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I have a df as shown below

        df = pd.DataFrame({'Session': ['s1', 's1', 's1', 's1', 's1', 's1', 's1',
                                       's1', 's1', 's1', 's1', 's1', 's1', 's1', 's1'],
                          'slot_num': [1, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9]})

df:

Session         slot_num
s1              1
s1              2
s1              3
s1              3
s1              4
s1              4
s1              5
s1              5
s1              6
s1              7
s1              7
s1              8
s1              8
s1              9
s1              9

From the above I would like to create a column called service_time randomly with mean exactly 20, maximum is 25 and minimum 2 as quickly as possible.

I tried below code but it is not giving the mean exactly 20.

Note: Service time should contain whole numbers only

# generate service time with mean = 20, min = 2 and max = 25
def gen_avg(n, expected_avg=20, a=2, b=25):
    l = np.random.randint(a, b, size=n)
    while True:
        if np.mean(l) == expected_avg:
            break
        while np.mean(l) > expected_avg:
            c = np.random.choice(np.where(l>expected_avg)[0])
            l[c] = np.random.randint(a, expected_avg+1)
        while np.mean(l) < expected_avg:
            c = np.random.choice(np.where(l<expected_avg)[0])
            l[c] = np.random.randint(expected_avg, b)
        return l

df['service_time'] = df.groupby('Session')['Session'].transform(lambda x: gen_avg(len(x)))

I tried below one as well but it is taking very long time

#https://stackoverflow.com/a/39435600/2901002
def gen_avg(n, expected_avg=20, a=5, b=25):
    while True:
        l = np.random.randint(a, b, size=n)
        avg = np.mean(l)

        if avg == expected_avg:
            return l

df['service_time'] = df.groupby('Session')['Session'].transform(lambda x: gen_avg(len(x)))
Danish
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1 Answers1

1

It is probably taking a long time because you are expecting the mean to be exactly equal to the expected_avg. Because it is a random variable in which one out of n observations can change the average, this is a problem, especially as n grows. If this is allowed, you could use a method such that the mean is sufficiently close, e.g. at most 5% away. Suppose that we call this tolerance. Try something like the following:

while abs((avg-expected_avg)/expected_avg) > tolerance:
  l = np.random.randint(a, b, size=n)
  avg = np.mean(l)
Mike Weltevrede
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