怎么理解Python交换：为什么a，b=b，a并不总是等同于b，a=a，b？

Question

我们都知道，在python中交换两个项目中a和b值的方法是：

a, b = b, a

它应该等同于：

b, a = a, b

但是，当我在敲代码时，我发现以下两种交换得到了不同的结果：

nums = [1, 2, 4, 3]
i = 2
nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
print(nums)
# [1, 2, 4, 3]

nums = [1, 2, 4, 3]
i = 2
nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
print(nums)
# [1, 2, 3, 4]

有没有人告诉我这到底是为什么？我之前一直认为在Python交换中，这两个赋值是同时且独立发生的。

Answer 1

assignment statement

An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.

Although the definition of assignment implies that overlaps between the left-hand side and the right-hand side are ‘simultaneous’ (for example a, b = b, a swaps two variables), overlaps within the collection of assigned-to variables occur left-to-right, sometimes resulting in confusion. For instance, the following program prints [0, 2]:

x = [0, 1]
i = 0
i, x[i] = 1, 2         # i is updated, then x[i] is updated
print(x)

---

assginment 先计算右侧得到一个 tuple ，然后将 tuple 中的值依次赋值给左侧的变量。

Answer 2

可以参考下这篇回答：https://stackoverflow.com/que...

简单地讲：

nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]

这种复制并不是同时发生的。它是先按从左往右的顺序计算右边两个变量的值，将结果临时保存在栈中，再按从左往右的顺序依次赋值给左边的变量。从实际运行的顺序来说，可以分成5步：

1. 计算nums[nums[i]-1]，将结果临时存入栈顶
2. 计算num[i],将结果存入栈顶
3. 将栈顶前面两个数据进行交换
4. 将栈顶结果取出，赋值给nums[i]
5. 将栈顶结果取出，赋值给nums[nums[i]-1]，注意此时经过第4步，nums[i]的结果已经改变了。

Answer 3

a, b = b, a
# 等同于
tmp1 = b
tmp2 = a
a = tmp1
b = tmp2

拆分成普通赋值就容易理解了

nums = [1, 2, 4, 3]
i = 2

# nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i]
num1 = nums[nums[i]-1]
num2 = nums[i]
nums[i] = num1
nums[nums[i]-1] = num2

print(nums)
# [1, 2, 4, 3]

nums = [1, 2, 4, 3]
i = 2

# nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
num1 = nums[i]
num2 = nums[nums[i]-1]
nums[nums[i]-1] = num1
nums[i] = num2

print(nums)
# [1, 2, 3, 4]

因为 nums[nums[i]-1] 的指向会受 nums[i] 变化的影响，所以 nums[nums[i]-1] 必须先赋值，然后再赋值 nums[i]