In this context (CREATE and APPEND), you do not want the value of the matrix, just its name.  So you would need to modify the code as follows: do k = 1 to 2;   call execute('create ' + CD +'  from ' + C + ';');   call execute('append from ' + C + ';') ;   call execute('close ' + CD + ';' ); end; and because of the call execute, you would have to use this code within a module.  Be warned this could get messy if you are creating hundreds of data sets,  why not save the original matrix from the beginning of this thread as a SAS data set? ... View more

If you are prepared to sacrifice readability of code, then you can create a character matrix y that contains the names of each matrix in your list. Then you can use the VALUE() function and the VALSET() call every time you want to refer to those matrices.   x = { 1 0.0036813 0.0030405 , 1 0.0030405 0.0032935 , 2 0.0015928 0.0017098 , 2 0.0017098 0.0020805 };   y = 'y_1' : 'y_2';  /* define matrix NAMES */   r = 1:2;   do k = 1 to 2;     call valset( y , x[r, 2:3]);      /* create the matrix y_k */     print (value( y ));              /* print y_k */     call valset( y , value(y )#2 ); /* multiply by 2 */     print (value( y ));              /* print again */     r = r + 2;   end; I can see Rick's point that it will be easier to store all the results in once place, however I would prefer to have a choice and for there to be an easier way to create a list of matrices, where each element can be a matrix of any size. ... View more

Try printing out the matrix x.  It has only 1 row and 12 columns because you have not used commas to determine where one row of the matrix ends and the next begins. ... View more

There is no equivalent of the by statement, instead you need to use a loop and matrix sub-setting to extract the data required.  For example if all 12 numbers above are in a 4x3 matrix called x, then the following would extract the 2x2 matrices that you want to work further with. r = 1:2;  /* row numbers to extract from x */ do boot = 1 to 2;   y = x[r, 2:3];   < do stuff here with the 2x2 matrix y >   r = r + 2;  /* increment row numbers for next boot */ end; ... View more

I think you can work entirely within IML, since it is essentially a 'multiplication' of the matrix from the ALLCOMB(n, k) function, with the matrix from the ALLPERM(k) function.  Again row order is likely to be different.   start permutations(n, k); c = allcomb(n, k); p = allperm( k ); nc = nrow( c ); np = nrow( p ); a = do(0, nc # k - 1, k)` @ j(np, k) + repeat(p, nc); return ( shape( c[a], nc # np, k) ); finish; print (permutations( 4, 3) );      [Edited January 2017 - at some point part of the return statement became a link and the code was no longer rendered correctly - converted to a SAS code box to fix the problem] ... View more

You can access and manipulate each matrix independently. Rick's point is that doing everything indirectly is going to look ugly, as I imagine the actual problem you are trying to solve is a lot more complicated than the three 5s, and you will end up with a program full of VALSET and VALUE which will be difficult for others to understand.  For example:   /* create an 'array' of 3 identity matrices of varying size */   y = 'y_1' : 'y_3';   do k = 1 to 3;     call valset( y , i(k) );     print (value(y ));   end;   /* scale each identity matrix in the array by its side,      for the 2nd matrix also add 5 to all elements */   do k = 1 to 3;     call valset( y , value(y ) # k );     if k = 2 then call valset (y , value(y ) + 5 );     print (value(y ));   end; ... View more

Just realised that once the matrices are defined as above, it's possible to refer to them using the value function.   So   x = 2 # value(yname[2]); would set x to 10. ... View more

The macro variable &mc_k will be resolved just once at compile time, so I presume it had the value 3 before you ran the code above (the macro variable here is not the same as the matrix mc_k).  There is no need to resort to the SAS macro language as you can use call execute to get what you want.  A line of IML code can be constructed in a character matrix and then you can have IML execute it, but note that this will only work from inside a module. start yarray;   yname = 'y_1' : 'y_3';   do k = 1 to 3;     call execute( yname + '=5;' );   end;   print y_1 y_2 y_3; finish; run yarray; In a real application the code quickly becomes difficult to read, since every time you refer to y_k, there will be a complicated call execute statement.  In the case where all the matrices are the same size, then it may be better to store them as sub-matrices of a single large matrix. I only wish there was a better way to have an array of matrices. ... View more

The function DET returns the determinant of a matrix,  so: if det(x)^=0 then ...... but you may want to exclude small determinants if whatever you want to do is likely to give stability problems with near-singular x, so something like: if abs(det(x))>s where s is some suitable small value. ... View more

I stand by H || K1, and think there are other problems with your code.   Test out the function TLAINT before you start calling it from QUAD, and check the value of 'C'.   Also move the module called EXPN, you can not have it inside a loop. ... View more

The cause of the error is that you are specifying a character matrix for the interval for integration, you are passing the names of the matrices instead of the values.  You need to use a 2 value numeric vector with the left and right hand limits.  So replace INTERVAL = {H K1}; with INTERVAL = H || K1; I think there are other errors in your code.  For example the loop "DO I = 1 TO NROW(HH);" has no matching END statement. ... View more

I have a function to do exactly this.   It creates the 'shift' that you mention by manipulating the row index numbers of a matrix in its sparse format (details of the format can be found in the documentation for the SPARSE function).  The result when converted back to normal form, with the FULL function, is a new matrix with one row for every antidiagonal.  Finally it is a simple matter to sum the rows of this new matrix. start AntiDiagSum( x );   nr = nrow( x );   nc = ncol( x );   s = sparse( x );   s[ , 2] = s[ , 2] + s[ , 3] - 1;   return( full( s , nr + nc - 1, nc )[ , +] ); finish; /* get example 3x5 matrix */ a = shape(1:15, 3, 5); print a; /* required sums are from the 2nd to 5th antidiagonals */ b = AntiDiagSum(a)[2:5]; print b; ... View more

Yes, I do like Rick's suggestion of the zero coefficients for the higher powers, plus there is a completely natural way of getting there using the gamma function. This, I imagine, is the mechanism behind the Wolfram Alpha and Google implementations of the n choose r function.  If I write my own version of the COMB function: start icomb(n, r);   return( round( gamma(n + 1) / (gamma(r + 1) # gamma(n - r + 1 + 1E-12)))); finish; all I need to do is add on a tiny amount to prevent any attempt at evaluating the gamma function exactly at zero, or at one of the negative integers, round the result and I have a function that can handle r>n.  Then I can have that elusive one line solution for the generation of the Pascal matrix. Start Pascal( n );   return( icomb( row(j(n,n)) - 1, col(j(n,n)) - 1 ) ); finish; print ( Pascal(12) ) [f=3.0]; As for utility of the modified function, I admit it is limited as I have been using the COMB function for some years and this is the first time I have tried using it with r>n, but my expectation was that it would perform like ICOMB above. ICOMB may not be robust and only work with smaller n, and my original Pascal function contains 3n^2 unnecessary multiplications in its desperate attempt to retain 'elegance' while circumventing the invalid argument issues, so thanks to Rick for the alternative suggestions.  Actually I prefer the shorter function with the loop rather than the fully vectorized version! ... View more