random – 生成伪随机数的详解 (6) Python语言的数学和数学模块(必读进阶学习教程)(参考资料)

random.gauss(mu, sigma)

高斯分布。 mu是均值，sigma是标准差。这比normalvariate()下面定义的函数稍快。

random.lognormvariate(mu, sigma)

记录正态分布。如果您取此分布的自然对数，您将得到一个均值为mu和标准差为sigma的正态分布。 mu可以是任意值，sigma必须大于零。

random.normalvariate(mu, sigma)

正态分布。 mu是均值，sigma是标准差。

random.vonmisesvariate(mu, kappa)

mu是平均角度，以 0 到 2* pi之间的弧度表示，kappa 是浓度参数，必须大于或等于零。如果 kappa等于零，则此分布在 0 到 2* pi范围内减少为均匀的随机角度。

random.paretovariate(alpha)

帕累托分布。 alpha是形状参数。

random.weibullvariate(alpha, beta)

威布尔分布。 alpha是尺度参数，beta是形状参数。

替代发电机

classrandom.Random([seed])

实现模块使用的默认伪随机数生成器的类 random。

classrandom.SystemRandom([seed])

使用该os.urandom()函数从操作系统提供的源生成随机数的类。并非在所有系统上都可用。不依赖于软件状态，序列不可重现。因此，该seed()方法没有任何作用并被忽略。getstate()和方法在调用时setstate()引发 NotImplementedError。

重现性注释

有时能够重现伪随机数生成器给出的序列是很有用的。通过重新使用种子值，只要多个线程没有运行，相同的序列就应该可以从运行到运行重现。

大多数 random 模块的算法和种子函数在不同的 Python 版本中都会发生变化，但有两个方面保证不会发生变化：

• 如果添加了新的播种方法，则会提供向后兼容的播种机。
• 当兼容的播种器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。

例子

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
7

>>> randrange(0, 101, 2)                 # Even integer from 0 to 100 inclusive
26

>>> choice(['win', 'lose', 'draw'])      # Single random element from a sequence
'draw'

>>> deck = 'ace two three four'.split()
>>> shuffle(deck)                        # Shuffle a list
>>> deck
['four', 'two', 'ace', 'three']

>>> sample([10, 20, 30, 40, 50], k=4)    # Four samples without replacement
[40, 10, 50, 30]

模拟：

>>> # 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
>>> # (a ten, jack, queen, or king).
>>> deck = collections.Counter(tens=16, low_cards=36)
>>> seen = sample(list(deck.elements()), k=20)
>>> seen.count('tens') / 20
0.15

>>> # 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(10000)) / 10000
0.4169

>>> # Probability of the median of 5 samples being in middle two quartiles
>>> def trial():
...     return 2500 <= sorted(choices(range(10000), k=5))[2] < 7500
...
>>> sum(trial() for i in range(10000)) / 10000
0.7958

使用有放回的重采样来估计大小为 5 的样本均值的置信区间的统计自举示例：

# http://statistics.about.com/od/Applications/a/Example-Of-Bootstrapping.htm
from statistics import mean
from random import choices

data = 1, 2, 4, 4, 10
means = sorted(mean(choices(data, k=5)) for i in range(20))
print(f'The sample mean of {mean(data):.1f} has a 90% confidence '
      f'interval from {means[1]:.1f} to {means[-2]:.1f}')

重采样排列检验的示例，以确定药物与安慰剂作用之间观察到的差异 的统计显着性或p 值：

# Example from "Statistics is Easy" by Dennis Shasha and Manda Wilson
from statistics import 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 = 10000
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.')

在单个服务器队列中模拟到达时间和服务交付：

from random import expovariate, gauss
from statistics import mean, median, stdev

average_arrival_interval = 5.6
average_service_time = 5.0
stdev_service_time = 0.5

num_waiting = 0
arrivals = []
starts = []
arrival = service_end = 0.0
for i in range(20000):
    if arrival <= service_end:
        num_waiting += 1
        arrival += expovariate(1.0 / average_arrival_interval)
        arrivals.append(arrival)
    else:
        num_waiting -= 1
        service_start = service_end if num_waiting else arrival
        service_time = gauss(average_service_time, stdev_service_time)
        service_end = service_start + service_time
        starts.append(service_start)

waits = [start - arrival for arrival, start in zip(arrivals, starts)]
print(f'Mean wait: {mean(waits):.1f}.  Stdev wait: {stdev(waits):.1f}.')
print(f'Median wait: {median(waits):.1f}.  Max wait: {max(waits):.1f}.')