如何在 JavaScript 中合并 TypedArrays？
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How can I merge TypedArrays in JavaScript?
提问by yomotsu
I'd like to merge multiple arraybuffers to create a Blob. however, as you know, TypedArraydosen't have "push" or useful methods...
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......
E.g.:
例如：
var a = new Int8Array( [ 1, 2, 3 ] );
var b = new Int8Array( [ 4, 5, 6 ] );
As a result, I'd like to get [ 1, 2, 3, 4, 5, 6 ].
结果，我想得到[ 1, 2, 3, 4, 5, 6 ].
回答by Prinzhorn
Use the setmethod. But note, that you now need twice the memory!
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);
回答by D?nu
I always use this function:
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;
}
The original function without checking for null or types
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;
}
回答by JerryCauser
for client-side ~ok solution:
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])
回答by hoogw
if I have multiple typed arrays
如果我有多个类型化数组
arrays = [ typed_array1, typed_array2,..... typed_array100]
I want concat all 1 to 100 sub array into single 'result' this function works for me.
我想将所有 1 到 100 个子数组连接到单个“结果”中，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;
}
回答by randomsock
As a one-liner, which will take an arbitrary number of arrays (myArrayshere) and of mixed types so long as the result type takes them all (Int8Arrayhere):
作为单行，它将采用任意数量的数组（myArrays此处）和混合类型，只要结果类型全部采用（Int8Array此处）：
let combined = Int8Array.from(Array.prototype.concat(...myArrays.map(a => Array.from(a))));
回答by mattdlockyer
I like @prinzhorn's answer but I wanted something a bit more flexible and compact:
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)
