CODEWITHSUNDEEP
pythonpython-3.xlistdictionary
Fat Wallets
Fat Wallets

Reputation: 101

Count occurance of number from given range

My objective is to count the frequency of number in num_lst to the range in num_range. And display the output to a dictionary where key is the range, value is the frequencies of numbers within the range from num_lst.

I've seen many posts and most are using numpy or pandas to solve it. However I want to find traditional ways to solve this without using np and pd. Can anyone give me the right direction.

num_range = [(0.0, 20.0), (20.0, 40.0), (40.0, 60.0), (60.0, 80.0), (80.0, 100.0)]
num_lst = [x for x in range(100)]

#Preferred output
frequency_dict ={(0.0, 20.0):20, (20.0, 40.0):20, 
                 (40.0, 60.0):20,(60.0, 80.0):20, 
                 (80.0, 100.0):20}

Upvotes: 1

Views: 105

Answers (1)

Tadhg McDonald-Jensen
Tadhg McDonald-Jensen

Reputation: 21453

if all the ranges use integers you can exploit set overlap, seta.intersection(setb) gives all elements in common between 2 sets, then the len of that is how many are in common:

num_range = [(0, 20), (20, 40), (40, 60), (60, 80), (80, 100)]
num_lst = set(range(100))

frequency_dict = {}
for a,b in num_range:
    frequency_dict[a,b] = len(num_lst.intersection(range(a,b)))

print(frequency_dict)

in more general, you can just use a nested loop over the range and see if it falls between each category:

num_range = [(0, 20), (20, 40), (40, 60), (60, 80), (80, 100)]
num_lst = range(100)

frequency_dict = dict.fromkeys(num_range, 0) # initial dictionary has 0 in all entries
for a,b in num_range:
    for i in num_lst:
        if a<=i<b:
            frequency_dict[a,b] += 1

print(frequency_dict)

or if you want it as a one liner with comprehensions:

frequency_dict = {(a,b):sum(a<=i<b for i in num_lst) for a,b in num_range}

And if this nested loop isn't fast enough for your liking, that is why so many people do it with numpy and pandas.

Upvotes: 1

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