如何在sas中动态改变观察值来找到最大值？ [英] how to find the maximum value across dynamically change observations in sas?

问题描述

所以我有以下三个变量的数据集：帐户，余额和时间。

 账户余额时间
 1 110 01/2006 
 1 111 02/2006 
 1 88 03/2006 
 1 61 04/2006 
 1 1203 05/2006 
 2 112 01/2006 
 2 111 02/2006 
 2 665 03/2006 
 2 61 04/2006 
 2 1243 05/2006 
 3 110 01/2006 
 3 111 02/2006 
 3 88 03/2006 
 3 61 04/2006 
 3 1203 05/2006 
  

每个帐户都有更多记录。所以开始时间可能在我写的东西之前，结束时间可能在我写的东西之后。

所以我的问题是：

我试图找到最初12个月的每个帐户的最大余额。例如，对于05/2006的帐户3，我试图找到最大值（2006年4月的帐户3余额，2006年3月的帐户3余额，2006年3月的帐户3余额，........ .....，帐户3余额在2006年4月）。

你的想法是什么？我所做的就是对数组使用滞后函数。然而，它并不那么高效。因为如果我需要前120个月，我会遇到麻烦。

谢谢。

Best

Xintong

解决方案

  options mprint; 
％宏观滞后; 
 data temp（drop = count i）; 
设定余额; 
 by acct time; 
 array x（*）lag_bal1  -  lag_bal3; 
％我= 1％到3; 
 lag_bal& i = lag& i。（余额）; 
％end; 
 if first.acct then count = 1; 
 do i = count to dim（x）; 
致电失踪（x（i））; 
 end; 
 count +1; 
 max_ball = max（balance_，lag_bal1-lag_bal3）; 
％修正; 
 
％滞后; 
  

so i have the following dataset with three variables: account, balance, and time.

account balance time
1       110     01/2006
1       111     02/2006
1       88      03/2006
1       61      04/2006
1       1203    05/2006
2       112     01/2006
2       111     02/2006
2       665     03/2006
2       61      04/2006
2       1243    05/2006
3       110     01/2006
3       111     02/2006
3       88      03/2006
3       61      04/2006
3       1203    05/2006

each account has more records. so the starting time might be before what I wrote and the ending time might be after what i wrote.

so my question is:

I am trying to find the maximum balance for each account in prievious 12 monthes. For example, for account 3 on 05/2006, i am trying to find max(account 3 balance at 04/2006, account 3 balance at 03/2006, account 3 balance at 03/2006,............., account 3 balance at 04/2006).

what is in your mind? what i did is to use lag function with array. however, it is NOT so efficient. because i will be in trouble if i need previous 120 months.

Thank you.

Best

Xintong

解决方案

options mprint;
%macro lag;
    data temp (drop=count i);
        set balance;
        by acct time;
    array x(*) lag_bal1 - lag_bal3;
        %do i=1 %to 3;
            lag_bal&i=lag&i.(balance);
        %end;
    if first.acct then count=1;
    do i=count to dim(x);
        call missing(x(i));
    end;
        count+1;
    max_ball=max(balance, of lag_bal1-lag_bal3);
%mend;

%lag;

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