Optimizing C code
Thanks both of you.
Inf - Inf
[1] NaN
So isn't the line 9 useless ? If either x[i] or y[i] are NA, then dev
will be NA and !ISNAN(dev) will detect it...
Sothe loop cool be
8. for(i = 0 ; i < taille ; i++) {
10. dev = (x[i] - y[i]);
11. if(!ISNAN(dev)) {
12. dist += dev * dev;
13. count++;
15. }
16. }
No ?
Christophe
Duncan Murdoch
Christophe
#define both_FINITE(a,b) (R_FINITE(a) && R_FINITE(b))
#define both_non_NA(a,b) (!ISNAN(a) && !ISNAN(b))
1. static double R_euclidean2(double *x, double *y, int taille)
2. {
3. double dev, dist;
4. int count, i;
5.
6. count= 0;
7. dist = 0;
8. for(i = 0 ; i < taille ; i++) {
9. if(both_non_NA(x[i], y[i])) {
10. dev = (x[i] - y[i]);
11. if(!ISNAN(dev)) {
12. dist += dev * dev;
13. count++;
14. }
15. }
16. }
17. if(count == 0)return NA_REAL;
18. if(count != taille) dist /= ((double)count/taille);
19. return sqrt(dist);
20.}
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.