Thanks. I prefer to stick with SAS function because I am more comfortable with SAS. Morever, With SAS function, it easier to do repetition to obtain sampling distribution. That is, it is easiet to write a program that will do compute such distance many times. The way we will have a third column containing the repl ... View more

Ksharp, awesome!! Thanks. You are indeed a genius. I am very grateful for help and patience. Thanks Jack ... View more

Thanks Rick. You are right. It seems that the issue comes from my using the means of each subgroups in HAVE1 to form distance with the mean of HAVE2. So that with standardization, the mean of each subgroup essentially supposed to be zero. With unit variance. The way to avoid this is to use the observations in the subgroubs instead of using the means. Similarly for HAVE2, using the observations instead of the mean. That is form the distance between the data sets in HAVE1 (each of the subgroups) and HAVE2 data set. The link below does that using MATLB: that is what I wanted to do. Except that in my own case I will be forming distance between subgroups and one data set. Thanks J http://people.revoledu.com/kardi/tutorial/Similarity/MahalanobisDistance.html ... View more

http://people.revoledu.com/kardi/tutorial/Similarity/MahalanobisDistance.html The above link contains exactly what I want to do. That is to compute distance between two matrix. In my case HAVE1 and HAVE2. The only difference is that HAVE1 have multiple data set in it, so I am computing distance between each of those replicate in HAVE1 and HAVE2. In the link above, the data were standardized and that didnt cause any problem. The site provide MATLAB code. But I want it in SAS. Thanks, J ... View more

below is the log:   Or could it be that it's because HAVE2  is only the mean values? The actual aim is to compute distance between of the datasets in each replicate in HAVE1 to the dataset HAVE2. I have only use the mean of HAVE2 becuase I thought that the center should be sufficient. It could be that it's because we are using only the covariance of HAVE1 instead of the pooled covariance of each of the replicate in HAVE1 and HAVE2.   Thanks Below is the SAS log.   270 proc iml; NOTE: IML Ready 271 use stdwant nobs nobs; 272 read all var {reps}; 273 read all var {x y} into data[c=vnames]; 274 close; NOTE: Closing WORK.STDWANT 275 use stdhave; 276 read all var {x y} into m; 277 close; NOTE: Closing WORK.STDHAVE 278 279 center=m[:,]; 280 level=unique(reps); 281 want=j(ncol(level),ncol(data)+2,.); 282 283 do i=1 to ncol(level); 284 want[i,1]=level[i]; 285 idx=loc(reps=level[i]); 286 want[i,2:ncol(data)+1]=data[idx,][:,]; 287 end; 288 289 xx=want[,2:ncol(data)+1]; 290 cov=cov(xx); 291 want[,ncol(data)+2]= mahalanobis(xx,center,cov); NOTE: Module MAHALANOBIS loaded from the storage SASHELP.IMLMLIB. ERROR: (execution) Matrix should be non-singular. operation : TRISOLV at offset 8 column 18 operands : *LIT1242, U, _TEM1001 *LIT1242 1 row 1 col (numeric) 2 U 2 rows 2 cols (numeric) 0 0 0 0 _TEM1001 2 rows 30 cols (numeric) statement : ASSIGN at offset 8 column 7 traceback : module MAHALANOBIS at offset 8 column 7 NOTE: Paused in module MAHALANOBIS. 292 293 names={reps}||vnames||{distance}; 294 create want from want[c=names]; ERROR: Matrix want has not been set to a value. statement : CREATE at line 294 column 1 295 append from want; ERROR: No data set is currently open for output. statement : APPEND at line 295 column 1 296 close; 297 298 quit; NOTE: Exiting IML. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.02 seconds cpu time 0.04 seconds   ... View more

The program worked as expected and computes the Mahalanobis distance when the numbers are not standardized.   However, when the numbers are standardized it gives an error and stops. The error is:   "ERROR: (execution) Matrix should be non-singular."   I suspect it's because one should be computing Euclidian distance when the numbers are standardized since standardized variables invlove identity matrix. Does anyone know why this is the case and how this may overcome in computing distance for standardized variables?    Any help is appreciated.   Thank,   J     ... View more

Thanks! Thanks!! Thanks!!!. It's perfect!!!!!!   Please, I have a few questions on some parts of the code to enable me generalise the code to handle any number of variables: This is because a data set can have any number of variable--not just x and y; So that I will use something like &varlist.   want=j(ncol(level),4,.); What is the 4 doing; I assume is the number of observation in each replicate, right? Or is it the 4th column? do i=1 to ncol(level); want[i,1]=level[i]; idx=loc(reps=level[i]); want[i,2:3]=data[idx,][:,]; What is the 2:3 doing? is it indexing column 2 to 3? If yes, in which data set? end; xx=want[,2:3]; what is the 2:3 doing? cov=cov(xx);   Thanks for you incredible help,   J want[,4]= mahalanobis(xx,center,cov); What is the 4 doing?   Thanks very much for your help. ... View more

Thanks very much Ksharp for sharing the program.    The program works, but it's computing dintance from each observation in HAVE1 to the mean of the variables in HAVE2.   But what I want to compute is the Mahalonobis distance from the means of the variables in each replicate in HAVE1 to the means of the variables in HAVE2, so that when the column containing distance is added back to HAVE1, it should be one value of distance for each replicate. This is because the distance is computed from the means of the variables in each replicate in HAVE1 which are unique to each replicate.   Thanks,   J   ... View more

So means of x and y in HAVE2 is the center of Mahalanobis distance ? Yes. HAVE2 is the center of Mahalanobis distance. ... View more

This is correct answer as well ... View more

All the answers are essentially correct. Thank you all very very much.  ... View more

Thanks Rick! ... View more

It's basicacally to obtain a random subset of possible permutations of the data set. The reason is that in some settings, computer memory might limit the enumeration of all permutations. Therefor, a sufficiently large number of permutaion of the data set should suffice.   Thanks! J ... View more

Incredible!!!    Let me evalute to this.  ... View more