Using regular expression to split into columns

Question

I'm in desperate need of help have a data that I would like to split into columns using regex(python), it must use regular expression

Jan  9 01:04:49 syntax sshd(pam_unix)[21354]: authentication failure; logname= uid=0 euid=0 tty=NODEVssh ruser= rhost=120-123-141-4.hinet-ip.hinet.com  

Jul 10 04:17:11 syntax su(pam_unix)[95367]: session opened for user abc by (uid=0)

May  1 14:06:19 syntax su(pam_unix)[95367]: session closed for user abc

Oct 23 18:08:26 syntax logrotate: ALERT exited abnormally with [1]

Jun 14 21:42:52 syntax su(pam_unix)[95367]: session opened for user cbx by (uid=0)

Supposed output

It's actually from a url, I made it into a pandas dataFrame and tried using re.split but it gives me error

*ValueError: 1 columns passed, passed data had 24 columns*

Hope can I get the output I need?

Answer 1

Use regex as follows

Data

 df=pd.DataFrame({'Text':['Jan  9 01:04:49 syntax sshd(pam_unix)[21354]: authentication failure; logname= uid=0 euid=0 tty=NODEVssh ruser= rhost=120-123-141-4.hinet-ip.hinet.com','Jul 10 04:17:11 syntax su(pam_unix)[95367]: session opened for user abc by (uid=0)','May  1 14:06:19 syntax su(pam_unix)[95367]: session closed for user ab']})

regex= ([A-Za-z]+\s+\d+\s+\d+:\d+:\d+)\s+|(?<=\])[:\s+]+|(?<=[x])\s+

df2=df.Text.str.split('([A-Za-z]+\s+\d+\s+\d+:\d+:\d+)\s+|(?<=\])[:\s+]+|(?<=[x])\s+', n=3, expand=True)

df2.rename(columns=({0:'DROP1',1:'Timestamp', 2:'A1', 3:'DROP', 4:'A2', 5:'DROP2',6:'A3'}),inplace=True)#Rename columns

df2.drop(columns=['DROP2','DROP1','DROP'],inplace=True)#Drop unwanted columns

Basically;

(?<=\])[:\s+]+ Split by space that comes after ]:

or -|

(?<=[x])\s+ Split by space that comes after x

or- |

([A-Za-z]+\s+\d+\s+\d+:\d+:\d+)\s+ Split the timestamp

Outcome