About mkeintz

mkeintz · ‎12-05-2016

Show the log please, including the error message. Can't diagnose what can't be seen.

mkeintz · ‎12-04-2016

data want (drop=nxt_: total_: I grpsize); do grpsize=1 to 1000; merge have (drop=count) have (firstobs=2 keep=studyid environment startdate rename=(studyid=nxt_i environment=nxt_e startdate=nxt_s)); total_days=sum(total_days,daysbwn); if not(studyid=nxt_i and environment=nxt_e and nxt_s=enddate+1 and environment='SCHOOL') then leave; end; daysbwn=total_days; if studyid^=lag(studyid) then count=1; else count+1; do I=1 to grpsize; set have (drop=count daysbwn mvmtdate enddate); if I=1 then output; end; run; This program has two loops in the data step The first loop gets the grpsize and calculates the totaldays. The second loop re-read the same records, but outputs only the first in the group. Notes: The MERGE in the first loop merges HAVE with itself. The "right-hand" HAVE starts at observation 2, providing a way to know when the current record is not part of the same group as the next record. During that loop TOTALDAYS is accumulated. The "do grpsie=1 to 1000; I use 1000 as a number that should exceed any record count for a single group. Inside the loop is an IF test the leaves the loop when the condiition is met. Between the loops DAYSBWN is set to totaldays. COUNT is reset to 1 when a new studyid is started, otherwise it's just incremented. In the second loop the SET statement does not include the variables daysbwn mvmtdate enddate which means those values will be taken from the end-of-group record read in by the first loop. But it does take all the other vars which means that they come from the first record in a group.

mkeintz · ‎12-04-2016

Perhaps you can find a way to allocate the full24 points to the winner of your outsourcing.

mkeintz · ‎12-04-2016

You could use, say, the 4-quarter trailing average for the jump-off date.

mkeintz · ‎12-04-2016

You have shown us the source table. Now show us explicitly what the result table would look like.

mkeintz · ‎12-03-2016

Aside from the missing semi-colon for the ARRAY TWO statement as mentioned by Tom, I have these observations on the messages you get. It says "syntax error. statement will be ignored". That means the "array two" statement is ignored. So the subsequent attempt to use the expression two{n} generates the "too many subscripts" messages. All due to the missing semicolon. But I actually would like you to consider an alternative based on these observations. You obviously want the cross-products of ONE put into the upper triangle of a 2-dimensional matrix. But you've defined that upper triangle as a one-dimensional array. Instead make the calculation easier by defining a 2-dimensional array ("array two{19,19"}. You can still calculate only the upper triangle. You don't need to use the unintuitive N to index array TWO. Just use a row index and a column index. Use the power of SAS to generate variable name lists to minimize your risk of typing error (e.g. g3g4-g3g15 generates all the names from g3g4, g3g5, g3g6 .... g3g14, g3g15). Use the names DUM1, DUM2 ... etc. for the lower triangle and diagonal elements of the matrix. Then you can drop them from output with a very simple statement. Note you can reuse the same name for multiple elements of the array. SAS doesn't care. data new1; set new; array ONE[*] E1-E4 G1-G15; array TWO[19,19] dum1 e1e2-e1e4 e1g1-e1g15 dum1-dum2 e2e3-e2e4 e2g1-e2g15 dum1-dum3 e3e4 e3g1-e3g15 dum1-dum4 e4g1-e4g15 dum1-dum5 g1g2-g1g15 dum1-dum6 g2g3-g2g15 dum1-dum7 g3g4-g3g15 dum1-dum8 g4g5-g4g15 dum1-dum9 g5g6-g5g15 dum1-dum10 g6g7-g6g15 dum1-dum11 g7g8-g7g15 dum1-dum12 g8g9-g8g15 dum1-dum13 g9g10-g9g15 dum1-dum14 g10g11-g10g15 dum1-dum15 g11g12-g11g15 dum1-dum16 g12g13-g12g15 dum1-dum17 g13g14-g13g15 dum1-dum18 g14g15 ; do r=1 to dim(one); do c=r+1 to dim(one); two{r,c}=one{r}*one{c}; end; end; drop dum1-dum18 r c; run; And out of curiosity why aren't you getting the squares as well as the cross-products?

