How to Remove repetitive characters from words of the given Pandas DataFrame using Regex?
Prerequisite: Regular Expression in Python
In this article, we will see how to remove continuously repeating characters from the words of the given column of the given Pandas Dataframe using Regex.
Here, we are actually looking for continuously occurring repetitively coming characters for that we have created a pattern that contains this regular expression (\w)\1+ here \w is for character, 1+ is for the characters that come more than once.
We are passing our pattern in the re.sub() function of re library.
Syntax: re.sub(pattern, repl, string, count=0, flags=0)
The ‘sub’ in the function stands for SubString, a certain regular expression pattern is searched in the given string(3rd parameter), and upon finding the substring pattern is replaced by repl(2nd parameter), count checks and maintains the number of times this occurs.
Now, Let’s create a Dataframe:
Python3
# importing required libraries import pandas as pd import re # creating Dataframe with column # as name and common_comments df = pd.DataFrame( { 'name' : [ 'Akash' , 'Ayush' , 'Diksha' , 'Priyanka' , 'Radhika' ], 'common_comments' : [ 'hey buddy meet me today ' , 'sorry bro i cant meet' , 'hey akash i love geeksforgeeks' , 'twiiter is the best way to comment' , 'geeksforgeeks is good for learners' ] }, columns = [ 'name' , 'common_comments' ] ) # printing Dataframe df |
Output:
Now, Remove continuously repetitive characters from words of the Dataframe common_comments column.
Python3
# define a function to remove # continuously repeating character # from the word def conti_rep_char(str1): tchr = str1.group( 0 ) if len (tchr) > 1 : return tchr[ 0 : 1 ] # define a function to check # whether unique character # is present or not def check_unique_char(rep, sent_text): # regular expression for # repetition of characters convert = re.sub(r '(\w)\1+' , rep, sent_text) # returning the converted word return convert df[ 'modified_common_comments' ] = df[ 'common_comments' ]. apply ( lambda x : check_unique_char(conti_rep_char, x)) # show Dataframe df |
Output:
Time Complexity : O(n) where n is the number of elements in the dataframe.
Space complexity : O(m * n) where m is the number of columns and n is the number of elements in the dataframe