Question

我一直在试图弄清楚如何编写一个函数来连接使用splat的任意数量的列表。

下面的代码是我最接近我想要的地方，但是我试图让函数输出一个如下所示的列表：

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

但我只是将它显示为三个嵌套列表。任何帮助将非常感激。谢谢。

m = [1, 2, 3]
n = [4, 5, 6]
o = [7, 8, 9]

# Update the below function to take
# an arbitrary number of arguments
def join_lists(*args):
    return args


print join_lists(m, n, o)

Answer 1

itertools救援！

def join_lists(*args):
    return list(itertools.chain(*args))

Answer 2

万一你想手动编码：）

>>> def joinlists(*args):
...     l = []
...     for arg in args:
...             l.extend(arg)
...     return l
... 
>>> joinlists(m, n, o)
[1, 2, 3, 4, 5, 6, 7, 8, 9]

或者： -

>>> def joinlists(*args):
...     l = []
...     for arg in args:
...         for item in arg:
...             l.append(item)
...     return l
... 
>>> joinlists(m, n, o)
[1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer 3

您可以使用reduce(lambda x,y: x+y, args)：

>>> m = [1, 2, 3]
>>> n = [4, 5, 6]
>>> o = [7, 8, 9]
>>>
>>> def join_lists(*args):
...     return reduce(lambda x,y: x+y, args)
...
>>> join_lists(m,n,o)
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>

或reduce(operator.add, args)：

>>> import operator
>>>
>>> m = [1, 2, 3]
>>> n = [4, 5, 6]
>>> o = [7, 8, 9]
>>>
>>> def join_lists(*args):
...     return reduce(operator.add, args)
...
>>> print join_lists(m,n,o)
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>

Answer 4

鉴于您仅限于“基本”功能（如最近的评论中所述），这里有一个简短的列表理解：

def join_lists(*args):
    return [i for L in args for i in L]

或者，作为嵌套循环展开（完全等效）：

def join_lists(*args):
    out = []
    for L in args:
        for i in L:
            out.append(i)
    return out

连接splatted列表

4 个答案: