Parse a file by multiple delimiters

[jgrabow1]

I have a file with multiple lines of words for example:
groovy
office space.
bypass
bad

I want to delimit each word by space/newline/punctuation so that I have a list with "groovy", "office", "space", "bypass", "bad".
So far I have this code written:

f = open('test.txt', 'r')
    lines = f.read().split()
    print(lines)
    f.close()
 

However, this only gets me spaces and new lines... not punctuation. Can anyone help me?

[dwblas]

You would then separate "isn't" into "isn" and "t". Sentences ending with punctuation will have a space following it so spaces should get you what you want. If not, please include a sample of the input that would require a split on punctuation, and the desired output.

Note that "office space" can be divided by splitting each line on a space. Previous post about iterating over a file.

[jgrabow1]

This is true, but you can assume isn't won't be a case. Also, office space can be split with spaces, but the period would still be there. I think I figured it out anyway on my own though, but if you still have a suggestion i'm open to it.

[dwblas]

Don't know how you can get "office space." using

lines = f.read().split() 

which is different from split("\n"), but anyway...the code is the same either way

You would only keep characters that are letters, which removes numbers, punctuation, and any other non-letter characters (removing punctuation instead of keeping letters is "bass-ackwards" as you automatically keep anything that you didn't think of). A solution using list comprehension.

import string
 
test_lines="""groo**vy
office space."""
words = test_lines.split()
## print while testing
print words
print string.letters
 
for word in words:
    new_word =[chr for chr in word if chr in string.letters]
    print "".join(new_word)