How to Remove Punctuation from a List of Strings in Python
This guide explains how to efficiently remove punctuation marks from strings within a list in Python. We'll cover using str.translate()
, regular expressions with re.sub()
, and basic looping, highlighting the strengths and weaknesses of each approach.
Removing Punctuation with str.translate()
(Recommended)
The str.translate()
method, combined with a pre-built translation table, is the most efficient and recommended way to remove punctuation:
import string
a_list = ['t.u.t.o.r.i.a.l', 're,fer,en,ce', '', 'c:om']
# Create the translation table ONCE, outside the loop/comprehension
translator = str.maketrans('', '', string.punctuation)
new_list = [item.translate(translator) for item in a_list if item] # Added check for empty strings
print(new_list) # Output: ['tutorial', 'reference', 'com']
string.punctuation
: This constant (from thestring
module) provides a string containing all common punctuation characters:!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~
str.maketrans('', '', string.punctuation)
: This creates a translation table. This is a highly optimized lookup table thattranslate()
uses. The arguments mean:- First argument (empty string): We're not replacing any characters with other characters.
- Second argument (empty string): We're not mapping any characters to other characters in a 1:1 fashion.
- Third argument (
string.punctuation
): These are the characters we want to delete.
item.translate(translator)
: This applies the translation table to each string in the list, efficiently removing all punctuation characters.- List comprehension filters out any empty string that might have resulted from removing punctuation, if the original list had empty strings.
Key Advantages of str.translate()
:
- Efficiency:
str.translate()
with a pre-built table is significantly faster than using regular expressions or looping within
. It's implemented in C and highly optimized. - Readability: Once you understand the
maketrans()
call, the code is very clear. - Correctness: It handles Unicode punctuation correctly.
Removing Punctuation with re.sub()
Regular expressions provide a flexible way to remove punctuation, but they are generally slower than str.translate()
. Use re.sub()
if you need to remove a specific, complex pattern of punctuation, not just all punctuation.
import re
a_list = ['t.u.t.o.r.i.a.l', 're,fer,en,ce', 'c:om']
new_list = [re.sub(r'[^\w\s]', '', item) for item in a_list]
print(new_list) # Output: ['tutorial', 'reference', 'com']
-
re.sub(r'[^\w\s]', '', item)
will remove any character which is not alphanumeric or whitespace character, i.e. it will remove all punctuation. -
re.sub(pattern, replacement, string)
: Substitutes all occurrences ofpattern
instring
withreplacement
. -
r'[^\w\s]'
: This regular expression matches any character that is not (^
inside[]
means "not") a word character (\w
- letters, numbers, and underscore) or whitespace (\s
). This effectively matches all punctuation.
Removing Punctuation with a for
Loop and string.punctuation
(Least Efficient)
You can use a nested for
loop and check each character against string.punctuation
, but this is the least efficient and least readable method:
import string
a_list = ['t.u.t.o.r.i.a.l', 're,fer,en,ce', '', 'c:om']
new_list = [''.join(char for char in item
if char not in string.punctuation)
for item in a_list if item != '']
print(new_list) # Output: ['tutorial', 'reference', 'com']
- We use a list comprehension which checks if a character is punctuation, and creates a list with the valid characters.
''.join(...)
joins the filtered list of characters back into a single string.