I have a DF column which has many strings in it. I need to remove all non-alpha numeric characters from that column: i.e:
df['strings'] = ["a#bc1!","a(b$c"]
Run code:
Print(df['strings']): ['abc','abc']
I've tried:
df['strings'].replace([',','.','/','"',':',';','!','@','#','$','%',"'","*","(",")","&",],"")
But this didn't work and I feel that there should be a more efficient way to do this using regex. Any help would be very appreciated.
Use str.replace.
df
strings
0 a#bc1!
1 a(b$c
df.strings.str.replace('[^a-zA-Z]', '')
0 abc
1 abc
Name: strings, dtype: object
To retain alphanumeric characters (not just alphabets as your expected output suggests), you'll need:
df.strings.str.replace('\W', '')
0 abc1
1 abc
Name: strings, dtype: object
[^0-9a-zA-Z.,-/ ] was what i was after personallySince you wrote alphanumeric, you need to add 0-9 in the regex. But maybe you only wanted alphabetic...
import pandas as pd
ded = pd.DataFrame({'strings': ['a#bc1!', 'a(b$c']})
ded.strings.str.replace('[^a-zA-Z0-9]', '')
But it's basically what COLDSPEED wrote
You can also use regex
import re
regex = re.compile('[^a-zA-Z]')
l = ["a#bc1!","a(b$c"]
print [regex.sub('', i) for i in l]
['abc', 'abc']
There has been an update with the Pandas str.replace() method since the final top answer was written in 2017, now you need to specify that the pattern that you are passing is a regular expression (regex) as now the default parameter boolean value is False, so the updated code from the answer above should be:
df
strings
0 a#bc1!
1 a(b$c
df.strings.str.replace('[^a-zA-Z]', '', regex=True)
#the regex parameter must be set to True for this to work post 2017.
# Additionally the regex parameter [^a-zA-Z] is saying to the computer:
# Match a single character not present in the list below, which is a-z and A-Z.
0 abc
1 abc
Name: strings, dtype: object
