x ** (alpha - 1) * math.exp(-x / beta)
pdf(x) =  --------------------------------------
            math.gamma(alpha) * beta ** alpha
random.gauss(mu=0.0, sigma=1.0)¶
Normal distribution, also called the Gaussian distribution.
mu is the mean,
and sigma is the standard deviation.  This is slightly faster than
the normalvariate() function defined below.
多线程注意事项：当两个线程同时调用此方法时，它们有可能将获得相同的返回值。 这可以通过三种办法来避免。 1) 让每个线程使用不同的随机数生成器实例。 2) 在所有调用外面加锁。 3) 改用速度较慢但是线程安全的 normalvariate() 函数。
在 3.11 版更改: mu and sigma now have default arguments.
关于再现性的说明¶
有时能够重现伪随机数生成器给出的序列是很有用处的。 通过重用一个种子值，只要没有运行多线程，相同的序列就应当可在多次运行中重现。
大多数随机模块的算法和种子函数都会在 Python 版本中发生变化，但保证两个方面不会改变：
如果添加了新的播种方法，则将提供向后兼容的播种机。
当兼容的播种机被赋予相同的种子时，生成器的 random() 方法将继续产生相同的序列。
基本示例:
>>> random()                             # Random float:  0.0 <= x < 1.0
0.37444887175646646
>>> uniform(2.5, 10.0)                   # Random float:  2.5 <= x <= 10.0
3.1800146073117523
>>> expovariate(1 / 5)                   # Interval between arrivals averaging 5 seconds
5.148957571865031
>>> randrange(10)                        # Integer from 0 to 9 inclusive
>>> randrange(0, 101, 2)                 # Even integer from 0 to 100 inclusive
>>> choice(['win', 'lose', 'draw'])      # Single random element from a sequence
'draw'
>>> deck = 'ace two three four'.split()
>>> shuffle(deck)                        # Shuffle a list
['four', 'two', 'ace', 'three']
>>> sample([10, 20, 30, 40, 50], k=4)    # Four samples without replacement
[40, 10, 50, 30]
>>> # Six roulette wheel spins (weighted sampling with replacement)
>>> choices(['red', 'black', 'green'], [18, 18, 2], k=6)
['red', 'green', 'black', 'black', 'red', 'black']
>>> # Deal 20 cards without replacement from a deck
>>> # of 52 playing cards, and determine the proportion of cards
>>> # with a ten-value:  ten, jack, queen, or king.
>>> dealt = sample(['tens', 'low cards'], counts=[16, 36], k=20)
>>> dealt.count('tens') / 20
>>> # Estimate the probability of getting 5 or more heads from 7 spins
>>> # of a biased coin that settles on heads 60% of the time.
>>> def trial():
...     return choices('HT', cum_weights=(0.60, 1.00), k=7).count('H') >= 5
>>> sum(trial() for i in range(10_000)) / 10_000
0.4169
>>> # Probability of the median of 5 samples being in middle two quartiles
>>> def trial():
...     return 2_500 <= sorted(choices(range(10_000), k=5))[2] < 7_500
>>> sum(trial() for i in range(10_000)) / 10_000
0.7958
statistical bootstrapping 的示例，使用重新采样和替换来估计一个样本的均值的置信区间:
# https://www.thoughtco.com/example-of-bootstrapping-3126155
from statistics import fmean as mean
from random import choices
data = [41, 50, 29, 37, 81, 30, 73, 63, 20, 35, 68, 22, 60




    
, 31, 95]
means = sorted(mean(choices(data, k=len(data))) for i in range(100))
print(f'The sample mean of {mean(data):.1f} has a 90% confidence '
      f'interval from {means[5]:.1f} to {means[94]:.1f}')
使用 重新采样排列测试 来确定统计学显著性或者使用  p-值 来观察药物与安慰剂的作用之间差异的示例:
# Example from "Statistics is Easy" by Dennis Shasha and Manda Wilson
from statistics import fmean as mean
from random import shuffle
drug = [54, 73, 53, 70, 73, 68, 52, 65, 65]
placebo = [54, 51, 58, 44, 55, 52, 42, 47, 58, 46]
observed_diff = mean(drug) - mean(placebo)
n = 10_000
count = 0
combined = drug + placebo
for i in range(n):
    shuffle(combined)
    new_diff = mean(combined[:len(drug)]) - mean(combined[len(drug):])
    count += (new_diff >= observed_diff)
print(f'{n} label reshufflings produced only {count} instances with a difference')
print(f'at least as extreme as the observed difference of {observed_diff:.1f}.')
print(f'The one-sided p-value of {count / n:.4f} leads us to reject the null')
print(f'hypothesis that there is no difference between the drug and the placebo.')
