About desireatem

desireatem · ‎11-06-2013

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.

desireatem · ‎11-06-2013

I start with a do... end; see below; do k = 1 to nrow(byGroup); byGroup = DM[loc(replicate=u ),]; InverseC=byGroup[,9];SurvivalM=byGroup[,8]; survival1=byGroup[,7];frequency=byGroup[,6];y=byGroup[,10];obs =byGroup[,4]; censored =byGroup[,3]; time= byGroup[,2]; replicate= byGroup[,1]; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big[(k-1)*10+(1:10),1:(sum(censored=1)+1)]=survival1*(tt); end;

desireatem · ‎11-06-2013

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. proc iml; use man; read all var{ Replicate time censored obs y Frequency SURVIVAL1 SurvivalM InverseC } into DM; close; /*close the link between the dataset and IML*/ InverseC=DM[,9];SurvivalM=DM[,8];survival1=DM[,7]; frequency=DM[,6]; y=DM[,5];obs =DM[,4]; censored=DM[,3]; time= DM[,2]; replicate = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ n = nrow(byGroup); u = unique(Replicate); BIG=J(20,10,-1) ; InverseC=J(nrow(survival1),1,1); survival1= J(nrow(survival1),1,1); t=J(nrow(survival1),1,1); do k = 1 to nrow(byGroup); byGroup = DM[loc(replicate=u ),]; InverseC=byGroup[,9];SurvivalM=byGroup[,8]; survival1=byGroup[,7];frequency=byGroup[,6];y=byGroup[,10];obs =byGroup[,4]; censored =byGroup[,3]; time= byGroup[,2]; replicate= byGroup[,1]; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big[(k-1)*10+(1:10),1:(sum(censored=1)+1)]=survival1*(tt); end; create GIB_1 from BIG[colname={COL1}]; append from BIG; quit;

desireatem · ‎11-06-2013

PLease help correct this, I wish to do it by replicates. This is the code. The data set and error message is below. CODE proc iml; use Complete_1; read all var _all_ into DM; close; frequency =DM[,7]; y =DM[,6]; obs=DM[,10]; Censored= DM[,4]; time = DM[,3]; x =DM[,2];replicate=DM[,1]; use GIB_&p; read all var _all_ into Col; timeS=time; locC=loc(censored=1); matrix=J(1,7,2.10); do im = 1 to 1; u = unique(Replicate); byIdx = loc(replicate=u[im]); /* the UNIQUE-LOC trick */ byGroup = DM[byIdx,]; frequency =byGroup[,7]; y =byGroupDM[,6]; obs=byGroup[,5]; Censored= byGroup[,4]; time = byGroup[,3]; x =byGroup[,2];replicate=byGroup[,1]; do j= 1 to frequency[1]; seed=-1; A=RANUNI(-1); i=max(loc(Col[,j]-A>0)); if i<nrow(time) then /** do nothing if the last is censored, that is censored=1*****/ do; timeS[locC ] = time+ (time[i+ 1]-time)*((Col[i,j]-A)/(Col[i,j]+col[i+1,j])); end; end; sim=repeat(im,nrow(time),1); temp=obs||censored||time||Y||timeS||X||Sim||replicate; Matrix=matrix//temp; end; matrix=matrix[2:nrow(matrix),]; create impute_t_&p var{obs censored time Y timeS X replicate}; append from matrix; quit; ods trace on/listing; proc sort data=impute_t_&p; by replicate; run; variables: col1=DM[,8] frequency =DM[,7]; y =DM[,6]; obs=DM[,10]; Censored= DM[,4]; time = DM[,3]; x =DM[,2];replicate=DM[,1] 1 1 0 0.005959 1 0 2.19394 3 8.00000 2 1 0 0.005959 1 0 2.19394 3 7.11111 3 1 0 0.005959 1 0 2.19394 3 6.22222 4 1 0 0.006302 0 1 2.10653 3 5.33333 5 1 0 0.001931 0 1 2.10633 3 4.44444 6 1 0 0.001931 0 1 2.10633 3 3.55556 7 1 0 0.000076 0 1 2.09604 3 2.22222 8 1 0 0.011368 0 1 2.11672 3 1.55556 9 1 0 0.001074 0 1 2.10064 3 0.88889 10 1 0 0.001074 0 1 2.10064 3 0.00000 11 2 0 0.005959 1 0 2.19394 3 8.00000 12 2 0 0.013170 0 1 2.13512 3 7.00000 13 2 0 0.006302 0 1 2.10653 3 6.50000 14 2 0 0.006302 0 1 2.10653 3 6.00000 15 2 0 0.006302 0 1 2.10653 3 5.00000 16 2 0 0.001931 0 1 2.10633 3 4.00000 17 2 0 0.001931 0 1 2.10633 3 2.50000 18 2 0 0.000076 0 1 2.09604 3 1.75000 19 2 0 0.005171 1 0 2.14700 3 1.00000 20 2 0 0.004562 1 0 2.12766 3 0.00000 This code gives me error NOTE: IML Ready NOTE: Closing WORK.COMPLETE_1 ERROR: (execution) Matrices do not conform to the operation. operation : || at line 1966 column 139 operands : obs, censored, time, Y, timeS, X, Sim, replicate obs 10 rows 1 col (numeric) 1 1 1 1 1 1 1 0 1 1 censored 10 rows 1 col (numeric) 0 0 0 0 0 0 0 1 0 0 time 10 rows 1 col (numeric) 0.0119728 0.0119728 0.0119728 0.0000702 0.0071248 0.0071248 0.0017914 0.0079639 0.0066853 0.0066853 y 10 rows 1 col (numeric) 2.1234929 2.1234929 2.1234929 2.1003696 2.1255796 2.1255796 2.1055103 2.176416 2.1040166 2.1040166 timeS 20 rows 1 col (numeric) X 10 rows 1 col (numeric) 0 0 0 0 0 0 1 0 0 0 sim 10 rows 1 col (numeric) 1 1 1 1 1 1 1 1 1 1 Replicate 10 rows 1 col (numeric) 1 1 1 1 1 1 1 1 1 1

