Pandas DataFrame Calculate time difference between 2 columns on specific time range
-1
I want to calculate time difference between two columns on specific time range.
I try df.between_time but it only works on index.
Ex. Time range: between 18.00 - 8.00
Data :
start stop
0 2018-07-16 16:00:00 2018-07-16 20:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00
2 2018-07-13 17:54:00 2018-07-13 21:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00
4 2018-07-20 00:21:00 2018-07-20 04:21:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00
Expect Result :
start stop time_diff
0 2018-07-16 16:00:00 2018-07-16 20:00:00 02:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00 0
2 2018-07-13 17:54:00 2018-07-13 21:54:00 03:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00 0
4 2018-07-20 00:21:00 2018-07-20 04:21:00 04:00:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00 14:00:00
Note: If time_diff > 1 days, I already deal with that case.
Question: Should I build a function to do this or there are pandas build-in function to do this? Any help or guide would be appreciated.
python pandas datetime
asked Nov 12 at 7:43
yolox
438
|
show 3 more comments
-1
I want to calculate time difference between two columns on specific time range.
I try df.between_time but it only works on index.
Ex. Time range: between 18.00 - 8.00
Data :
start stop
0 2018-07-16 16:00:00 2018-07-16 20:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00
2 2018-07-13 17:54:00 2018-07-13 21:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00
4 2018-07-20 00:21:00 2018-07-20 04:21:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00
Expect Result :
start stop time_diff
0 2018-07-16 16:00:00 2018-07-16 20:00:00 02:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00 0
2 2018-07-13 17:54:00 2018-07-13 21:54:00 03:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00 0
4 2018-07-20 00:21:00 2018-07-20 04:21:00 04:00:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00 14:00:00
Note: If time_diff > 1 days, I already deal with that case.
Question: Should I build a function to do this or there are pandas build-in function to do this? Any help or guide would be appreciated.
python pandas datetime
asked Nov 12 at 7:43
yolox
438
What isdf.dtypes?
– timgeb
Nov 12 at 7:49
1
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
1
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59
|
show 3 more comments
-1
-1
-1
1
I want to calculate time difference between two columns on specific time range.
I try df.between_time but it only works on index.
Ex. Time range: between 18.00 - 8.00
Data :
start stop
0 2018-07-16 16:00:00 2018-07-16 20:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00
2 2018-07-13 17:54:00 2018-07-13 21:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00
4 2018-07-20 00:21:00 2018-07-20 04:21:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00
Expect Result :
start stop time_diff
0 2018-07-16 16:00:00 2018-07-16 20:00:00 02:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00 0
2 2018-07-13 17:54:00 2018-07-13 21:54:00 03:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00 0
4 2018-07-20 00:21:00 2018-07-20 04:21:00 04:00:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00 14:00:00
Note: If time_diff > 1 days, I already deal with that case.
Question: Should I build a function to do this or there are pandas build-in function to do this? Any help or guide would be appreciated.
python pandas datetime
asked Nov 12 at 7:43
yolox
438
I want to calculate time difference between two columns on specific time range.
I try df.between_time but it only works on index.
Ex. Time range: between 18.00 - 8.00
Data :
start stop
0 2018-07-16 16:00:00 2018-07-16 20:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00
2 2018-07-13 17:54:00 2018-07-13 21:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00
4 2018-07-20 00:21:00 2018-07-20 04:21:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00
Expect Result :
start stop time_diff
0 2018-07-16 16:00:00 2018-07-16 20:00:00 02:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00 0
2 2018-07-13 17:54:00 2018-07-13 21:54:00 03:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00 0
4 2018-07-20 00:21:00 2018-07-20 04:21:00 04:00:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00 14:00:00
Note: If time_diff > 1 days, I already deal with that case.
Question: Should I build a function to do this or there are pandas build-in function to do this? Any help or guide would be appreciated.
python pandas datetime
python pandas datetime
asked Nov 12 at 7:43
yolox
438
asked Nov 12 at 7:43
yolox
438
edited Nov 12 at 9:18
asked Nov 12 at 7:43
yolox
438
asked Nov 12 at 7:43
yolox
438
asked Nov 12 at 7:43
yolox
438
438
What isdf.dtypes?
