Python, change email to name format-CodePudding

How can I parse the below email options to just expected output. These are not in a dataframe, they are separate strings. I have a loop that loops through each string.

example input

[email protected]
[email protected]
[email protected]
[email protected]  
[email protected]
[email protected]

expected output:

Louis Stevens
Louis Stevens
Louis Stevens
Louis Stevens
Mike Williams
Lebron James

Thanks

CodePudding user response：

Assuming s the input Series, and using str.replace:

import re

s.str.replace(r'^([a-z] )\.(?:.\.)?([a-z] ).*', r'\1 \2', regex=True, flags=re.I)

Output:


0    Louis Stevens
1    Louis Stevens
2    Louis Stevens
3    Louis Stevens
4    Mike Williams
5     Lebron James
dtype: object

For individual strings:

import re

s = '[email protected]'

out = re.sub(r'^([a-z] )\.(?:.\.)?([a-z] ).*', r'\1 \2', s, flags=re.I)

CodePudding user response：

Remove everything after @ with regex @.*:

s = pd.Series("""[email protected]
[email protected]
[email protected]
[email protected]""".splitlines())

s.str.replace('@.*', '', regex=True)
#0      Louis.Stevens
#1    Louis.a.Stevens
#2      Louis.Stevens
#3     Louis.Stevens2
#dtype: object

CodePudding user response：

Use regex's findall to extract the alpha numerics at the start of sentence and the alphanumeric immediately before @ . Then proceed and replace digits with nothing. Code below

    email
0        [email protected]
1      [email protected]
2  [email protected]
3       [email protected]
4       [email protected]
5       [email protected]






  df= df.assign(email_new =df['email'].str.findall('^\w |\w (?=\@)').str.join(' ').str.replace('\d','', regex=True))

              

                           email      email_new
0        [email protected]  Louis Stevens
1      [email protected]  Louis Stevens
2  [email protected]  Louis Stevens
3       [email protected]  Louis Stevens
4       [email protected]  Mike Williams
5       [email protected]   Lebron James