Add/subtract days/month to a Stata date

[Francois Durant]

Hi,

I would like to substract 1 months to a certain date

date
28jan2016
29jan2016
01feb2016
02feb2016
03feb2016
04feb2016
05feb2016
08feb2016
09feb2016
10feb2016
11feb2016
12feb2016

I would like to create a variable "date1" that substract 1 month to date. Can anyone help? Thanks!

[Nick Cox]

This isn't trivial. The previous month often has fewer days (true of March, May, June, etc.), so what should happen for e.g. 31mar2016?

[Francois Durant]

how can i subtract days then? I'd like to subtract 30 days. Thanks

[Nick Cox]

gen wanted = whatever - 30

[Francois Durant]

Easier than I thought. Thanks!

[Francois Durant]

I'm circling back on an old discussion. I would like to subtract or add months to a daily date in way that the day remains unchanged unless this is the end of the month, in which case, I would like to create a date with the last day of that month.

For instance if I want to subtract 1 month:
feb 11st 2020 becomes Jan 11st 2020
and
March 31st becomes Feb 28th in non leap years and Feb 29th in leap years.

Please advise. Thanks.

[cu dao huy]

I'm circling back on an old discussion. I would like to subtract or add months to a daily date in way that the day remains unchanged unless this is the end of the month, in which case, I would like to create a date with the last day of that month.

For instance if I want to subtract 1 month:
feb 11st 2020 becomes Jan 11st 2020
and
March 31st becomes Feb 28th in non leap years and Feb 29th in leap years.

Please advise. Thanks.

HI Francois,

I think -egenmore- package has -eomd- command useful for your question. Try to install this package and use example command

 egen neededvar = eomd(date), f(%dd_m_y) lag(1)

.

Goodluck,
Cu,

[Francois Durant]

Elegant and efficient solution, exactly what I needed. Thanks a lot cu dao huy .

[Nick Cox]

Note also

The Stata Journal (2019) 19, Number 3, pp. 719–728

DOI: 10.1177/1536867X19874247

Speaking Stata: The last day of the month

Nicholas J. Cox

Abstract. I discuss three related problems about getting the last day of the month in a new variable. Commentary ranges from the specifics of date and other functions to some generalities on developing code. Modular arithmetic belongs in every Stata user’s coding toolbox.

Keywords: dm0100, dates, days, weeks, months, functions, modulus, remainders, rotations

[Francois Durant]

Thanks for bringing that up, I will take a look.

[Rebecca Frausel]

Hi all,

Adding again to this old conversation to contribute my solution. My problem was that I have a set of intake dates, and I need to calculate target dates that are 3-, 6-, 9-, and 12-months after the intake date. This brute force solution worked for me and hopefully it will help others. Note, though, that I was extremely lucky that none of my intake dates fell on the 29th, 30th, or 31st, which would make this approach more complicated. If people have ideas for how to improve this code or how to account for days ending in 29-31, I would appreciate it!

Data:

Intake_Date
19jan2023
19jan2023
26jan2023
26jan2023
22feb2023
22feb2023
22feb2023
02mar2023
02mar2023
02mar2023
10mar2023
10mar2023
10mar2023
10mar2023
16mar2023
16mar2023
16mar2023
16mar2023
28mar2023
28mar2023
28mar2023
28mar2023
04apr2023
04apr2023
25apr2023
25apr2023
26may2023
26may2023
02aug2023
02aug2023
02aug2023
02aug2023
04aug2023
13sep2023
13sep2023
13sep2023
08nov2023
21nov2023
21nov2023
09jan2024
09jan2024
09jan2024
18jan2024
18jan2024
18jan2024
15apr2024
15apr2024
02jul2024
15jul2024

Code to calculate 3-, 6-, 9-, and 12-months following intake date:

* adding in target survey dates

gen month = month(Intake_Date)
gen day = day(Intake_Date)
gen year = year(Intake_Date)

tab day
* latest day of intake is 28th - phew! If any youth intake on 29th, 30th, or 31st, additional adjustmnets are needed

* 3 months first

gen target_month3 = month+3

* fix Oct, Nov, Dec
replace target_month3=1 if target_month3==13
replace target_month3=2 if target_month3==14
replace target_month3=3 if target_month3==15

* also need to add a year
gen target_year3 = year + 1 if target_month3<=3

gen target_survey_date3 = mdy(target_month3,day,year)
replace target_survey_date3 = mdy(target_month3,day,target_year3) if target_year3!=.

format target_survey_date3 %td

* now 6 months - add 3 months to target_month3 which is already clean

gen target_month6 = target_month3+3

* fix Oct, Nov, Dec
replace target_month6=1 if target_month6==13
replace target_month6=2 if target_month6==14
replace target_month6=3 if target_month6==15

* also need to add a year
gen target_year6 = year + 1 if target_month6<=3 | target_year3!=.

gen target_survey_date6 = mdy(target_month6,day,year)
replace target_survey_date6 = mdy(target_month6,day,target_year6) if target_year6!=.

format target_survey_date6 %td

* now 9 months - add 3 months to target_month6 which is already clean

gen target_month9 = target_month6+3

* fix Oct, Nov, Dec
replace target_month9=1 if target_month9==13
replace target_month9=2 if target_month9==14
replace target_month9=3 if target_month9==15

* also need to add a year
gen target_year9 = year + 1 if target_month9<=3 | target_year6!=.

gen target_survey_date9 = mdy(target_month9,day,year)
replace target_survey_date9 = mdy(target_month9,day,target_year9) if target_year9!=.

format target_survey_date9 %td

* for 12 months, just add 1 to year

gen target_year12 = year + 1

gen target_survey_date12 = mdy(month,day,target_year12)

format target_survey_date12 %td

* Let's reorder to put target dates at end

order target_survey_date3 target_survey_date6 target_survey_date9 target_survey_date12, after(target_year12)

* for sanity's sake, get difference between intake and survey dates
gen difference3 = target_survey_date3 - Intake_Date
gen difference6 = target_survey_date6 - Intake_Date
gen difference9 = target_survey_date9 - Intake_Date
gen difference12 = target_survey_date12 - Intake_Date

sum difference*
* everything looks reasonable