Python, given a word, generate a list with one letter changing for each letter in alphabet

Question

Here is my problem:

Return the number of times that the string code appears anywhere in the given string, except we'll accept any letter for the d, so cope and cooe count.

I solved it by creating a list with all 26 possible co_e combinations and then used two for loops to iterate through the list and search for matches in the given string.

Is there another way I could have created that list? or is there a way to ignore the third character in word and only search for matching characters 1, 2 and 4?

Here is my solution:

if len(inputstring)<4:
    return 0
wordlist=['coae','cobe','coce','code'...etc]
count=0

for i in wordlist:
    for j in range(len(inputstring)):
        if inputstring[j:j+4] == i:
            count+=1
return count

Answer 1

Just use regular expressions!

import re
# creating regular expression with "any" letter at 3 position
r = re.compile("co[a-z]e")
# and taking all the matches:
matches = r.findall("some of my code, nope to come, cofe and cole, or cope and dome, rope and hope!")
count = len(matches)
print(count)

Answer 2

You can use in to test if a substring appears:

>>> 'code' in 'recoded'
True

This removes the need to loop over the string.

You can generate your test strings by looping over all letters in the alphabet; Python already has the latter available for you at string.ascii_lowercase and a list comprehension:

import string

possibilities = ['co{l}e'.format(l) for l in string.ascii_lowercase]

count = 0
for possibility in possibilities:
    if possibility in inputstring:
        count += 1

You could also just test for co appearing, and see if there is a letter e further along. You can use the str.find() method to find the position of an occurrence and search from there; str.find() takes a starting position to search for the next match:

count = 0
start = 0
while True:
    position = inputstring.find('co')
    if position == -1:
        # not found, end the search
        break
    if len(inputstring) > position + 2 and inputstring[position + 2] == 'e':
        count += 1
    start = position + 1

However, most experienced programmers will use regular expressions to find such matches:

import re

count = len(re.findall(r'co[a-z]e', inputstring))

Here the expression uses [a-z] to match a single character as a class, anything in that series (so letters from a to z) would match. The re.findall() function returns a list of all matches found in the input string, so all you have to do then is take the len() length of that list to get a count.

Answer 3

Simple and beautiful. I hope it is what you want

import re

list=['coae','cobe','coce','code','coco']
count=0

for i in list:
  if re.match(r'co[a-z]e', i):
    count+=1

print(count)