一个很好的函数是:

function chunk(arr,times){
    if(times===null){var times = 10} //Fallback for users wanting to use the default of ten
   var tempArray = Array() //Array to be populated with chunks
    for(i=0;i<arr.length/times;i++){
     tempArray[i] = Array() //Sub-Arrays        //Repeats for each chunk         
   for(j=0;j<times;j++){
        if(!(arr[i*times+j]===undefined)){tempArray[i][j] = arr[i*times+j]//Populate Sub-  Arrays with chunks
    }
     else{
       j = times //Stop loop
       i = arr.length/times //Stop loop
  }
    }
     }
   return tempArray //Return the populated and chunked array
   }

用法如下:

chunk(array,sizeOfChunks)

我对它做了注释，这样你就能理解发生了什么。

(格式有点不对，我在移动设备上编程)

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;
}

基于数组的ES6一行方法。原型缩减和推送方法:

const doChunk = (list, size) => list.reduce((r, v) =>
  (!r.length || r[r.length - 1].length === size ?
    r.push([v]) : r[r.length - 1].push(v)) && r
, []);

console.log(doChunk([0,1,2,3,4,5,6,7,8,9,10,11,12], 5));
// [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12]]

in coffeescript:

b = (a.splice(0, len) while a.length)

demo 
a = [1, 2, 3, 4, 5, 6, 7]

b = (a.splice(0, 2) while a.length)
[ [ 1, 2 ],
  [ 3, 4 ],
  [ 5, 6 ],
  [ 7 ] ]

array.slice()方法可以根据需要从数组的开头、中间或结尾提取切片，而不需要改变原始数组。

const chunkSize = 10;
for (let i = 0; i < array.length; i += chunkSize) {
    const chunk = array.slice(i, i + chunkSize);
    // do whatever
}

最后一个块可能小于chunkSize。例如，当给定一个包含12个元素的数组时，第一个块将有10个元素，第二个块只有2个。

注意，chunkSize为0将导致无限循环。

results = []
chunk_size = 10
while(array.length > 0){
   results.push(array.splice(0, chunk_size))
}