python中是否有一个适用于数组的二元OR运算符? [英] is there a binary OR operator in python that works on arrays?

问题描述

我是从 matlab 背景到 python 的，我只是想知道 python 中是否有一个简单的运算符可以执行以下功能:

I have come from a matlab background to python and i am just wondering if there is a simple operator in python that will perform the following function:

a = [1, 0, 0, 1, 0, 0]
b = [0, 1, 0, 1, 0, 1]
c = a|b
print c
[1, 1, 0, 1, 0, 1]

或者我是否必须编写一个单独的函数来做到这一点?

or would i have to write a separate function to do that?

推荐答案

您可以使用列表推导式.如果您使用的是 itertools 中的 izip蟒蛇 2.

You could use a list comprehension. Use izip from itertools if you're using Python 2.

c = [x | y for x, y in zip(a, b)]

或者，@georg 在评论中指出您可以导入按位或运算符并将其与 map 一起使用.这仅比列表理解略快.map 在 Python 2 中不需要用 list() 包裹.

Alternatively, @georg pointed out in a comment that you can import the bitwise or operator and use it with map. This is only slightly faster than the list comprehension. map doesn't need wrapped with list() in Python 2.

import operator
c = list(map(operator.or_, a, b))

性能

列表理解:

$ python -m timeit -s "a = [1, 0, 0, 1, 0, 0]; b = [0, 1, 0, 1, 0, 1]" \
> "[x | y for x, y in zip(a, b)]"

1000000 loops, best of 3: 1.41 usec per loop

地图:

$ python -m timeit -s "a = [1, 0, 0, 1, 0, 0]; b = [0, 1, 0, 1, 0, 1]; \
> from operator import or_" "list(map(or_, a, b))"
1000000 loops, best of 3: 1.31 usec per loop

NumPy

$ python -m timeit -s "import numpy; a = [1, 0, 0, 1, 0, 0]; \
> b = [0, 1, 0, 1, 0, 1]" "na = numpy.array(a); nb = numpy.array(b); na | nb"

100000 loops, best of 3: 6.07 usec per loop

NumPy(其中 a 和 b 已经转换为 numpy 数组):

NumPy (where a and b have already been converted to numpy arrays):

$ python -m timeit -s "import numpy; a = numpy.array([1, 0, 0, 1, 0, 0]); \
> b = numpy.array([0, 1, 0, 1, 0, 1])" "a | b"

1000000 loops, best of 3: 1.1 usec per loop

结论:除非您需要 NumPy 进行其他操作，否则不值得转换.

Conclusion: Unless you need NumPy for other operations, it's not worth the conversion.

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