– timgeb
Nov 12 at 7:49
1
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
1
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59
|
show 3 more comments
What isdf.dtypes?
– timgeb
Nov 12 at 7:49
1
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
1
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59
What is
df.dtypes?– timgeb
Nov 12 at 7:49
What is
df.dtypes?– timgeb
Nov 12 at 7:49
1
1
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
1
1
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59
|
show 3 more comments
1 Answer
1
active
oldest
votes
1
I think this can be a solution
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00
answered Nov 12 at 8:38
J. Doe
935421
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row ispd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00')which is16:00:00but withrule18it becomes15:00:00
– J. Doe
Nov 12 at 9:39
because if time exceed08:00:00it should be convert to08:00:00. Thank you for providing the example dataframe.
– yolox
Nov 12 at 10:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
I think this can be a solution
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00
answered Nov 12 at 8:38
J. Doe
935421
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row ispd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00')which is16:00:00but withrule18it becomes15:00:00
– J. Doe
Nov 12 at 9:39
because if time exceed08:00:00it should be convert to08:00:00. Thank you for providing the example dataframe.
– yolox
Nov 12 at 10:02
add a comment |
1
I think this can be a solution
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00
answered Nov 12 at 8:38
J. Doe
935421
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row ispd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00')which is16:00:00but withrule18it becomes15:00:00
– J. Doe
Nov 12 at 9:39
because if time exceed08:00:00it should be convert to08:00:00. Thank you for providing the example dataframe.
– yolox
Nov 12 at 10:02
add a comment |
1
1
1
I think this can be a solution
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00
answered Nov 12 at 8:38
J. Doe
935421
I think this can be a solution
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00
answered Nov 12 at 8:38
J. Doe
935421
answered Nov 12 at 8:38
J. Doe
935421
answered Nov 12 at 8:38
J. Doe
935421
answered Nov 12 at 8:38
J. Doe
935421
935421
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row ispd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00')which is16:00:00but withrule18it becomes15:00:00
– J. Doe
Nov 12 at 9:39
because if time exceed08:00:00it should be convert to08:00:00. Thank you for providing the example dataframe.
– yolox
Nov 12 at 10:02
add a comment |
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row ispd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00')which is16:00:00but withrule18it becomes15:00:00
– J. Doe
Nov 12 at 9:39
because if time exceed08:00:00it should be convert to08:00:00. Thank you for providing the example dataframe.
– yolox
Nov 12 at 10:02
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
Thanks for the answer but I think the last row should be 14:00:00.
– yolox
Nov 12 at 8:49
why do you think that ?
– J. Doe
Nov 12 at 9:09
why do you think that ?
– J. Doe
Nov 12 at 9:09
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
If it in my case I guess it should be like this pd.to_datetime('2018-07-21 08:00:00') - pd.to_datetime('2018-07-20 18:00:00') give 14:00:00
– yolox
Nov 12 at 9:19
in your example the last row is
pd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00') which is 16:00:00 but with rule18 it becomes 15:00:00– J. Doe
Nov 12 at 9:39
in your example the last row is
pd.to_datetime('2018-07-21 09:00:00') - pd.to_datetime('2018-07-20 17:00:00') which is 16:00:00 but with rule18 it becomes 15:00:00– J. Doe
Nov 12 at 9:39
because if time exceed
08:00:00 it should be convert to 08:00:00. Thank you for providing the example dataframe.– yolox
Nov 12 at 10:02
because if time exceed
08:00:00 it should be convert to 08:00:00. Thank you for providing the example dataframe.– yolox
Nov 12 at 10:02
add a comment |
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What is
df.dtypes?– timgeb
Nov 12 at 7:49
1
@timgeb I convert it to datetime by using pd.to_datetime
– yolox
Nov 12 at 7:51
1
Ok. I don't understand how you get to those expected valules. For example, the times in the first row are 4 hours apart, how do you get 2?
– timgeb
Nov 12 at 7:53
@timegeb if time exceeds 18.00 only use 18.00 to calculate. Sorry for not clear question.
– yolox
Nov 12 at 7:57
Ok and in the second row both times are between 8 and 18. Why is their difference 0?
– timgeb
Nov 12 at 7:59