让我们说我有一个Javascript数组看起来如下:
["Element 1","Element 2","Element 3",...]; // with close to a hundred elements.
什么样的方法适合将数组分成许多更小的数组,假设最多有10个元素?
让我们说我有一个Javascript数组看起来如下:
["Element 1","Element 2","Element 3",...]; // with close to a hundred elements.
什么样的方法适合将数组分成许多更小的数组,假设最多有10个元素?
当前回答
一个有效的解决方案是通过indexChunk将解决方案与slice和push连接起来,解决方案被分割成块:
function splitChunks(sourceArray, chunkSize) { if(chunkSize <= 0) throw "chunkSize must be greater than 0"; let result = []; for (var i = 0; i < sourceArray.length; i += chunkSize) { result[i / chunkSize] = sourceArray.slice(i, i + chunkSize); } return result; } let ar1 = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 ]; console.log("Split in chunks with 4 size", splitChunks(ar1, 4)); console.log("Split in chunks with 7 size", splitChunks(ar1, 7));
其他回答
下面的ES2015方法不需要定义函数,直接在匿名数组上工作(例如块大小为2):
[11,22,33,44,55].map((_, i, all) => all.slice(2*i, 2*i+2)).filter(x=>x.length)
如果你想为此定义一个函数,你可以这样做(改进K._对Blazemonger的回答的评论):
const array_chunks = (array, chunk_size) => array
.map((_, i, all) => all.slice(i*chunk_size, (i+1)*chunk_size))
.filter(x => x.length)
我是这样解决的:
const chunks = [];
const chunkSize = 10;
for (let i = 0; i < arrayToSplit.length; i += chunkSize) {
const tempArray = arrayToSplit.slice(i, i + chunkSize);
chunks.push(tempArray);
}
使用发电机
函数*块(arr, n) { 对于(设I = 0;I < arrr .length;I += n) { 加勒比海盗。Slice (i, i + n); } } let someArray = [0,1,2,3,4,5,6,7,8,9] console.log([…块(someArray, 2)]) / /[[0, 1],[2、3],[4,5],[6、7],[8 9]]
可以像这样用Typescript输入:
function* chunks<T>(arr: T[], n: number): Generator<T[], void> {
for (let i = 0; i < arr.length; i += n) {
yield arr.slice(i, i + n);
}
}
这里有一个更具体的案例,有人可能会觉得有价值。我还没看到这里提到过。
如果你不想要固定/均匀的数据块大小,而是想要指定拆分数组的下标怎么办?在这种情况下,你可以使用这个:
const splitArray = (array = [], splits = []) => {
array = [...array]; // make shallow copy to avoid mutating original
const chunks = []; // collect chunks
for (const split of splits.reverse()) chunks.push(array.splice(split)); // go backwards through split indices and lop off end of array
chunks.push(array); // add last remaining chunk (at beginning of array)
return chunks.reverse(); // restore chunk order
};
然后:
splitArray([1, 2, 3, 4, 5, 6, 7, 8, 9], [4, 6])
// [ [1, 2, 3, 4] , [5, 6] , [7, 8, 9] ]
请注意,如果你给它非升序/重复/负/非整数/等分割索引,这将会发生有趣的事情。您可以为这些边缘情况添加检查(例如array .from(new Set(array))来消除重复。
对源数组进行突变:
let a = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ], aa = [], x
while((x = a.splice(0, 2)).length) aa.push(x)
// aa == [ [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ], [ 9 ] ]
// a == []
不改变源数组:
let a = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ], aa = []
for(let i = 0; i < a.length; i += 2) aa.push(a.slice(i, i + 2))
// aa == [ [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ], [ 9 ] ]
// a == [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]