CODEWITHSUNDEEP
pythonregexstringpandas
Pranshu
Pranshu

Reputation: 23

Combine multiple regex expressions in pandas.DataFrame.str.replace?

I've got a column in dataframe I want to clean up by removing the brackets.

1                          Auburn (Auburn University)[1]
2                 Florence (University of North Alabama)
3        Jacksonville (Jacksonville State University)[2]
4             Livingston (University of West Alabama)[2]
5               Montevallo (University of Montevallo)[2]
6                              Troy (Troy University)[2]
7      Tuscaloosa (University of Alabama, Stillman Co...
8                      Tuskegee (Tuskegee University)[5]
10         Fairbanks (University of Alaska Fairbanks)[2]
12            Flagstaff (Northern Arizona University)[6]

I used unitowns['City'].str.replace('\(.*\)','').str.replace('\[.*\]','') to get the intended result as follows-

1                            Auburn 
2                          Florence 
3                      Jacksonville 
4                        Livingston 
5                        Montevallo 
6                              Troy 
7                        Tuscaloosa 
8                          Tuskegee 
10                        Fairbanks 
12                        Flagstaff

Is there a way to combine these expressions? This code does not seem to work -> unitowns['City'].str.replace('(\(.*\)) | (\[.*\])','')

Upvotes: 2

Views: 2237

Answers (1)

cs95
cs95

Reputation: 402473

Option 1
str.extract/str.findall
Rather than removing irrelevant content, why not extract the relevant ones instead?

df.City.str.extract(r'(.*?)(?=\()', expand=False)

Or,

df.City.str.findall(r'(.*?)(?=\()').str[0]


0          Auburn 
1        Florence 
2    Jacksonville 
3      Livingston 
4      Montevallo 
5            Troy 
6      Tuscaloosa 
7        Tuskegee 
8       Fairbanks 
9       Flagstaff 
Name: City, dtype: object

You may also want to get rid of leading/trailing spaces post extraction. You can call str.strip on the result - 

df.City = df.City.str.extract(r'(.*?)(?=\()', expand=False).str.strip()

Or, 

df.City = df.City.str.findall(r'(.*?)(?=\()').str[0].str.strip()

Regex Details

(      # capture group
.*?    # non-greedy matcher
)
(?=    # lookahead
\(     # opening parenthesis
)

Option 2
str.split
If your city names only consist of one word, str.split would also work. 

df.City.str.split('\s', 1).str[0]

0          Auburn
1        Florence
2    Jacksonville
3      Livingston
4      Montevallo
5            Troy
6      Tuscaloosa
7        Tuskegee
8       Fairbanks
9       Flagstaff
Name: City, dtype: object

Option 3
str.replace
Condensing your chained calls, you can use - 

df['City'].str.replace(r'\(.*?\)|\[.*?\]', '').str.strip()

0          Auburn
1        Florence
2    Jacksonville
3      Livingston
4      Montevallo
5            Troy
6      Tuscaloosa
7        Tuskegee
8       Fairbanks
9       Flagstaff
Name: City, dtype: object

Upvotes: 5

Related Questions