mkeintz · ‎12-03-2016

You want rolling J-month cumulative returns for each permno. It looks to me like you are usuing the CRSP monthly returns data, not the daily returns. Given that CRSP data is almost always sorted by permno/date, you can use a DATA step to process the data much faster. This program untested program makes the following assumptions. data sorted by permno/date the data are monthly RET is never missing The second example works even when there are "holes" (i.e. missing months between the beginning and end of a permno). Your data has no more than 241 months for any given permno data umd (drop=month0_date m); set commonstocks; by permno; /* define array to hold cumulative log(1+ret) starting month 0 up to 20 years*/ array sumlog_retplus1 {0:240} _temporary_ (); if first.permno then do; month0_date=date; do m=0 to 240; sumlog_retplus1{m}=0; end; end; retain month0_date; m=intck('month',month0_date,date); if m=0 then sumlog_retplus1{m}=log(1+ret); else sumlog_retplus1{m}=log(1+ret)+sumlog_retplus1{m-1}; if m<&J then delete; else if m=&J then cmret=exp(sumlog_retplus1{m})-1; else cmret=exp(sumlog_retplus1{m}-sumlog_retplus1{m-&J})-1; run; The second version accomodates holes in the monthly series: data umd (drop=month0_date m m2 lastm); set commonstocks; by permno; /* define array to hold cumulative log(1+ret) starting month 0 up to 20 years*/ array sumlog_retplus1 {0:240} _temporary_ (); if first.permno then do; month0_date=date; do m=0 to 240; sumlog_retplus1{m}=0; end; end; retain month0_date; m=intck('month',month0_date,date); lastm=lag(m); if m=0 then sumlog_retplus1{m}=log(1+ret); else if m=lastm+1 then sumlog_retplus1{m}=log(1+ret)+sumlog_retplus1{m-1}; else do m2=lastm+1 to m-1; else sumlog_retplus1{m2}=sumlog_retplus1{m2-1}; end; if m<&J then delete; else if m=&J then cmret=exp(sumlog_retplus1{m})-1; else cmret=exp(sumlog_retplus1{m}-sumlog_retplus1{m-&J})-1; run;

mkeintz · ‎12-03-2016

@Ksharp Nice, suggest this tweak for the second do group: _w=ifc(value>200,' ',w); /* instead of if value le 200 then do;_w=w;yes=1;end;*/ do until(last.flag); set temp; by flag notsorted; new=coalescec(_w,w); /* new=ifc(yes,_w,w);*/ output; end;

mkeintz · ‎12-02-2016

You don't want the second SET statement to prematurely end the data step, which it would do if it attempts to read beyond the end-of-file. So the SET statement inside the loop has an "end=eof" parameter, which sets the dummy variable EOF to 1 only if the last record is in hand. So I use "if eof=0 then set" to avoid reading beyond end of file. But of course the loop containing the data set might not have the UNTIL expression satisfied at the end-of-file. So the "if eof=0 then set" is followed by ELSE setting next_w to 'ZZZZZZ' and next_v>200 thereby satisfying the until expression and getting out of the loop.

mkeintz · ‎12-02-2016

One of your GROUP BY variables is product. So should you expect more that one distinct product within such a group?

mkeintz · ‎12-02-2016

No need to sort and lag- and then re-sort to original order. Just use a suitably define DO loop containing a SET statement to read ahead to find the desired value data have; input w :$20. value; datalines; abate 200 abbreviations 50 abdomen 20 abdominal 110 abelcet 100 aberrant 100 abiraterone 100 abl1 100 ablation 210 abnormal 30 abo 780 abscission 260 absence 100 absenteeism 140 absorption 130 abstinence 150 abstracting 110 academia 210 academic 30 academy 600 acceleration 180 access 130 accessibility 100 accident 200 accidents 230 accountability 50 accounting 100 accreditation 230 acculturation 30 ace 600 acetates 920 acetazolamide 310 acetic 110 acetone 210 acetyl 50 acetylaminofluorene 200 acetylatio 230 acetylation 200 bupropion 550 burden 560 burkitt's 230 burn 300 burns 140 businesses 550 butanoic 800 butterflies 500 butyrates 100 butyric 450 butyryl 800 bypass 330 bystander 300 c 300 c2 130 c2h2 150 c3h 250 c57bl 720 run; data want (drop=next_:); set have; length nw $20; retain nw ' '; if value<=200 and nw<w then do until (next_v>200 and next_w>=w); if eof=0 then set have (rename=(w=next_w value=next_v)) end=eof; else do; next_w='ZZZZZZ'; next_v=201; end; nw=lag(next_w); end; else if value>200 then nw=w; run; xxx

mkeintz · ‎12-02-2016

Why does w=accident/value=200 get nw=accident, yet w=abate/value=200 get nw=abl1?

mkeintz · ‎12-02-2016

If you get fewer "001.00" cases than expected, then perhaps the original visit coding has errors or incosistencies. I think you have to find a patient with no visit="001.00" to see whether there are other codes indicating baseline visit.

mkeintz · ‎12-02-2016

Instead of re-activating a zombie topic, start a new one, using appropriate title.

mkeintz · ‎12-02-2016

But proc reg will use ALL the non-missing values to make estimates. I think the OP just wants to use the nearest non-missing value on each end of the missing sequence.

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