Reputation: 603
I want to create a new column which will show the timedelta in days between two dates as shown in the following pandas dataframe:
>>> hg[['not inc','date']]
not inc date
0 False 2012-02-29 00:00:00
1 False 2012-03-16 00:00:00
2 False 2012-04-04 00:00:00
3 True 2012-05-08 00:00:00
4 False 2012-05-12 00:00:00
5 False 2012-05-26 00:00:00
6 False 2012-06-09 00:00:00
7 False 2012-10-13 00:00:00
8 False 2012-11-10 00:00:00
9 True 2013-03-19 00:00:00
10 False 2013-04-01 00:00:00
11 False 2013-04-25 00:00:00
12 False 2013-05-04 00:00:00
13 False 2013-05-18 00:00:00
14 False 2013-06-01 00:00:00
15 True 2013-08-20 00:00:00
16 False 2013-08-31 00:00:00
17 False 2013-09-21 00:00:00
18 False 2013-10-05 00:00:00
19 False 2013-10-19 00:00:00
20 False 2013-11-09 00:00:00
21 True 2014-01-21 00:00:00
22 False 2014-02-08 00:00:00
23 False 2014-02-22 00:00:00
24 False 2014-03-08 00:00:00
25 False 2014-03-29 00:00:00
26 False 2014-04-19 00:00:00
27 True 2014-07-21 00:00:00
28 True 2014-08-01 00:00:00
29 False 2014-08-09 00:00:00
30 False 2014-08-30 00:00:00
31 False 2014-09-13 00:00:00
32 True 2014-09-26 00:00:00
33 False 2014-10-04 00:00:00
34 True 2015-01-08 00:00:00
35 True 2015-01-20 00:00:00
36 False 2015-01-31 00:00:00
37 False 2015-02-14 00:00:00
I want the date difference to start by subtracting 2012-01-02 and to be an int.
This is what I have tried but with no success because prevdate does not update to the new row's date but keeps referring to the original starting position of datetime(2012,01,02). I am using iterrows through the rows of the dataframe.
>>>for index, row in hg.iterrows():
prevdate = datetime(2012,01,02)
dsince = row['date']-prevdate
prevdate = row['date']
print dsince
Result (Also I don't know how to modify the value into int):
58 days, 0:00:00
74 days, 0:00:00
93 days, 0:00:00
127 days, 0:00:00
131 days, 0:00:00
145 days, 0:00:00
159 days, 0:00:00
285 days, 0:00:00
313 days, 0:00:00
442 days, 0:00:00
455 days, 0:00:00
479 days, 0:00:00
488 days, 0:00:00
502 days, 0:00:00
516 days, 0:00:00
596 days, 0:00:00
607 days, 0:00:00
628 days, 0:00:00
642 days, 0:00:00
656 days, 0:00:00
677 days, 0:00:00
750 days, 0:00:00
768 days, 0:00:00
782 days, 0:00:00
796 days, 0:00:00
817 days, 0:00:00
838 days, 0:00:00
931 days, 0:00:00
942 days, 0:00:00
950 days, 0:00:00
971 days, 0:00:00
985 days, 0:00:00
998 days, 0:00:00
1006 days, 0:00:00
1102 days, 0:00:00
1114 days, 0:00:00
1125 days, 0:00:00
1139 days, 0:00:00
To make things a little more complicated I would like to only create another column with the date difference between rows which have False for 'not inc' column.
Thank you.
Upvotes: 1
Views: 776
Reputation: 5468
Assuming your date column is already cast as a datetime64
:
In [61]: hg = pd.DataFrame({"not inc":[False , False, False, True, False],"date":pd.to_datetime(pd.Series(["2012-02-29", "2012-03-16", "2012-04-04", "2012-05-08", "2012-05-12"]))})
In [63]: hg.dtypes
Out[63]:
date datetime64[ns]
not inc bool
dtype: object
Temporarily filter out the rows you don't want to include:
In [64]: included = hg[hg["not inc"] == False]
Use shift
to get a series of the dates you want to subtract, filling in your start date at the beginning:
In [66]: prev_dates = included.date.shift().fillna(pd.datetime(2012,1,2))
In [67]: prev_dates
Out[67]:
0 2012-01-02
1 2012-02-29
2 2012-03-16
4 2012-04-04
Name: date, dtype: datetime64[ns]
Subtract the dates and recast the timedelta as an int:
In [68]: delta = included.date - prev_dates
In [69]: delta = delta.astype("timedelta64[D]")
In [70]: delta
Out[70]:
0 58
1 16
2 19
4 38
Name: date, dtype: float64
Then concat
the new Series to your original dataframe.
In [71]: delta.name = "delta"
In [72]: hg = pd.concat((hg, delta), axis=1)
In [73]: hg
Out[73]:
date not inc delta
0 2012-02-29 False 58
1 2012-03-16 False 16
2 2012-04-04 False 19
3 2012-05-08 True NaN
4 2012-05-12 False 38
Upvotes: 1
Reputation: 3281
Put the line prevdate = datetime(2012,01,02)
before the loop.
prevdate = datetime(2012,01,02)
for index, row in hg.iterrows():
dsince = (row['date'] - prevdate).days
prevdate = row['date']
print dsince
If it doesn't work, convert prevdate
and row['date']
to dates.
Upvotes: 0