我有一个这样的清单:
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
我希望它看起来像这样
[['a', 'b', 'c'],['d', 'e', 'f'],['g', 'h', 'i']]
最有效的方法是什么?
编辑: 怎么样呢?
[['a', 'b', 'c'],['d', 'e', 'f'],['g', 'h', 'i']]
- >
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
答案 0 :(得分:12)
你可以通过简单的列表理解来做你想做的事。
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> [a[i:i+3] for i in range(0, len(a), 3)]
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
如果您希望填充最后一个子列表,可以在列表理解之前执行此操作:
>>> padding = 0
>>> a += [padding]*(3-len(a)%3)
将这些组合成一个单一的功能:
def group(sequence, group_length, padding=None):
if padding is not None:
sequence += [padding]*(group_length-len(sequence)%group_length)
return [sequence[i:i+group_length] for i in range(0, len(sequence), group_length)]
走另一条路:
def flatten(sequence):
return [item for sublist in sequence for item in sublist]
>>> a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> flatten(a)
[1, 2, 3, 4, 5, 6, 7, 8, 9]
答案 1 :(得分:4)
如果您可以使用numpy,请尝试x.reshape(-1, 3)
In [1]: import numpy as np
In [2]: x = ['a','b','c','d','e','f','g','h','i']
In [3]: x = np.array(x)
In [4]: x.reshape(-1, 3)
Out[4]:
array([['a', 'b', 'c'],
['d', 'e', 'f'],
['g', 'h', 'i']],
dtype='|S1')
如果数据足够大,则此代码更有效。
<强>更新强>
附加cProfile结果以解释更高效
import cProfile
import numpy as np
a = range(10000000*3)
def impl_a():
x = [a[i:i+3] for i in range(0, len(a), 3)]
def impl_b():
x = np.array(a)
x = x.reshape(-1, 3)
print("cProfile reuslt of impl_a()")
cProfile.run("impl_a()")
print("cProfile reuslt of impl_b()")
cProfile.run("impl_b()")
输出
cProfile reuslt of impl_a()
5 function calls in 15.614 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.499 0.499 15.614 15.614 <string>:1(<module>)
1 14.968 14.968 15.114 15.114 impla.py:6(impl_a)
1 0.000 0.000 0.000 0.000 {len}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
1 0.146 0.146 0.146 0.146 {range}
cProfile reuslt of impl_b()
5 function calls in 3.142 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 3.142 3.142 <string>:1(<module>)
1 0.000 0.000 3.142 3.142 impla.py:9(impl_b)
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
1 0.000 0.000 0.000 0.000 {method 'reshape' of 'numpy.ndarray' objects}
1 3.142 3.142 3.142 3.142 {numpy.core.multiarray.array}
答案 2 :(得分:3)
您可以将grouper
recipe from itertools
与list comprehension:
from itertools import izip_longest # or zip_longest for Python 3.x
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args) # see note above
in_ = [1, 2, 3, 4, 5, 6, 7, 8, 9]
out = [list(t) for t in grouper(in_, 3)]
答案 3 :(得分:2)
我的解决方案:
>>> list=[1,2,3,4,5,6,7,8,9,10]
>>> map(lambda i: list[i:i+3], range(0,len(list),3))
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
答案 4 :(得分:1)
使用itertools,更具体地说,使用Recipes提到的函数grouper
:
from itertools import izip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print [list(x) for x in grouper(a, 3)]
打印
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
答案 5 :(得分:1)
我已将所有已回答的方法运行到基准测试并找到最快的方法。
样本量 999999(1&lt; = x&lt; = 258962)
Python: Python 2.7.5 | Anaconda 1.8.0(32位)(IPython)
操作系统:Windows 7 32位 @ Core i5 / 4GB RAM
import random as rd
lst = [rd.randrange(1,258963) for n in range(999999)]
>>> %timeit x = [lst[i:i+3] for i in range(0, len(lst), 3)]
10 loops, best of 3: 114 ms per loop
>>> %timeit array = np.array(lst)
10 loops, best of 3: 127 ms per loop
>>> %timeit array.reshape(-1,3)
1000000 loops, best of 3: 679 ns per loop
>>> %timeit out = [list(t) for t in grouper(lst, 3)]
10 loops, best of 3: 158 ms per loop
所以,似乎在IPython(Anaconda)上, list-comprehension比itertools / izip_longest / grouper方法快快30%
P.S。我想,这个结果在CPython运行时会有所不同,我也希望补充一下。