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

There is no need to count or groupby. Just note the indices where a change occurs and subtract consecutive indicies.

w = "111000222334455555"
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]

print(dw)  # digits
['1', '0', '2', '3', '4']
print(cw)  # counts
[3, 3, 3, 2, 2, 5]

w = 'XXYXYYYXYXXzzzzzYYY'
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]
print(dw)  # characters
print(cw)  # digits

['X', 'Y', 'X', 'Y', 'X', 'Y', 'X', 'z', 'Y']
[2, 1, 1, 3, 1, 1, 2, 5, 3]