import itertools
a = [['a','b'], ['c']]
print(list(itertools.chain.from_iterable(a)))

这给了

['a', 'b', 'c']

这就是所谓的扁平化，有很多实现。

这个怎么样，尽管它只适用于1级深嵌套:

>>> x = [["a","b"], ["c"]]
>>> for el in sum(x, []):
...     print el
...
a
b
c

从这些链接中，显然最完整的-fast-elegant-etc实现如下:

def flatten(l, ltypes=(list, tuple)):
    ltype = type(l)
    l = list(l)
    i = 0
    while i < len(l):
        while isinstance(l[i], ltypes):
            if not l[i]:
                l.pop(i)
                i -= 1
                break
            else:
                l[i:i + 1] = l[i]
        i += 1
    return ltype(l)

x = [["a","b"], ["c"]]

result = sum(x, [])

您所描述的就是所谓的将列表扁平化，有了这些新知识，您将能够在谷歌上找到许多解决方案(没有内置的扁平化方法)。以下是其中一个，来自http://www.daniel-lemire.com/blog/archives/2006/05/10/flattening-lists-in-python/:

def flatten(x):
    flat = True
    ans = []
    for i in x:
        if ( i.__class__ is list):
            ans = flatten(i)
        else:
            ans.append(i)
    return ans

import itertools
a = [['a','b'], ['c']]
print(list(itertools.chain.from_iterable(a)))

这给了

['a', 'b', 'c']

对于一级扁平化，如果你关心速度，在我尝试过的所有条件下，这比之前的任何答案都快。(也就是说，如果您需要结果作为列表。如果你只需要在运行中迭代它，那么链的例子可能更好。)它的工作原理是预先分配一个最终大小的列表，并按片复制部分(这是一种比任何迭代器方法都低级别的块复制):

def join(a):
    """Joins a sequence of sequences into a single sequence.  (One-level flattening.)
    E.g., join([(1,2,3), [4, 5], [6, (7, 8, 9), 10]]) = [1,2,3,4,5,6,(7,8,9),10]
    This is very efficient, especially when the subsequences are long.
    """
    n = sum([len(b) for b in a])
    l = [None]*n
    i = 0
    for b in a:
        j = i+len(b)
        l[i:j] = b
        i = j
    return l

带注释的排序时间列表:

[(0.5391559600830078, 'flatten4b'), # join() above. 
(0.5400412082672119, 'flatten4c'), # Same, with sum(len(b) for b in a) 
(0.5419249534606934, 'flatten4a'), # Similar, using zip() 
(0.7351131439208984, 'flatten1b'), # list(itertools.chain.from_iterable(a)) 
(0.7472689151763916, 'flatten1'), # list(itertools.chain(*a)) 
(1.5468521118164062, 'flatten3'), # [i for j in a for i in j] 
(26.696547985076904, 'flatten2')] # sum(a, [])