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Start your free trialMaba Bah
2,744 PointsHow would I return a list of the words?
I understand this would return a "matchobject". How can I just get the words and return them in to a list?
import re
# EXAMPLE:
# >>> find_words(4, "dog, cat, baby, balloon, me")
# ['baby', 'balloon']
def find_words(count, string):
words = re.findall(r'\w{count,}', string)
return words
2 Answers
Jeff Muday
Treehouse Moderator 28,720 PointsThe other issue (and why you have to use str(count)
is the regular expression is not a "normal string" but a RAW string. So the presence of a \
character signals in a regular string to interpret it in an escape sequence like newline \n
or other escape sequences. Thus the typical "coeercion" may not work in the way you expect.
Here's another solution which would work too. Basically uses a %s
which marks a string substitution of count which is intrepreted as a string.
def find_words(count, mystring):
matches = re.findall(r'\w{%s,}' % count, mystring)
return matches
Jeff Muday
Treehouse Moderator 28,720 PointsConceptually you got it correct, nice work! The regular expression can't interpret count
as an integer-- so we have to take the regular expression "apart", and EMBED count
as a string INSIDE the regular expression. See below.
Good luck with your Python journey. I have always found regular expressions to be powerful... But!!! one of the more difficult topics in programming.
def find_words(count, mystring):
matches = re.findall(r'\w{' + str(count) + ',}', mystring)
return matches
Maba Bah
2,744 PointsIt works! Although I understand how it works but not the why. What if I coerced "count" into an integer beforehand, for example "count = int(count)". Would I still be able to plug in "count" in the curly braces?
Maba Bah
2,744 PointsMaba Bah
2,744 PointsThank you so much for clarifying! The above example is much easier to digest!