desireatem · ‎11-04-2013

Hello the code below is within a macro and it will create the data set below data tes_&p; set tes_&p; if _n_ = 10 then censored = 0; drop alpha1 seed i beta1 beta2 lambdaT lambdaC er c c1 c2 t R y1; run; proc surveyselect data=tes_&p method = urs sampsize = 10 OUTHITS reps=4 seed=12345 out=test_&p; run; OUTPUT which i do not like: 1 1 0.008272 0 1 2.49670 8 0.90000 2 1 0.008272 0 1 2.50739 8 0.85000 3 1 0.008272 1 0 2.48662 8 0.82500 4 1 0.008272 1 0 2.53422 8 0.81250 5 1 0.008272 1 0 2.56987 8 0.80625 6 1 0.012790 1 0 2.56987 8 0.80000 7 1 0.015102 1 0 2.56987 8 0.60000 8 1 0.026568 1 0 2.56987 8 0.50000 9 1 0.031404 1 0 2.56987 8 0.40000 10 1 0.056710 1 0 2.83734 8 0.40000 11 2 0.007388 0 1 2.50739 4 0.90000 12 2 0.007388 0 1 2.50739 4 0.80000 13 2 0.012790 0 1 2.50739 4 0.60000 14 2 0.012790 0 1 2.50739 4 0.50000 15 2 0.012790 1 0 2.48662 4 0.45000 16 2 0.012790 1 0 2.53422 4 0.40000 17 2 0.015102 1 0 2.53422 4 0.40000 18 2 0.026568 0 1 2.53611 4 0.40000 19 2 0.026568 0 1 2.53611 4 0.40000 20 2 0.056710 1 0 2.83734 4 0.40000 21 3 0.000713 0 1 2.49670 7 0.90000 22 3 0.006193 1 0 2.48662 7 0.80000 23 3 0.015102 1 0 2.48662 7 0.66667 24 3 0.015102 1 0 2.48662 7 0.60000 25 3 0.015102 1 0 2.53422 7 0.56667 26 3 0.026568 1 0 2.53422 7 0.55000 27 3 0.026568 1 0 2.73459 7 0.54167 28 3 0.031404 0 1 2.55807 7 0.53333 29 3 0.032303 1 0 2.83734 7 0.53333 30 3 0.056710 0 1 2.49329 7 0.53333 31 4 0.000713 0 1 2.49670 4 0.90000 32 4 0.006193 0 1 2.49670 4 0.80000 33 4 0.007388 0 1 2.50739 4 0.70000 34 4 0.008272 1 0 2.48662 4 0.64167 35 4 0.012790 1 0 2.53422 4 0.58333 36 4 0.015102 0 1 2.53611 4 0.38889 37 4 0.026568 0 1 2.55807 4 0.29167 38 4 0.031404 1 0 2.56987 4 0.24306 39 4 0.031404 1 0 2.83734 4 0.19444 40 4 0.056710 0 1 2.49329 4 0.19444 BUT this is not what I want, i want to loop it such that rather than having a data set of 40 with replicate from 1 to 4. I want the simple random sample to select one data set at a time and loop 4 times. That is what I want are 4 data set for this simple random sample like test_&p_1 , test_&p_2, test_&p_3, test_&p_4. This might be wrong notation but I am just using this for someone to uderstand what I want to do. Poorly written like this, within a macro do i=1 to 4; proc surveyselect data=tes_&p method = urs sampsize = 10 OUTHITS reps=1 seed=12345 out=test_&p; run end test_&p_1 1 1 0.008272 0 1 2.49670 8 0.90000 2 1 0.008272 0 1 2.50739 8 0.85000 3 1 0.008272 1 0 2.48662 8 0.82500 4 1 0.008272 1 0 2.53422 8 0.81250 5 1 0.008272 1 0 2.56987 8 0.80625 6 1 0.012790 1 0 2.56987 8 0.80000 7 1 0.015102 1 0 2.56987 8 0.60000 8 1 0.026568 1 0 2.56987 8 0.50000 9 1 0.031404 1 0 2.56987 8 0.40000 10 1 0.056710 1 0 2.83734 8 0.40000 test_&p_2 1 3 0.000713 0 1 2.49670 7 0.90000 2 3 0.006193 1 0 2.48662 7 0.80000 3 3 0.015102 1 0 2.48662 7 0.66667 4 3 0.015102 1 0 2.48662 7 0.60000 5 3 0.015102 1 0 2.53422 7 0.56667 6 3 0.026568 1 0 2.53422 7 0.55000 7 3 0.026568 1 0 2.73459 7 0.54167 8 3 0.031404 0 1 2.55807 7 0.53333 9 3 0.032303 1 0 2.83734 7 0.53333 10 3 0.056710 0 1 2.49329 7 0.53333 ETC THANKS t

