I am working with a string of text that I want to search through and only find 4 letters words. It works, except it also finds 4+ letter words as well.
import re
test ="hello, how are you doing tonight?"
total = len(re.findall(r'[a-zA-Z]{3}', text))
print (total)
It finds 15, although I am not sure how it found that many. I thought I might have to use \b to pick the beginning and the end of the word, but that didn't seem to work for me.
Try this
re.findall(r'\b\w{4}\b',text)
The regex matches:
\b, which is a word boundary.
It matches the beginning or end of a word.
\w{4} matches four word characters (a-z, A-Z, 0-9 or _).
\b is yet another word boundary.
**As a side note, your code contains typos, the second parameter of the re.findall should be the name of your string variable, which is test. Also, your string does not contain any 4 letter words so the suggested code will give the output of 0.
Here's a way without regex:
from string import punctuation
s = "hello, how are you doing tonight?"
[i for i in s.translate(str.maketrans('', '', punctuation)).split(' ') if len(i) > 4]
# ['hello', 'doing', 'tonight']
You can use re.findall to locate all letters, and then filter based off of length:
import re
test ="hello, how are you doing tonight?"
final_words = list(filter(lambda x:len(x) == 4, re.findall('[a-zA-Z]+', test)))
