Count consecutive characters

Question

How would I count consecutive characters in Python to see the number of times each unique digit repeats before the next unique digit?

At first, I thought I could do something like:

word = '1000'  counter=0 print range(len(word))   for i in range(len(word)-1):     while word[i]==word[i+1]:         counter +=1         print counter*"0"     else:         counter=1         print counter*"1" 

So that in this manner I could see the number of times each unique digit repeats. But this, of course, falls out of range when i reaches the last value.

In the example above, I would want Python to tell me that 1 repeats 1, and that 0 repeats 3 times. The code above fails, however, because of my while statement.

I know you can do this with just built-in functions and would prefer a solution that way.

Answer 1

Consecutive counts:

Ooh nobody's posted itertools.groupby yet!

s = "111000222334455555"  from itertools import groupby  groups = groupby(s) result = [(label, sum(1 for _ in group)) for label, group in groups] 

After which, result looks like:

[("1": 3), ("0", 3), ("2", 3), ("3", 2), ("4", 2), ("5", 5)] 

And you could format with something like:

", ".join("{}x{}".format(label, count) for label, count in result) # "1x3, 0x3, 2x3, 3x2, 4x2, 5x5" 

Total counts:

Someone in the comments is concerned that you want a total count of numbers so "11100111" -> {"1":6, "0":2}. In that case you want to use a collections.Counter:

from collections import Counter  s = "11100111" result = Counter(s) # {"1":6, "0":2} 

Your method:

As many have pointed out, your method fails because you're looping through range(len(s)) but addressing s[i+1]. This leads to an off-by-one error when i is pointing at the last index of s, so i+1 raises an IndexError. One way to fix this would be to loop through range(len(s)-1), but it's more pythonic to generate something to iterate over.

For string that's not absolutely huge, zip(s, s[1:]) isn't a a performance issue, so you could do:

counts = [] count = 1 for a, b in zip(s, s[1:]):     if a==b:         count += 1     else:         counts.append((a, count))         count = 1 

The only problem being that you'll have to special-case the last character if it's unique. That can be fixed with itertools.zip_longest

import itertools  counts = [] count = 1 for a, b in itertools.zip_longest(s, s[1:], fillvalue=None):     if a==b:         count += 1     else:         counts.append((a, count))         count = 1 

If you do have a truly huge string and can't stand to hold two of them in memory at a time, you can use the itertools recipe pairwise.

def pairwise(iterable):     """iterates pairwise without holding an extra copy of iterable in memory"""     a, b = itertools.tee(iterable)     next(b, None)     return itertools.zip_longest(a, b, fillvalue=None)  counts = [] count = 1 for a, b in pairwise(s):     ...