How to count all occurrences of a word in a string using python

Question

I\'m trying to find the number of times \'bob\' occurs in a string of characters like \'abdebobdfhbobob\'.

My code (that I found through another stackoverflow question)

Answer 1

We can just check all possible candidates:

def count_substrings(sub, main):
    n = len(sub)
    return sum(sub == main[i : i+n] for i in range(len(main) - n + 1))

s = 'abdebobdfhbobob'
sub = 'bob'
print('The number of times %s occurs is: %d' % (sub, count_substrings(sub, s)))  # 3

Answer 2

If you don't want to use regular expressions you can create all triplets from the string using zip and then use list.count:

>>> word = 'bob'
>>> triplets = (''.join(k) for k in zip(*[s[i:] for i in range(len(word))]))
>>> triplets.count(word)
3

The triplets are created by zipping these strings:

     ▼     ▼ ▼
'abdebobdfhbobob'
'bdebobdfhbobob'
'debobdfhbobob'
     ▲     ▲ ▲

If you don't mind working with tuples:

>>> word = 'bob'
>>> triplets = zip(*[s[i:] for i in range(len(word))])
>>> triplets.count(tuple(word))
3

Tip: If you're going to count other words as well, use a collections.Counter.

Answer 3

Based on documentation, str.count() return the number of non-overlapping occurrences of substring sub in the range [start, end]. You can use a positive lookahead based regular expression in order to find the overlapped strings:

>>> import re
>>> s = 'abdebobdfhbobob'
>>> len(re.findall(r'(?=bob)', s))
3

If you don't want to use regex you can use a generator expression within the sum() function that will iterate over the all sub-strings with length 3 and count the number of those that are equal to 'bob':

>>> sum(s[i:i+3] == 'bob' for i in range(len(s)-2))
3

Answer 4

why do not you make it easy?

bobc=0
for i in range (0,len(s)-2):
    if s[i:i+3]=='bob':
        bobc+=1
        i=+1
print('Number of bob:'+str(bobc))