> ## Content Index > Fetch the complete content index at: https://datascientyst.com/llms.txt > Use this file to discover other available public pages before exploring further. # How to Replace Regex Groups in Pandas - URL: https://datascientyst.com/replace-regex-groups-in-pandas/ - Published: 2021-11-03T14:32:42.000Z - Updated: 2021-11-03T14:32:42.000Z - Author: John D K - Tags: Regex In this short tutorial, we'll look at how to **match and replace regex groups in Pandas.** Here you can find the short answer: ```python df_e['Date'].str.replace(r'(\d{2})/(\d{2})/(\d{4})', r"\3-\2-\1", regex=True) ``` ## How to match and replace regex groups in Pandas Let's check first for the regex pattern `r'(\d{2})/(\d{2})/(\d{4})'`. It has 3 capturing groups:: - 1st capturing group `(\d{2})` - `\d` \- matches a digit (equivalent to \[0-9\]) - {2} matches the previous token exactly 2 times - 2nd capturing group `(\d{2})` - `\d` \- matches a digit (equivalent to \[0-9\]) - {2} matches the previous token exactly 2 times - 4nd capturing group `(\d{4})` - `\d` \- matches a digit (equivalent to \[0-9\]) - {4} matches the previous token exactly 4 times Between the capturing groups there are separators - `/` which will be matched but not captured. Now in the replacement part - `r"\3-\2-\1"` \- the groups are presented by numbers - starting from 1: - 1st group is `\1` - 2nd group is `\2` - 3rd group is `\3` So if the initial value in Pandas column is: ``` 0 01/02/1965 1 01/04/1965 2 01/05/1965 3 01/08/1965 4 01/09/1965 ``` After the replacement we will end with: ``` 0 1965-02-01 1 1965-04-01 2 1965-05-01 3 1965-08-01 4 1965-09-01 ``` Another example is: ```python df_e['Date'].str.replace(r'(\d{2})/(\d{2})/\d{2}(\d{2})', r"\2 \1 '\3", regex=True) ``` result will be: ``` 0 02 01 '65 1 04 01 '65 2 05 01 '65 3 08 01 '65 4 09 01 '65 ``` ![](https://datascientyst.com/content/images/2021/11/replace-regex-groups-in-pandas.png) ## How to match and replace regex groups - string patterns Let's have string data which has a different number of words. Suppose we like to extract only the last word from all rows: ``` 0 Noida 1 Noida 2 Work From Home 3 Work From Home 4 Work From Home ... ``` We can use regex groups to match the last word by: ```python df['location'].str.replace(r'(.*) (.*) (.*)', r"\3", regex=True) ``` result: ``` 0 Noida 1 Noida 2 Home 3 Home 4 Home ``` If you need to work with special characters and complex expressions - you may need to test a lot - in order to avoid problems. Let's have data like: ``` 0 Software Testing 1 Java, SQL, Unix, Oracle, MS SQL Server, Hibern... 2 English Proficiency (Spoken), English Proficie... 3 HTML, CSS, Flask, Python, Django 4 HTML, CSS, JavaScript, ReactJS, Redux ``` For example if you expect: ```python df['skills'].str.replace(r'(.*)(\(.*\))(.*)', r"\1\3", regex=True) ``` The regex above to remove all words surrounded by `(` and `)` \- this will not happen: ``` 0 Software Testing 1 Java, SQL, Unix, Oracle, MS SQL Server, Hibern... 2 English Proficiency (Spoken), English Proficie... 3 HTML, CSS, Flask, Python, Django 4 HTML, CSS, JavaScript, ReactJS, Redux ``` because the regex are greedy by default and will try to match as much as possible. You can check what is captured as group 2: ```python df['skills'].str.replace(r'(.*)(\(.*\))(.*)', r"\2", regex=True) ``` So as you can if there are multiple matches - the regex will try to match everything until the end - or matching the last one possible: ``` 0 Software Testing 1 (Java) 2 (Written) 3 HTML, CSS, Flask, Python, Django 4 HTML, CSS, JavaScript, ReactJS, Redux ``` For non greedy matching in Pandas regex groups you can add `?`. So: - `(.*)` \- greedy match - `(.*)` \- non greedy match ```python df['skills'].str.replace(r'(.*?)(\(.*?\))(.*)', r"\2", regex=True) ``` result: ``` 0 Software Testing 1 (Java) 2 (Spoken) 3 HTML, CSS, Flask, Python, Django 4 HTML, CSS, JavaScript, ReactJS, Redux ``` Note that the third line changed from `(Written)` to `(Spoken)`. ## How to match and replace regex groups - numeric patterns Suppose we have ranged amounts like: ``` 0 8000 /month 1 10000 /month 2 1000-2000 /month 3 1000-2000 /month 4 1000-2000 /month ``` And we need to get only the upper amount from this range and leave the rest as it is. This can be achieved by: ```python df['stipend'].str.replace(r'(\d+)-(\d+)(.*)', r"\2\3", regex=True) ``` result: ``` 0 8000 /month 1 10000 /month 2 2000 /month 3 2000 /month 4 2000 /month ``` ## Resources - [Notebook](https://github.com/softhints/Pandas-Tutorials/blob/master/regex/replace-regex-groups-in-pandas.ipynb?ref=datascientyst.com) - [pandas.Series.str.replace](https://pandas.pydata.org/docs/reference/api/pandas.Series.str.replace.html?ref=datascientyst.com) - [pandas.Series.replace](https://pandas.pydata.org/docs/reference/api/pandas.Series.replace.html?ref=datascientyst.com)