在python中获取一个随机布尔值? [英] Get a random boolean in python?

问题描述

我正在寻找在 python 中获取随机布尔值的最佳方法(快速而优雅)(掷硬币).

I am looking for the best way (fast and elegant) to get a random boolean in python (flip a coin).

目前我使用的是 random.randint(0, 1) 或 random.getrandbits(1).

For the moment I am using random.randint(0, 1) or random.getrandbits(1).

是否有我不知道的更好的选择?

Are there better choices that I am not aware of?

推荐答案

Adam 的回答相当快，但我发现 random.getrandbits(1) 快了很多.如果你真的想要一个布尔值而不是一个 long 那么

Adam's answer is quite fast, but I found that random.getrandbits(1) to be quite a lot faster. If you really want a boolean instead of a long then

bool(random.getrandbits(1))

仍然是 random.choice([True, False])

两种方案都需要import random

如果最大速度不是优先考虑的，那么 random.choice 肯定会更好读.

If utmost speed isn't to priority then random.choice definitely reads better.

请注意，由于属性查找，random.choice() 比 choice() 慢(在 from random import choice 之后).

Note that random.choice() is slower than just choice() (after from random import choice) due to the attribute lookup.

$ python3 --version
Python 3.9.7
$ python3 -m timeit -s "from random import choice" "choice([True, False])"
1000000 loops, best of 5: 376 nsec per loop
$ python3 -m timeit -s "from random import choice" "choice((True, False))"
1000000 loops, best of 5: 352 nsec per loop
$ python3 -m timeit -s "from random import getrandbits" "getrandbits(1)"
10000000 loops, best of 5: 33.7 nsec per loop
$ python3 -m timeit -s "from random import getrandbits" "bool(getrandbits(1))"
5000000 loops, best of 5: 89.5 nsec per loop
$ python3 -m timeit -s "from random import getrandbits" "not getrandbits(1)"
5000000 loops, best of 5: 46.3 nsec per loop
$ python3 -m timeit -s "from random import random" "random() < 0.5"
5000000 loops, best of 5: 46.4 nsec per loop

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