The following my code is suited for  only TWO levels of paygrade as you showed in your data. If you need to apply to more pattern ,that is another story.   data have; infile cards expandtabs; input id date :date9. paygrade; format date date9.; datalines; 23456 01JAN2021 7 23456 01FEB2021 7 23456 01MAR2021 7 23456 01APR2021 7 23456 05APR2021 7 23456 10APR2021 7 23456 15APR2021 6 23456 01MAY2021 6 23456 01JUN2021 6 12345 01JAN2021 3 12345 01FEB2021 3 12345 01MAR2021 3 12345 01APR2021 3 12345 05APR2021 3 12345 10APR2021 2 12345 15APR2021 3 12345 01MAY2021 3 12345 01JUN2021 3 34567 01JAN2021 5 34567 01FEB2021 5 34567 01MAR2021 5 34567 01APR2021 4 34567 05APR2021 5 34567 10APR2021 5 34567 15APR2021 4 34567 01MAY2021 4 34567 01JUN2021 4 ; proc freq data=have noprint order=data; table id*paygrade/out=level; run; data level; set level; by id notsorted; if first.id then level=-1; level+1; keep id paygrade level; run; data temp; if _n_=1 then do; if 0 then set level; declare hash h(dataset:'level'); h.definekey('id','paygrade'); h.definedata('level'); h.definedone(); end; set have; call missing(level); rc=h.find(); drop rc; run; data temp2; set temp; by id level notsorted; if first.level; keep id level; run; data want; do until(last.id); set temp2; by id notsorted; length pattern $ 80; pattern=cats(pattern,level); end; keep id pattern; run;   ... View more

Calling @GraphGuy  (a.k.a Robert.Allision ) ,since you are asking some special feature with 3D chart. Or you could try PROC SGPLOT + GTL Annotation Facility (sganno= option) to get this 3D chart, but that need to write many code to finish it.   P.S. Why not use PROC SGPLOT + VBAR to simplify this question as Paige and ballardw pointed out ? ... View more

@PaigeMiller I tried it before. One time I posted the data step without the running man icon, I found multiple blanks compressed into 1. If you don't believe it, you could try it. ... View more

I think you should post your code or data with this button:   Otherwise it would eat your multiple blanks, like there are four blanks in your data,but display it only one blank. data SRT; input ProjID $ Manager & $15. Budget CompletionPercent; datalines; P001 James Lee 100000 85 P002 Sophia Ray 120000 90 P003 Michael Chen 75 P004 Emma Stone 95000 . P005 Olivia Park 110000 88 ; run; proc print data=SRT; run;   ... View more

