I am doing multilevel analyses using beta regression because my dependent variable is a proportion. The range of proportion values includes 0 and 1; however, prior to analysis 0 was recoded to 0.005 and 1 was recoded to 0.995. Most of the analyses executed without error but four did not. The attached file shows the syntax for each analysis and the noted error (the same for three of the four analyses). As this my first experience with beta regression, please pitch your reply to the level of somebody with no experience with beta regression and an incomplete understanding of all glimmix options and keywords and their possible interactions. A low level slow walk would be appreciated. Thanks Gene Maguin ... View more

Q1: I have a dataset that has become disordered in the sense that items from a given measure/scale have become scattered across the dataset. I want to re-order the dataset so that common items are grouped together and within the group are ordered sequentially. I had asked about this previously and after some discussion among persons replying the recommendation was to use a retain statement.(Why re-order? Because i need to something with all variables associated with a scale or group of scales and i can define an array referencing all variables in the dataset and for a given scale(s) operate on those items by using element range in the Do statement.) So (from the log):   17 * reorganize the dataset to bring all items in a measure together; 18 * group measures together into an AOD group, dsa1 group, and an all group; 19 data work5; 20 retain Year--Ddr2 Dalc5a--Dalc5f Dyaq1-Dyaq9 Ddmq1-Ddmq12 ERROR: Variable Year cannot be found on the list of previously defined variables. ERROR: Variable Dalc5a cannot be found on the list of previously defined variables. 21 Dsa1-Dsa13 Drr1-Drr31 Dcs1-Dcs15 Dcs16 Dsbi1-Dsbi47 22 Doa1-Doa40 Dnsa1--Dnsa15 Dncs1-Dncs16 Dnsb1-Dnsb47 ERROR: Variable Dnsa1 cannot be found on the list of previously defined variables. 23 Dmood--Dmis4 Dass1--Dass16 Dass17 Dpb1-Dpb19 DCovid1-DCovid6 DCovid7 ERROR: Variable Dmood cannot be found on the list of previously defined variables. ERROR: Variable Dass1 cannot be found on the list of previously defined variables. 24 Dscsev Dscs1-Dscs6 DcovSt1-DcovSt4; 25 drop i dot YB miss1-miss10 j; 26 set PNA.CleanFile; 27 run;   Each of the variables identified as "cannot be found" are very definitely in the set dataset PNA.CleanFile. What am i doing wrong and how do i correct it?   Q2: How do i find questions i have previously submitted (and replies to those questions)?         ... View more

In spss: if (any(variablename,value1,value2, ..., valuen)) newvar=17.   Is there a sas equivalent? What is it?   secondary question how would i even go about finding the answer to this question in the sas help system? Gene Maguin ... View more

I hope you can see the log output. Ok. Two questions. (1)  The log output sas that missing values were generated on line 1187. If you go the bottom of  the posted code you will see 1187 pulled out by itself and the column locations are marked by |&| following the noted column. So i know that there was one record with a missing value for A2WEIGHT and so '.' should have been the result for vbacwk{i}, i=1 to 7 and it was. My question is why was three missing value notes reported. I've seen these messages but i've nerver understood them except for the fact that a missing value was assigned to the result variable for some record in the dataset. Please help me understand.   (2) Now look at lines 1186, 87, and 88. My intention was that if the else if statement is true,  both 1187 and 1188 are executed. However, if the else if is false, both 1187 and 1188 are not executed. These lines execute correctly, as near as i can tell, but why shouldn't i have had to write this instead 1186      else if (valcwk{i} ge 1 and valhwk{i} ge 1) then do 1187         vbacwk{i}=((valcwk{i}/2)*(9.0/A2WEIGHT))-(0.02*valhwk{i}); 1188         if (vbacwk{i} le 0.0 and vbacwk{i} gt -1.) then vbacwk{i}=0.0;                 end; When i tried that i got this 2335 vbacwk{i}=((valcwk{i}/2)*(9.0/A2WEIGHT))-(0.02*valhwk{i}); - 73 76 ERROR 73-322: Expecting an =. ERROR 76-322: Syntax error, statement will be ignored.   Why isn't a do required? I want conditionally execute both 1187 and 1188. Why the error? Thanks, Gene Maguin     1173 * COMPUTE WOMEN'S DAILY BAC BASED ON TYPICAL DAILY DRINKS AND WEIGHT; 1174 * NOT SURE OF THE EXACT SOURCE BUT SOMEBODY WIDMARK IS NAMED; 1175 * [(c/2)*(GC/w)] -(B60*t), where c = total standard drinks consumed, 1176 GC = gender constant (9.0 for women, 7.5 for men), w = weight in pounds, 1177 B60 = hourly rate of metabolism of alcohol (estimated at .02), and 1178 t = total drinking hours; 1179 * PER CRAIG (ABOUT 08.26.2021) COMPUTE BAC AS .p IF DRINKS > 1 AND HOURS=0 1180 * OR IF DRINKS = 0 AND HOURS > 1; 1181 * RECODE TO .s IF BAC LE 0.0; >>>> array valcwk{i} defined above this section. 1182 array valhwk{7} A2ALCH1-A2ALCH7; 1183 array vbacwk{7} A2BAC1-A2BAC7; 1184 do i=1 to 7; 1185 if (valcwk{i} eq 0 and valhwk{i} eq 0) then vbacwk{i}=.s; 1186 else if (valcwk{i} ge 1 and valhwk{i} ge 1) then 1187 vbacwk{i}=((valcwk{i}/2)*(9.0/A2WEIGHT))-(0.02*valhwk{i}); 1188 if (vbacwk{i} le 0.0 and vbacwk{i} gt -1.) then vbacwk{i}=0.0; 1189 else if (valcwk{i} eq .p or valhwk{i} eq .p) then vbacwk{i}=.p; 1190 else if (valcwk{i} eq 0 and valhwk{i} gt 0) then vbacwk{i}=.p; 1191 else if (valcwk{i} gt 0 and valhwk{i} eq 0) then vbacwk{i}=.p; 1192 end; 1193 run; NOTE: Missing values were generated as a result of performing an operation on missing values. Each place is given by: (Number of times) at (Line):(Column). 1 at 1187:32 1 at 1187:37 1 at 1187:48 1187 vbacwk{i}=((valcwk{i}/2)|&|*(9.0|&|/A2WEIGHT))|&|-(0.02*valhwk{i}); ... View more