desireatem · ‎11-04-2013

I made some changes, howver the problem with this code is where I mark bold, any corrections NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: (execution) Matrices do not conform to the operation. operation : + at line 7396 column 81 operands : _TEM1002, _TEM1003 _TEM1002 1 row 7 cols (numeric) 0 20 40 60 80 100 120 _TEM1003 1 row 20 cols (numeric) statement : ASSIGN at line 7396 column 69 NOTE: Exiting IML. NOTE: The data set WORK.GIB_1 has 140 observations and 11 variables. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.11 seconds cpu time 0.06 seconds proc iml; use KMM_&p; read all var{replicate time censored obs y frequency survival1 } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ survival1=DM[,7]; frequency=DM[,6]; y=DM[,5];obs =DM[,4]; censored=DM[,3]; time= DM[,2]; replicate = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ *n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 20x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; u = unique(replicate); do k = 1 to ncol(u); BIG=J(140,11,-1) ; byGroup = DM[loc(replicate=u ),]; *byIdx = loc(replicate=u ); /* the UNIQUE-LOC trick */ *byGroup = DM[byIdx,]; /* extract BY group */ /* extract BY group */ obs=byGroup[,6];y=byGroup[,5];freqency =byGroup[,4]; survival1 =byGroup[,3]; Censored= byGroup[,2]; time = byGroup[,1]; do i = 2 to nrow(byGroup); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to nrow(byGroup); if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to nrow(byGroup); if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; do i = 1 to ncol(frequency); call symputx("j", i); m = num(symget("j")); m=1*m; print m; %let mm=m; end; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big[(u-1)*20+(1:20),1]=k; Big[(u-1)*20+(1:20),2:(sum(censored=1)+1)]=survivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit;

desireatem · ‎11-04-2013

I have been reading since morning, got more than 70 hits but no one can help, Can you help me with this Rick? I made some changes to the code NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: (execution) Matrices do not conform to the operation. operation : + at line 7396 column 81 operands : _TEM1002, _TEM1003 _TEM1002 1 row 7 cols (numeric) 0 20 40 60 80 100 120 _TEM1003 1 row 20 cols (numeric) statement : ASSIGN at line 7396 column 69 NOTE: Exiting IML. NOTE: The data set WORK.GIB_1 has 140 observations and 11 variables. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.11 seconds cpu time 0.06 seconds proc iml; use KMM_&p; read all var{replicate time censored obs y frequency survival1 } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ survival1=DM[,7]; frequency=DM[,6]; y=DM[,5];obs =DM[,4]; censored=DM[,3]; time= DM[,2]; replicate = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ *n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 20x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; u = unique(replicate); do k = 1 to ncol(u); BIG=J(140,11,-1) ; byGroup = DM[loc(replicate=u ),]; *byIdx = loc(replicate=u ); /* the UNIQUE-LOC trick */ *byGroup = DM[byIdx,]; /* extract BY group */ /* extract BY group */ obs=byGroup[,6];y=byGroup[,5];freqency =byGroup[,4]; survival1 =byGroup[,3]; Censored= byGroup[,2]; time = byGroup[,1]; do i = 2 to nrow(byGroup); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to nrow(byGroup); if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to nrow(byGroup); if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; do i = 1 to ncol(frequency); call symputx("j", i); m = num(symget("j")); m=1*m; print m; %let mm=m; end; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big[(u-1)*20+(1:20),1]=k; Big[(u-1)*20+(1:20),2:(sum(censored=1)+1)]=survivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit; Wrong output:

