BookmarkSubscribeRSS Feed
🔒 This topic is solved and locked. Need further help from the community? Please sign in and ask a new question.
jmoody77
Calcite | Level 5

Hi - 

I'm using IML to do an agent based model, treating the matrix space as the substrate agents move over. 

 

As part of the input for agents I'd like to smooth the data around all other agents' values.    So for every ij cell in the matrix, the goal is to return the average of all neighboring cells (including index).  The code listed below works; but I'm curious if there's a way to do this that doesn't involve looping over all cells.

 

Thanks in advance!

Jim

 

proc iml;

/* function to pull out the subset of cells adjacent 
to selected node, including index */
start nrhd(inmat,idx);
  rc=ndx2sub(dimension(inmat),idx); /* give xy coord */
  rs=(rc[1]-1):(rc[1]+1); /* get nbrhd rows*/
  rs=xsect(rs,1:nrow(inmat)); /* limit by bndry */

  cs=(rc[2]-1):(rc[2]+1); /* get nbrhd cols */
  cs=xsect(cs,1:ncol(inmat));
  nbhd=inmat[rs,cs]; 
 return(nbhd);
 free rs cs;
finish;

/* silly example */
mat=shape(1:25,5,5);
newmat=mat>100; /*silly way to get empty matrix. I know.  being lazy.*/
do i=1 to 25;
 nbrs_i=nrhd(mat,i);
 val=mat[nbrs_i][:]; 
 newmat[i]=val;
end;


print mat newmat;

quit;
1 ACCEPTED SOLUTION

Accepted Solutions
Rick_SAS
SAS Super FREQ

This computation is an example of a 2D convolution. Usually computations like this are performed by "padding" the boundary of the matrix with missing values (or sometimes zeros) so that you can average the neighbors without worrying about whether a cell is in the first or last row/column.

 

There are lots of different averaging schemes. It looks like you want to compute an evenly weighted average of each 3x3 set of cells.  Other schemes might put more weight on the center of the square or less weight in the corners.  To accommodate different weighting schemes, researchers often use a "mask" of weights to indicate how much weight to give to each element in the 3x3 grid. They also have to decide how to average the boundary cells. For your example, the weight matrix would be j(3,3,1) and wouldn't affect the result, so I have omitted it.

 

I think your solution is fine. But since you asked for an alternative way to solve the problem, you might consider using the padding idea, as follows:

 

proc iml;
mat=shape(1:30,5,6);
nr = nrow(mat);
nc = ncol(mat);
avg=j(nr, nc, 0);

pad = j(nr+2, nc+2, .);   /* add row/col of missing values on boundary */
pad[ 2:nr+1, 2:nc+1 ] = mat;
do i = 1 to nr;
  do j = 1 to nc;
     avg[i,j] = ( pad[i:i+2, j:j+2] )[:];   /* average */
  end;
end;
print avg;

By the way, these ideas a used in studying automata, such as Conway's Game of Life and Wolfram's Rule 30, which I have explored in SAS/IML.

 

View solution in original post

8 REPLIES 8
Rick_SAS
SAS Super FREQ

This computation is an example of a 2D convolution. Usually computations like this are performed by "padding" the boundary of the matrix with missing values (or sometimes zeros) so that you can average the neighbors without worrying about whether a cell is in the first or last row/column.

 

There are lots of different averaging schemes. It looks like you want to compute an evenly weighted average of each 3x3 set of cells.  Other schemes might put more weight on the center of the square or less weight in the corners.  To accommodate different weighting schemes, researchers often use a "mask" of weights to indicate how much weight to give to each element in the 3x3 grid. They also have to decide how to average the boundary cells. For your example, the weight matrix would be j(3,3,1) and wouldn't affect the result, so I have omitted it.

 

I think your solution is fine. But since you asked for an alternative way to solve the problem, you might consider using the padding idea, as follows:

 

proc iml;
mat=shape(1:30,5,6);
nr = nrow(mat);
nc = ncol(mat);
avg=j(nr, nc, 0);

pad = j(nr+2, nc+2, .);   /* add row/col of missing values on boundary */
pad[ 2:nr+1, 2:nc+1 ] = mat;
do i = 1 to nr;
  do j = 1 to nc;
     avg[i,j] = ( pad[i:i+2, j:j+2] )[:];   /* average */
  end;
end;
print avg;

By the way, these ideas a used in studying automata, such as Conway's Game of Life and Wolfram's Rule 30, which I have explored in SAS/IML.

 

Rick_SAS
SAS Super FREQ

PS. Your program doesn't actually work because you are returning inmat[rs,cs].  You should be returning the subscripts, not the matrix values.

jmoody77
Calcite | Level 5

Yup...good eye; thanks.  the nrhd() function pulls out the values of the input "mat" matrix -- i.e. returns the submatrix w. i in the center. The example here only worked because the cell values are the index values!.  Pure coincidence on how I constructed the example matrix.

 

In case anyone is interested, this modification fixes it:

do i=1 to 25;
    nbrs_i=nrhd(mat,i);
	*val=mat[nbrs_i][:]; /* this is wrong */
	val=nbrs_i[:]; /* this is right */
	newmat[i]=val;
  end;

 

jmoody77
Calcite | Level 5

thanks much; appreciate it!

Rick_SAS
SAS Super FREQ

The "pad" method seems to be substantially faster. On a 600x400 matrix, the "Nbhd" method takes about 4.5 seconds on my system whereas the "Pad" method takes about 0.2 seconds (about 25 times faster).

jmoody77
Calcite | Level 5
Very cool. Thank you!
Rick_SAS
SAS Super FREQ

Do you have other questions? If not, you can close this thread by choosing a response to mark as the "Accepted Solution."

jmoody77
Calcite | Level 5
ahh...sorry; done!

sas-innovate-2024.png

Available on demand!

Missed SAS Innovate Las Vegas? Watch all the action for free! View the keynotes, general sessions and 22 breakouts on demand.

 

Register now!

Multiple Linear Regression in SAS

Learn how to run multiple linear regression models with and without interactions, presented by SAS user Alex Chaplin.

Find more tutorials on the SAS Users YouTube channel.

From The DO Loop
Want more? Visit our blog for more articles like these.
Discussion stats
  • 8 replies
  • 863 views
  • 0 likes
  • 2 in conversation