54

我想合并多个数组缓冲区来创建一个 Blob。但是，如您所知， TypedArray没有“推送”或有用的方法......

例如：

var a = new Int8Array( [ 1, 2, 3 ] );
var b = new Int8Array( [ 4, 5, 6 ] );

结果，我想得到[ 1, 2, 3, 4, 5, 6 ].

4

8 回答 8

97

使用set方法。但请注意，您现在需要两倍的内存！

var a = new Int8Array( [ 1, 2, 3 ] );
var b = new Int8Array( [ 4, 5, 6 ] );

var c = new Int8Array(a.length + b.length);
c.set(a);
c.set(b, a.length);

console.log(a);
console.log(b);
console.log(c);
于 2012-12-28T15:07:28.837 回答
11

对于客户端〜ok解决方案：

const a = new Int8Array( [ 1, 2, 3 ] )
const b = new Int8Array( [ 4, 5, 6 ] )
const c = Int8Array.from([...a, ...b])
于 2019-09-15T16:41:59.193 回答
9

我总是使用这个功能：

function mergeTypedArrays(a, b) {
    // Checks for truthy values on both arrays
    if(!a && !b) throw 'Please specify valid arguments for parameters a and b.';  

    // Checks for truthy values or empty arrays on each argument
    // to avoid the unnecessary construction of a new array and
    // the type comparison
    if(!b || b.length === 0) return a;
    if(!a || a.length === 0) return b;

    // Make sure that both typed arrays are of the same type
    if(Object.prototype.toString.call(a) !== Object.prototype.toString.call(b))
        throw 'The types of the two arguments passed for parameters a and b do not match.';

    var c = new a.constructor(a.length + b.length);
    c.set(a);
    c.set(b, a.length);

    return c;
}

不检查 null 或类型的原始函数

function mergeTypedArraysUnsafe(a, b) {
    var c = new a.constructor(a.length + b.length);
    c.set(a);
    c.set(b, a.length);

    return c;
}
于 2016-02-25T17:15:21.100 回答
2

如果我有多个类型化数组

            arrays = [ typed_array1, typed_array2,..... typed_array100]

我想将所有 1 到 100 个子数组连接成单个“结果”，这个函数对我有用。

  single_array = concat(arrays)


function concat(arrays) {
  // sum of individual array lengths
  let totalLength = arrays.reduce((acc, value) => acc + value.length, 0);

  if (!arrays.length) return null;

   let result = new Uint8Array(totalLength);

      // for each array - copy it over result
      // next array is copied right after the previous one
      let length = 0;
      for(let array of arrays) {
            result.set(array, length);
            length += array.length;
      }

      return result;
   }
于 2020-01-24T19:29:36.703 回答
1

作为一个单行，它将采用任意数量的数组（myArrays此处）和混合类型，只要结果类型将它们全部（Int8Array此处）：

let combined = Int8Array.from(Array.prototype.concat(...myArrays.map(a => Array.from(a))));
于 2019-06-08T15:32:13.733 回答
0

对于喜欢单线的人：

  const binaryData = [
    new Uint8Array([1, 2, 3]),
    new Int16Array([4, 5, 6]),
    new Int32Array([7, 8, 9])
  ];

  const mergedUint8Array = new Uint8Array(binaryData.map(typedArray => [...new Uint8Array(typedArray.buffer)]).flat());
于 2020-09-04T09:40:18.277 回答
0 
function concat (views: ArrayBufferView[]) {
    let length = 0
    for (const v of views)
        length += v.byteLength
        
    let buf = new Uint8Array(length)
    let offset = 0
    for (const v of views) {
        const uint8view = new Uint8Array(v.buffer, v.byteOffset, v.byteLength)
        buf.set(uint8view, offset)
        offset += uint8view.byteLength
    }
    
    return buf
}
于 2021-11-17T03:20:15.573 回答
-1

我喜欢@prinzhorn 的回答，但我想要一些更灵活和紧凑的东西：

var a = new Uint8Array( [ 1, 2, 3 ] );
var b = new Float32Array( [ 4.5, 5.5, 6.5 ] );

const merge = (tArrs, type = Uint8Array) => {
  const ret = new (type)(tArrs.reduce((acc, tArr) => acc + tArr.byteLength, 0))
  let off = 0
  tArrs.forEach((tArr, i) => {
    ret.set(tArr, off)
    off += tArr.byteLength
  })
  return ret
}

merge([a, b], Float32Array)
于 2019-07-11T16:09:30.520 回答