Bob Counter in Python

Question

Using Python 2.7, I was attempting to count the number of occurances of \'bob\' in the phrase \'bobbbobobboobobookobobbobbboj.\' To do this, I wrote the code below:

Answer 1

If you want to do it the Pythonic way, then you'll want to read up on the string find function.

If you want to learn how make a string find function, then you'll want to read up on string searching algorithms.

Answer 2

let's see if we can help you out.

Your code is

s='bobbbobobboobobookobobbobbboj'

 for i in s:

     if(i=='BOB' or i=='bob'):
        b=b+1

It is important to think about this- a string like "s" is a list of characters. When you do for i in s, you are looping through each individual character.

on the first iteration, i == 'b', on the second one it equals 'o' and so on.

What you want is something that checks sections of code. a way of doing that would be

for i in range(len(s)):
    if s[i:i+4] == "bob":

The reason this works is that range returns a list of numbers in order, so it will go 0, 1, 2, 3... the [i:i+4] part cuts out a section of the string based on the how far into the string it is. For your string s[0:2] would be "bob" (it starts with 0).

I left several problems... for example, if you let it run to the end you'll have a problem (if s is 10 characters long and you try to do s[9:12] you will get an error) ... but that should help you get going

Answer 3

s = ('bobobobobobob')
count = 0

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

print ('Number of times bob occurs is: ' + str(count))

Answer 4

>>> s = 'bobbbobobboobobookobobbobbboj'
>>> term = 'bob'
sum(1 for i, j in enumerate(s) if s[i:i+len(term)] == term)
6

Answer 5

def bobcount(x):
    bobcounter = 0
    for i in range(len(x)):
        if (x[i:i+3]=="bob") or (x[i:i+3]=="BOB"):
            bobcounter = bobcounter + 1
    return bobcounter

This is an extension of Paul's answer. The result of the code is:

bobcount("bobobobBOBBOB")
5

Answer 6

Guess i am a little bit late, but you can try this:

from re import findall

    def bobcount(word):
        return len(findall(r'(bob)', word.lower()))