SAS code query

[Mubatsiri]

Good Day

I am running the following code:

data ass3;

input income1 income2 house stand double prop;

cards;

  521.502    118.348    735.779     920.53      0     1215.86

   14.116    457.801    413.522     690.15      0      917.67

  308.237    205.341    567.238     903.11      1     1201.16

  449.589    157.470    496.226     659.05      1     1099.73

   47.286    555.871    414.292     769.92      0     1077.25

   12.702    400.744    283.223     539.62      1      973.28

  303.539    360.630    671.121     934.58      0     1206.01

  325.548    369.610    284.473     707.85      0     1071.03

  328.079     17.192    492.552     699.23      0      834.60

  479.735     34.212    767.408    1097.32      0     1102.11

   70.381    319.148    373.140     760.19      0      774.12

  232.232    255.517    238.515     577.39      0      828.64

   56.125    326.705    589.865     930.42      0     1020.10

  510.569     36.773    461.059     920.65      0     1044.39

   15.890    353.851    345.385     655.05      1      932.65

  298.906    126.398    531.592    1093.24      0     1036.68

  280.401    105.089    497.296     727.87      0      867.00

  188.411    419.229    383.097     903.32      0     1114.49

   11.004    462.602    351.969     575.44      0      833.46

  408.952    119.757    650.882     950.26      0     1044.39

  114.999    253.868    439.853     849.32      0      831.60

  200.932    141.234    400.907     571.64      0      773.70

  276.907    350.366    554.191     948.33      1     1273.24

  271.076    109.235    734.862     970.72      1     1124.66

  357.141    324.151    507.147     686.02      0     1146.86

   74.029    403.535    372.881     520.79      0      836.79

  112.752    195.755    550.987    1048.71      0     1023.95

  189.496    273.100    400.458     550.31      0      834.02

  283.516    395.697    445.404     600.35      0     1064.84

  255.701    154.743    535.123    1078.51      0     1075.30

  ;

run;

proc iml;

use ass3;

read all into EKT3;

print EKT3;

JI = EKT3[,1]+EKT3[,2];   /* to create the joint the joint income variable */

rat = EKT3[,3]/EKT3[,4];  /* to create the ratio variable as defined */

n=30;

p= 4; /* number of indepndenr varibles excluding intercept */

k=5;       /* number of variables */

m = 2;    /* number of linear restrictions */

/*part a */

In = J(nrow(EKT3),1,1);

X = In || JI || EKT3[ ,4] || rat || EKT3[ ,5];

Y = EKT3[ ,6];

bhat = inv(X`*X)*X`*y;

print bhat;

bnew = J(5,n,.);

/*part b */

do i = 1 to n;

if i = 1 then do;

Xnew = X[2:n,];

ynew = EKT3[2:n,];

end;

if  i = n then do ;

Xnew = X[1:n-1,];

ynew = EKT3[1:n-1,];

end;

else do;

Xnew = X[1:i-1,] || X[i+1:n,];

ynew = EKT3[1:i+1,] || EKT3[i+1:n,];

end;

bhat2 = ginv(Xnew`*Xnew)*Xnew`*ynew;

bnew[,i]= bhat2;

end;

print bnew;

quit;

I keep getting the error message

ERROR: (execution) Invalid subscript or subscript out of range.

operation : [ at line 1588 column 13

operands  : X, *LIT1024, _TEM1001,

X     30 rows      5 cols    (numeric)

*LIT1024      1 row       1 col     (numeric)

         1

_TEM1001      1 row       1 col     (numeric)

         0

statement : ASSIGN at line 1588 column 5

1594  print bnew;

1595  quit;

         0

I have tried to fix the error but it is somehow still not running. Any advise on how I can get the code to run smoothly?

[RW9]

Move your question to the SAS/IML section and you will get more coverage of people who use that technology.

[Rick_SAS]

In the following statements

else do;

Xnew = X[1:i-1,] || X[i+1:n,];

...

