], [ 'e' ] ]

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;
}

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/

# in coffeescript
# assume "ar" is the original array
# newAr is the new array of arrays

newAr = []
chunk = 10
for i in [0... ar.length] by chunk
   newAr.push ar[i... i+chunk]

# or, print out the elements one line per chunk
for i in [0... ar.length] by chunk
   console.log ar[i... i+chunk].join ' '