Get the time spent since midnight in dataframe - pandas

Question

I have a dataframe which has a column of type Timestamp. I want to find the time elapsed (in seconds) since midnight as a new column. How to do it in a simple way ?
Eg :
Input :
samples['time']
2018-10-01 00:00:01.000000000
2018-10-01 00:00:12.000000000
type(samples['time'].iloc[0])
<class 'pandas._libs.tslib.Timestamp'>
Output :
samples['time_elapsed']
1
12

Answer 1

Current answers either too complicated or specialized.
samples = pd.DataFrame(data=['2018-10-01 00:00:01', '2018-10-01 00:00:12'], columns=['time'], dtype='datetime64[ns]')
samples['time_elapsed'] = ((samples['time'] - samples['time'].dt.normalize()) / pd.Timedelta('1 second')).astype(int)
print(samples)
time time_elapsed
0 2018-10-01 00:00:01 1
1 2018-10-01 00:00:12 12
normalize() removes the time component from the datetime (moves clock back to midnight).
pd.Timedelta('1 s') sets the unit of measurement, i.e. number of seconds in the timedelta.
.astype(int) casts the decimal number of seconds to int. Use round functionality if that is preferred.

Answer 2

Note that the date part in each row may be other (not from one
and the same day), so you can not take any "base date" (midnight)
for the whole DataFrame, as it can be seen in one of other solutions.
My intention was also not to "contaminate" the source DataFrame
with any intermediate columns, e.g. the time (actually date and time)
as string converted to "true" DateTime.
Then my proposition is:
convert the DateTime string to DateTime,
take the time part from it,
compute the number of seconds from hour / minute / second
part.
All the above steps in a dedicated function.
So to do the task, define a function:
def secSinceMidnight(datTimStr):
tt = pd.to_datetime(datTimStr).time()
return tt.hour * 3600 + tt.minute * 60 + tt.second
Then call:
samples['Secs'] = samples.time.apply(secSinceMidnight)
For source data:
samples = pd.DataFrame(data=[
[ '2018-10-01 00:00:01' ], [ '2018-10-01 00:00:12' ],
[ '2018-11-02 01:01:10' ], [ '2018-11-04 03:02:15' ] ],
columns = ['time']);
when you print the result, you will see:
time Secs
0 2018-10-01 00:00:01 1
1 2018-10-01 00:00:12 12
2 2018-11-02 01:01:10 3670
3 2018-11-04 03:02:15 10935

Answer 3

Doing this in Pandas is very simple!
midnight = pd.Timestamp('2018-10-01 00:00:00')
print(pd.Timestamp('2018-10-01 00:00:01.000000000') - midnight).seconds
>
1
And by extension we can use an apply on a Pandas Series:
samples = pd.DataFrame(['2018-10-01 00:00:01.000000000', '2018-10-01 00:00:12.000000000'], columns=['time'])
samples.time = pd.to_datetime(samples.time)
midnight = pd.Timestamp('2018-10-01 00:00:00')
samples['time_elapsed'] = samples['time'].apply(lambda x: (x - midnight).seconds)
samples
>
time time_elapsed
0 2018-10-01 00:00:01 1
1 2018-10-01 00:00:12 12
Note that the answers here use an alternative method: comparing the timestamp to itself converted to a date. This zeros all time data and so is the equivalent of midnight of that day. This method might be slightly more performant.

Answer 4

I ran into the same problem in my one of my projects and here's how I solved it (assuming your time column has already been converted to Timestamp):
(samples['time'] - samples['time'].dt.normalize()) / pd.Timedelta(seconds=1)
The beauty of this approach is that you can change the last part to get seconds, minutes, hours or days elapsed:
... / pd.Timedelta(seconds=1) # seconds elapsed
... / pd.Timedelta(minutes=1) # minutes elapsed
... / pd.Timedelta(hours=1) # hours elapsed
... / pd.Timedelta(days=1) # days elapsed

Answer 5

datetime = samples['time']
(datetime - datetime.dt.normalize()).dt.total_seconds()

Answer 6

We can do :
samples['time'].dt.hour * 3600 +
samples['time'].dt.minute * 60 +
samples['time'].dt.second