The ELSE clause is being executed when i=1, which implies that te vector 1:i-1 evaluates to {1 0}, which is causing the "subscript out of range" error.

It looks like your loop is attempting to omit the i_th row. You can use the SETDIF function to generate the vector

rows = setdif(1:n, i);  /* rows except the i_th */

For a detailed explanation, see the article

"Negative indexing" in SAS/IML: Excluding elements from an array

[Mubatsiri]

Thanks Rick it worked

[Ksharp]

You missed ELSE before " if  i = n " ,

and you want // (vertical operator) not || (horizontal)  right ?

and at the end of code , you need to consider about

bnew[, i]=bhat2;

which bhat2 is not column vector.

Code: Program

data ass3;
   input income1 income2 house stand double prop;
   cards;
  521.502 118.348 735.779 920.53 0 1215.86
  14.116 457.801 413.522 690.15 0 917.67
  308.237 205.341 567.238 903.11 1 1201.16
  449.589 157.470 496.226 659.05 1 1099.73
  47.286 555.871 414.292 769.92 0 1077.25
  12.702 400.744 283.223 539.62 1 973.28
  303.539 360.630 671.121 934.58 0 1206.01
  325.548 369.610 284.473 707.85 0 1071.03
  328.079 17.192 492.552 699.23 0 834.60
  479.735 34.212 767.408 1097.32 0 1102.11
  70.381 319.148 373.140 760.19 0 774.12
  232.232 255.517 238.515 577.39 0 828.64
  56.125 326.705 589.865 930.42 0 1020.10
  510.569 36.773 461.059 920.65 0 1044.39
  15.890 353.851 345.385 655.05 1 932.65
  298.906 126.398 531.592 1093.24 0 1036.68
  280.401 105.089 497.296 727.87 0 867.00
  188.411 419.229 383.097 903.32 0 1114.49
  11.004 462.602 351.969 575.44 0 833.46
  408.952 119.757 650.882 950.26 0 1044.39
  114.999 253.868 439.853 849.32 0 831.60
  200.932 141.234 400.907 571.64 0 773.70
  276.907 350.366 554.191 948.33 1 1273.24
  271.076 109.235 734.862 970.72 1 1124.66
  357.141 324.151 507.147 686.02 0 1146.86
  74.029 403.535 372.881 520.79 0 836.79
  112.752 195.755 550.987 1048.71 0 1023.95
  189.496 273.100 400.458 550.31 0 834.02
  283.516 395.697 445.404 600.35 0 1064.84
  255.701 154.743 535.123 1078.51 0 1075.30
  ;
run;

proc iml ;
   use ass3;
   read all into EKT3;
   print EKT3;
   JI=EKT3[, 1]+EKT3[, 2];

   /* to create the joint the joint income variable */
   rat=EKT3[, 3]/EKT3[, 4];

   /* to create the ratio variable as defined */
   n=30;
   p=4;

   /* number of indepndenr varibles excluding intercept */
   k=5;

   /* number of variables */
   m=2;

   /* number of linear restrictions */
   /*part a */
   In=J(nrow(EKT3), 1, 1);
   X=In || JI || EKT3[ , 4] || rat || EKT3[ , 5];
   Y=EKT3[ , 6];
   bhat=inv(X`*X)*X`*y;
   print bhat;
   bnew=J(5, n, .);

   /*part b */
   do i=1 to n;

   if i=1 then
   do;
   Xnew=X[2:n, ];
   ynew=EKT3[2:n, ];
   end;
   else if i=n then
   do;
   Xnew=X[1:n-1, ];
   ynew=EKT3[1:n-1, ];
   end;
   else
   do;
   Xnew=X[1:i-1, ]//X[i+1:n, ];
   ynew=EKT3[1:i-1, ]//EKT3[i+1:n, ];
   end;
   bhat2=ginv(Xnew`*Xnew)*Xnew`*ynew;
   bnew[, i]=bhat2[1];
   end;
   print bnew;
   quit;