0

I have a pandas DataFrame with the following column (first column = index):

0         14:43:45:921

1         14:43:45:923

2         14:43:45:925

I would like to modify this column, or add another column with the time starting at 0:

0         00:00:00.000

1         00:00:00.002

2         00:00:00.004

So far, I've tried the following code:

df['time'].apply(pd.Timedelta)

This is giving me the following error:

expected hh:mm:ss format

To me, the problem is

a) convert the time format HH:MM:SS:fff to HH:MM:SS.fff and

b) get the timedelta function to work.

Does anyone have any suggestions? Thanks!

Tamil Selvan
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M195
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1 Answers1

1

Use to_datetime:

s = pd.to_datetime(df['time'], format='%H:%M:%S:%f')

Or Series.str.replace with to_timedelta:

s = pd.to_timedelta(df['time'].str.replace('(:)(\d+)$', r'.\2'))

And then subtract first value:

df['new'] = s.sub(s.iat[0])
print (df)
           time                    new
0  14:43:45:921        0 days 00:00:00
1  14:43:45:923 0 days 00:00:00.002000
2  14:43:45:925 0 days 00:00:00.004000

If need times:

df['new'] = s.sub(s.iat[0])
df['new1'] = df['new'].apply(lambda x: (pd.datetime.min + x).time())

print (df)
           time                    new             new1
0  14:43:45:921        0 days 00:00:00         00:00:00
1  14:43:45:923 0 days 00:00:00.002000  00:00:00.002000
2  14:43:45:925 0 days 00:00:00.004000  00:00:00.004000


print (type(df.at[0, 'new']))
<class 'pandas._libs.tslibs.timedeltas.Timedelta'>

print (type(df.at[0, 'new1']))
<class 'datetime.time'>
jezrael
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