python - String instead of integer in finding "bob" -

my question code:

count = 0  char in s:      if char.startswith("bob"):          count += 1  print ("number of times bob occurs is: " + str(count)) 

i have solution followed:

count = 0  in range(len(s)):       if s[i: i+3] == "bob"          count += 1  print ("number of times bob occurs is: " + str(count)) 

my question: instead of taking out solution using integer "for in range(len(s))", want alternative solution using character/string. tell me why above solution returns "0" in finding "bob"? thanks.

what wrong first piece of code become apparent print statement in loop. statement for char in s loops through each character in s , no character starts word bob.

if want for in something type loop, can do:

count = 0 word in s.split():     if word == "bob":         count += 1 print ("number of times bob occurs is: " + str(count)) 

this work if wan match occurrences of bob occur word alone. if want match bob anywhere, use string.count method:

count = s.count("bob") 

or alternatively, regex:

import re count = len(re.findall("bob", s)) 

if want overlapping answers. doing in loop more concisely. don't think there simpler way count overlapping occurrences this.

[s[i:i+3] in range(len(s))].count('bob')