网民不要很cccccccc
问题描述
l1=[0,2,1] index=1 from ipdb import set_trace; set_trace() l1[index], l1[l1[index]] = l1[l1[index]], l1[index] print(l1)
为什么l1相同？l1[1]和l1[2]不会交换。
推荐答案
您可以更改顺序，它会起作用：
l1=[0,2,1] index=1 l1[l1[index]], l1[index] = l1[index], l1[l1[index]] print(l1)
输出：
[0, 1, 2]
让我们首先看一下代码的反汇编：
import dis
def switch():
l1=[0,2,1]
index=1
l1[index], l1[l1[index]] = l1[l1[index]], l1[index]
return l1
dis.dis(switch)
2 0 LOAD_CONST 1 (0)
2 LOAD_CONST 2 (2)
4 LOAD_CONST 3 (1)
6 BUILD_LIST 3
8 STORE_FAST 0 (l1)
3 10 LOAD_CONST 3 (1)
12 STORE_FAST 1 (index)
5 14 LOAD_FAST 0 (l1)
16 LOAD_FAST 0 (l1)
18 LOAD_FAST 1 (index)
20 BINARY_SUBSCR
22 BINARY_SUBSCR
24 LOAD_FAST 0 (l1)
26 LOAD_FAST 1 (index)
28 BINARY_SUBSCR
30 ROT_TWO
32 LOAD_FAST 0 (l1)
34 LOAD_FAST 1 (index)
36 STORE_SUBSCR
38 LOAD_FAST 0 (l1)
40 LOAD_FAST 0 (l1)
42 LOAD_FAST 1 (index)
44 BINARY_SUBSCR
46 STORE_SUBSCR
6 48 LOAD_FAST 0 (l1)
50 RETURN_VALUE
在这种类型的赋值中，首先计算表达式的右侧(请参见Evaluation Order)。因此，首先，指令集(14 - 18)加载l1[index]，即1，并将其推送到堆栈。然后，24-26加载l1[l1[index]]，即2并将其推送到堆栈。因此，堆栈现在可以容纳[2,1]。ROT_TWO(30)交换堆栈并将其设置为[1, 2]，这是我们需要的顺序。
现在，在32-36中，堆栈的顶部，即1分配给l1[index]，所以现在，l1[index] == 1，即l1[1] = 1。
然后38-42，堆栈中剩余的元素2弹出为l1[l1[index]]，但现在l1[index]的值是1，所以您实际上是在做l1[1] = 1。让我们来看看：
l1[index], l1[l1[index]] = l1[l1[index]], l1[index] loaded == 2, 1 after stack swapping == 1, 2 l1[1] == 1 l1[1] == 2 # So you have modified only index 1, and then overwritten it with its original value.
类似以下内容：
14 LOAD_FAST 0 (l1) ˉˉ|
16 LOAD_FAST 0 (l1) ˉˉ| 2 | 1 ---------->
18 LOAD_FAST 1 (index)__| __| ↓
20 BINARY_SUBSCR |
22 BINARY_SUBSCR |
24 LOAD_FAST 0 (l1) ˉˉ| 2 ------------------------>
26 LOAD_FAST 1 (index)__| | ↓
28 BINARY_SUBSCR | |
30 ROT_TWO | |
32 LOAD_FAST 0 (l1) ˉˉ| ↓ |
34 LOAD_FAST 1 (index)__| l1[1] = 1 <-------- |
36 STORE_SUBSCR | |
38 LOAD_FAST 0 (l1) | ˉˉ| |
40 LOAD_FAST 0 (l1) ˉˉ| ↓ | |
42 LOAD_FAST 1 (index)__| l1[1] == 1 __| l1[1] = 2 <---
44 BINARY_SUBSCR
46 STORE_SUBSCR
如果我们在我的解决方案中遵循相同的推理：
l1[l1[index]], l1[index] = l1[index], l1[l1[index]] loaded = 1, 2 after stack swapping == 2, 1 l1[2] = 2 l1[1] = 1 # Here, as you have not changed the value of `l1[index]` in the first assignment, the order remains.现在，您可以对l1 = [0, 1, 2]遵循相同的逻辑。虽然不需要解释，但因为l1[index]和l1[l1[index]]是一样的：
l1 = [0, 1, 2] l1[index], l1[l1[index]] = l1[l1[index]], l1[index] loaded = 1, 1 after stack swapping == 1, 1 l1[1] == 1 l1[1] == 1 ------------------------------------------------------------------ l1[l1[index]], l1[index] = l1[index], l1[l1[index]] loaded = 1, 1 after stack swapping == 1, 1 l1[1] = 1 l1[1] = 1 # Here both have same value, so it does not modify.
因此，当您通过将列表元素作为索引传递来访问索引时，最好避免这种赋值。而是Explicit：
l1 = [0, 2, 1] index1 = 1 index2 = l1[index1] l1[index1], l1[index2] = l1[index2], l1[index1] print(l1) # [0, 1, 2]
