Finding all occurrences + substrings of a word

Question

I have the 'main' word, "LAUNCHER", and 2 other words, "LAUNCH" and "LAUNCHER". I want to find out (using regex), which words are in the 'main' word. I'm using findAll, with the regex: "(LAUNCH)|(LAUNCHER)" , but this will only return LAUNCH and not both of them. How do i fix this?

import re
mainword = "launcher"
words = "(launch|launcher)"
matches = re.findall(words,mainword)
for match in matches:
  print(match)

Answer 1

you can try something like this:

import re
mainword = "launcher"
words = "(launch|launcher)"
for x in (re.findall(r"[A-Za-z@#]+|\S", words)):
    if x in mainword:
        print (x)

result:

launch

launcher

Answer 2

If you're not required to use regular expressions, this would be done more efficiently with the IN operator and a simple loop or list comprehension:

mainWord = "launcher"
words    = ["launch","launcher"]
matches  = [ word for word in words if word in mainWord ] 

# case insensitive...
matchWord = mainWord.lower()
matches   = [ word for word in words if word.lower() in matchWord ]

Even if you do require regex, a loop would be needed because re.findAll() never matches overlapping patterns :

import re
pattern   = re.compile("launcher|launch")
mainWord  = "launcher"
matches   = []
startPos  = 0
lastMatch = None
while startPos < len(mainWord):
    if lastMatch : match = pattern.match(mainWord,lastMatch.start(),lastMatch.end()-1) 
    else         : match = pattern.match(mainWord,startPos)
    if not match: 
        if not lastMatch : break
        startPos  = lastMatch.start() + 1
        lastMatch = None
        continue
    matches.append(mainWord[match.start():match.end()])
    lastMatch = match

print(matches)

note that, even with this loop, you need to have the longer words appear before shorter ones if you use the | operator in the regular expression. This is because | is never greedy and will match the first word, not the longest one.