Python 扩展列表中任意数量的子列表

我试图在列表中展开子列表。我为列表中的一层子列表编写了它。有没有更好的方法来编写此代码，以便为任意数量的嵌套列表进行缩放？见以下示例：

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]

def expand(x):
    # expands a list of lists with no further sublists 
    a = []
    for i in x:
        # could also here do a += i and avoid the second for loop
        for j in i:
            a.append(j)
    return a

print expand(c)

返回所需的[1,2,3,4,5,6]

为了进一步澄清，我想知道如何更好地扩展

`e=[c，a，b]'和进一步的嵌套迭代

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

# first method
res = [x for lst in c for x in lst]

# second method
from itertools import chain
res = list(chain(*c))

编辑：他想要迭代嵌套列表

编辑：他想要迭代嵌套列表

使用递归：

def flat_list(l):
    for a in l:
        if isinstance(a, list):
            for x in flat_list(a):
                yield x
        else:
            yield a

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]
e = [c, a, b]

print list(flat_list(c))
print list(flat_list(e))

输出为：

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

使用递归：

def flat_list(l):
    for a in l:
        if isinstance(a, list):
            for x in flat_list(a):
                yield x
        else:
            yield a

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]
e = [c, a, b]

print list(flat_list(c))
print list(flat_list(e))

输出为：

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

一个解决方案：

def flatten(ls):
        result = []
        stack = [ls]
        while stack:
            if isinstance(stack[-1], list):
                try: stack.append(stack[-1].pop(0))
                except IndexError: stack.pop() # remove now-empty sublist
            else:
                result.append(stack.pop())
        return result

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]
e = [c, a, b]

print flatten(e)

输出：

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

或：

使用总和：

>>> e
[[[1, 2, 3], [4, 5, 6]], [1, 2, 3], [4, 5, 6]]
>>> sum(e[0], [])
[1, 2, 3, 4, 5, 6]

使用reduce：

>>> reduce(lambda x,y: x+y, e[0])
[1, 2, 3, 4, 5, 6]
>>>

一个解决方案：

def flatten(ls):
        result = []
        stack = [ls]
        while stack:
            if isinstance(stack[-1], list):
                try: stack.append(stack[-1].pop(0))
                except IndexError: stack.pop() # remove now-empty sublist
            else:
                result.append(stack.pop())
        return result

a = [1, 2, 3]
b = [4, 5, 6]
c = [a, b]
e = [c, a, b]

print flatten(e)

输出：

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

或：

使用总和：

>>> e
[[[1, 2, 3], [4, 5, 6]], [1, 2, 3], [4, 5, 6]]
>>> sum(e[0], [])
[1, 2, 3, 4, 5, 6]

使用reduce：

>>> reduce(lambda x,y: x+y, e[0])
[1, 2, 3, 4, 5, 6]
>>>

---

+1对于itertools方法。我相信itertools和collections是python的stdlib.+1的itertools方法中功能最强大但使用不足的两个模块。我相信itertools和collections是python的stdlib中功能最强大但使用不足的两个模块。