Morgan Law in Stata

[Silvana Builes]

Hi!

I have three binary variables hh_d_floor, hh_d_wall and hh_d_roof. The three of them take values of 0 and 1.

I am trying to run the following instruction:

gen housing = 1 if (hh_d_floor == 1| hh_d_wall == 1| hh_d_roof == 1)**The household has inadequate (1) housing materials in any of the three components: floor, roof, or walls.
replace housing=0 if (hh_d_floor==0) & (hh_d_wall==0) & (hh_d_roof==0) **The household has adequate (0) housing materials if it has all of the three components: floor, roof, or walls.

The first line is giving me an error....How could I express the union of threeevents in stata?
and... is the intersection defined with &?

Thanks!

[Silvana Builes]

I solved it


gen housing = 1 if hh_d_floor == 1 | hh_d_wall == 1 | hh_d_roof == 1
replace housing=0 if hh_d_floor==0 & hh_d_wall==0 & hh_d_roof==0

[William Liu]

By the way, Boolean expressions can be directly used without the if conditional (not to be confused with the -if- command). A more efficient version of your code would be:

gen housing = (hh_d_floor == 1| hh_d_wall == 1| hh_d_roof == 1)

Also, there is nothing wrong with your original code. The source of the error isn't because of the parentheses.

[Nick Cox]

gen housing = inlist(1, hh_d_floor, hh_d_roof, hh_d_wall)

is another way to do it.

[Hemanshu Kumar]

Side note: in all these methods, be mindful of missing values and their consequences for the boolean evaluations.