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

Unique method:- In case if you are just looking for counting consecutive 1's Using Bit Magic: The idea is based on the concept that if we AND a bit sequence with a shifted version of itself, we’re effectively removing the trailing 1 from every sequence of consecutive 1s.

  11101111   (x)
& 11011110   (x << 1)
----------
  11001110   (x & (x << 1)) 
    ^    ^
    |    |

trailing 1 removed So the operation x = (x & (x << 1)) reduces length of every sequence of 1s by one in binary representation of x. If we keep doing this operation in a loop, we end up with x = 0. The number of iterations required to reach 0 is actually length of the longest consecutive sequence of 1s.

Answer 2

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'

Answer 3

You only need to change len(word) to len(word) - 1. That said, you could also use the fact that False's value is 0 and True's value is 1 with sum:

sum(word[i] == word[i+1] for i in range(len(word)-1))

This produces the sum of (False, True, True, False) where False is 0 and True is 1 - which is what you're after.

If you want this to be safe you need to guard empty words (index -1 access):

sum(word[i] == word[i+1] for i in range(max(0, len(word)-1)))

And this can be improved with zip:

sum(c1 == c2 for c1, c2 in zip(word[:-1], word[1:]))