Question

我的示例代码如下：

ID  DX  S_DT
1   VV  12/20/2013
1   VV  12/21/2013
2   VV  3/11/2006
2   VV  6/20/2008
2   VV  6/4/2009
2   VV  6/5/2009
3   VV  11/12/2001
3   VV  11/13/2001
3   VV  1/1/2014

我希望我的最终数据看起来像这样，除非将粗体日期设置为零：

ID  d_dt1        d_dt2           d_dt3          d_dt4
1   12/20/2013  **12/21/2013**      
2   3/11/2006   6/20/2008       **6/4/2009**    **6/5/2009**
3   11/12/2001  **11/13/2001**  1/1/2014

这是我的代码：

data data_out;
SET data_in;

by ID S_dt;

retain d num_d prev_d_dt d_dt1-d_dt4;

array d_dt(4) d_dt1-d_dt4;

array dur(3) dur1-dur3;
if first.ID then do;

d=0;
num_d=0;
prev_d_dt=.;
do i = 1 to 4;
d_dt(i)=.;
end;
end;
if (DX='VV' and prev_d_dt ne S_dt ) then do; 
d=1;
num_d+1;
prev_d_dt=S_dt;
d_dt(num_D)=S_dt;
end;
end;
end;
if (last.ID) then do;
current_ddt=.;
do  i = 1 to 3;
dur(i)=d_dt(i+1) - d_dt(i);
if ( dur(i) >= 270) then do;
current_ddt=d_dt(i+1);
dur(i) = d_dt(i+1) - d_dt(i);
end;
if (dur(i) < 270) then do;
d_dt(i+1)=.;
current_ddt=d_dt(i+2);
dur(i+1)=d_dt(i+2) - d_dt(i);
end;
end;
output;
end;
keep patid num_d d_dt1-d_dt4 dur1-dur3 current_ddt;
format d_dt1-d_dt4 current_ddt mmddyy10.;
run;

我无法将这些不需要的日期设置为零。

Answer 1

一种方法是转置然后使用数组进行计算。

data  have;
input ID  DX $  S_DT:mmddyy10.;
format s_dt mmddyy10.;
datalines;
1   VV  12/20/2013
1   VV  12/21/2013
2   VV  3/11/2006
2   VV  6/20/2008
2   VV  6/4/2009
2   VV  6/5/2009
3   VV  11/12/2001
3   VV  11/13/2001
3   VV  1/1/2014
;

/* sorting the data*/

proc sort data = have;
by id s_dt;
run;

/* transposing the data*/

proc transpose data =have out =have1 (drop =_name_) prefix =date;
by id ;
 var s_dt;
run;

 /* picking all the date variables and counting number of variables
  for array processing in next step*/

  proc sql;
  select name , count(name)-1 into :date SEPARATED by ' ',
  :cnt
 from dictionary.columns
 where memname = 'HAVE1'
 and upcase(name) like 'DATE%';

  %put &date &cnt;
 /* data processing*/

data have2;
set have1;
array date(*) &date;
dateno =date1;
format dateno mmddyy10.;
do i = 1 to &cnt;
if date(i) -date(i+1) = -1 then do ;
call missing(date(i));
call missing (date(i+1));
end;
end;
drop date1 i;
run;

 /* putting variables in the order*/

 data have_final;
 retain id &date;
  set have2(rename =dateno=date1);
 run;

SAS会循环并删除不需要的记录

1 个答案: