Extracting hh:mm:ss from a column in a dataframe

Question

I am trying to extract hh:mm:ss from a column

Column1 Column 2 Column 3
1        2     12:30:11
2        3     1904-01-01 13:40:20
3        4     20:03:04

In column three there are some entries starting with 1904-01-01. I just want to get hh:mm:ss from the column.

does this help? -- https://stackoverflow.com/q/16138744/8279585 — samkart, Feb 22 '22 at 12:45

score 1 · Accepted Answer · answered Feb 22 '22 at 12:44

If values are strings filter last 8 values:

df['Column 3'] = df['Column 3'].str[-8:]
print (df)
   Column1 Column 2  Column 3
1                 2  12:30:11
2                 3  13:40:20
3                 4  20:03:04

Or convert to datetimes and get HH:MM:SS format:

df['Column 3'] = pd.to_datetime(df['Column 3']).dt.strftime('%H:%M:%S')

score 0 · Answer 2 · answered Feb 22 '22 at 13:01

A safe way is to use a regex, thus if you have NaN or non-aligned strings (e.g. trailing space), you'll get exactly what you want:

df['Column 3'] = df['Column 3'].str.extract('(\d{2}:\d{2}:\d{2})')

output:

   Column1  Column 2  Column 3
0        1         2  12:30:11
1        2         3  13:40:20
2        3         4  20:03:04

Extracting hh:mm:ss from a column in a dataframe

2 Answers2