Write a program that prints the number of times the string contains a substring

Question

s = "bobobobobobsdfsdfbob"
count = 0 
for x in s : 
    if x == "bob" : 
       count += 1
print count

i want to count how many bobs in string s, the result if this gives me 17 what's wrong with my code i'm newbie python.

Answer 1

When you are looping overt the string, the throwaway variable will hold the characters, so in your loop x is never equal with bob .

If you want to count the non-overlaping strings you can simply use str.count :

In [52]: s.count('bob')
Out[52]: 4

For overlapping sub-strings you can use lookaround in regex:

In [57]: import re

In [59]: len(re.findall(r'(?=bob)', s))
Out[59]: 6

Answer 2

you can use string.count

for example:

s = "bobobobobobsdfsdfbob"
count = s.count("bob")
print(count)

Answer 3

I'm not giving the best solution, just trying to correct your code.

Understanding what for each (aka range for) does in your case

for c in "Hello":
    print c  

Outputs:

H
e
l
l
o

In each iteration you are comparing a character to a string which results in a wrong answer.
Try something like
(For no overlapping, ie no span)

s = "bobobobobobsdfsdfbob"
w = "bob"
count = 0
i = 0
while i <= len(s) - len(w):
    if s[i:i+len(w)] == w:
        count += 1
        i += len(w)
    else:
        i += 1
print (count)

Output:

Count = 4

Overlapping

s = "bobobobobobsdfsdfbob"
w = "bob"
count = 0
for i in range(len(s) - len(w) + 1):
    if s[i:i+len(w)] == w:
         count += 1
print (count)

Output:
Count = 6