Re: st: generate variable

[Phil Schumm <pschumm@uchicago.edu>]

On May 27, 2006, at 1:05 PM, peter harper wrote:
> Thanks Nick. I tried several things given below, but I still cannot  
> get "18825 changes made" in total, they are the people without work  
> limiting disability. Can I please ask what precisely needs to be  
> issued to get the "18825 or 18826 changes made" please.
>
>  ge disabledworklim=.
> (22451 missing values generated)
> . replace disabledworklim=1 if e6a==1&e6b==1
> (997 real changes made)
> . replace disabledworklim=2 if (e6a==2)&(e6a==1-e6b==1)
> (17174 real changes made)
> . replace disabledworklim=2 if (e6a==1&2)-(e6b==1)
> (1700 real changes made)
> . replace disabledworklim=2 if (e6a==1&e6a==2)&-(e6b==1)
> (0 real changes made)
> . replace disabledworklim=2 if (e6a==1&e6a==2)-(e6b==1)
> (997 real changes made)
> . replace disabledworklim=2 if (e6a<=e6a==1)&(e6a==1<=e6a==2)
> (0 real changes made)
> . replace disabledworklim=2 if (e6b<=e6a==1)&(e6a==1<=e6a==2)
> (0 real changes made)

Take Nick's advice, and use parentheses to make your meaning clear.   
For example, consider your expression:


    (e6a==1&e6a==2)&-(e6b==1)


You have three operators here (in order of precedence, high to low):  
-, ==, and &.  This means your expression is equivalent to


    ( (e6a==1) & (e6a==2) ) & ( - (e6b==1) )
      --------   --------     --------------
         a           b               c
    -----------------------
               d


Now, a will equal 1 if e6a equals 1 (or 0 otherwise), b will equal 1  
if e6a equals 2 (or 0 otherwise), and c will equal -1 if e6b equals 1  
(or 0 otherwise).  Moving outward, d will equal 1 if a and b are both  
non-zero; this obviously cannot happen, so d will automatically equal  
0.  Finally, the entire expression will be true (i.e., non-zero) only  
if both d and c are non-zero, and since d is automatically 0, the  
entire expression will always be false.

Similarly, consider your expression:


    (e6b<=e6a==1)&(e6a==1<=e6a==2)


Here, the operators (high to low precedence) are: <=, ==, and &.   
This means that your expression is equivalent to


    ( (e6b<=e6a) == 1) & ( ( e6a == (1<=e6a) ) == 2 )
      ----------                    --------
          a                             b
    ------------------     -------------------
             c                      d
                         ----------------------------
                                        e


Starting with the inner-most evaluations, a will equal 1 if e6b is  
less than or equal to e6a (or 0 otherwise), and b will equal 1 if e6a  
is greater than or equal to 1 (including missing).  As a result, c  
will equal 1 if e6b is less than or equal to e6a (or 0 otherwise),  
and d will equal 1 if e6a equals b, which, by the way, can only  
happen if e6a equals either 1 or 0.  Moving further outward, e will  
equal 1 only if d equals 2, which, since d itself is a logical  
expression and will therefore only ever be 0 or 1, can never happen.   
Thus, just like before, the entire expression will always be false.

If you have questions about the above, look carefully at the sections  
in the Stata manual ([U] I believe) on operators and expressions; you  
should find what you need there.  In addition, be very careful when  
using the comparison operators with variables that can have missing  
values.  The expression


    x <= y


will always be true if y is missing, regardless of what x is.


-- Phil

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