Calculate average value of dictionaries

Question:-
We have a list of annual rainfall recordings of cities. Each element in the list is of the form (c,r) where c is the city and r is the annual rainfall for a particular year. The list may have multiple entries for the same city, corresponding to rainfall recordings in different years.
Write a Python function rainaverage(l) that takes as input a list of rainfall recordings and computes the avarage rainfall for each city. The output should be a list of pairs (c,ar) where c is the city and ar is the average rainfall for this city among the recordings in the input list. Note that ar should be of type float. The output should be sorted in dictionary order with respect to the city name.
Here are some examples to show how rainaverage(l) should work.
>>> rainaverage([(1,2),(1,3),(2,3),(1,1),(3,8)])
[(1, 2.0), (2, 3.0), (3, 8.0)]

>>> rainaverage([('Bombay',848),('Madras',103),('Bombay',923),('Bangalore',201),('Madras',128)])
[('Bangalore', 201.0), ('Bombay', 885.5), ('Madras', 115.5)]

Solution:-
First, try to calculate the number of occurrences of each element in the list and store in a dictionary. To the same with the sum of all the values; finally, divide sum/value to get the average.



def rainaverage(l):
    count = {}
    sum1 = {}
    ar = {}
    for i in l:
        count[i[0]] = 0
        sum1[i[0]] = 0
    for j in l:
        count[j[0]]+=1
    for k in l:
        sum1[k[0]] += k[1]

    for m in count.keys():
        ar[m] = sum1[m]/count[m]
    x = sorted(ar.items())
    return x

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