Python-建模概率 [英] Python - modelling probability

问题描述

我有一个简单的问题.我需要一种使函数在p％的情况下生成0并在所有其他情况下生成1的函数的方法.我尝试使用random.random()这样操作:

I have a simple problem. I need a way to make a function which generates 0s in p percent cases and 1s in all other cases. I tried doing it with random.random() like this:

p = 0.40

def generate():
    x = random.random()
    if x < p:
        return 0
    else:
        return 1

但是，这似乎不是一个好方法.还是它?

However, this doesn't seem like a good way. Or it is?

推荐答案

您当前的方法非常好，您可以通过多次尝试进行多次试验来验证这一点.例如，我们期望通过1000000次尝试获得大约600000个True结果:

Your current method is perfectly fine, you can verify this by performing a few trials with a lot of attempts. For example we would expect approximately 600000 True results with 1000000 attempts:

>>> sum(generate() for i in range(1000000))
599042
>>> sum(generate() for i in range(1000000))
599670
>>> sum(generate() for i in range(1000000))
600011
>>> sum(generate() for i in range(1000000))
599960
>>> sum(generate() for i in range(1000000))
600544

看对了.

如评论中所述，您可以稍微缩短方法:

As noted in comments, you can shorten your method a bit:

p = 0.40

def generate():
    return random.random() >= p

如果要使用1和0而不是True和False，则可以使用int(random.random() >= p)，但这绝对是不必要的.

If you want 1 and 0 instead of True and False you can use int(random.random() >= p), but this is almost definitely unnecessary.

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