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

Answer 1

Totals (without sub-groupings)

#!/usr/bin/python3 -B

charseq = 'abbcccffffdd'
distros = { c:1 for c in charseq  }

for c in range(len(charseq)-1):
    if charseq[c] == charseq[c+1]:
        distros[charseq[c]] += 1

print(distros)

I'll provide a brief explanation for the interesting lines.

distros = { c:1 for c in charseq  }

The line above is a dictionary comprehension, and it basically iterates over the characters in charseq and creates a key/value pair for a dictionary where the key is the character and the value is the number of times it has been encountered so far.

Then comes the loop:

for c in range(len(charseq)-1):

We go from 0 to length - 1 to avoid going out of bounds with the c+1 indexing in the loop's body.

if charseq[c] == charseq[c+1]:
    distros[charseq[c]] += 1

At this point, every match we encounter we know is consecutive, so we simply add 1 to the character key. For example, if we take a snapshot of one iteration, the code could look like this (using direct values instead of variables, for illustrative purposes):

# replacing vars for their values
if charseq[1] == charseq[1+1]:
    distros[charseq[1]] += 1

# this is a snapshot of a single comparison here and what happens later
if 'b' == 'b':
    distros['b'] += 1

You can see the program output below with the correct counts:

➜  /tmp  ./counter.py
{'b': 2, 'a': 1, 'c': 3, 'd': 4}