# [SOLVED] Count consecutive characters

## Issue

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.

## Solution

A solution “that way”, with only basic statements:

``````word="100011010" #word = "1"
count=1
length=""
if len(word)>1:
for i in range(1,len(word)):
if word[i-1]==word[i]:
count+=1
else :
length += word[i-1]+" repeats "+str(count)+", "
count=1
length += ("and "+word[i]+" repeats "+str(count))
else:
i=0
length += ("and "+word[i]+" repeats "+str(count))
print (length)
``````

Output :

``````'1 repeats 1, 0 repeats 3, 1 repeats 2, 0 repeats 1, 1 repeats 1, and 0 repeats 1'
#'1 repeats 1'
``````