CODEWITHSUNDEEP
pythonpandas
valid
valid

Reputation: 63

caluclate duration time without date in a pandas dataframe

I've got a dataframe with a column (t1) giving the time someone usually goes to bed and a column (t2) giving the time someone usually gets up, looking like that:

ID       t1           t2
p001   22:30:00    7:00:00
...
p055   01:20:00    9:00:00
...

I want to create a new column giving the hours spent in bed, computed by those two times. A simple subtraction, like that

Df['hoursinbed'] = Df['t2']- Df['t1']

of course does not work since the starting time is usually on the day before.

I've also tried this:

t1 = pd.to_datetime(Df['t1'])
t2 = pd.to_datetime(Df['t2'])

print (pd.Timedelta(t2-t1).seconds / 60.0)

But probably this works only if the start- and end-time includes a date?

Does anyone have an idea how to solve this? TIA :)

Upvotes: 0

Views: 119

Answers (1)

NYC Coder
NYC Coder

Reputation: 7594

You can do this, add one day to t2 if t1 > 12 hours and then find the difference:

df['t1'] = pd.to_datetime(df['t1'])
df['t2'] = pd.to_datetime(df['t2'])

df['t2'] = np.where(df['t1'].dt.hour > 12, df['t2'] + pd.DateOffset(1), df['t2'])
df['timeinbed'] = df['t2'] - df['t1']

df['t1'] = df['t1'].dt.strftime('%H:%M')
df['t2'] = df['t2'].dt.strftime('%H:%M')

print(df)

     ID     t1     t2       timeinbed
0  p001  22:30  07:00 0 days 08:30:00
1  p055  01:20  09:00 0 days 07:40:00

Upvotes: 1

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