如何在Python中重新分配列表中的项目？ [英] How to re-assign items in a list in Python?

问题描述

我想在Python中重新分配列表中的每个项目。

I want to re-assign each item in a list in Python.

In [20]: l = [1,2,3,4,5]
In [21]: for i in l:
   ....:     i = i + 1
   ....:     
   ....:     

但是列表根本没有变化。

But the list didn't change at all.

In [22]: l
Out[22]: [1, 2, 3, 4, 5]

我想知道为什么会这样。任何机构都可以详细解释列表吗？谢谢。

I want to know why this happened. Could any body explain the list iterating in detail? Thanks.

推荐答案

你不能这样做，你只是将绑定的值更改为名称我。在 for 循环的每次迭代中， i 被绑定到列表中的值。它不是一个指针，通过更改 i 的值，您正在更改列表中的值。相反，正如我之前所说，它只是一个名称而你只是在改变名称所指的值。在这种情况下， i = i + 1 ，将 i 绑定到值 i + 1 。因此，您实际上并没有影响列表本身，要做到这一点，您必须通过索引进行设置。

You can't do it like that, you are merely changing the value binded to the name i. On each iteration of the for loop, i is binded to a value in the list. It is not a pointer in the sense that by changing the value of i you are changing a value in the list. Instead, as I said before, it is simply a name and you are just changing the value that name refers to. In this case, i = i + 1, binds i to the value i + 1. So you aren't actually affecting the list itself, to do that you have to set it by index.

>>> L = [1,2,3,4,5]
>>> for i in range(len(L)):
        L[i] = L[i] + 1

>>> L
[2, 3, 4, 5, 6]

有些pythonistas可能更喜欢迭代如下：

Some pythonistas may prefer to iterate like this:

for i, n in enumerate(L): # where i is the index, n is each number
    L[i] = n + 1

然而，您可以轻松地获得相同的结果列表理解：

However you can easily achieve the same result with a list comprehension:

>>> L = [1,2,3,4,5]
>>> L = [n + 1 for n in L]
>>> L
[2, 3, 4, 5, 6]

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