desireatem · ‎11-03-2013

Hi Rick, I am confused while I am still getting this error: Thought this occurs when you try to assign or print a matrix that has not been assigned.: NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: (execution) Matrix has not been set to a value. operation : [ at line 11381 column 173 operands : DM, byIidx, DM 700 rows 6 cols (numeric) byIidx 0 row 0 col (type ?, size 0) statement : ASSIGN at line 11381 column 161 ERROR: Matrix Big has not been set to a value. statement : CREATE at line 11388 column 94 ERROR: No data set is currently open for output. statement : APPEND at line 11388 column 134 NOTE: Exiting IML. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.00 seconds cpu time 0.00 seconds

desireatem · ‎11-03-2013

Thanks, I will look at and keep working on

desireatem · ‎11-03-2013

Thanks for the help, I defined u, yet still having issues; LOG FILE: NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: (execution) Argument should be a scalar. operation : J at line 5393 column 113 operands : u, *LIT1010 u 1 row 7 cols (numeric) 1 2 3 4 5 6 7 *LIT1010 1 row 1 col (numeric) statement : ASSIGN at line 5393 column 105 ERROR: (execution) Matrix has not been set to a value. operation : [ at line 5393 column 189 operands : DM, idx, DM 700 rows 6 cols (numeric) idx 0 row 0 col (type ?, size 0) statement : ASSIGN at line 5393 column 177 ERROR: Matrix Big has not been set to a value. CODE proc iml; use KMM_&p; read all var{time censored survival1 frequency y replicate } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ replicate=DM[,6]; y=DM[,5];frequency =DM[,4]; survival1 =DM[,3]; Censored= DM[,2]; time = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 100x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; u = unique(replicate); do k = 1 to ncol(u); byIdx = loc(replicate=u ); /* the UNIQUE-LOC trick */ byGroup = DM[idx,]; /* extract BY group */ y=byGroup[,5];freqency =byGroup[,4]; survival1 =byGroup[,3]; Censored= byGroupDM[,2]; time = timeDM[,1]; do i = 2 to nrow(DM); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; do i = 1 to ncol(frequency); call symputx("j", i); m = num(symget("j")); m=1*m; print m; %let mm=m; end; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big=SurvivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit;

desireatem · ‎11-02-2013

I do not want to duplicate messages but this has a different heading. How can I fixed it? LOG FILE: NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: Matrix Big has not been set to a value. statement : CREATE at line 913 column 93 ERROR: No data set is currently open for output. statement : APPEND at line 913 column 133 NOTE: Exiting IML. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.00 seconds cpu time 0.01 seconds CODE proc iml; use KMM_&p; read all var{time censored survival1 frequency y replicate } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ replicate=DM[,6]; y=DM[,5];frequency =DM[,4]; survival1 =DM[,3]; Censored= DM[,2]; time = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 100x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; do k = 1 to ncol(u); byIdx = loc(replicate=u ); /* the UNIQUE-LOC trick */ byGroup = DM[idx,]; y=byGroup[,5];freqency =byGroup[,4]; survival1 =byGroup[,3]; Censored= byGroupDM[,2]; time = timeDM[,1]; do i = 2 to nrow(DM); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; do i = 1 to ncol(frequency); call symputx("j", i); m = num(symget("j")); m=1*m; print m; %let mm=m; end; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big=SurvivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit;