多服务器队列的到达时间和服务交付模拟:
from heapq import heapify, heapreplace
from random import expovariate, gauss
from statistics import mean, quantiles
average_arrival_interval = 5.6
average_service_time = 15.0
stdev_service_time = 3.5
num_servers = 3
waits = []
arrival_time = 0.0
servers = [0.0] * num_servers  # time when each server becomes available
heapify(servers)
for i in range(1_000_000):
    arrival_time += expovariate(1.0 / average_arrival_interval)
    next_server_available = servers[0]
    wait = max(0.0, next_server_available - arrival_time)
    waits.append(wait)
    service_duration = max(0.0, gauss(average_service_time, stdev_service_time))
    service_completed = arrival_time + wait + service_duration
    heapreplace(servers, service_completed)
print(f'Mean wait: {mean(waits):.1f}   Max wait: {max(waits):.1f}')
print('Quartiles:', [round(q, 1) for q in quantiles(waits)])
Statistics for Hackers Jake Vanderplas 撰写的视频教程，使用一些基本概念进行统计分析，包括模拟、抽样、洗牌和交叉验证。
Economics Simulation
a simulation of a marketplace by
Peter Norvig that shows effective
use of many of the tools and distributions provided by this module
(gauss, uniform, sample, betavariate, choice, triangular, and randrange).
A Concrete Introduction to Probability (using Python)
a tutorial by Peter Norvig covering
the basics of probability theory, how to write simulations, and
how to perform data analysis using Python.
These recipes show how to efficiently make random selections
from the combinatoric iterators in the itertools module:
def random_product(*args, repeat=1):
    "Random selection from itertools.product(*args, **kwds)"
    pools = [tuple(pool) for pool in args] * repeat
    return tuple(map(random.choice, pools))
def random_permutation(iterable, r=None):
    "Random selection from itertools.permutations(iterable, r)"
    pool = tuple(iterable)
    r = len(pool) if r is None else r
    return tuple(random.sample(pool, r))
def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(range(n), r))
    return tuple(pool[i] for i in indices)
def random_combination_with_replacement(iterable, r):
    "Choose r elements with replacement.  Order the result to match the iterable."
    # Result will be in set(itertools.combinations_with_replacement(iterable, r)).
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.choices(range(n), k=r))
    return tuple(pool[i] for i in indices)
默认的 random() 返回在 0.0 ≤ x < 1.0 范围内 2⁻⁵³ 的倍数。 所有这些数值间隔相等并能精确表示为 Python 浮点数。 但是在此间隔上有许多其他可表示浮点数是不可能的选择。 例如，0.05954861408025609 就不是 2⁻⁵³ 的整数倍。
以下规范程序采取了一种不同的方式。 在间隔上的所有浮点数都是可能的选择。 它们的尾数取值来自 2⁵² ≤ 尾数 < 2⁵³ 范围内整数的均匀分布。 指数取值则来自几何分布，其中小于 -53 的指数的出现频率为下一个较大指数的一半。
from random import Random
from math import ldexp
class FullRandom(Random):
    def random(self):
        mantissa = 0x10_0000_0000_0000 | self.getrandbits(52)
        exponent = -53
        x = 0
        while not x:
            x = self.getrandbits(32)
            exponent += x.bit_length() - 32
        return ldexp(mantissa, exponent)
该类中所有的 实值分布 都将使用新的方法:
>>> fr = FullRandom()
>>> fr.random()
0.05954861408025609
>>> fr.expovariate(0.25)
8.87925541791544
该规范程序在概念上等效于在 0.0 ≤ x < 1.0 范围内对所有 2⁻¹⁰⁷⁴ 的倍数进行选择的算法。 所有这样的数字间隔都相等，但大多必须向下舍入为最接近的 Python 浮点数表示形式。 （2⁻¹⁰⁷⁴ 这个数值是等于 math.ulp(0.0) 的未经正规化的最小正浮点数。）
生成伪随机浮点数值 为 Allen B. Downey 所撰写的描述如何生成相比通过 random() 正常生成的数值更细粒度浮点数的论文。
    This page is licensed under the Python Software Foundation License Version 2.
    Examples, recipes, and other code in the documentation are additionally licensed under the Zero Clause BSD License.
    See History and License for more information.

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    最后更新于 9月 01, 2023.
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