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

Here is an implementation based on Brandon's answer to a linked question.

def MIT(full_string, substring):
    results = []
    sub_len = len(substring)
    for i in range(len(full_string)):
    # range returns a list of values from 0 to (len(full_string) - 1)
        if full_string[i:i+sub_len] == substring:
        # this is slice notation;
        # it means take characters i up to (but not including) i
        # + the length of th substring
            results.append(i)
    return results

full_string = "bobBOBBobBoj"
lower_substring = "bob"
upper_substring = "BOB"
lower_occurrence_array = MIT(full_string, lower_substring)
upper_occurrence_array = MIT(full_string, upper_substring)
number_of_lower_occurrences = len(lower_occurrence_array)
number_of_upper_occurrences = len(upper_occurrence_array)
number_of_occurrences = number_of_lower_occurrences + number_of_upper_occurrences

print ("Number of times bob occurs is: %s" %number_of_occurrences)

The result is

Number of times bob occurs is: 2

The main gist is already there with Paul's answer but I hope this helps.

Answer 2

start=0
count=0 
while start<=len(s):
    n=s.find('b',start,len(s))
    prnt=(s[start:start+3])
    if prnt =='bob':
       start=n+2 
       count+=1
    else:
        start+=1        
print count