Remove periods from string python

Many times while working with Python strings, we have a problem in which we need to remove certain characters from strings. This can have applications in data preprocessing in the Data Science domain and also in day-day programming. Let’s discuss certain ways in which we can perform this task using Python.

Method 1: Remove Punctuation from a String with Translate

The first two arguments for string.translate method is empty strings, and the third input is a Python list of the punctuation that should be removed. This instructs the Python method to eliminate punctuation from a string. This is one of the best ways to strip punctuation from a string.

Python3

import string

test_str = 'Gfg, is best: for ! Geeks ;'

test_str = test_str.translate

    (str.maketrans('', '', string.punctuation))

print(test_str)

Output:

Gfg is best for  Geeks 

Method 2: Remove Punctuation from a String with Python loop

This is the brute way in which this task can be performed. In this, we check for the punctuations using a raw string that contain punctuations and then we construct a string removing those punctuations.

Python3

test_str = "Gfg, is best : for ! Geeks ;"

print("The original string is : " + test_str)

punc =

for ele in test_str:

    if ele in punc:

        test_str = test_str.replace(ele, "")

print("The string after punctuation filter : " + test_str)

Output: 

The original string is : Gfg, is best : for ! Geeks ;
The string after punctuation filter : Gfg is best  for  Geeks 

Method 3: Remove Punctuation from a String with regex 

The part of replacing with punctuation can also be performed using regex. In this, we replace all punctuation with an empty string using a certain regex.

Python3

import re

test_str = "Gfg, is best : for ! Geeks ;"

print("The original string is : " + test_str)

res = re.sub(r'[^\w\s]', '', test_str)

print("The string after punctuation filter : " + res)

Output : 

The original string is : Gfg, is best : for ! Geeks ;
The string after punctuation filter : Gfg is best  for  Geeks 

Method 4:  Using for loop, punctuation string and not in operator

Python3

test_str = "Gfg, is best : for ! Geeks ;"

print("The original string is : " + test_str)

punc =

res=" "

for ele in test_str:

    if ele not in punc:

        res+=ele

print("The string after punctuation filter : " + res)

Output

The original string is : Gfg, is best : for ! Geeks ;
The string after punctuation filter :  Gfg is best  for  Geeks 

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Auxiliary Space: O(n)

From an efficiency perspective, you're not going to beat

s.translate(None, string.punctuation)

For higher versions of Python use the following code:

s.translate(str.maketrans('', '', string.punctuation))

It's performing raw string operations in C with a lookup table - there's not much that will beat that but writing your own C code.

If speed isn't a worry, another option though is:

exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)

This is faster than s.replace with each char, but won't perform as well as non-pure python approaches such as regexes or string.translate, as you can see from the below timings. For this type of problem, doing it at as low a level as possible pays off.

Timing code:

import re, string, timeit

s = "string. With. Punctuation"
exclude = set(string.punctuation)
table = string.maketrans("","")
regex = re.compile('[%s]' % re.escape(string.punctuation))

def test_set(s):
    return ''.join(ch for ch in s if ch not in exclude)

def test_re(s):  # From Vinko's solution, with fix.
    return regex.sub('', s)

def test_trans(s):
    return s.translate(table, string.punctuation)

def test_repl(s):  # From S.Lott's solution
    for c in string.punctuation:
        s=s.replace(c,"")
    return s

print "sets      :",timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000)
print "regex     :",timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000)
print "translate :",timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000)
print "replace   :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000)

This gives the following results:

sets      : 19.8566138744
regex     : 6.86155414581
translate : 2.12455511093
replace   : 28.4436721802

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