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05-18-2010 09:42 AM

I want to know if there is any SAS/IML function which returns the row and column location of an element with a M x N or N x N Matrix.

For example, I want to know the location of the smallest element in the matrix A. I can find the element using

>>a=loc(min(A));

This returns the elements value, but not the position within the matrix.

Even when I try

>>b=ncol(min(A));

>>c=nrow(min(A));

SAS/IML only return column=1 row=1, since it still recognizes the elements as a value, separate from the Matrix.

Appreciate any help you could give.

For example, I want to know the location of the smallest element in the matrix A. I can find the element using

>>a=loc(min(A));

This returns the elements value, but not the position within the matrix.

Even when I try

>>b=ncol(min(A));

>>c=nrow(min(A));

SAS/IML only return column=1 row=1, since it still recognizes the elements as a value, separate from the Matrix.

Appreciate any help you could give.

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Solution

06-30-2017
03:51 PM

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05-18-2010 11:43 AM

Put the condition that you are searching for inside the LOC statement, as in

idx = loc( A = min(A) ); /* indices where A equals min(A) */

or, in general,

idx = loc( A = value ); /* indices where A equals 'value' */

The previous statements return ALL indices that satisfy the condition.

It turns out that for "index of minimum" and "index of maximum" there is a subscript reduction operator that returns the FIRST index that satisfies the condition. See

http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/workmatrix_sect14.htm

idx = loc( A = min(A) ); /* indices where A equals min(A) */

or, in general,

idx = loc( A = value ); /* indices where A equals 'value' */

The previous statements return ALL indices that satisfy the condition.

It turns out that for "index of minimum" and "index of maximum" there is a subscript reduction operator that returns the FIRST index that satisfies the condition. See

http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/workmatrix_sect14.htm

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Solution

06-30-2017
03:51 PM

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05-18-2010 11:43 AM

idx = loc( A = min(A) ); /* indices where A equals min(A) */

or, in general,

idx = loc( A = value ); /* indices where A equals 'value' */

The previous statements return ALL indices that satisfy the condition.

It turns out that for "index of minimum" and "index of maximum" there is a subscript reduction operator that returns the FIRST index that satisfies the condition. See

http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/workmatrix_sect14.htm