列表迭代列表如何在 python 中工作？

Question

我有一个这样的列表：-

lst = [[1, 2, 3, 4, 5, 6], [4, 5, 6], [7], [8, 9]]

如果我运行这些，我会得到这样的输出。我不知道它们是如何工作的。

>>>[j for i in lst for j in i]
[1, 2, 3, 4, 5, 6, 4, 5, 6, 7, 8, 9]

>>>[j for j in i for i in lst]
[8, 8, 8, 8, 9, 9, 9, 9]

谁能解释一下它们是如何给出这样的输出的。这两次迭代有什么不同？

Answer 1

在第一个 LC 结束时，i 被分配给 [8,9]：

>>> lis = [[1, 2, 3, 4, 5, 6], [4, 5, 6], [7], [8, 9]]
>>> [j for i in lis for j in i]
[1, 2, 3, 4, 5, 6, 4, 5, 6, 7, 8, 9]
>>> i
[8, 9]

现在在第二个 LC 中，您正在迭代这个 i：

>>> [j for j in i for i in lis]
[8, 8, 8, 8, 9, 9, 9, 9]

两个 LC （大致）等同于：

>>> lis = [[1, 2, 3, 4, 5, 6], [4, 5, 6], [7], [8, 9]]
>>> for i in lis:
...     for j in i:
...         print j,
...         
1 2 3 4 5 6 4 5 6 7 8 9
>>> i
[8, 9]
>>> for j in i:
...     for i in lis:
...         print j,
...         
8 8 8 8 9 9 9 9

---

This has been fixed in py3.x:

特别是循环控制变量不再泄漏到 周围的范围。

演示（py 3.x）：

>>> lis = [[1, 2, 3, 4, 5, 6], [4, 5, 6], [7], [8, 9]]
>>> [j for i in lis for j in i]
[1, 2, 3, 4, 5, 6, 4, 5, 6, 7, 8, 9]
>>> i
Traceback (most recent call last):
NameError: name 'i' is not defined

>>> j
Traceback (most recent call last):
NameError: name 'j' is not defined

Answer 2

[j for i in list for j in i]

这类似于

result = []
for i in list:
    for j in i:
        result.append(j)

一般情况下，[p for a in b if c == d for e in f if etc] 之类的列表理解将被翻译为喜欢

reuslt = []
for a in b: if c == d: for e in f: if etc: result.append(p)

---

[j for j in i for i in list]

通常这甚至不会运行。可能您之前已将i 定义为[8, 9]。

>>> [j for j in i for i in list]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'i' is not defined

这相当于

result = []
for j in i: 
    for i in list:
        result.append(j)

所以如果一开始没有定义i，循环将不会工作。