I define a 7-element array and put a value in one or more of the elements but never two of them. (Why you say. Because LEC16 (see code) by design never includes a value for i1=2 or 5.) I later drop those two elements. But, i get a warning that those two elements were never addressed and the elements are not dropped. I understand the warning as useful but why weren't the elements dropped? Here's the code for one of the arrays. The drop command that follows is just drop  LEC16_2 LEC16_5; * LEC_16; array vlec16{7} A3LEC16_1-A3LEC16_7; nlen=lengthn(LEC_16); if (nlen ge 1) then do; nlen=1+(nlen-1)/2; do i=1 to nlen; i1=1+2*(i-1); val=substr(LEC_16,i1,1); vlec16{input(val,BEST32.)}=1; end; end; ... View more

What i want is an if-else if (i won't need an else statement for this problem) structure so that if a statement or set of statements within an if/else if is executed the rest of the structure is not executed. in the log output everything seems to go ok until the fifth else if block where for unknown to me reasons an if-then is expected. If i take out the 'else's, it will run but isn't that equivalent to a set of if-then statements, all of which are executed? (also, you'll see a vdu{1} referenced. That is defined above this block of syntax and i do not believe is the problem).   625 array voderr(12) ODERR1-ODERR12; 626 if (A3ODRG90 eq 1 or A3ODRG90 eq .p) then 627 do i=1 to 12; 628 voderr{i}=0; 629 nval=0; 630 k=(i-1)*9; 631 do j=1 to 7; 632 if (vdu{k+j} eq 1) then nval=sum(of nval,1); 633 end; 634 * j=8 IS DNU, j=9 IS PNA; 635 * possible cases; 636 * nval=0, DNU=1, PNA=. no error. A3ODxxW1-A3ODxxW7=0, A3ODxxW9=0; 637 if (nval eq 0 and vdu{k+8} eq 1 and vdu{k+9} eq .) then do; 638 do j=1 to 7; 639 vdu{k+j}=0; 640 end; 641 vdu{k+9}=0; 642 voderr{i}=0; 643 end; 644 * nval=0, DNU=., PNA=1 no error. A3ODxxW1-A3ODxxW7=.p, A3ODxxW8=0; 645 else if (nval eq 0 and vdu{k+8} eq . and vdu{k+9} eq 1) then do; 646 do j=1 to 7; 647 vdu{k+j}=.p; 648 end; 649 vdu{k+8}=0; 650 voderr{i}=0; 651 end; 652 * nval=0, DNU=., PNA=. no error. A3ODxxW1-A3ODxxW9=.p; 653 else if (nval eq 0 and vdu{k+8} eq . and vdu{k+9} eq .) then do; 654 do j=1 to 7; 655 vdu{k+j}=.p; 656 end; 657 vdu{k+8}=0; 658 vdu{k+9}=0; 659 voderr{i}=0; 660 end; 661 * nval>0, DNU=., PNA=. no error. A3ODxxW1-A3ODxxW9: recode . to 0; 662 else if (nval gt 0 and vdu{k+8} eq . and vdu{k+9} eq .) then do; 663 do j=1 to 7; 664 if (vdu{k+j} ne 1) then vdu{k+j}=0; 665 end; 666 vdu{k+8}=0; 667 vdu{k+9}=0; 668 voderr{i}=0; 669 end; 670 * nval=0, DNU=1, PNA=1 error. do what? TYPE 1 ERROR; 671 else if (nval eq 0 and vdu{k+8} eq 1 and vdu{k+9} eq 1) 672 then voderr{i}=1; /* ERROR TYPE 1: e.g., 000000011 */; 673 * nval>0, DNU=1, PNA=. error. do what? TYPE 2 ERROR; 674 else if (nval ge 1 and vdu{k+8} eq 1 and vdu{k+9} eq .) ---- 160 ERROR 160-185: No matching IF-THEN clause. 675 then voderr{i}=2; /* ERROR TYPE 2: e.g., 10000001. */; 676 * nval>0, DNU=., PNA=1 error. do what? TYPE 3 ERROR; 677 else if (nval ge 1 and vdu{k+8} eq . and vdu{k++9} eq 1) ---- 160 ERROR 160-185: No matching IF-THEN clause. 678 then voderr{i}=3; /* ERROR TYPE 3: e.g., 1000000.1 */; 679 * nval>0, DNU=1, PNA=1 error. do what? TYPE 4 ERROR; 680 else if (nval ge 1 and vdu{k+8} eq 1 and vdu{k++9} eq 1) ---- 160 ERROR 160-185: No matching IF-THEN clause. 681 then voderr{i}=4; /* ERROR TYPE 4: e.g., 100000011 */; 682 end; 683 ANYODERR=sum(of ODERR1-ODERR12); 684 run;   ... View more