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

ThomasH已经补充了一个很好的答案，但我想展示发生了什么:

>>> a = [[1, 2], [3, 4]]
>>> [x for x in b for b in a]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

>>> [x for b in a for x in b]
[1, 2, 3, 4]
>>> [x for x in b for b in a]
[3, 3, 4, 4]

我猜Python是从左到右解析列表理解的。这意味着，发生的第一个for循环将首先执行。

第二个“问题”是b被“泄露”出了列表理解。在第一个成功的列表理解b ==[3,4]。

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]

假设你有一个充满句子的文本，你想要一个单词数组。

# Without list comprehension
list_of_words = []
for sentence in text:
    for word in sentence:
       list_of_words.append(word)
return list_of_words

我喜欢将列表理解理解为横向扩展代码。

试着把它分解成:

# List Comprehension 
[word for sentence in text for word in sentence]

例子:

>>> text = (("Hi", "Steve!"), ("What's", "up?"))
>>> [word for sentence in text for word in sentence]
['Hi', 'Steve!', "What's", 'up?']

这也适用于生成器

>>> text = (("Hi", "Steve!"), ("What's", "up?"))
>>> gen = (word for sentence in text for word in sentence)
>>> for word in gen: print(word)
Hi
Steve!
What's
up?

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]

此外，您可以对当前访问的输入列表成员和该成员中的元素使用相同的变量。然而，这甚至可能使它更加难以理解。

input = [[1, 2], [3, 4]]
[x for x in input for x in x]

首先计算输入中的x，导致输入的一个成员列表，然后，Python遍历x中的x的第二部分，在此期间x的值被它正在访问的当前元素覆盖，然后第一个x定义我们想要返回的内容。

这个记忆技巧对我帮助很大:

[< currend_value > < outer_loop1 > < inner_loop2 > < inner_loop3 >…]< OPTIONAL_IF >)

现在你可以考虑返回+外循环 作为唯一的正义秩序

知道了上面，列表综合的顺序即使是3个循环看起来也很简单:

c=[111, 222, 333]
b=[11, 22, 33]
a=[1, 2, 3]

print(
  [
    (i, j, k)                            # <RETURNED_VALUE> 
    for i in a for j in b for k in c     # in order: loop1, loop2, loop3
    if i < 2 and j < 20 and k < 200      # <OPTIONAL_IF>
  ]
)
[(1, 11, 111)]

因为上面只是一个:

for i in a:                         # outer loop1 GOES SECOND
  for j in b:                       # inner loop2 GOES THIRD
    for k in c:                     # inner loop3 GOES FOURTH
      if i < 2 and j < 20 and k < 200:
        print((i, j, k))            # returned value GOES FIRST

对于迭代一个嵌套的列表/结构，技术是相同的: 对于a这个问题:

a = [[1,2],[3,4]]
[i2    for i1 in a      for i2 in i1]
which return [1, 2, 3, 4]

对于另一个嵌套级别

a = [[[1, 2], [3, 4]], [[5, 6], [7, 8, 9]], [[10]]]
[i3    for i1 in a      for i2 in i1     for i3 in i2]
which return [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

等等

这个flat_nlevel函数递归调用嵌套的list1来转换到一个级别。试试这个

def flatten_nlevel(list1, flat_list):
    for sublist in list1:
        if isinstance(sublist, type(list)):        
            flatten_nlevel(sublist, flat_list)
        else:
            flat_list.append(sublist)

list1 = [1,[1,[2,3,[4,6]],4],5]

items = []
flatten_nlevel(list1,items)
print(items)

输出:

[1, 1, 2, 3, 4, 6, 4, 5]

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()]