The count() function returns the number of times a substring occurs in a string, but it fails in case of overlapping strings.
Let's say my input is:
^_^_^-_-
I want to find how many times ^_^ occurs in the string.
mystr=input()
happy=mystr.count('^_^')
sad=mystr.count('-_-')
print(happy)
print(sad)
Output is:
1
1
I am expecting:
2
1
How can I achieve the desired result?
New Version
You can solve this problem without writing any explicit loops using regex. As @abhijith-pk's answer cleverly suggests, you can search for the first character only, with the remainder being placed in a positive lookahead, which will allow you to make the match with overlaps:
def count_overlapping(string, pattern):
regex = '{}(?={})'.format(re.escape(pattern[:1]), re.escape(pattern[1:]))
# Consume iterator, get count with minimal memory usage
return sum(1 for _ in re.finditer(regex, string))
Using [:1] and [1:] for the indices allows the function to handle the empty string without special processing, while using [0] and [1:] for the indices would not.
Old Version
You can always write your own routine using the fact that str.find allows you to specify a starting index. This routine will not be very efficient, but it should work:
def count_overlapping(string, pattern):
count = 0
start = -1
while True:
start = string.find(pattern, start + 1)
if start < 0:
return count
count += 1
Usage
Both versions return identical results. A sample usage would be:
>>> mystr = '^_^_^-_-'
>>> count_overlapping(mystr, '^_^')
2
>>> count_overlapping(mystr, '-_-')
1
>>> count_overlapping(mystr, '')
9
>>> count_overlapping(mystr, 'x')
0
Notice that the empty string is found len(mystr) + 1 times. I consider this to be intuitively correct because it is effectively between and around every character.
you can use regex for a quick and dirty solution :
import re
mystr='^_^_^-_-'
print(len(re.findall('\^(?=_\^)',mystr)))
You need something like this
def count_substr(string,substr):
n=len(substr)
count=0
for i in range(len(string)-len(substr)+1):
if(string[i:i+len(substr)] == substr):
count+=1
return count
mystr=input()
print(count_substr(mystr,'121'))
Input: 12121990
Output: 2
string[i:i+len(substr)] creates a new string object every single time.