是否可以为列表理解中的每个项目返回2个(或更多)项目?
我想要的(示例):
[f(x), g(x) for x in range(n)]
应返回[f(0), g(0), f(1), g(1), ..., f(n-1), g(n-1)]
因此,替换这段代码的内容如下:
result = list()
for x in range(n):
result.add(f(x))
result.add(g(x))
发布于 2012-08-09 00:29:44
>>> from itertools import chain
>>> f = lambda x: x + 2
>>> g = lambda x: x ** 2
>>> list(chain.from_iterable((f(x), g(x)) for x in range(3)))
[2, 0, 3, 1, 4, 4]
计时:
from timeit import timeit
f = lambda x: x + 2
g = lambda x: x ** 2
def fg(x):
yield f(x)
yield g(x)
print timeit(stmt='list(chain.from_iterable((f(x), g(x)) for x in range(3)))',
setup='gc.enable(); from itertools import chain; f = lambda x: x + 2; g = lambda x: x ** 2')
print timeit(stmt='list(chain.from_iterable(fg(x) for x in range(3)))',
setup='gc.enable(); from itertools import chain; from __main__ import fg; f = lambda x: x + 2; g = lambda x: x ** 2')
print timeit(stmt='[func(x) for x in range(3) for func in (f, g)]',
setup='gc.enable(); f = lambda x: x + 2; g = lambda x: x ** 2')
print timeit(stmt='list(chain.from_iterable((f(x), g(x)) for x in xrange(10**6)))',
setup='gc.enable(); from itertools import chain; f = lambda x: x + 2; g = lambda x: x ** 2',
number=20)
print timeit(stmt='list(chain.from_iterable(fg(x) for x in xrange(10**6)))',
setup='gc.enable(); from itertools import chain; from __main__ import fg; f = lambda x: x + 2; g = lambda x: x ** 2',
number=20)
print timeit(stmt='[func(x) for x in xrange(10**6) for func in (f, g)]',
setup='gc.enable(); f = lambda x: x + 2; g = lambda x: x ** 2',
number=20)
2.69210777094
3.13900787874
1.62461071932
25.5944058287
29.2623711793
25.7211849286
发布于 2012-08-09 00:50:19
双重列表理解:
[f(x) for x in range(5) for f in (f1,f2)]
演示:
>>> f1 = lambda x: x
>>> f2 = lambda x: 10*x
>>> [f(x) for x in range(5) for f in (f1,f2)]
[0, 0, 1, 10, 2, 20, 3, 30, 4, 40]
发布于 2012-08-09 00:34:14
sum( ([f(x),g(x)] for x in range(n)), [] )
这相当于[f(1),g(1)] + [f(2),g(2)] + [f(3),g(3)] + ...
你也可以把它想象成:
def flatten(list):
...
flatten( [f(x),g(x)] for x in ... )
注意:正确的方法是使用itertools.chain.from_iterable
或双重列表理解。(它不需要在每个+上重新创建列表,因此具有O(N)性能,而不是O(N^2)性能。)当我想要一个快速的一行,或者我很着急,或者当组合的词条数量有限时(例如<= 10),我仍然会使用sum(..., [])
。这就是为什么我仍然在这里提到它,但有这个警告。你也可以使用元组:((f(x),g(x)) for ...), ()
(或者根据卡奇克的评论,有一个生成器fg(x),它产生一个二元组)。
https://stackoverflow.com/questions/11868964
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