This apparently simple question is bugging my head for a while, thought somebody might be of help.
I have a simple string
s = 'AAABCAA'
How to find the number of repetitions of first letter 'A'? Answer should be 3.
I have tried:
from collections import Counter
c = Counter(s)
But, this gives 'A' = 5, instead of 3.
You could use a for loop with a break statement.
s = 'AAABCAA'
counter=0
firstletter=s[0]
for each in s:
if each==firstletter:
counter+=1
else:
break
print(counter)
This just returns 3.
Alternatively, you could return index of the first element of the string which is not the same as the first character of your string:
import numpy as np
s = 'AAABCAA'
firstletter=s[0]
checklist=[(each==firstletter)*1 for each in s]
print(np.where(np.asarray(checklist)==0)[0][0])
In this case, with list comprehension ([(each==firstletter)*1 for each in s]) we produce a list:
[1, 1, 1, 0, 0, 1, 1]
The value is 1 wherever the character in that spot is identical to the first character of the string.
Then np.where(np.asarray(checklist)==0)[0][0] gives you the index of the first 0 (ie the first character not identical to starting character) of this newly created list.
You can use the function groupby() to find all letter groups and then you can use next() to get the first group from the iterator:
from itertools import groupby
s = 'AAABCAA'
sum(1 for _ in next(groupby(s))[1])
# 3
Alternatively you can use the function takewhile():
from itertools import takewhile
sum(1 for _ in takewhile(lambda x: x == s[0], s))
# 3
And finally you can use regex:
import re
len(re.search(r'^(\w)\1+', s, flags=re.MULTILINE).group(0))
# 3
Here's a short solution that uses list comp. Of course, readability won't be the goal here :)
repetitions = lambda str, letter: [i + 1 for i, num in enumerate(str) if num == letter][-1]
Examples:
str = 'BBBBC'
letter = 'B'
repetitions(str, letter) # 4
str = 'AABC'
letter = 'A'
repetitions(str, letter) # 2
If you are looking for patterns in strings in general, use a suffix tree.
