[英]Is there a more efficient way to call functions with different arguments multiple times?
[英]Efficient way to call multiple reduce functions on an iterable in Python?
我想在 Python (2.7) 中的可迭代對象上運行幾個 reduce 函數。 一個例子是對整數的可迭代調用min
和max
。 但是當然你不能在同一個迭代器上調用reduce(min, it)
和reduce(max, it)
,因為它在第一次調用后就耗盡了。 因此,您可能會考慮執行以下操作:
reduce(lambda a, b: (min(a[0], b[0]), max(a[1], b[1])), ((x, x) for x in it))
你認為,嘿,這很漂亮,所以你把它概括為這樣的:
from itertools import izip
def multireduce(iterable, *funcs):
""":Return: The tuple resulting from calling ``reduce(func, iterable)`` for each `func` in `funcs`."""
return reduce(lambda a, b: tuple(func(aa, bb) for func, aa, bb in izip(funcs, a, b)), ((item,) * len(funcs) for item in iterable))
(你喜歡單元測試,所以你包括這樣的東西:)
import unittest
class TestMultireduce(unittest.TestCase):
def test_multireduce(self):
vecs = (
((1,), (min,), (1,)),
(xrange(10), (min, max), (0, 9)),
(xrange(101), (min, max, lambda x, y: x + y,), (0, 100, (100 * 101) // 2))
)
for iterable, funcs, expected in vecs:
self.assertSequenceEqual(tuple(multireduce(iterable, *funcs)), expected)
但是當你嘗試它時,你會發現它真的很慢:
%timeit reduce(min, xrange(1000000)) ; reduce(max, xrange(1000000))
10 loops, best of 3: 140 ms per loop
%timeit reduce(lambda a, b: (min(a[0], b[0]), max(a[1], b[1])), ((x, x) for x in xrange(1000000)))
1 loop, best of 3: 682 ms per loop
%timeit multireduce(xrange(1000000), min, max)
1 loop, best of 3: 1.99 s per loop
哎喲。 那么你來到 Stack Overflow 尋找 Python 優化智慧......
嗯,就是這樣,這有點違背了迭代的意義......
def multireduce(iterable, *funcs):
""":Return: The tuple resulting from calling ``reduce(func, iterable)`` for each `func` in `funcs`."""
return tuple(imap(reduce, funcs, tee(iterable, len(funcs))))
但是對於我的測試用例來說速度非常快:
%timeit multireduce(xrange(1000000), min, max)
10 loops, best of 3: 166 ms per loop
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