I think you should use IMPAR statement to get rid of that ERROR information: options nodate ; proc datasets lib=work kill noprint; run; %let n=10; %let fb=0.89; %let CLcr = 100; %let ka=2.202; %let clnr=0.364; %let weight=70; %let dose=400; data covar; *Create listing of covariates; do i=1 to &n; num=i; weight=50+rand("uniform")*50; *Weight range 50-100 kg; CLCR=20+rand("uniform")*10; *CLcr uniform distribtuion 20-30 ml/min; output; end; run; */; data param; weight=&weight; clcr=&clcr; * clearance renal; CLr=CLCR*0.643/90; *covariate adjusted CLr; vCLr=(1.68*CLr)**2; *1.68 is the RSD for CLr; CLrphi=sqrt(vCLr+CLr**2); *Phi - used to calculate log parameters; LogCLrmu=log(CLr**2/clrphi); *Mean for simulations; logclrsig=sqrt(log(clrphi**2/CLr**2)); *sd for logCLtmu; * V; V=48.226/72*weight; vV=(0.696*V)**2; * 0.696 is the cov; Vphi=sqrt(vV+V**2); logVmu=log(V**2/Vphi); *Mean for log of V; logVsig=sqrt(log(Vphi**2/V**2)); *sd for log of Vc run; data s1; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='MEAN'; _NAME_="logCLr"; logclr=logclrmu; logv=logvmu; keep _TYPE_ _NAME_ logclr logv; data s4; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='COV'; _NAME_="logclr"; logclr=logclrsig**2; logv=0.5*logclrsig*logVsig; keep _TYPE_ _NAME_ logclr logv; data s5; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='COV'; _NAME_="logv"; logclr=0.5*logclrsig*logVsig ; logv=logvsig**2; keep _TYPE_ _NAME_ logclr logv; run; data s (type=cov); set s4 s5 s1; run; Title2 'Parmeter Means and Correlation Matrix'; proc simnormal data=s outsim=SimMVN numreal = &n /* number of realizations = size of sample */ seed = 12345; /* random number seed */ var logclr logv ; run; data f_parm; set SimMVN ; CLr=exp(logclr); V=exp(logv); CLs=clr+&clnr; ke=CLs/V; Dose=&dose; ka=&ka; run; title2 'Patameter Values'; proc means data=f_parm; var cls v; run; proc univariate data=f_parm; var cls v; run; proc sgplot data=flu_parm; histogram cls; run; proc sgplot data=flu_parm; histogram v; run; data sim2; set SimMVN ; CLr=exp(logclr); V=exp(logv); CLs=clr+&clnr; ke=CLs/V; Dose=&dose; ka=&ka; length Condition $12; Condition='400 mg Q 24 h'; dose=400; tau=24; if CLs < 0.388 or CLs > 3.14 or V < 6.6 or v > 167 then delete; C_0=dose*ka/(v*(ka-ke))*(exp(-ke*0)/(1-exp(-ke*24))-exp(-ka*0)/(1-exp(-ka*24))); C_2=dose*ka/(v*(ka-ke))*(exp(-ke*2)/(1-exp(-ke*24))-exp(-ka*2)/(1-exp(-ka*24))); C_4=dose*ka/(v*(ka-ke))*(exp(-ke*4)/(1-exp(-ke*24))-exp(-ka*4)/(1-exp(-ka*24))); C_6=dose*ka/(v*(ka-ke))*(exp(-ke*6)/(1-exp(-ke*24))-exp(-ka*6)/(1-exp(-ka*24))); C_8=dose*ka/(v*(ka-ke))*(exp(-ke*8)/(1-exp(-ke*24))-exp(-ka*8)/(1-exp(-ka*24))); C_10=dose*ka/(v*(ka-ke))*(exp(-ke*10)/(1-exp(-ke*24))-exp(-ka*10)/(1-exp(-ka*24))); C_12=dose*ka/(v*(ka-ke))*(exp(-ke*12)/(1-exp(-ke*24))-exp(-ka*12)/(1-exp(-ka*24))); C_16=dose*ka/(v*(ka-ke))*(exp(-ke*16)/(1-exp(-ke*24))-exp(-ka*16)/(1-exp(-ka*24))); C_20=dose*ka/(v*(ka-ke))*(exp(-ke*20)/(1-exp(-ke*24))-exp(-ka*20)/(1-exp(-ka*24))); C_22=dose*ka/(v*(ka-ke))*(exp(-ke*22)/(1-exp(-ke*24))-exp(-ka*22)/(1-exp(-ka*24))); C_23=dose*ka/(v*(ka-ke))*(exp(-ke*23)/(1-exp(-ke*24))-exp(-ka*23)/(1-exp(-ka*24))); C_24=dose*ka/(v*(ka-ke))*(exp(-ke*24)/(1-exp(-ke*24))-exp(-ka*24)/(1-exp(-ka*24))); AUC24_sim=dose/cls; format c_0 best4. c_2 best4. c_4 best4. c_6 best4. c_8 best4. c_10 best4. c_12 best4. c_16 best4. c_20 best4. c_22 best4. c_23 best4. c_24 best4.; format auc24_sim best4.; run; proc means data=sim2 n median min max; var C_0 c_2 c_4 c_6 c_8 c_10 c_12 c_16 c_20 c_22 c_23 c_24; run; proc corr data=sim2; var cls v; run; proc optmodel; set OBS; num init {OBS}; num ke {OBS}; num V {OBS}; num ka {OBS}; num dose {OBS}; num tau{OBS}; num c_23{OBS}; num auc24_sim {OBS}; read data sim2 into OBS=[_N_] ke V ka dose tau c_23 auc24_sim; num simno; print ke v ka tau c_23 auc24_sim ; var ki init 0.01; impvar Vi = Dose[simno]*ka[simno]/(C_23[simno]*(ka[simno]-ki))*(exp(-ki*23)/(1-exp(-ki*tau[simno]))-exp(-ka[simno]*23)/(1-exp(-ka[simno]*tau[simno]))); min f1=(ke[simno]-ki)**2/(ke[simno]*0.5)**2 + (V[simno]-Vi)**2/(V[simno]*0.5)**2; /* Declare parameters to store solution values */ num ki_sol {OBS}; num vi_sol {OBS}; cofor {b in OBS} do; simno = b; solve; ki_sol[b] = ki.sol; vi_sol[b] = vi.sol; end; create data outki from [simno]=OBS ki_sol vi_sol; quit; ... View more

