Given a binary string count the frequency of all binary number with various lenghts-CodePudding

Given a string like '0100011' and a length = 2 we have to count the frequency of pairs in the string: 00,01,10,11.

00 -> 2 , 01 -> 2 , 10 ->1 , 11->1

The problem is if I just use the count('00') function, it doesn't work since it only finds one occurrence: " _ _ 000 _ _".

I've been wondering what the right approach to this kind of problem would be, since for the moment i've tried to create some sort of substring checker with 2 pointer(index and length) which goes back and forth to be sure it doesn't miss any combination but it doesn't feel right

CodePudding user response：

If you want to find the number of occurrences of a single binary substring, you can use string slicing to get each consecutive pair of letters:

data = "0100011"
target = "00"

sum(data[start:start len(target)] == target for start in range(len(data) - len(target)))

This outputs:

If you want to find the number of occurrences of all binary strings of a particular length, you could use a collections.Counter instead:

Counter(data[start:start len(target)] for start in range(len(data) - len(target)))

This outputs:

Counter({'01': 2, '00': 2, '10': 1})

(Note that if you try to look up the frequency of a binary string that doesn't occur, the Counter will simply return 0.)

CodePudding user response：

here is how I did it:

length = 2
s = '0100011'

counter = {}

for i in range(2**length):
    b = f'{i:0{length}b}'
    for j in range(len(s)-1):
        if b == s[j:j length]:
            counter[b] = counter.get(b, 0)   1

print(counter)

will produce:

{'00': 2, '01': 2, '10': 1, '11': 1}

for length == 3:

{'000': 1, '001': 1, '010': 1, '011': 1, '100': 1}

...

for length == 7, obviously:

{'0100011': 1}