Python 值交换什么都不做

Question

我目前正在编写一个列表排序函数，我试图将列表的最小值与列表的第一个元素交换:

foo = [4, 7, 2, 9]

foo[0], foo[foo.index(min(foo))] = foo[foo.index(min(foo))], foo[0]

我期待的结果:

foo = [2, 7, 4, 9]

但是我得到了

foo = [4, 7, 2, 9]

一切都没有改变。有帮助吗？

Answer 1

让我们用一些临时变量来分解它，看看发生了什么:

>>> foo = [4, 7, 2, 9]
>>> tup = foo[foo.index(min(foo))], foo[0]
>>> print tup
(2, 4)
>>> foo[0] = tup[0]
>>> print foo
[2, 7, 2, 9]
>>> dx = foo.index(min(foo))
>>> print dx
0
>>> foo[dx] = tup[1] # foo[dx] equivalent to foo[foo.index(min(foo))]
>>> print foo
[4, 7, 2, 9]

赋值的效果是将 2 赋值给 foo[0]。然后将 4 分配给 foo 最小值的索引，该值被计算为 foo[0] 因为你刚刚分配给它。

这是另一种通过记下计算顺序来查看它的方法:

foo[0], foo[foo.index(min(foo))] = foo[foo.index(min(foo))], foo[0]
                                                 1_______
                                       2__________________
                                   3_______________________  4_____
                                   5_______________________________
6_____
                      7_______
            8__________________
        9_______________________
1 find min
2 find index
3 get item by index
4 get item by index
5 make tuple
6 assign to foo at index 0 from left hand side of tuple from step 5)
7 find min
8 find index
9 assign to foo at index from step 8 from right hand side of tuple from step 5