Separating dates

[Beenish Amir]

I am working on a dataset which has date in the following format, I want separate columns for month, date and year can anyone help me with the relevant command please? Thank you in advance.

dataex interviewdate

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* Example generated by -dataex-. To install: ssc install dataex
clear
input long interviewdate
 .
29
29
 .
 .
 .
 .
24
 .
 .
27
 .
 .
 .
 .
 .
47
47
 .
37
14
37
 .
 .
29
29
 .
 .
 .
15
 .
32
 .
 .
 .
 .
15
 .
15
 .
 .
15
 .
 .
42
42
42
 .
 .
 .
 .
42
42
 .
 .
32
 7
 .
 7
 .
 .
20
20
 .
20
20
20
15
 .
 .
 .
 .
 .
 .
13
13
 .
13
13
 .
13
 .
 .
 .
 7
 .
 .
 7
 7
 .
 7
 .
19
19
19
19
19
19
19
19
end
label values interviewdate interviewdate
label def interviewdate 7 "11/3/2004", modify
label def interviewdate 13 "16/02/04", modify
label def interviewdate 14 "16/03/04", modify
label def interviewdate 15 "17/02/04", modify
label def interviewdate 19 "18/03/04", modify
label def interviewdate 20 "19/02/04", modify
label def interviewdate 24 "21/02/04", modify
label def interviewdate 27 "23/02/04", modify
label def interviewdate 29 "24/02/04", modify
label def interviewdate 32 "25/02/04", modify
label def interviewdate 37 "28/02/04", modify
label def interviewdate 42 "4/3/2004", modify
label def interviewdate 47 "6/3/2004", modify

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[Carole J. Wilson]

The only thing that is a little tricky is that you have two digit and four digit years. I am assuming that any two digit years (like "04") should be in this century ("2004").

*create a new var date that is the label of interview date
decode interviewdate, gen(date)

*split date into parts
split date, parse("/") gen(d_) destring

*fix the 2 digit years; change to 4 digit years
replace d_3=real(string(2)+string(d_3, "%03.0f")) if d_3<1000

*rename the variables
rename d_1 day
rename d_2 month
rename d_3 year

[Beenish Amir]

The command worked perfectly, thank you so much.

[Nick Cox]

Carole gives excellent advice as you asked. For most Stata purposes, however, you'd be better off with a Stata daily date.

gen ddate = mdy(month, day, year) 
format ddate %td

[Beenish Amir]

The command Carole gave worked with the first dataset but in the second data set it isn't working I am sharing the interview date example below
dataex interview date

----------------------- copy starting from the next line -----------------------
[CODE]
* Example generated by -dataex-. To install: ssc install dataex
clear
input str9 interviewdate
"15-Jan-05"
""
"15-Jan-05"
""
"15-Jan-05"
"15-Jan-05"
""
"15-Jan-05"
"15-Jan-05"
""
"15-Jan-05"
"15-Jan-05"
"15-Jan-05"
"15-Jan-05"
""
"15-Jan-05"
"15-Jan-05"
""
"15-Jan-05"
""
"15-Jan-05"
""
"15-Jan-05"
"15-Jan-05"
"15-Jan-05"
"12-Jan-05"
""
"12-Jan-05"
""
"12-Jan-05"
"12-Jan-05"
""
"12-Jan-05"
""
""
"12-Jan-05"
""
""
"12-Jan-05"
"12-Jan-05"
""
"12-Jan-05"
""
"12-Jan-05"
"12-Jan-05"
""
""
""
"12-Jan-05"
"12-Jan-05"
""
""
""
"12-Jan-05"
"12-Jan-05"
""
""
""
"12-Jan-05"
""
"12-Jan-05"
""
""
"12-Jan-05"
"12-Jan-05"
""
""
"10-Jan-05"
""
""
"10-Jan-05"
""
""
"10-Jan-05"
"10-Jan-05"
""
""
--more--

[Nick Cox]

It really isn't fair comment that a solution designed for a quite different problem doesn't work with the data shown in #5.

Just read please

help datetime

to learn how to convert string dates into Stata numeric daily dates. You can't hope to deal with dates successfully in Stata without a minimal understanding of those principles.

[Carole J. Wilson]

I agree with Nick. If you are going to be working with dates, it is essential to understand the way that Stata represents dates. Once you have accomplished the task of turning your string that is in HRF to a SIF (the datetime help file will explain these terms) then you can then use the functions:

gen day=day(new_date_var)
gen month=month(new_date_var)
gen year=year(new_date_var)

[Beenish Amir]

Okay, thanks I will read the suggested material and run the commands.