data CP; input time A B; cards; 1  .   . 2  0  .1 2  .4  .7 3   .3  .5 5   .9   0 6    .3  .33 ; data CPn; set CP; n = _n_; run; proc sort data=CPn; by descending n; run; data want; set CPn; Asum + A; Bsum + B; run; proc sort data=want out=want(drop=n); by n; run; PG ... View more

Alternatively by  SQL PROC SQL;    CREATE TABLE want AS    SELECT coalesce(t2.a,t1.a) AS a,           t1.b       FROM WORK.A t1            FULL JOIN WORK.B t2 ON (t1.a = t2.a); QUIT; Thanks, Jag ... View more

For any value of U, use the FROOT function to find the numerical value of X that solves the equation. If you are running a version of prior to SAS/IML 12.1, use the bisection module instead of the FROOT function. See Finding the root (zero) of a function in SAS ... View more

This question is a duplicate of https://communities.sas.com/message/192463#192463 , where it is answered. ... View more

See How to numerically integrate a function in SAS - The DO Loop ... View more

This is within a simulation, I run it for just one iteration, that is &p=1. The code has append and quit.  Your help is highly need Sir Here is the log file: data KMM_&p; 570  set testt_&p; 571  set KM_&p; 572  if censored=1 then SurvivalM=Survival1; else SurvivalM=Survival1; 573  if censored=1 then InverseC=1/survival1; else InverseC=0; 574  keep time  censored  y frequency survival1 obs replicate SurvivalM InverseC; 575  run; 576 577 578  proc iml; 579  use KMM_&p; 580  read all var{ Replicate time censored obs y Frequency SURVIVAL1 SurvivalM InverseC } into 580! DM; 581  close; 581!  /*close the link between the dataset and IML*/ 582    InverseC=DM[,9];SurvivalM=DM[,8];survival1=DM[,7]; frequency=DM[,6]; y=DM[,5];obs 582! =DM[,4]; censored=DM[,3]; time= DM[,2]; replicate = DM[,1];   /*Survivalival is second 582! column*/  /*time is first column*/ 583 584  n = nrow(byGroup); 585  u = unique(Replicate); 586  BIG=J(20,10,-1) ; 587  InverseC=J(nrow(survival1),1,1); 588  survival1= J(nrow(survival1),1,1); 589  t=J(nrow(survival1),1,1); 590  do k = 1 to nrow(byGroup); 591       byGroup = DM[loc(replicate=u ),]; 592         *byIdx = loc(Replicate=u ); /* the UNIQUE-LOC trick */ 593          *byGroup = DM[byIdx,];           /* extract BY group */        /* extract BY group 593!  */ 594      InverseC=byGroup[,9];SurvivalM=byGroup[,8]; survival1=byGroup[,7]; 594! frequency=byGroup[,6];y=byGroup[,10];obs =byGroup[,4]; censored =byGroup[,3]; time= 594! byGroup[,2]; replicate= byGroup[,1]; 595   t=InverseC[(loc(InverseC^=0))]; print t; 596   tt=t(t); *print tt; 597  Big[(k-1)*10+(1:10),1:(sum(censored=1)+1)]=survival1*t(t); 598  end; 599 600 601  create GIB_&p from BIG[colname={COL1}]; 602  append from BIG; 603  quit; 604 605 606 OTE: There were 20 observations read from the data set WORK.TESTT_1. NOTE: There were 20 observations read from the data set WORK.KM_1. NOTE: The data set WORK.KMM_1 has 20 observations and 9 variables. NOTE: DATA statement used (Total process time):       real time           0.00 seconds       cpu time            0.00 seconds NOTE: IML Ready NOTE: Closing WORK.KMM_1 NOTE: Exiting IML. NOTE: The data set WORK.GIB_1 has 20 observations and 10 variables. NOTE: PROCEDURE IML used (Total process time):       real time           0.01 seconds       cpu time            0.01 seconds NOTE: Line generated by the invoked macro "SURV". 26   (InverseC^=0))]; print t;  tt=t(t); *print tt; 26 ! Big[(k-1)*10+(1:10),1:(sum(censored=1)+1)]=survival1*t(t); end;   create GIB_&p from 26 ! BIG[colname={COL1}]; append from BIG; quit;            ;          end; *print InverseC;                                                                        ---                                                                        180 ERROR 180-322: Statement is not valid or it is used out of proper order. ... View more