This was my attempt , I wanted to take a window of 3 letters and walk through the string, But not getting the expected answer of 2 . I am getting into some infinite loop. Wondering why .
def find_bob(s1):
check_list = 'bob'
count = 0
n=0
cmp_chars=0
while n<=len(s1):
while cmp_chars == s1[n:3]:
if cmp_chars == check_list:
count += 1
continue
return count
s1= 'azcbobobegghakl'
#check_list='bob'
val1 = find_bob(s1)
You can use str.find() to find your bobs by using the start param, e.g.:
def find_bob(s):
check_list = 'bob'
c, n = 0, s.find(check_list)
while n != -1:
c += 1
n = s.find(check_list, n+1)
return c
In []:
find_bob('azcbobobegghakl')
Out[]:
2
You can use a regex with a zero width lookahead and just count the matches:
>>> import re
>>> s1= 'azcbobobegghakl'
>>> len([m for m in re.finditer(r'(?=bob)', s1)])
2
Wondering why...
n=0
cmp_chars=0
while n<=len(s1):
while cmp_chars == s1[n:3]:
if cmp_chars == check_list:
count += 1
continue
n starts at zero and does not change within the loop so the while condition is always True.
while cmp_chars == s1[n:3]: Not sure what to make of this, cmp_chars starts at zero and will never equal a string so this loop condition is always False.
Try
while n<=len(s1):
if s1[n:n+3] == check_list:
count += 1
n = n +1
