涉及列表索引的 Python 元组分配失败

2022-01-20 00:00:00 python list tuples

问题描述

说出清单

s = [1, 2]

我想使用元组分配来编辑列表.为什么这样做很好

and I want to use tuple assignments to edit the list. Why it is fine to do

s[s[0]], s[0] = s[0], s[s[0]] # s == [2, 1]

但不是

s[0], s[s[0]] = s[s[0]], s[0] # get error, index out of range


解决方案

可以在这里.基本上,当您有多个分配时,= 右侧的任何内容都会首先被完全评估.

A good explanation of what is happening under the hood can be found here. Essentially when you have multiple assignments anything to the right of the = will be fully evaluated first.

让我们以您的第一个示例为例,了解正在发生的事情.

Let’s take your first example and walk through what is happening.

s[s[0]], s[0] = s[0], s[s[0]] # s == [2, 1]

这相当于下面的临时值存储评估的右手边

This is equivalent to the following where temp values store the evaluated right hand sides

# assigns right hand side to temp variables

temp1 = s[0]    # 1
temp2 = s[s[0]] # 2

# then assign those temp variables to the left-hand side

s[s[0]] = temp1 # s[1] = 1 —-> s= [1, 1]
s[0] = temp2    # s[0] = 2 —-> s = [2, 1]

如果您遇到索引错误,则会发生以下情况...

In the case you get an index error the following happens...

# assigns right hand side to temp variables

temp1 = s[s[0]] # 2
temp2 = s[0]    # 1

# then assign those temp variables to the left-hand side

s[0] = temp1 # s[0] = 2 —-> s= [2, 1]
s[s[0]] = temp2  # s[2] -> Index error (highest index of s is 1)

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