Python count occurence of a string without overlapping from a string

Question

I was trying to find the occurrence of every 2 consecutive characters from a string.
The result will be in a dictionary as key = 2 characters and value = number of occurrence.

I tried the following :

seq = "AXXTAGXXXTA"

d = {seq[i:i+2]:seq.count(seq[i:i+2]) for i in range(0, len(seq)-1)}

The problem is that the result of XX should be 3 not 2 .

Asked By: BouhaaCode

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Source

Answer 1

You can use collections.Counter.

from collections import Counter

seq = "AXXTAGXXXTA"

Counter((seq[i:i+2] for i in range(len(seq)-1)))

Output:

Counter({'AX': 1, 'XX': 3, 'XT': 2, 'TA': 2, 'AG': 1, 'GX': 1})

Or without additional libraries. You can use dict.setdefault.

seq = "AXXTAGXXXTA"

d = {}
for i in range(len(seq)-1):
    key = seq[i:i+2]
    d[key] = d.setdefault(key, 0) + 1
print(d)

Answered By: I'mahdi

Python count occurence of a string without overlapping from a string

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