我觉得这样更容易理解

[row[i] for row in a for i in range(len(a))]

result: [1, 2, 3, 4]

哎呀，我想我找到了答案:我没有足够关心哪个循环是内部的，哪个是外部的。列表推导式应该是这样的:

[x for b in a for x in b]

为了得到想要的结果，一个当前值可以作为下一个循环的迭代器。

用你自己的建议来回答你的问题:

>>> [x for b in a for x in b] # Works fine

当你要求列表理解答案时，让我也指出优秀的itertools.chain():

>>> from itertools import chain
>>> list(chain.from_iterable(a))
>>> list(chain(*a)) # If you're using python < 2.6

在我第一次尝试的时候，我从来没有写过双表理解。读一读PEP202，你会发现原因是它的实施方式和你在英语中读到的完全相反。好消息是，这是一个逻辑上合理的实现，因此一旦您理解了结构，就很容易得到正确的实现。

设a, b, c, d是依次嵌套的对象。对我来说，扩展列表理解的直观方法是模仿英语:

# works
[f(b) for b in a]
# does not work
[f(c) for c in b for b in a]
[f(c) for c in g(b) for b in a]
[f(d) for d in c for c in b for b in a]

换句话说，你将从下往上阅读，即。

# wrong logic
(((d for d in c) for c in b) for b in a)

然而，这不是Python实现嵌套列表的方式。相反，实现将第一个块视为完全独立的块，然后从上到下(而不是从下到上)将for和in链接到单个块中，即。

# right logic
d: (for b in a, for c in b, for d in c)

请注意，嵌套最深的层(对于c中的d)距离列表(d)中的最后一个对象最远。这是因为Guido自己:

形式[…]x……y……]巢，最后一个索引变化最快，就像嵌套for循环一样。

使用Skam的文本示例，这变得更加清楚:

# word: for sentence in text, for word in sentence
[word for sentence in text for word in sentence]

# letter: for sentence in text, for word in sentence, for letter in word
[letter for sentence in text for word in sentence for letter in word]

# letter:
#     for sentence in text if len(sentence) > 2, 
#     for word in sentence[0], 
#     for letter in word if letter.isvowel()
[letter for sentence in text if len(sentence) > 2 for word in sentence[0] for letter in word if letter.isvowel()]

迭代器的顺序似乎与直觉相反。

例如:[str(x) for i in range(3) for x in foo(i)]

让我们分解它:

def foo(i):
    return i, i + 0.5

[str(x)
    for i in range(3)
        for x in foo(i)
]

# is same as
for i in range(3):
    for x in foo(i):
        yield str(x)

如果你想保持多维数组，就应该嵌套数组括号。参见下面的示例，其中每个元素都添加了一个。

>>> a = [[1, 2], [3, 4]]

>>> [[col +1 for col in row] for row in a]
[[2, 3], [4, 5]]

>>> [col +1 for row in a for col in row]
[2, 3, 4, 5]