Convolution of Two Vectors

[vioravis]

Are there any functions in Base SAS (I do not have SAS enterprise miner) to compute the convolution of two vectors???

Please guide. Thank you.

[art297]

I don't think there is in base sas, but there is if you license IML (e.g., http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/viewer.htm#langref_sect138.htm ).

[vioravis]

Thanks art297. I do not have IML. Looking for a way to implement it in Base SAS itself.

Is there a way to implement the vector operations shown below in Base SAS??

In the example below, data table test1 contains three variable Variable1, Variable2 and Variable3.

DATA test1;

set test1;

DO i = 1 to num;  

set test1 nobs point = i;

   variable1=  sum(variable2[1:i-1]*variable3[(num-i+2)]:num)];

END;

RUN;

Thank you.

[art297]

Of course there is, but you would have to show a small example with some data and what should go into the calculations.  e.g., when i eq 1 what value should the calculation sum(variable2[1:i-1] get?

What is the maximum number of records you will have?  If it isn't too big, an easy solution might be to just preload the entire table into an array.

[vioravis]

Hi art297, my dataset has close to 10,000 observations.  The following is a sample dataset:

DATA test;

input X Y Z;

cards;

1 10 0

2 20 0

3 30 0

4 40 0

5 50 0

;

run;

I have also given some sample calculations below:

Iteration 1:

SUM(X[1:1]*Y[5–2+2:5]) = SUM(X[1:1]*Y[5:5]) = SUM([1]*[50]) = 50

Iteration 2:

SUM(X[1:2]*Y[5–3+2:5]) = SUM(X[1:2]*Y[4:5]) = SUM([1,2]*[40,50]) = SUM(40+100) = 140

Iteration 3:

SUM(X[1:3]*Y[5–4+2:5]) = SUM(X[1:3]*Y[3:5]) = SUM([1,2,3]*[30,40,50]) = SUM(30+80+150) = 260

… so on

The final dataset after all the calculations will look like this

DATA test;

input X Y Z;

cards;

1 10 0

2 20 50

3 30 140

4 40 260

5 50 400

;

run;

Thank you.

[vioravis]

In my previous post, Z corresponds to Variable 1, X corresponds to Variable 2 and Y corresponds to Variable 3.

Thanks.

[Tom]

This may not be the fastest way, but it should work for any size vectors.

DATA test;

input X Y expZ;

cards;

1 10 0

2 20 50

3 30 140

4 40 260

5 50 400

;

run;

data want ;
  set test nobs=nobs ;
  z=0;
  i=_n_-1;
  do p1=1 to i;
    p2=nobs -i + p1 ;
    set test (keep=x rename=(x=x1)) point=p1;
    set test (keep=y rename=(y=y1)) point=p2;
    z1=(x1*y1);
    z=sum(z,z1);
   *  put (_n_ i p1 p2 x1 y1 z1 z) (=);
  end;
run;
proc print width=min;
run;

Obs    X     Y    expZ      z    i    x1    y1     z1
 1     1    10       0      0    0     .     .      .
 2     2    20      50     50    1     1    50     50
 3     3    30     140    140    2     2    50    100
 4     4    40     260    260    3     3    50    150
 5     5    50     400    400    4     4    50    200

[art297]

The following doesn't look as tidy as Tom's suggested code but, with 10000 records, took 6 seconds on my machine, while Tom's approach took over a minute:

/*create some test data*/

DATA test (drop=i);

  input X Y Z;

  do i=1 to 2000;

    output;

  end;

cards;

1 10 0

2 20 0

3 30 0

4 40 0

5 50 0

;

data _null_;

if 0 then set test nobs=nobs;

CALL SYMPUT('NUMREC',nobs);      /*** put # of records into NUMREC macro var ***/

stop;                            /*** stop, got number of records ***/

run;

/* now read values of x and y into arrays, and then

   reread test and do calculations*/

data want (drop=_:);

    array _xval(&numrec); /* create two arrays with same number of

                                        elements as there are records in test

                                    */

    array _yval(&numrec);

    i=0;

    do until (eof1);      /* load the array with x and y values */

      set test end=eof1;

      i+1;

      _xval(i)=x;

      _yval(i)=y;

    end;

    _rec=-1;

    do until (eof2);      /* read in each record separately */

      set test end=eof2;

      _rec+1;

      z=0;

      do _i=1 to _rec;

        z+_xval(_i)*_yval(&numrec-_rec+_i);

      end;

      output;

    end;

run;

[vioravis]

Thanks a lot Tom and Art. I am looking at Art's solution since it is faster (For a record with 8000+ records it takes around 1.9 secs). Speed is very critical since this is going to be called multiple times for optimization. Any faster solution is welcome.

I am trying out one more enhancement to the code with an additional variable A as follows (A is just another variable in the dataset):

DO i = 1 to num;  

set test nobs point = i;

   Z =  sum(X[1:i-1]*Y[(num-i+2)]:num)];

  X = A + Z;

END;

RUN;

I tried the following but it isn't giving the expected solution:

data _null_;

if 0 then set test nobs=nobs;

CALL SYMPUT('NUMREC',nobs);      /*** put # of records into NUMREC macro var ***/

stop;                            /*** stop, got number of records ***/

run;

/* now read values of x and y into arrays, and then

   reread test and do calculations*/

data want (drop=_:);

    array _xval(&numrec); /* create two arrays with same number of

                                        elements as there are records in test

                                    */

    array _yval(&numrec);

   array _Aval(&numrec);

    i=0;

    do until (eof1);      /* load the array with x and y values */

      set test end=eof1;

      i+1;

      _xval(i)=x;

      _yval(i)=y;

      _Aval(i) = A;

    end;

    _rec=-1;

    do until (eof2);      /* read in each record separately */

      set test end=eof2;

      _rec+1;

      z=0;

      do _i=1 to _rec;

        z+_xval(_i)*_yval(&numrec-_rec+_i);

        _xval(_i) = _Aval(_i) + z;

      end;

      output;

    end;

run;

Can someone please point out the correct approach???

Thank you

[art297]

I haven't looked at all of your code, but noticed that you never declared the new array.  Make sure that you add one.  e.g.

    array _xval(&numrec); /* create THREE arrays with same number of

                                        elements as there are records in test

                                    */

    array _yval(&numrec);

    array _Aval(&numrec);

[vioravis]

Art,

That was a typo in the code I pasted here. In the actual code have declared it. It still doesn't work.

Thank you.

[art297]

Was this also just a typo?:

        -xval(_i) = _Aval(_i) + z;

Or should it have been _xval(_i) = _Aval(_i) + z;

And, of course, that will use the new values in all of the later calculations.  Is that what you want?

[vioravis]

Oops...sorry. That was another typo.

New values are to be used in the later calculation. But only the current i is replace in each iteration.

Thank you.

[art297]

Then, I suggest that you post an expansion of your original sample data, including the new variable, and what you want as an output.

[vioravis]

Art,

Please find below the new data and the expected output.

DATA test;

input X Y A Z;

cards;

1 10 0.1 0

2 20 0.2 0

3 30 0.3 0

4 40 0.4 0

5 50 0.5 0

;

run;

DATA results;

input X Y A Z;

cards;

1.0 10 0.1    0

50.2 20 0.2   50

2550.3 30 0.3 2550

129553.4 40 0.4  129553

6581208.5 50 0.5 6581208

;

run;

Thank you