Sas 聚合以覆盖现有数据

我使用了下面的代码，它工作得非常好，以获得以下结果：

data No_int_weeksPaid; 
set no_internet4;
keep account_number week0-week61;
by account_number;
array week{62} week0-week61;
do i = 1 to 62;
  if i > subscription_start and i <= (subscription_end+1) then 
    week{i} = weeks_paid ;
  else
    week{i} = 0;
end;
drop i;
run;

我希望将所有帐户#都放在一行上，并覆盖这些值，以便得到如下结果：

Account#   Week0   week1 week2  week3 week4
 1          0        1     1      1     1
 1          0        0     0      5     5
 2          1        1     1      1     1
 2          0        2     2      2     2
 2          0        0     0      4     4

 Account#   Week0   week1 week2  week3 week4
 1          0        1     1      5     5
 2          1        2     2      4     4

---

我原以为by语句会有帮助，但不，像这样的东西应该行得通。输出if last.account\u编号，并使用retain跨行保留值。我使用coalesce将非missing设置为零，您可以通过几种不同的方法来实现

data No_int_weeksPaid; 
set no_internet4;
keep account_number week0-week61;
retain week0-week61;  **CHANGED**
by account_number;
array week{62} week0-week61;
do i = 1 to 62;
  if i > subscription_start and i <= (subscription_end+1) then 
    week{i} = weeks_paid ;
  else 
    week{i} = coalesce(week[i],0);  **CHANGED**
end;
drop i;
if last.account_number then output; **CHANGED**
run;

data No_int_weeksPaid；
不设互联网4；
保持账号为第0周至第61周；
保留第0周至第61周**改变**
按账号；
阵列周{62}周0-61周；
i=1到62；
如果i>subscription\u start并且i假设我了解您想要做什么，请尝试以下操作：

data have;
  input Account  Week0   week1 week2  week3 week4;
datalines;
 1          0        1     1      1     1
 1          0        0     0      5     5
 2          1        1     1      1     1
 2          0        2     2      2     2
 2          0        0     0      4     4
run;

data want;
  set have(rename=(Week0=oWeek0 Week1=oWeek1 Week2=oWeek2
                   Week3=oWeek3 Week4=oWeek4));
    by account;

  retain Week0 Week1 Week2 Week3 Week4;
  array new{*} Week0 Week1 Week2 Week3 Week4;
  array old{*} oWeek0 oWeek1 oWeek2 oWeek3 oWeek4;

  keep Account Week0 Week1 Week2 Week3 Week4;

  if First.account then
     do i=1 to dim(new);
        new{i} = old{i};
        end;

  else do i=1 to dim(new);
     if old{i} ne 0 then new{i} = old{i};
     end;

  if last.Account;
run;

我看到的唯一“规则”是希望保留变量的最后一个非零值。只要您的原始数据按显示顺序排列，它就应该可以工作。
如果您希望将其作为
过程sql
，可能最容易构建一个快速宏来为您执行迭代：

%MACRO Week(W) ;
  %DO N=1 %TO &W ;
    max(Week&N) as Week&N, 
  %END ;
  0 as _null_
%END ;

proc sql ;
  create table output as
  select Account, %Week(61)
  from No_int_weeksPaid
  group by Account
  ;
quit ;

不，不是这样。我希望合并给定帐号的所有行，以便无论何时开始，week的值都会覆盖现有的行。对于accnt#1，第二行在第三周有一个5，这将覆盖第一行的剩余1。