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Consider the pattern %at% where % can have zero or any number of occurrences of an alphabetical character. For example, the string `"BAT, FAT, LATER, HABIT" has three words that match the pattern.

How can this be done in Python?

import re

str_list = ['BAT', 'FAT', 'LATER', 'HABIT']    
counter = 0

for strr in str_list:
    if re.search('at', strr, re.IGNORECASE):
        counter += 1   
print(counter)

OUTPUT:

3

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mylist = ['BAT', 'FAT', 'LATER', 'HABIT']

count = 0
for pattern in mylist:
    if 'at' in pattern.lower():
        count = count + 1    
print(count)

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Use Counter to make the codes shorter:

from collections import Counter
str_list = ['BAT', 'FAT', 'LATER', 'HABIT']
my_counter = Counter('at' in word.lower() for word in str_list)
# 3
print(my_counter[True])

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Comments
  • Still not crystal, so in your example HAT CAT MATTER, we should be able to just print uhm 3? since AT came thrice?
  • @santosh Have you tried to code it?
  • @planet260 i am not able to get the logic
  • @user5173426 yes AT came thrice so i should print 3
  • Okay, does it have to be a pattern or just a searching for a substring?