更新时间:2023-02-26 16:20:29
无numpy( chain.from_iterable
,它是itertools.chain
的备用构造函数:
Without numpy ( ndarray.flatten
) one way would be using chain.from_iterable
which is an alternate constructor for itertools.chain
:
>>> list(chain.from_iterable([[1,2,3],[1,2],[1,4,5,6,7]]))
[1, 2, 3, 1, 2, 1, 4, 5, 6, 7]
或者作为另一种Python方式,您可以使用列表理解:
Or as another yet Pythonic approach you can use a list comprehension :
[j for sub in [[1,2,3],[1,2],[1,4,5,6,7]] for j in sub]
另一种非常适合短列表的功能方法也可以是Python2中的reduce
和Python3中的functools.reduce
(不要将其用于长列表):
Another functional approach very suitable for short lists could also be reduce
in Python2 and functools.reduce
in Python3 (don't use it for long lists):
In [4]: from functools import reduce # Python3
In [5]: reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]])
Out[5]: [1, 2, 3, 1, 2, 1, 4, 5, 6, 7]
要使其更快一点,可以使用内置的operator.add
代替lambda
:
To make it slightly faster you could can use operator.add
, which is built-in, instead of lambda
:
In [6]: from operator import add
In [7]: reduce(add ,[[1,2,3],[1,2],[1,4,5,6,7]])
Out[7]: [1, 2, 3, 1, 2, 1, 4, 5, 6, 7]
In [8]: %timeit reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]])
789 ns ± 7.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [9]: %timeit reduce(add ,[[1,2,3],[1,2],[1,4,5,6,7]])
635 ns ± 4.38 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
基准:
:~$ python -m timeit "from itertools import chain;chain.from_iterable([[1,2,3],[1,2],[1,4,5,6,7]])"
1000000 loops, best of 3: 1.58 usec per loop
:~$ python -m timeit "reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]])"
1000000 loops, best of 3: 0.791 usec per loop
:~$ python -m timeit "[j for i in [[1,2,3],[1,2],[1,4,5,6,7]] for j in i]"
1000000 loops, best of 3: 0.784 usec per loop
@Will答案的基准测试使用sum
(对于短列表来说它是快速的,但对于长列表来说它不是很快):
A benchmark on @Will's answer that used sum
(its fast for short list but not for long list) :
:~$ python -m timeit "sum([[1,2,3],[4,5,6],[7,8,9]], [])"
1000000 loops, best of 3: 0.575 usec per loop
:~$ python -m timeit "sum([range(100),range(100)], [])"
100000 loops, best of 3: 2.27 usec per loop
:~$ python -m timeit "reduce(lambda x,y :x+y ,[range(100),range(100)])"
100000 loops, best of 3: 2.1 usec per loop