Python交换值在更改交换顺序时给出了不同的结果

发布于 2025-02-06 01:36:16 字数 540 浏览 4 评论 0原文

输入：
[1,2,3,4]

预期输出：

[2，1，3，4]

给出了预期结果

a = [1,2,3,4]
a[a[0]], a[0] = a[0],a[a[0]]
#Output -> [2, 1, 3, 4]

以下代码在更改交换后，。订单给出了不正确的输出

a = [1,2,3,4]
a[0],a[a[0]] = a[a[0]],a[0]
# Output -> [2, 2, 1, 4]

，我刚刚切换了作业的顺序。为什么会有所作为？

编辑：那么，为什么下面的两个案例给出相同的输出：

a = 1 b= 2 
a,b = b,a #Case1
b,a = a,b #Case2
print(a,b) #=> prints 2,1 for both

如果问题不够描述，请让我知道我会添加更多评论。我试图使其尽可能简单。谢谢

原文

Input:
[1,2,3,4]

Expected Output:

[2, 1, 3, 4]

The below code gives the expected result

a = [1,2,3,4]
a[a[0]], a[0] = a[0],a[a[0]]
#Output -> [2, 1, 3, 4]

The below code ie, after changing swapping order gives the incorrect output

a = [1,2,3,4]
a[0],a[a[0]] = a[a[0]],a[0]
# Output -> [2, 2, 1, 4]

I just switched the order of the assignments. Why does that make a difference?

Edit:
Then why do the two cases below give the same output:

a = 1 b= 2 
a,b = b,a #Case1
b,a = a,b #Case2
print(a,b) #=> prints 2,1 for both

If the question is not descriptive enough please let me know I will add more comments.
I have tried to make it as simple as possible.
Thanks

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执笏见 2025-02-13 01:36:16

这很重要，因为您基本上执行两个单独的作业。这些作业之一必须首先执行。因此，如果您首先分配a [0]，则a [a [0]]已经可以与a [0] 确定索引。如果您首先分配a [a [0]]，则该索引的行为按预期行为，并且之后只有a [0]更改的值。

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时间你老了 2025-02-13 01:36:16

Python首先评估整个右侧，然后从左到右分配到左侧。在第一种情况下：

tmp1=a[0]      # that is tmp1=1
tmp2=a[a[0]]   # that is tmp2=a[1] that is tmp2=2
a[a[0]]=tmp1   # that is a[1]=1
a[0]=tmp2      # that is a[0]=2

在第二种情况下：

tmp1=a[a[0]]   # that is tmp1=a[1], that is tmp1=2
tmp2=a[0]      # that is tmp2=1
a[0]=tmp1      # that is a[0]=2
a[a[0]]=tmp2   # that is a[2]=1

请参见 https：https：// docss：// docs .python.org/3/reference/expressions.html#evaluation-order

Python evaluates first the whole right-hand side, then assigns to the left hand side from left to right. In the first case:

tmp1=a[0]      # that is tmp1=1
tmp2=a[a[0]]   # that is tmp2=a[1] that is tmp2=2
a[a[0]]=tmp1   # that is a[1]=1
a[0]=tmp2      # that is a[0]=2

In the second case:

tmp1=a[a[0]]   # that is tmp1=a[1], that is tmp1=2
tmp2=a[0]      # that is tmp2=1
a[0]=tmp1      # that is a[0]=2
a[a[0]]=tmp2   # that is a[2]=1

See https://docs.python.org/3/reference/expressions.html#evaluation-order

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暮年 2025-02-13 01:36:16

两者都会返回不同的结果，您的代码含义上述

a = [1,2,3,4]
a[a[0]], a[0] = a[0],a[a[0]]
# code above access a[with index a[0]] = 1
# simply means that a[1], a[0] = a[0], a[1]
# and equal with [2, 1, 3, 4]
#Output -> [2, 1, 3, 4]

代码返回您的预期结果，但是不正确的结果呢？

a = [1,2,3,4]
a[0],a[a[0]] = a[a[0]],a[0]
# this means a[0], a[1] = a[1], a[0]
# Output -> [2, 2, 1, 4]

解释：
您定义第一个a [0] = a [a [0]]与a [0] = a [1]（当前值为2）或在简单的单词中a [0] = 2。
接下来，您调用a [a [0]] = a [0]与a [2] = a [0]（您已经将A [0]定义为2）< /code>或简单词a [2] = 2 这为您的结果提供了[2,2,1,4]的结果，

用于交换值时的其他情况

a = 1 b= 2 
a,b = b,a #Case1
b,a = a,b #Case2
print(a,b) #=> prints 2,1 for both

，使用变量的简单案例，并且像常见的交换案例一样工作，但是您使用列表的示例不正确，这更棘手，就像我已经很棘手以前解释过

Both will return different result, your code means like this

a = [1,2,3,4]
a[a[0]], a[0] = a[0],a[a[0]]
# code above access a[with index a[0]] = 1
# simply means that a[1], a[0] = a[0], a[1]
# and equal with [2, 1, 3, 4]
#Output -> [2, 1, 3, 4]

Above code return what your expected result, but how about the incorrect one?

a = [1,2,3,4]
a[0],a[a[0]] = a[a[0]],a[0]
# this means a[0], a[1] = a[1], a[0]
# Output -> [2, 2, 1, 4]

Explanation:
You define the first a[0] = a[a[0]] which equal with a[0] = a[1] (which current value is 2) or in simple word a[0] = 2.
Next, you call a[a[0]] = a[0] which equal with a[2] = a[0] (you already define a[0] as 2) or in simple word a[2] = 2
And this give you the result as [2, 2, 1, 4]

For the additional case in swapping value

a = 1 b= 2 
a,b = b,a #Case1
b,a = a,b #Case2
print(a,b) #=> prints 2,1 for both

This's simple case using variable, and work like common swapping case, but your incorrect sample using list, which are more tricky as I already explained before

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