Comparing Sample Mean Vs Random Assortments In Python

Given df A B C Date 2010-01-17 -0.9304 3.7477 0.0000 2010-01-24 -3.6348 1.5733 -3.6348 2010-01-31 -1.8950 0.4957 -1.8950 201

Solution 1:

Well, there's a shortcut:

Since we have equal number of elements in both columns A, B. We could put them in a list and take the 10000 random samples from that list and compare them with the mean of C

sample = df['C'].values
a = df['A'].values
b = df['B'].values
population = np.concatenate((a,b), axis=0)

def mean_diff(s, p):
    m = np.mean(s)
    len_ = len(s)
    result= np.mean([m > np.mean(npr.choice(p, len_, True))
                            for _ inrange(10000)])
    returnresult

mean_diff(sample, population)