Is there an easy way to get the number of repeating character in a word?

Original question: order of repetition does not matter

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x))  # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values())  # 6

New question: repetitions must be sequential

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")
res = sum(sum(1 for _ in j) - 1 for i, j in g)  # 5

To avoid the nested sum calls, you can use itertools.islice:

from itertools import groupby, islice

g = groupby("aooooaooaoo")
res = sum(1 for _, j in g for _ in islice(j, 1, None))  # 5

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
    if word.count(char)>1:
       sum+=word.count(char)
    print('{}|{}'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

You could use a regular expression if you want:

import re

rx = re.compile(r'(\w)\1+')

repeating = sum(x[1] - x[0] - 1
                for m in rx.finditer("loooooveee")
                for x in [m.span()])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

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The expression is

(\w)\1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.
See a demo on regex101.com for the repeating pattern.

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If you want to match any character (that is, not only word characters), change your expression to:

(.)\1+

See another demo on regex101.com.