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
下面是一个使用ImmutableJS的解决方案，其中items是一个不可变列表，size是所需的分组大小。
const partition = ((items, size) => {
return items.groupBy((items, i) => Math.floor(i/size))
})
其他回答
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));
使用来自lodash的chunk
lodash.chunk(arr,<size>).forEach(chunk=>{
console.log(chunk);
})
使用ES6的拼接版本
让[列表,chunkSize] =[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12、13、14、15),6]; list =[…数组(Math.ceil(list. ceil))length / chunkSize))]。map(_ => list.splice(0,chunkSize)) console.log(列表);
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 ]
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)
