SAS Support Communities

@PaigeMiller  Thank you so much for your help! ... View more

@Tom It is not a mistake. Once SAS produce the desired format, it will replace the other software.  ... View more

@Quentin Thanks for your comments. The desired format was produced by another software. I need to replicate the results using SAS. ... View more

@Reeza  Thanks much for looking into my question! The desired is when you view an portrait orientation table after 90 degree counterclockwise rotation.   I opened the PDF generated using the code in Adobe. In view-->rotation view-->counterclockwise, I got the desired format.  I am wondering how to get the desired format using SAS.   Thanks! ... View more

@PaigeMiller  Thanks again for your time.  I learned a lot from your code. As I mentioned in my previous message. My correlation matrix was from another source, which saved as .csv.  I used SAS to read in. If I directly apply your code, I can only get the correlations of the last row. That's why I need to figure out each step of your code, so that I can apply it or modify it as needed. I very much appreciate your help and expertise! Thanks!   ... View more

@PaigeMiller  Thanks so much for your code! Since I got my correlation matrix from another source, I need to read it into SAS. I applied your code, but the result I got only contains the correlation between the first and the last variable.   So I run your example code step by step to check what appended in each step. My trial is as following proc corr data=sashelp.class noprob nosimple outp=try; var age height weight; /* ods output pearsoncorr=corrs;*/ run; data try1; set try; if _TYPE_="CORR"; run; data try1; set try1; drop _TYPE_; run; data try2; set try1; length variable2 $ 32; array r _numeric_; do j=_n_+1 to dim(r); corr=r(j); variable2=vname(r(j)); end; run; But the result table I got is like the following. The corr column only contains the correlations from the last row.  I can't figure out how to fix that. Could you please give me direction? Thank you!         ... View more

@Ksharp  As always, thanks a bunch! a big thumb up to your expertise! ... View more

@Reeza @Rick_SAS @Ksharp  Thank you all so much for your expertise! I very much appreciate your time and help! ... View more

@Reeza  Thanks much for providing the information! My sas code is as below proc logistic data=d1 outmodel=outp_d1; where first=1; class sex (ref='0')/param=ref; model pass (event="1")= sex time passrate/expb; run; in my code above, male is coded as 0. By using the code above, I got the following output, what does 1 (in bold) following sex mean? Does it mean 0.3299 is the parameter estimate for female? My results is a little bit counter intuitive. I'd like to check with you expert. Thank you for taking time to answer my question.     Analysis of Maximum Likelihood Estimates Parameter   DF Estimate Standard Error Wald Chi-Square Pr > ChiSq Exp(Est) Intercept   1 -3.7746 0.1433 693.8620 <.0001 0.023 sex 1 1 0.3299 0.0439 56.5428 <.0001 1.391 time   1 -0.00713 0.000960 55.2062 <.0001 0.993 pass   1 7.3763 0.2057 1285.7808 <.0001 1597.655 ... View more

@Rick_SAS  I tried as you suggested. The probabilities are still a little higher by coping and pasting the coefficients than using the proc logistic inmodel procedure. Is there anything we can do about proc logistic inmodel procedure? Or is there anything wrong of my proc logistic inmodel as below? Thank you!!   proc logistic inmodel=outp_d1; score clm data=pre11 out=prob_first; where first=1 ; run ... View more