如何使用另一个数组扩展现有的 JavaScript 数组，而无需创建新数组 [英] How to extend an existing JavaScript array with another array, without creating a new array

问题描述

似乎没有办法用另一个数组扩展现有的 JavaScript 数组，即模拟 Python 的 extend 方法.

我想实现以下目标:

<预><代码>>>>a = [1, 2][1, 2]>>>b = [3, 4, 5][3, 4, 5]>>>这里的东西>>>一种[1, 2, 3, 4, 5]

我知道有一个 a.concat(b) 方法，但它创建了一个新数组，而不是简单地扩展第一个数组.我想要一种算法，当 a 明显大于 b(即不复制 a 的算法)时，它可以有效工作.

_{注意:这不是如何将某些内容附加到数组中? -- 这里的目标是将一个数组的全部内容添加到另一个数组中，并就地"完成，即不复制所有元素扩展数组.}

解决方案

.push 方法可以接受多个参数.您可以使用 spread operator 来传递所有第二个数组的元素作为 .push 的参数:

<预><代码>>>>a.push(...b)

如果您的浏览器不支持 ECMAScript 6，您可以使用 .apply 代替:

<预><代码>>>>a.push.apply(a, b)

或者，如果您认为它更清楚:

<预><代码>>>>Array.prototype.push.apply(a,b)

请注意，如果数组 b 太长(问题从大约 100,000 个元素开始，取决于浏览器)，所有这些解决方案都将失败并出现堆栈溢出错误.
如果您不能保证 b 足够短，您应该使用另一个答案中描述的基于循环的标准技术.

There doesn't seem to be a way to extend an existing JavaScript array with another array, i.e. to emulate Python's extend method.

I want to achieve the following:

>>> a = [1, 2]
[1, 2]
>>> b = [3, 4, 5]
[3, 4, 5]
>>> SOMETHING HERE
>>> a
[1, 2, 3, 4, 5]

I know there's a a.concat(b) method, but it creates a new array instead of simply extending the first one. I'd like an algorithm that works efficiently when a is significantly larger than b (i.e. one that does not copy a).

_{Note: This is not a duplicate of How to append something to an array? -- the goal here is to add the whole contents of one array to the other, and to do it "in place", i.e. without copying all elements of the extended array.}

解决方案

The .push method can take multiple arguments. You can use the spread operator to pass all the elements of the second array as arguments to .push:

>>> a.push(...b)

If your browser does not support ECMAScript 6, you can use .apply instead:

>>> a.push.apply(a, b)

Or perhaps, if you think it's clearer:

>>> Array.prototype.push.apply(a,b)

Please note that all these solutions will fail with a stack overflow error if array b is too long (trouble starts at about 100,000 elements, depending on the browser).
If you cannot guarantee that b is short enough, you should use a standard loop-based technique described in the other answer.

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