couting a set of word in python

Question

Assume s is a string of lower case characters.

Write a program that prints the number of times the string 'bob' occurs in s. For example, if s = 'azcbobobegghakl', then your program should print

Number of times bob occurs is: 2

This is my answer, but i dont know what's wrong with my code. Please help

s = "azcbobobegghakl"
coutBob=0
i=0
for char in range (len(s)):
    if char[i:i+3]=="bob":
        coutBob+=1
    else:
        i=i+1
print ("Number of times bob occurs is: " + str(coutBob))

This seems AWFULLY familiar...... https://stackoverflow.com/questions/19617125/python-slicing-bob-in-s — L_Church, Jan 22 '18 at 16:40
Possible duplicate of [Python Slicing 'bob' in s](https://stackoverflow.com/questions/19617125/python-slicing-bob-in-s) — ti7, Jan 22 '18 at 16:46
Possible duplicate of [string count with overlapping occurrences](https://stackoverflow.com/questions/2970520/string-count-with-overlapping-occurrences) — Nathan Vērzemnieks, Jan 22 '18 at 18:06

score 0 · Answer 1 · answered Jan 22 '18 at 16:37

0

You need to subscript the string s, not the index:

for i in range(len(s)):
    if s[i:i+3]=="bob":
        coutBob+=1

answered Jan 22 '18 at 16:37

Mureinik

297,002
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score 0 · Answer 2 · answered Feb 19 '18 at 18:21

0

I think this will help you.

b = list(s)
i = 0
j = 0
for i in range(0,len(b)-2):
    if b[i]=='b' and b[i+1]=='o' and b[i+2]=='b':
       j = j + 1
print ("Number of times bob occurs is: %d"%j)

answered Feb 19 '18 at 18:21

Shashank Singh

719
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couting a set of word in python

2 Answers2