Hi ... here's another idea.  Just sort by CLAM and IND and then check if the first IND within a CLAM is an N or a Y.  If it's an N, keeps the Ns.  If it's a Y, keep the Ys. data have; input CLAM  LNNO : $3. IND : $1.; datalines; 12345    001           Y 12345    002           Y 12345    003           Y 23456    001           Y 23456    002           N 34567    001           Y 34567    002           N 34567    003           N 78968    001           Y 78968    002           Y 78968    003           Y ; run; proc sort data=have; by clam ind; run; data want; do until (last.clam);    set have;    by clam;    if first.clam and ind eq 'N' then hold = 1;    if ind eq 'Y' and missing(hold) or       ind eq 'N' and ^missing(hold) then output; end; drop hold; run; proc print data=want noobs; run; CLAM    LNNO    IND 12345    001      Y 12345    002      Y 12345    003      Y 23456    002      N 34567    002      N 34567    003      N 78968    001      Y 78968    002      Y 78968    003      Y ... View more

Hi ... you don't have to use an ARRAY.  Here's another idea (with only five questions, but it'll work with 20) ... data test; input q1-q5; datalines; 4 4 4 4 4 . 2 3 1 4 4 . . . 4 . . . . . 1 2 3 4 5 . 4 4 4 . ; run; data test; set test; questcat = ifn(verify(catt(of q:),'.4') or sum(of q:) eq . , 9 , 3); run; proc print data=test noobs; run; q1    q2    q3    q4    q5    questcat 4     4     4     4     4        3 .     2     3     1     4        9 4     .     .     .     4        3 .     .     .     .     .        9 1     2     3     4     5        9 .     4     4     4     .        3 More info at ... "Searching for Variable Values with CAT Functions:  An Alternative to Arrays and Loops" http://www.nesug.org/Proceedings/nesug09/ff/ff04.pdf ... View more

Hi ... OK, upon a little more thought,  I think that this works with ties.  If there are multiple large values, this allows up to 10 variables  (Y1-Y10) to hold the 5 largest values and variable names (Z1-Z10) ... * some data ... restricted range of X1-X20 leads to some duplicate values within an observation; data x; array x(20); do j=1 to 10; do k= 1 to 20;    x(k) = ceil(100*ranuni(999)); end; output; end; keep x1-x20; run; data y; set x; array x(20); array y(10); array z(10) $32; do j = 1 by 1 until(count eq 5 or j eq 10);    y(j) = largest(j, of x:);    var = whichn(y(j),of x:);    z(j) = vname(x(var));    if x(var) ne hold then do; count = sum(count,1); hold=x(var); x(var) = 1e6 ; end; end; keep y: z:; run; ... View more

Hi ... not sure how you deal with TIES (multiple variables with same value within an observation), but if there are not any duplicate large values within an observation, this should work ...  * some data; data x; array x(20); do j=1 to 10; do k= 1 to 20;    x(k) = ceil(10000*ranuni(999)); end; output; end; keep x1-x20; run; * use WHICHN and VNAME to get the variable names ... put into variables z1-z5); data y; set x; array x(20); array y(5); array z(5) $32; do j = 1 to 5;    y(j) = largest(j, of x:);    var = whichn(y(j),of x:);    z(j) = vname(x(var)); end; keep y: z:; run; y1      y2      y3      y4      y5     z1     z2     z3     z4     z5 8968    7881    7777    7443    6810    x16    x19    x1     x9     x5 9990    9956    9568    9384    9257    x9     x3     x18    x4     x16 9297    8066    7714    7355    7072    x11    x9     x10    x18    x4 9969    9903    8598    8386    8184    x10    x7     x1     x20    x18 9947    9512    9398    8740    7714    x11    x17    x18    x1     x4 9609    9528    8581    7187    7028    x10    x4     x9     x5     x7 9550    9403    8798    8024    5612    x9     x6     x12    x17    x2 9727    9605    9173    8484    8360    x14    x18    x13    x3     x2 9279    9091    9010    8755    8178    x2     x10    x6     x14    x17 9898    8788    8753    8444    8188    x13    x20    x16    x19    x17 I guess if there were ties , you could resort to  transposing the data and using PROC RANK.  You could end up with more than 5 largest for each observation (or do something more elaborate with the above ARRAY based code). ... View more

Hi ... here's a way to add a blank line.  The OPTIONS MISSING is used to keep periods out of that blank line in the columns for the numeric variables.  You'll get a WARNING in the LOG,  ... WARNING: A table has been extended with null columns to perform the UNION ALL set operation Image of HTML output is attached. data x; input SUBSYS : $6. CAP ALLOC AVAL; datalines; 2107#1 67.94 49.26 18.68 2107#2 67.94 50.93 17.01 2107#3 69.39 39.54 29.85 2107#4 67.24 24.94 42.30 ; run; ods listing close; ods results off; ods html file='z:\table.html' style=barrettsblue; options missing=' '; proc sql; select  SUBSYS, CAP, ALLOC, AVAL format=6.2, 100*alloc/cap as PERCENT format=4.1 from x union all select distinct ' ' as subsys from x union all select 'TOTAL' as subsys, sum(cap), sum(alloc) , sum(aval), 100*sum(alloc)/sum(cap) from x; quit; options missing='.'; ods html close; ods results; ods listing; ... View more

Hi ... there was a PROC SQL suggestion, so ... You use PROC SQL just like PROC PRINT, just don't create a table. Since one of the values of AVAL ends in a zero, a format (6.2) is used to assure two decimal places in the table (the width also accomodates the SUM). Otherwise ... select  * , 100*alloc/cap as PERCENT format=4.1 from x would suffice.  PROC SQL uses the variable names as entered, so they are all in CAPS. data x; input SUBSYS : $6. CAP ALLOC AVAL; datalines; 2107#1 67.94 49.26 18.68 2107#2 67.94 50.93 17.01 2107#3 69.39 39.54 29.85 2107#4 67.24 24.94 42.30 ; run; proc sql; select  SUBSYS, CAP, ALLOC, AVAL format=6.2, 100*alloc/cap as PERCENT format=4.1 from x union all select 'TOTAL' as subsys, sum(cap), sum(alloc) , sum(aval), 100*sum(alloc)/sum(cap) from x; quit; SUBSYS       CAP     ALLOC    AVAL  PERCENT ------------------------------------------- 2107#1     67.94     49.26   18.68     72.5 2107#2     67.94     50.93   17.01     75.0 2107#3     69.39     39.54   29.85     57.0 2107#4     67.24     24.94   42.30     37.1 TOTAL     272.51    164.67  107.84     60.4 ... View more

Hi.  Another question ... what is the LENGTH of CAU, 3 or 4?  Since ICD10 cause of death codes have four characters, it is possible thaty CAU has a length of 4.  Then Art's solution ... where cau in ('I00', 'I01', 'I02', 'I03', 'I04', 'I05', 'I06', 'I07', 'I08','I09') or cau=('I11') or cau=('I13') or('I20'<cau<'I51'); should be changed (I agree with "other Art" about the I20 to I59 range needing the =s) ... where cau in : ('I00', 'I01', 'I02', 'I03', 'I04', 'I05', 'I06', 'I07', 'I08','I09') or       cau eq : 'I11' or cau eq : 'I13' or       'I20' <= cau <= : 'I51'); or the "other Art' ... where ('I00'<=cau<= : 'I09') or cau in : ('I11','I13') or ('I20'<=cau<= : 'I51'); ... View more