Python中random.shuffle()是否采用均匀分布？技术问询

Great question—this is a common point of confusion since the docs don’t spell it out explicitly, but let’s clear it up:

• random.shuffle() absolutely produces uniformly random permutations of your list. That means every possible ordering of the list has an equal probability of being chosen (specifically, 1 divided by the factorial of the list length, 1/n! for a list of size n).

Here’s why that’s the case:

1. It uses the Fisher-Yates (Knuth) Shuffle algorithm under the hood. This algorithm works by iterating through the list from last to first, and at each step swapping the current element with a randomly selected element from the unshuffled portion (from start to current index). Each selection is made using the random module’s core uniform random number generator (typically the Mersenne Twister), which picks each candidate element with equal probability.
2. The math checks out: For each position in the list, every remaining element has an equal chance to end up there. When you chain these independent uniform choices together, the end result is a uniform distribution over all possible permutations of the list.

While the official docs don’t explicitly state "uniform distribution," this is a standard property of the Fisher-Yates shuffle, and Python’s implementation adheres to this behavior. If you run repeated tests with a small list (like [1,2,3]), you’ll see that each of the 6 possible permutations appears roughly equally often over many trials.

内容的提问来源于stack exchange，提问作者HT121