How To Add Integer (seconds) To Hh:mm:ss Format In Python?
I have the following data frame in python where am trying to compute column 'New time' by adding 'Duration' (which is in seconds) to the 'start_time' Serial start_date start_t
Solution 1:
You can use to_timedelta and output is timedelta too:
df['New time'] = pd.to_timedelta(df['start_time']) +
pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
Serial start_date start_time Duration(seconds) New time
0 A 5/22/2017 10:37:24 216 10:41:00
1 A 5/22/2017 10:37:26 213 10:40:59
2 A 5/22/2017 10:37:29 3 10:37:32
3 A 5/22/2017 10:39:55 60 10:40:55
4 A 5/22/2017 10:51:50 380 10:58:10
5 A 5/22/2017 10:51:57 339 10:57:36
But if seconds is more, output is changed, because there are also days:
print (df)
Serial start_date start_time Duration(seconds)
0 A 5/22/2017 10:37:24 216
1 A 5/22/2017 10:37:26 213000
2 A 5/22/2017 10:37:29 3
3 A 5/22/2017 10:39:55 60
4 A 5/22/2017 10:51:50 380
5 A 5/22/2017 10:51:57 339
df['New time'] = pd.to_timedelta(df['start_time']) +
pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
Serial start_date start_time Duration(seconds) New time
0 A 5/22/2017 10:37:24 216 0 days 10:41:00
1 A 5/22/2017 10:37:26 213000 2 days 21:47:26
2 A 5/22/2017 10:37:29 3 0 days 10:37:32
3 A 5/22/2017 10:39:55 60 0 days 10:40:55
4 A 5/22/2017 10:51:50 380 0 days 10:58:10
5 A 5/22/2017 10:51:57 339 0 days 10:57:36
Also is posible add datetime:
df['New date'] = pd.to_datetime(df['start_date']) + \
pd.to_timedelta(df['start_time']) + \
pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
Serial start_date start_time Duration(seconds) New date
0 A 5/22/2017 10:37:24 216 2017-05-22 10:41:00
1 A 5/22/2017 10:37:26 213 2017-05-22 10:40:59
2 A 5/22/2017 10:37:29 3 2017-05-22 10:37:32
3 A 5/22/2017 10:39:55 60 2017-05-22 10:40:55
4 A 5/22/2017 10:51:50 380 2017-05-22 10:58:10
5 A 5/22/2017 10:51:57 339 2017-05-22 10:57:36
df['New date'] = pd.to_datetime(df['start_date']) + \
pd.to_timedelta(df['start_time']) + \
pd.to_timedelta(df['Duration(seconds)'], unit='s')
print (df)
Serial start_date start_time Duration(seconds) New date
0 A 5/22/2017 10:37:24 216 2017-05-22 10:41:00
1 A 5/22/2017 10:37:26 213000 2017-05-24 21:47:26
2 A 5/22/2017 10:37:29 3 2017-05-22 10:37:32
3 A 5/22/2017 10:39:55 60 2017-05-22 10:40:55
4 A 5/22/2017 10:51:50 380 2017-05-22 10:58:10
5 A 5/22/2017 10:51:57 339 2017-05-22 10:57:36
---
And if need convert timedelta to string in format HH:MM:SS and lost days (if exist):
df['New time'] = pd.to_timedelta(df['start_time']) +
pd.to_timedelta(df['Duration(seconds)'], unit='s')
df['New time'] = pd.to_datetime(df['New time']).dt.strftime('%H:%M:%S')
print (df)
Serial start_date start_time Duration(seconds) New time
0 A 5/22/2017 10:37:24 216 10:41:00
1 A 5/22/2017 10:37:26 213000 21:47:26
2 A 5/22/2017 10:37:29 3 10:37:32
3 A 5/22/2017 10:39:55 60 10:40:55
4 A 5/22/2017 10:51:50 380 10:58:10
5 A 5/22/2017 10:51:57 339 10:57:36
Solution 2:
Here is a snippet that may help you solve the problem yourself:
from datetime import datetime, timedeltamy_date= datetime.strptime('5/22/2017 10:37:24', '%m/%d/%Y %H:%M:%S')
my_time_diff = timedelta(seconds=216)
my_new_date = my_date + my_time_diff
print(my_new_date.strftime('%m/%d/%Y %H:%M:%S'))
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