我有一个方法，可以在给定的x时间内创建 2 个不同的实例(M, N) (math.random * x)该方法将创建对象M，其余时间对象N。
我已经编写了模拟随机数的单元测试，因此我可以确保该方法按预期运行。但是，我不确定如何（以及是否）测试概率是否准确，例如，如果x = 0.1我希望十分之一的情况返回实例 M。
如何测试此功能？
拆分测试。第一个测试应该允许您定义随机数生成器返回的内容（我假设您已经有了）。这部分测试只是满足"如果随机数生成器返回一些值，我是否得到预期的结果"。
第二个测试应该使用一些统计分析函数运行随机数生成器（例如计算它返回每个值的频率）。
我建议用一个返回"create M"和"create N"（或者可能只是 0 和 1）的包装器包装真正的生成器。这样，您可以将实现与使用它的位置分开（创建两个不同实例的代码不需要知道生成器是如何初始化的，或者如何将实际结果转换为"create X"。
我将以Python的形式执行此操作。
首先描述您的功能：
def binomial_process(x):
'''
given a probability, x, return M with that probability,
else return N with probability 1-x
maybe: return random.random() > x
'''
然后测试此功能：
import random
def binom(x):
return random.random() > x
然后编写测试函数，首先是一个设置函数，将来自昂贵过程的数据放在一起：
def setUp(x, n):
counter = dict()
for _ in range(n):
result = binom(x)
counter[result] = counter.get(result, 0) + 1
return counter
然后实际测试：
import scipy.stats
trials = 1000000
def test_binomial_process():
ps = (.01, .1, .33, .5, .66, .9, .99)
x_01 = setUp(.01, trials)
x_1 = setUp(.1, trials)
x_33 = setUp(.1, trials)
x_5 = setUp(.5, trials)
x_66 = setUp(.9, trials)
x_9 = setUp(.9, trials)
x_99 = setUp(.99, trials)
x_01_result = scipy.stats.binom_test(x_01.get(True, 0), trials, .01)
x_1_result = scipy.stats.binom_test(x_1.get(True, 0), trials, .1)
x_33_result = scipy.stats.binom_test(x_33.get(True, 0), trials, .33)
x_5_result = scipy.stats.binom_test(x_5.get(True, 0), trials)
x_66_result = scipy.stats.binom_test(x_66.get(True, 0), trials, .66)
x_9_result = scipy.stats.binom_test(x_9.get(True, 0), trials, .9)
x_99_result = scipy.stats.binom_test(x_99.get(True, 0), trials, .99)
setups = (x_01, x_1, x_33, x_5, x_66, x_9, x_99)
results = (x_01_result, x_1_result, x_33_result, x_5_result,
x_66_result, x_9_result, x_99_result)
print 'can reject the hypothesis that the following tests are NOT the'
print 'results of a binomial process (with their given respective'
print 'probabilities) with probability < .01, {0} trials each'.format(trials)
for p, setup, result in zip(ps, setups, results):
print 'p = {0}'.format(p), setup, result, 'reject null' if result < .01 else 'fail to reject'
然后编写你的函数（好的，我们已经这样做了）：
def binom(x):
return random.random() > x
并运行测试：
test_binomial_process()
在最后一个输出中给我：
can reject the hypothesis that the following tests are NOT the
results of a binomial process (with their given respective
probabilities) with probability < .01, 1000000 trials each
p = 0.01 {False: 10084, True: 989916} 4.94065645841e-324 reject null
p = 0.1 {False: 100524, True: 899476} 1.48219693752e-323 reject null
p = 0.33 {False: 100633, True: 899367} 2.96439387505e-323 reject null
p = 0.5 {False: 500369, True: 499631} 0.461122365668 fail to reject
p = 0.66 {False: 900144, True: 99856} 2.96439387505e-323 reject null
p = 0.9 {False: 899988, True: 100012} 1.48219693752e-323 reject null
p = 0.99 {False: 989950, True: 10050} 4.94065645841e-324 reject null
为什么我们不能在p=0.5时拒绝？让我们看看scipy.stats.binom_test上的帮助：
Help on function binom_test in module scipy.stats.morestats: binom_test(x, n=None, p=0.5, alternative='two-sided') Perform a test that the probability of success is p. This is an exact, two-sided test of the null hypothesis that the probability of success in a Bernoulli experiment is `p`. Parameters ---------- x : integer or array_like the number of successes, or if x has length 2, it is the number of successes and the number of failures. n : integer the number of trials. This is ignored if x gives both the number of successes and failures p : float, optional The hypothesized probability of success. 0 <= p <= 1. The default value is p = 0.5 alternative : {'two-sided', 'greater', 'less'}, optional Indicates the alternative hypothesis. The default value is 'two-sided'.
因此，.5 是检验的默认原假设，在这种情况下不拒绝原假设是有意义的。