desireatem · ‎11-02-2013

Hello Rick, I have modified my code as you said. This is the code, but it does not creat GIB. I do not know the dimension so I do not know how to set a default matrix; proc iml; use KMM_&p; read all var{time censored survival1 frequency y replicate } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ replicate=DM[,6]; y=DM[,5];frequency =DM[,4]; survival1 =DM[,3]; Censored= DM[,2]; time = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 100x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; do k = 1 to ncol(u); byIdx = loc(replicate=u ); /* the UNIQUE-LOC trick */ byGroup = DM[idx,]; y=byGroup[,5];freqency =byGroup[,4]; survival1 =byGroup[,3]; Censored= byGroupDM[,2]; time = timeDM[,1]; do i = 2 to nrow(DM); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; do i = 1 to ncol(frequency); call symputx("j", i); m = num(symget("j")); m=1*m; print m; %let mm=m; end; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big=SurvivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit; LOCK FILE NOTE: IML Ready NOTE: Closing WORK.KMM_1 ERROR: Matrix Big has not been set to a value. statement : CREATE at line 913 column 93 ERROR: No data set is currently open for output. statement : APPEND at line 913 column 133 NOTE: Exiting IML. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.00 seconds cpu time 0.01 seconds

desireatem · ‎11-01-2013

I used the blog,this is my code,runs in my simulations but I do not know if it is doing the right. I just want it to run like a by statement; I bold it. Does what is withing the bold do statement works by a by statement? I do not think correct because the matrix GIB is supposed to be of varying dimension but that snot what I got or may be the way I created my matrix is wrong. proc iml; use KMM_&p; read all var{time censored survival1 frequency y replicate } into DM; /*read the 3 variables into matrix*/ /*have to print within DM*/ close; /*close the link between the dataset and IML*/ replicate=DM[,6]; y=DM[,5];frequency =DM[,4]; survival1 =DM[,3]; Censored= DM[,2]; time = DM[,1]; /*Survivalival is second column*/ /*time is first column*/ n = nrow(DM); /*number of rows in the dataset = obs*/ /*THE N DOESNT NEED TO BE FIXED*/ /*Segment equals the number of consecutive patterns*/ /*J creates a matrix in IML , it creates a 20x1 matrix of 1s, the 3rd 1 is the value in matrix*/ segment = J(nrow(DM),1,1); *print segment; u = unique(replicate); s = j(1, ncol(u)); do i=1 to ncol(u); do i = 2 to nrow(DM); if ((Censored=1)=(Censored[i-1]=1)) then segment=segment[i-1]; /**/ /* coming up with unique segments for censored*****/; else if ((Censored=1)=1-(Censored[i-1]=1)) then segment=segment[i-1]+1; *print time survival1 censored frequency segment; *print segment; end; SurvivalM=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then SurvivalM=survival1; else if censored=0 then SurvivalM=survival1; /*****Create a 1 by N -matrix of Survival ****/; end; *print SurvivalM; InverseC=J(nrow(survival1),1,1); do i = 1 to n; if censored=1 then InverseC=1/survival1; else if censored=0 then InverseC=0; /*****survivalC is the imputed censored survival and set the real Survivalval to 1**/; end; *print InverseC; m=frequency[1]; m=1*m; print m; %let mm=m; t=InverseC[(loc(InverseC^=0))]; print t; tt=t(t); print tt; Big=SurvivalM*(tt); end; create GIB_&p from BIG[colname={COL1}]; append from BIG; quit;

desireatem · ‎11-01-2013

Hello,I wish to perform this procedure by replicate. I have a by replicate but it gave me the error message below. Any help? proc iml; use Complete_&p; read all var _all_ into DM; close; frequency =DM[,7]; y =DM[,6]; obs=DM[,5]; Censored= DM[,4]; time = DM[,3]; x =DM[,2];replicate=DM[,1]; use GIB_&p; read all var _all_ into Col; by replicate; timeS=time; locC=loc(censored=1); matrix=J(1,8,2.5); do im=1 to 1; do j= 1 to frequency[1]; seed=-1; A=RANUNI(-1); i=max(loc(Col[,j]-A>0)); if i<nrow(time) then /** do nothing if the last is censored, that is censored=1*****/ do; timeS[locC ] = time+ (time[i+1]-time)*((Col[i,j]-A)/(Col[i,j]+col[i+1,j])); end; end; sim=repeat(im,nrow(time),1); temp=obs||censored||time||Y||timeS||X||sim||replicate; Matrix=matrix//temp; end; matrix=matrix[2:nrow(matrix),]; create impute_t_&p var{obs censored time Y timeS X sim replicate}; append from matrix; quit; ods trace on/listing; NOTE: IML Ready NOTE: Closing WORK.COMPLETE_1 NOTE: Line generated by the invoked macro "SURV". 30 by replicate; -- 180 ERROR 180-322: Statement is not valid or it is used out of proper order. NOTE: Exiting IML. NOTE: The data set WORK.IMPUTE_T_1 has 500 observations and 8 variables. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE IML used (Total process time): real time 0.98 seconds cpu time 0.04 seconds

desireatem · ‎10-31-2013

Thanks PGStats

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