为什么random.random（）会占用两个random.randint（）值？

Question

I've developed a simple application that generates test data series and I've built it to be able to be repeatable, using a random seed. I noticed the following and wanted to know why this happens:

>>> random.seed(1)
>>> [random.randint(0,10) for _ in range(0,10)]
[2, 9, 1, 4, 1, 7, 7, 7, 10, 6]
>>> random.seed(1)
>>> random.random()
0.13436424411240122
>>> [random.randint(0,10) for _ in range(0,10)]
[1, 4, 1, 7, 7, 7, 10, 6, 3, 1]

Note how a single call to random() uses up two values for randint(). I'm guessing this has something to do with the amount of random information required to generate a float vs. an int in the given range, but is there some way of keeping track of 'how many random values have been used so far?', ie how far along in the sequence of semi-random values the system is?

I ended up writing my own function, always using a single call to random.random() in its logic. So I'm not asking for a solution, just some background / an explanation.

Answer 1

Your guess is accurate, at least for the latest version of CPython. There are a couple of relevant places in the code for the random module where you can see why this is happening. I would consider all of this an implementation detail, but you can see that essentially, groups of 32 random bits are generated as needed to get random numbers. For this reason, getting 53 random bits (to represent the fractional part of a double) uses twice as many random bits as getting 4 random bits.

• random.randint Python code
• random._randbelow Python code
• random.getrandbits C code
• random.random C code

With respect to being able to tell how much random data has been generated, the provided functions don't seem to give you an easy and reliable way of doing that.