Removing punctuation from only the beginning and end of each element in a list in python

Question

I'm fairly new to python (and this community), this is a question branching off of a question asked and answered from a long time ago from here

With a list like:

['hello', '...', 'h3.a', 'ds4,']

Creating a new list x with no punctuation (and deleting empty elements) would be:

x = [''.join(c for c in s if c not in string.punctuation) for s in x]
x = [s for s in x if s]
print(x)

Output:

['hello', 'h3a', 'ds4']

However, how would I be able to remove all punctuation only from the beginning and end of each element? I mean, to instead output this:

['hello', 'h3.a', 'ds4']

In this case, keeping the period in the h3a but removing the comma at the end of the ds4.

Answer 1

You could use regular expressions. re.sub() can replace all matches of a regex with a string.

import re
X = ['hello', '.abcd.efg.', 'h3.a', 'ds4,']
X_rep = [re.sub(r"(^[^\w]+)|([^\w]+$)", "", x) for x in X] 
print(X_rep)
# Output: ['hello', 'abcd.efg', 'h3.a', 'ds4']

Explanation of regex: Try it

• (^[^\\w]+) :
  • ^ : Beginning of string
  • [^\\w]+ : One or more non-word characters
• | : The previous expression, or the next expression
• ([^\\w]+$) :
  • [^\\w]+ : One or more non-word characters
  • $ : End of string

Answer 2

x = ['hello', '...', 'h3.a', 'ds4,']
x[0] = [''.join(c for c in s if c not in string.punctuation) for s in x][0]
x[(len(x)-1)] = [''.join(c for c in s if c not in string.punctuation) for s in x][(len(x)-1)]
x = [s for s in x if s]
print(x)