OK. Try this one. But I noticed you have some error when running your code. 1874! Dose[simno]*ka[simno]/(C_23[simno]*(ka[simno]-ki))*(exp(-ki*23)/(1-exp(-ki*tau[simno]))-exp(-ka[simno]*23)/(1-exp(-ka[sim 1874! no]*tau[simno]))); ERROR: 符号“simno”在第 1874 行第 18 列没有值。   options nodate ; proc datasets lib=work kill noprint; run; %let n=10; %let fb=0.89; %let CLcr = 100; %let ka=2.202; %let clnr=0.364; %let weight=70; %let dose=400; data covar; *Create listing of covariates; do i=1 to &n; num=i; weight=50+rand("uniform")*50; *Weight range 50-100 kg; CLCR=20+rand("uniform")*10; *CLcr uniform distribtuion 20-30 ml/min; output; end; run; */; data param; weight=&weight; clcr=&clcr; * clearance renal; CLr=CLCR*0.643/90; *covariate adjusted CLr; vCLr=(1.68*CLr)**2; *1.68 is the RSD for CLr; CLrphi=sqrt(vCLr+CLr**2); *Phi - used to calculate log parameters; LogCLrmu=log(CLr**2/clrphi); *Mean for simulations; logclrsig=sqrt(log(clrphi**2/CLr**2)); *sd for logCLtmu; * V; V=48.226/72*weight; vV=(0.696*V)**2; * 0.696 is the cov; Vphi=sqrt(vV+V**2); logVmu=log(V**2/Vphi); *Mean for log of V; logVsig=sqrt(log(Vphi**2/V**2)); *sd for log of Vc run; data s1; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='MEAN'; _NAME_="logCLr"; logclr=logclrmu; logv=logvmu; keep _TYPE_ _NAME_ logclr logv; data s4; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='COV'; _NAME_="logclr"; logclr=logclrsig**2; logv=0.5*logclrsig*logVsig; keep _TYPE_ _NAME_ logclr logv; data s5; set param; length _TYPE_ $4 _NAME_ $7 ; _TYPE_='COV'; _NAME_="logv"; logclr=0.5*logclrsig*logVsig ; logv=logvsig**2; keep _TYPE_ _NAME_ logclr logv; run; data s (type=cov); set s4 s5 s1; run; Title2 'Parmeter Means and Correlation Matrix'; proc simnormal data=s outsim=SimMVN numreal = &n /* number of realizations = size of sample */ seed = 12345; /* random number seed */ var logclr logv ; run; data f_parm; set SimMVN ; CLr=exp(logclr); V=exp(logv); CLs=clr+&clnr; ke=CLs/V; Dose=&dose; ka=&ka; run; title2 'Patameter Values'; proc means data=f_parm; var cls v; run; proc univariate data=f_parm; var cls v; run; proc sgplot data=flu_parm; histogram cls; run; proc sgplot data=flu_parm; histogram v; run; data sim2; set SimMVN ; CLr=exp(logclr); V=exp(logv); CLs=clr+&clnr; ke=CLs/V; Dose=&dose; ka=&ka; length Condition $12; Condition='400 mg Q 24 h'; dose=400; tau=24; if CLs < 0.388 or CLs > 3.14 or V < 6.6 or v > 167 then delete; C_0=dose*ka/(v*(ka-ke))*(exp(-ke*0)/(1-exp(-ke*24))-exp(-ka*0)/(1-exp(-ka*24))); C_2=dose*ka/(v*(ka-ke))*(exp(-ke*2)/(1-exp(-ke*24))-exp(-ka*2)/(1-exp(-ka*24))); C_4=dose*ka/(v*(ka-ke))*(exp(-ke*4)/(1-exp(-ke*24))-exp(-ka*4)/(1-exp(-ka*24))); C_6=dose*ka/(v*(ka-ke))*(exp(-ke*6)/(1-exp(-ke*24))-exp(-ka*6)/(1-exp(-ka*24))); C_8=dose*ka/(v*(ka-ke))*(exp(-ke*8)/(1-exp(-ke*24))-exp(-ka*8)/(1-exp(-ka*24))); C_10=dose*ka/(v*(ka-ke))*(exp(-ke*10)/(1-exp(-ke*24))-exp(-ka*10)/(1-exp(-ka*24))); C_12=dose*ka/(v*(ka-ke))*(exp(-ke*12)/(1-exp(-ke*24))-exp(-ka*12)/(1-exp(-ka*24))); C_16=dose*ka/(v*(ka-ke))*(exp(-ke*16)/(1-exp(-ke*24))-exp(-ka*16)/(1-exp(-ka*24))); C_20=dose*ka/(v*(ka-ke))*(exp(-ke*20)/(1-exp(-ke*24))-exp(-ka*20)/(1-exp(-ka*24))); C_22=dose*ka/(v*(ka-ke))*(exp(-ke*22)/(1-exp(-ke*24))-exp(-ka*22)/(1-exp(-ka*24))); C_23=dose*ka/(v*(ka-ke))*(exp(-ke*23)/(1-exp(-ke*24))-exp(-ka*23)/(1-exp(-ka*24))); C_24=dose*ka/(v*(ka-ke))*(exp(-ke*24)/(1-exp(-ke*24))-exp(-ka*24)/(1-exp(-ka*24))); AUC24_sim=dose/cls; format c_0 best4. c_2 best4. c_4 best4. c_6 best4. c_8 best4. c_10 best4. c_12 best4. c_16 best4. c_20 best4. c_22 best4. c_23 best4. c_24 best4.; format auc24_sim best4.; run; proc means data=sim2 n median min max; var C_0 c_2 c_4 c_6 c_8 c_10 c_12 c_16 c_20 c_22 c_23 c_24; run; proc corr data=sim2; var cls v; run; proc optmodel; set OBS; num init {OBS}; num ke {OBS}; num V {OBS}; num ka {OBS}; num dose {OBS}; num tau{OBS}; num c_23{OBS}; num auc24_sim {OBS}; read data sim2 into OBS=[_N_] ke V ka dose tau c_23 auc24_sim; num simno; print ke v ka tau c_23 auc24_sim ; var ki init 0.01; var Vi; Vi = Dose[simno]*ka[simno]/(C_23[simno]*(ka[simno]-ki))*(exp(-ki*23)/(1-exp(-ki*tau[simno]))-exp(-ka[simno]*23)/(1-exp(-ka[simno]*tau[simno]))); min f1=(ke[simno]-ki)**2/(ke[simno]*0.5)**2 + (V[simno]-Vi)**2/(V[simno]*0.5)**2; /* Declare parameters to store solution values */ num ki_sol {OBS}; num vi_sol {OBS}; cofor {b in OBS} do; simno = b; solve; ki_sol[b] = ki.sol; vi_sol[b] = vi.sol; end; create data outki from [simno]=OBS ki_sol vi_sol; quit;   ... View more

