在使用Python进行列表拼接时,使用sum()
函数是否符合Pythonic的风格?
>>> sum(([n]*n for n in range(1,5)),[])
[1, 2, 2, 3, 3, 3, 4, 4, 4, 4]
In [28]: [t for n in range(1,5) for t in [n]*n ]
Out[28]: [1, 2, 2, 3, 3, 3, 4, 4, 4, 4]
或者使用 itertools.chain
:
In [29]: from itertools import chain
In [32]: list(chain.from_iterable([n]*n for n in range(1,5)))
Out[32]: [1, 2, 2, 3, 3, 3, 4, 4, 4, 4]
或者作为一种纯生成器的方法,您可以使用repeat
而不是乘以列表:
In [33]: from itertools import chain, repeat
# In python2.X use xrange instead of range
In [35]: list(chain.from_iterable(repeat(n, n) for n in range(1,5)))
Out[35]: [1, 2, 2, 3, 3, 3, 4, 4, 4, 4]
如果你对NumPy感兴趣,或者想要一种超快的方法,这里有一个:
In [46]: import numpy as np
In [46]: np.repeat(np.arange(1, 5), np.arange(1, 5))
Out[46]: array([1, 2, 2, 3, 3, 3, 4, 4, 4, 4])
不,对于大型列表,这种方法会变得非常缓慢。列表推导式是更好的选择。
用列表推导式、总和和itertools.chain.from_iterable计时展平列表的代码:
import time
from itertools import chain
def increasing_count_lists(upper):
yield from ([n]*n for n in range(1,upper))
def printtime(func):
def clocked_func(*args):
t0 = time.perf_counter()
result = func(*args)
elapsed_s = time.perf_counter() - t0
print('{:.4}ms'.format(elapsed_s*1000))
return result
return clocked_func
@printtime
def concat_list_sum(upper):
return sum(increasing_count_lists(upper), [])
@printtime
def concat_list_listcomp(upper):
return [item for sublist in increasing_count_lists(upper) for item in sublist]
@printtime
def concat_list_chain(upper):
return list(chain.from_iterable(increasing_count_lists(upper)))
并运行它们:
>>> _ = concat_list_sum(5)
0.03351ms
>>> _ = concat_list_listcomp(5)
0.03034ms
>>> _ = concat_list_chain(5)
0.02717ms
>>> _ = concat_list_sum(50)
0.2373ms
>>> _ = concat_list_listcomp(50)
0.2169ms
>>> _ = concat_list_chain(50)
0.1467ms
>>> _ = concat_list_sum(500)
167.6ms
>>> _ = concat_list_listcomp(500)
8.319ms
>>> _ = concat_list_chain(500)
12.02ms
itertools.chain
或itertools.chain.from_iterable
。 - chapelo