## Statistical programming, matrix languages, and more

Occasional Contributor
Posts: 11

Hi all. Basically im doing coding for boxplot for purpose outlier detection. I dont know what went wrong with my coding, no results no errors. I have tried FLOOR func to replace LOC, nothing happen still. code LEFT meaning, the data that clear from outlier as LOWFEN and UPPFEN are the fences to detect outlier. So how can i print the LEFT(data without outlier). my data contain 3 group.

PROC IML;
RESET NONAME;

START BP(A) GLOBAL (NY,NCOL,RMEAN,VARX,N);
RMEAN=J(NCOL(NY),1,0);
Q1=RMEAN;
Q3=RMEAN;
LOWFEN=RMEAN;
UPPFEN=RMEAN;
LEFT=RMEAN;
F=1;
M=0;

DO J=1 TO NCOL(NY);

SAMP=NY[J];
L=M+SAMP;
TEMP=A[F:L];

Q=quartile (TEMP);
Q1[J]=Q[2,1];
Q3[J]=Q[4,1];
LEFT[J]=TEMP[LOC(LOWFEN[J]<TEMP & TEMP<UPPFEN[J])];

M=L;
F=F+SAMP;

END;

PRINT LEFT;

FINISH;

********cubaan menggunakan data yg dijana*******;

NY = {11 11 11};
A = {5,8,7,3,9,4,3,29,5,6,7,
3,2,6,4,14,4,7,6,9,3,4,
9,9,8,7,10,11,27,12,15,17,16};

QUIT;

SAS Super FREQ
Posts: 508

You did not print anything.  That is why nothing is printed.  You defined BP and two vectors, then you quit.

SAS Super FREQ
Posts: 4,274

First, you need to call the BP module before you quit:

RUN BP(A);

Second, in your module, you have the statement

LEFT[J]=TEMP[LOC(LOWFEN[J]<TEMP & TEMP<UPPFEN[J])];

The left-hand side of the assignment is a scalar. However, the right-hand side is *in general) a vector. You cannot assign a vector into a single element, so you will get

ERROR: (execution) Matrices do not conform to the operation.

which means that you are attempting a vector operation that doesn't make sense.

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