Pandas split after month day time from rest of string

You can also pass the exact form of date\time:

df['data'].str.extract('(\w* \d* \d*:\d*:\d*) (.*)')

output:

                 0        1
0  Oct 22 12:56:52  server1
1  Oct 22 12:56:52  server2
2  Oct 22 12:56:53  server2
3  Oct 22 12:56:54  server2
4  Oct 22 12:56:56    comp2

Using positive lookbehind to split on {semicolon}{two numbers}{space}:

Details:

• (?<=) is positive lookbehind (check if anything is before the string)
• :\d{2} is pattern: {semicolon}{two numbers}
• \s is white space

Conclusion: we split on a whitespace but only if its preceeded by the pattern described above.

s = df['data'].str.split('(?<=:\d{2})\s')

df['date'] = s.str[0]
df['machine'] = s.str[1]
df = df.drop(columns='data')

Or as piRSquared & jezrael suggest in the comments, in a one-liner:

df['date'], df['machine'] = zip(*df.pop('data').str.split('(?<=:\d{2})\s'))

Output

              date  machine
0  Oct 22 12:56:52  server1
1  Oct 22 12:56:52  server2
2  Oct 22 12:56:53  server2
3  Oct 22 12:56:54  server2
4  Oct 22 12:56:56    comp2

comprehension

This depends on the data format always being 15 characters.
Also, since we're going to have to drop a column 'data' anyway, I thought it would be better to simply create a dataframe from scratch.

pd.DataFrame([[s[:15], s[16:]] for s in df.data], columns=['date', 'machine'])

              date  machine
0  Oct 22 12:56:52  server1
1  Oct 22 12:56:52  server2
2  Oct 22 12:56:53  server2
3  Oct 22 12:56:54  server2
4  Oct 22 12:56:56    comp2

rsplit

Depends on 'machine' name never having spaces.

This works because the string accessor provided by pandas.Series.str is an iterable and can be used in an assignment statement similar to x, y = (1, 2)

Also note that I unapologetically took the idea to use pop in this instance from @jezrael

df['date'], df['machine'] = df.pop('data').str.rsplit(n=1).str

df

              date  machine
0  Oct 22 12:56:52  server1
1  Oct 22 12:56:52  server2
2  Oct 22 12:56:53  server2
3  Oct 22 12:56:54  server2
4  Oct 22 12:56:56    comp2

If the format remains the same:

df['date'] = df['data'].str[:-8]
df['machine'] = df['data'].str[-8:]

print(df)

                      data             date   machine
0  Oct 22 12:56:52 server1  Oct 22 12:56:52   server1
1  Oct 22 12:56:52 server2  Oct 22 12:56:52   server2

Try this:

r"(?<=[\S ][\d]{2}:[\d]{2}:[\d]{2} )[\S ]+"

I'm not the best with regex there is definitely a better approach but this works