I don't understand what you are trying to do. You want to impute missing value with median or just bring median into original dataset ? 1) if you want to impute missing value with median, try this: proc stdize data=have missing=median reponly; by cluster; run; 2)if you want to bring median value back into orginal dataset, try this: proc summary data=have; by cluster; var var; output out=median median=median_var; run; data want; merge have median; by cluster; run; 3)Last, if you just simply impute missing WITHOUT median ,try PROC MI. https://blogs.sas.com/content/iml/2020/12/02/score-external-logistic-model.html ... View more

You could use libname() funtion in data step to get job done. data _null_; response =libref("data"); put response=; if response ne 0 then rc=libname('data','c:\temp'); run; ... View more

If this random slope is one slope for one subject. Try this: data trial; call streaminit(123); do idno=1 to 10; do time=1 to 4; globalcog=rand('normal'); brain=rand('normal'); output; end; end; run; ods output SolutionR= SolutionR; proc mixed data=trial; class idno time ; model GLOBALCOG = time brain/s chisq RESIDUAL; /*random intercept /subject=idno solution ;*/ random brain/subject=idno solution ; /*repeated time/type=un subject=idno r rcorr;*/ run;   ... View more

I think your code should look like this: data trial; call streaminit(123); do idno=1 to 10; do time=1 to 4; globalcog=rand('normal'); brain=rand('normal'); output; end; end; run; ods output SolutionR= SolutionR; proc mixed data=trial; class idno time; model GLOBALCOG = time brain/s chisq RESIDUAL; random intercept /subject=idno solution ; repeated time/type=un subject=idno r rcorr; run;   ... View more

I think you are using thest two libname statements in many times in your code. You should clear them out before reassigning them. Like: LIBNAME DATA "&path2./&subfolder./data"; LIBNAME BROK spde "&path2./&subfolder./temp/master"; .....other code...... LIBNAME _ALL_ CLEAR; .......other code....... LIBNAME DATA "&path1./data"; LIBNAME BROK spde "&path1./master"; .....other code...... LIBNAME _ALL_ CLEAR; Check your code carefully and inserting "LIBNAME _ALL_ CLEAR;" at right place; ... View more

No. You can't expect percent is conforming to normal distribution when true percent is near zero or one, @Rick_SAS talk this topic at other session. But from the graph you posted, it is near 0.5 and have bell shape , so I think use Normal or Lognormal dist is decent. ... View more

You could use dist=binomial and link=identity to directly build a model for a ratio/percent, but you need to restructure your data into Y=0 1: https://support.sas.com/kb/37/228.html proc genmod data=test descending; class a; model y = a / dist=binomial link=identity; lsmeans a / diff cl; run; Or try Possion Regression as StatDave said. But you also need to change your data structure. https://support.sas.com/kb/24/188.html proc genmod data=insure; class car age; model c = car age / dist=poisson link=log offset=ln; run; ... View more