Hello The simple random code with replacement usually produce output with hits number, for instant id 22 is chosen three times. However I do not want the output like that , I want all the three hit live in the second output. That I want the sample size to stay 10, even if it is not independent like in output two below. proc surveyselect data = hsb25 method = SRS rep = 1   sampsize = 10 seed = 12345 out = hsbs1;   id _all_; run; proc print data = hsbs1 noobs; run; proc surveyselect data=hsb25  method = urs sampsize = 10    rep=1 seed=12345 out=hsbs2 out=outhits;    id id read write math science socst; run; proc print data = hsbs2 noobs; run; OUTPUT ONE             Rep    id    re    wri   math    sc    soc     Hits                   1        108     34       33      41        36        36        1     1         50     50       59      42        53        61        1     1        102     52       41      51        53        56        1     1        136     65       59      70        63        51        1     1        171     60       54      60        55        66        1     1         22     42       39      39        56        46        3     1         47     47       46      49        33        41        1     1         56     55       45      46        58        51        1 Hello The code above gives the output of Simple random sampling with replacement.  The sample size is 10. However I am not happy with the way the data is reported. I want the Hits= 3 to be reported three times. I wish to ask if any knows how to help me report the data like this: OUTPUT TWO       Rep    id    re    wri   math    sc    soc     Hits     1        108     34       33      41        36        36        1     1         50     50       59      42        53        61        1     1        102     52       41      51        53        56        1     1        136     65       59      70        63        51        1     1        171     60       54      60        55        66        1     1         22     42       39      39        56        46        1     1         22     42       39      39        56        46        1     1         22     42       39      39        56        46        1     1         47     47       46      49        33        41        1     1         56     55       45      46        58        51        1 ... View more

I ve my data set I call  "TES" I want to simultaneously draw sample with replacement from it then do a regression on all the samples . I want to draw 25 different samples with replacement. The code below is not correct. Please help. do sample=1 to 25 proc surveyselect data=tes_  method = urs sampsize = 100    rep=1 seed=12345 out=test; run; end; Proc reg data=test; by sample; model y=t; run; ... View more

Hello, in the following data generation; I want to change the censored value for the highest observed time to be always 0 '. That is after sorting by time from minumum time to maximum, I want censored=0 for the largest time data test; seed=-1;   alpha1 = 2.50;   beta1 = 9.50;   do i = 1 to 100;    lambdaT = 0.025; *baseline hazard;    lambdaC= .03;  *light=0.15; *heavy=0.03;    er=0+sqrt(0.0001)*rannor(1);     t = rand("WEIBULL", 0.75, lambdaT);  * time of event;     c = rand("WEIBULL", 1.25, lambdaC) ;* time of censoring;     time = min(t, c);    * which came first?;     censored = (c lt t);       obs=(t lt c); * creating observation variable from censored when       y= alpha1 + beta1*t + er;     output;   end; run; ... View more

Hello, in the following data generation; I want to change the last observation for variable 'censored' to '0' when ever it is '1'. That is the 100th observation should have censored=0 , even if it is supposed to e 1 data test; seed=-1;   alpha1 = 2.50;   beta1 = 9.50;   do i = 1 to 100;    lambdaT = 0.025; *baseline hazard;    lambdaC= .03;  *light=0.15; *heavy=0.03;    er=0+sqrt(0.0001)*rannor(1);     t = rand("WEIBULL", 0.75, lambdaT);  * time of event;     c = rand("WEIBULL", 1.25, lambdaC) ;* time of censoring;     time = min(t, c);    * which came first?;     censored = (c lt t);       obs=(t lt c); * creating observation variable from censored when       y= alpha1 + beta1*t + er;     output;   end; run; ... View more

data kaplan; input time survival1; cards; 4 .6 5 .55 6 .3 7 .1 9  0 ; run; proc iml; use kaplan; read all  var{TIME SURVIVAL1} into DM; close; SURVIVAL1 = DM[,2]; TIME = DM[,1]; n = nrow(DM);     segment = J(nrow(DM),1,1); *print segment; SS=J(nrow(survival1),1,1); do i = 1 to n ; if 1=n then SS=0; else if  i< n-1 then SS=SURVIVAL1-SURVIVAL1[i+1]; end; TT=J(nrow(time),1,1); do i = 1 to n; if 1=n then TT=0;                     else if i < n-1  then TT=time[i+1]-time; end; create kaplann var{time  survival1 SS TT }; append; quit; data kaplan_sum; set kaplann; sur_time=SS*TT; drop time  survival1; run; proc print data=kaplan_sum; run; This output is wrong, there is supposed to be just 4 observations. The 4th and fifth observation for SS is wrong. SS= (.6-.55)+(.55-.3)+(.3-.1)+(.1-0) TT= (5-4)+(6-5)+((7-6)+(9-7)) This is wrong. i   SS  TT  TT*SS 1 0.05 1 0.05 2 0.25 1 0.25 3 0.20 1 0.20 4 1.00 1 1.00 5 1.00 1 1.00 This is how the correct output is supposed to look like; SS= (.6-.55)+(.55-.3)+(.3-.1)+(.1-0) TT= (5-4)+(6-5)+((7-6)+(9-7))    i    SS   TT       SS*TT 1 0.05 1         0.05 2 0.25 1        0.25 3 0.20 1        0.20 4 0.1 2        0.2 ... View more

The first pick  the first row from the data set kaplan. How come the statement _n_=n does not pick the last row.? How do I pick the last row? data KS_; set Kaplan; if _n_ =1 then keep=1; if keep ne 1 then delete; drop keep survival; run; data KSm_&p(rename=survival1=(survival_max time=time_max)); set Kaplan_&p; if _n_ =n then keep=1; if keep ne 1 then delete; drop keep survival; run; ... View more