Extracting month and day from a variable

[Van Ha]

Hi All, I have a date variable with observations recorded as "ddmmyyyy" (e.g. 07may2001). I want to drop the year in this variable OR keep the day and month (e.g. 07may). I couldn't find anything about this online and in the manual u24.pdf (stata.com). Any tips on how I can approach this? Thanks in advance.

[Clyde Schechter]

Well, descriptions of data sets are usually insufficient to support help with code, and yours is a good example. From what you say, it's anybody's guess whether your variable "recorded as ddmmyyyy" is a string variable, or a numeric variable that reads just as ddmmyyyy without any datetime display format applied, or is a numeric variable that reads as ddmmyyyy after applying a datetime display format. Since the solution to your problem would differ depending on which this is, you'll need to post back showing your data in the helpful way that is recommended in the Forum FAQ (at #12). Specifically, use the -dataex- command.

If you are running version 18, 17, 16 or a fully updated version 15.1 or 14.2, -dataex- is already part of your official Stata installation. If not, run -ssc install dataex- to get it. Either way, run -help dataex- to read the simple instructions for using it. -dataex- will save you time; it is easier and quicker than typing out tables. It includes complete information about aspects of the data that are often critical to answering your question but cannot be seen from tabular displays or screenshots. It also makes it possible for those who want to help you to create a faithful representation of your example to try out their code, which in turn makes it more likely that their answer will actually work in your data.

When asking for help with code, always show example data. When showing example data, always use -dataex-.

[Van Ha]

Thanks for the response! Here is a sample of the variable:

* Example generated by -dataex-. For more info, type help dataex
clear
input float hosp_date_num
20948
20844
20842
20828
20819
21564
18407
18616
18695
18732
18750
18802
18912
18945
18955
19024
19083
19172
19133
19195
19207
19208
19340
19357
19344
19344
19212
18414
18641
18495
18932
18778
19229
18680
18739
18338
18296
18257
18276
18388
18411
18496
18746
18753
18793
18823
18851
18942
15836
15835
15831
15822
15800
15801
15860
15884
16020
16021
16026
16133
16133
16272
16274
16274
16553
16616
16586
16666
16622
16611
16553
16620
16680
16713
16689
16709
16650
16657
16652
16674
16705
16650
16541
16700
16658
16586
16705
16677
16617
16577
16761
16703
16766
16772
16683
16777
16745
16704
16787
16751
end
format %td hosp_date_num

[Clyde Schechter]

Extracting the day and month:

gen month = month(hosp_date_num)
gen day = day(hosp_date_num)

This gives you two separate variables, one for month and the other for day.

If you want to combine those into a single string variable that looks like "m/d" you could run

egen month_day = concat(month day), punct("/")

This gives you a genuine month_day single variable. It contains no year information at all. It is a string: you can't calculate anything with it, and you can't sort it into chronological order. But it has only the month and day as content so that the values corresponding to 3mar2024 and 3mar2005 would not only look the same, they would be equal if you were to test for equality.

If, however, all you want is to not see the year, you can do this:

clonevar hosp_date_num2 = hosp_date_num
format hosp_date_num2 %tdDDMon

But be warned that the year is really still there--Stata is just hiding it at your request. So, with this approach, hosp_date_num2 will look like 3mar for both 3mar2024 and 3mar2005, but the two values would be unequal if you were to test them.

[Van Ha]

Thanks Clyde! Is there a way that we can convert the variable from string to date? I want to use it with another date variable to calculate a change in time...

[Clyde Schechter]

There is no Stata date variable that contains day and month but not year. And it makes no sense to speak of calculating a change in time with such a variable if it did exist. Since 13mar2024 and 13mar2023 would be "the same" but for their year, change in time between 13mar and, say 25dec would differ depending on which year 13mar represented.

Added: Even if we knew they referred to the same year, the change in time between 1feb and 1mar depends on which year is involved, as some years are leap years and others not.