Hi,
I just did a practice of array and happened to find that one arithmetic result is not logical.
**** data preparation ;
data visits;
infile datalines;
input patid visdt0 visdt1 visdt2 visdt3 visdt4;
datalines;
1001 0.1 1.1 2.1 3.1 4.1
1002 0.2 1.2 2.2 3.2 4.2
;
run;
**** foolproof dataset;
data work.checkvisits;
set work.visits (keep=patid visdt0 visdt1 visdt2 visdt3 visdt4);
array visdt(1:5) visdt0 visdt1 visdt2 visdt3 visdt4;
do i=2 to 5;
x = visdt(i) - visdt(i-1);
if x gt 1 then do;
* there should be no output since all differences;
* should equal to 1 ;
put visdt(i-1)= visdt(i)= x=;
output;
end;
end;
drop i x;
run;
**** print ;
proc print data=checkvisits;
run;
and the log:
```log
visdt1=1.2 visdt2=2.2 x=1
NOTE: There were 2 observations read from the data set WORK.VISITS.
NOTE: The data set WORK.CHECKVISITS has 1 observations and 6 variables.
NOTE: DATA statement used (Total process time):
real time 0.00 seconds
cpu time 0.00 seconds
```
The environment is SAS 9.2 on Linux. Please advise, thanks.
Numerical precision.
Although reference is 9.4 it does apply to 9.2 as well
Numerical precision.
Although reference is 9.4 it does apply to 9.2 as well
Thanks so much Reeza, I adjusted a bit the program and found two remainings do not match the real value probably due to truncation in binary numbers.
13 **** foolproof dataset; 14 data work.checkvisits; 15 set work.visits; 16 array visdt(1:5) visdt0 visdt1 visdt2 visdt3 visdt4; 17 do i=2 to 5; 18 * the expeced result should be 0 ; 19 x = visdt(i) - visdt(i-1) - 1; 20 put visdt(i-1)= visdt(i)= x=; 21 if x gt 0 then do; 22 output; 23 end; 24 end; The SAS System 25 drop i x; 26 run; visdt0=0.1 visdt1=1.1 x=0 visdt1=1.1 visdt2=2.1 x=0 visdt2=2.1 visdt3=3.1 x=0 visdt3=3.1 visdt4=4.1 x=-4.44089E-16 visdt0=0.2 visdt1=1.2 x=0 visdt1=1.2 visdt2=2.2 x=2.220446E-16 visdt2=2.2 visdt3=3.2 x=0 visdt3=3.2 visdt4=4.2 x=0 NOTE: There were 2 observations read from the data set WORK.VISITS. NOTE: The data set WORK.CHECKVISITS has 1 observations and 6 variables. NOTE: DATA statement used (Total process time): real time 0.00 seconds cpu time 0.00 seconds
Those are the exact types of issues that arise with numerical precision for decimals. If you do something like
x = round( visdt(i) - visdt(i-1), 0.00001) -1;
you may be happier.
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