CODEWITHSUNDEEP
pythonstringalgorithmcount
Paul Van Doren
Paul Van Doren

Reputation: 35

Counting Instances of Consecutive Duplicate Letters in a Python String

I'm trying to figure out how I can count the number of letters in a string that occur 3 times. The string is from raw_input().

For example, if my input is:

abceeedtyooo

The output should be: 2

This is my current code:

print 'Enter String:'
x = str(raw_input (""))

print  x.count(x[0]*3)

Upvotes: 3

Views: 5252

Answers (3)

Pavan Rajput
Pavan Rajput

Reputation: 39

We can count the chars in the string through 'for' loop

s="abbbaaaaaaccdaaab"
st=[]
count=0
for i in set(s):
    print(i+str(s.count(i)),end='')

Output: a10c2b4d1

Upvotes: 1

jfs
jfs

Reputation: 414265

To count number of consecutive duplicate letters that occur exactly 3 times:

>>> from itertools import groupby
>>> sum(len(list(dups)) == 3 for _, dups in groupby("abceeedtyooo"))
2

Upvotes: 3

hlt
hlt

Reputation: 6317

To count the chars in the string, you can use collections.Counter:

>>> from collections import Counter
>>> counter = Counter("abceeedtyooo")
>>> print(counter)
Counter({'e': 3, 'o': 3, 'a': 1, 'd': 1, 'y': 1, 'c': 1, 'b': 1, 't': 1})

Then you can filter the result as follows:

>>> result = [char for char in counter if counter[char] == 3]
>>> print(result)
['e', 'o']

If you want to match consecutive characters only, you can use regex (cf. re):

>>> import re
>>> result = re.findall(r"(.)\1\1", "abceeedtyooo")
>>> print(result)
['e', 'o']
>>> result = re.findall(r"(.)\1\1", "abcaaa")
>>> print(result)
['a']

This will also match if the same character appears three consecutive times multiple times (e.g. on "aaabcaaa", it will match 'a' twice). Matches are non-overlapping, so on "aaaa" it will only match once, but on "aaaaaa" it will match twice. Should you not want multiple matches on consecutive strings, modify the regex to r"(.)\1\1(?!\1)". To avoid matching any chars that appear more than 3 consecutive times, use (.)(?<!(?=\1)..)\1{2}(?!\1). This works around a problem with Python's regex module that cannot handle (?<!\1).

Upvotes: 2

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