from datetime import datetime
start='12:25:03'
format = '%H:%M:%S'
startDateTime = datetime.strptime(start, format)
end='12:30:40'
endDateTime = datetime.strptime(end, format)
diff = endDateTime - startDateTime
print(diff)
0:05:37
Above code works fine, but when I apply the same to entire column using lambda function I get result in different format, I would like to get values of Diff column in hh:mm:ss format.
t1 - Object type
t2 - Object type
Diff - timedelta64[ns] type
df["Diff"] = df.apply(lambda x: datetime.strptime(x["t1"], format) - datetime.strptime(x["t2"], format), axis = 1)
df.head()
t1 t2 Diff
0 01:27:19 01:28:58 -1 days +23:58:21
1 01:49:57 01:50:40 -1 days +23:59:17
2 03:35:24 03:36:14 -1 days +23:59:10
related: Format timedelta to string
You can write your own formatter for the timedelta objects, e.g.
def formatTimedelta(td):
"""
format a timedelta object to string, in HH:MM:SS format (seconds floored).
negative timedeltas will be prefixed with a minus, '-'.
"""
total = td.total_seconds()
prefix, total = ('-', total*-1) if total < 0 else ('', total)
h, r = divmod(total, 3600)
m, s = divmod(r, 60)
return f"{prefix}{int(h):02d}:{int(m):02d}:{int(s):02d}"
which would give you for the example df
df
t1 t2
0 01:27:19 01:28:58
1 01:49:57 01:50:40
2 03:35:24 03:36:14
# to datetime
df['t1'] = pd.to_datetime(df['t1'])
df['t2'] = pd.to_datetime(df['t2'])
# calculate timedeltas and format
df['diff0'] = (df['t1']-df['t2']).apply(formatTimedelta)
df['diff1'] = (df['t2']-df['t1']).apply(formatTimedelta)
df['diff0']
0 -00:01:39
1 -00:00:43
2 -00:00:50
Name: diff0, dtype: object
df['diff1']
0 00:01:39
1 00:00:43
2 00:00:50
Name: diff1, dtype: object