列表理解:为每个项目返回两个(或更多)项目 [英] List comprehension: Returning two (or more) items for each item

问题描述

是否可以为列表理解中的每个项目返回2个(或更多)项目?

我想要的东西(示例):

What I want (example):

[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)]

所以，有一些东西可以代替这段代码:

So, something to replace this block of code:

result = list()
for x in range(n):
    result.add(f(x))
    result.add(g(x))

推荐答案

>>> 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]

时间:

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

2.69210777094

3.13900787874

3.13900787874

1.62461071932

1.62461071932

25.5944058287

25.5944058287

29.2623711793

29.2623711793

25.7211849286

25.7211849286

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