Question

跟我一起忍受！这是一个我已经工作了一段时间的项目，尝试了许多事情，我似乎无法让它发挥作用。

以下是我用来创建新表的现有代码。现有的输出表（claims.simulation_ISLPx）也在下面。

我想做的是采用＆＃34;阵列＆＃34;成员（5,6,8,10,12）和5个成员组的过程的5次迭代，6个成员组的6次迭代，以及12个成员组的12次迭代。

＆＃34;流程＆＃34;应该从U（0,1）中抽取一个随机值，然后用claim.simulationISLPx表中的给定值对其进行线性插值，以确定该组中的平均声明（它需要总声明并除以5,6， 8,10或12）。

我不知道添加到我的代码中的内容。任何帮助将非常感激。

/ *从零膨胀对数正态函数输入并允许付费公式* /

%let P_0 = .25; 
%let Mean = 8.9; 
%let Std_Dev = 1.8;
%let ISL = 50000;
%let Deductible = 1750;
%let COINS = .80; 
%let OOPM = 2000; 
%let Min_Paid = 253.08;
%let Ind_Cap_Claim = 2500000;
%let Iterations = 10;
%let Distribution = 'Lognormal';
%let Member_Count = (5,6,8,10,12,15,20,25,30,40,50,60,70,80,90,100,125,150,175,200,250,300,400,500)

libname claims 



data simulation; 
    do i = 1 to &Iterations; 
        Px = rand('Uniform',&P_0,1);
        Px1 = rand('Uniform',0,1);/*generate random numbers from 0 to 1*/
            if Px>= %sysevalf(&P_0) then 
        Allowed_Claims = quantile(&Distribution,Px1,%sysevalf(&Mean),%sysevalf(&Std_Dev));  /*inverse of cdf*/
        output; 
    end;
run; 

proc sql;
   create table claims.simulation_ISLPx as 
   select Allowed_Claims, Px1
      from simulation
      order by Allowed_Claims,Px1;
quit;

proc sql; 
alter table claims.simulation_ISLPx add Paid_Claims_NoISL float; 
    update claims.simulation_ISLPx
        set Paid_Claims_NoISL = min(max(0,Allowed_Claims - %sysevalf(&OOPM), min(Allowed_Claims, 
                                %sysevalf(&Min_Paid) + max(Allowed_Claims - %sysevalf(&Deductible)*%sysevalf(&COINS),0))),
                                %sysevalf(&Ind_Cap_Claim));
alter table claims.simulation_ISLPx add Paid_Claims_AfterISL float; 
    update claims.simulation_ISLPx
        set Paid_Claims_AfterISL = min(Paid_Claims_NoISL, %sysevalf(&ISL)); 
alter table claims.simulation_ISLPx add Total_Cost float; 
    update claims.simulation_ISLPx
        set Total_Cost = Paid_Claims_NoISL - Paid_Claims_AfterISL;
quit;

proc sql;
select * from claims.simulation_ISLPx;
run;
quit;

以下是我的想法

%let Member_Count= [5,6,8,10,12,15,20,25,30,40,50,60,70,80,90,100,125,150,175,200,250,300,400,500]; 


data simulation_interpolation;  
    do i = 1 to &Member_Count[0]; 
    xyz = rand('uniform',0,1);  
        if xyz >= 0 then Px = xyz;
        output;
        end;
        run;
    proc print data= simulation_interpolation;
    run;

Answer 1

有几种方法可以通过宏字符串进行迭代。下面是我将根据您的示例使用的一种方法，尽管有更有效的方法。这会将Member_Count宏字符串转换为数据集，该数据集用于生成模拟。然后使用PROC SUMMARY来平均每个组的模拟。

%let Member_Count= [5,6,8,10,12,15,20,25,30,40,50,60,70,80,90,100,125,150,175,200,250,300,400,500]; 

*** GET MACRO STRING AND CONVERT INTO A DATASET - ONE RECORD PER ITEM ***;
data start;
    *** COMPRESS BRACKETS FROM MACRO STRING ***;
    temp = compress("&member_count.", '[]');
    *** COUNT THE NUMBER OF COMMAS IN STRING AND ADD 1 ***;
    n = count(temp, ',') + 1;
    *** LOOP OVER STRING - GET MEMBER_COUNT, CONVERT TO NUMERIC, AND OUTPUT ***;
    do i = 1 to n;
        member_count = input( scan(temp, i), best8.);
        output;
    end;
run;

data simulation_interpolation;  
    set start(keep=member_count);
    do i = 1 to member_count ;
        xyz = rand('uniform',0,1);  
        if xyz >= 0 then Px = xyz;
        output;
    end;
run;

proc print data= simulation_interpolation;
run;

*** USE PROC SUMMARY TO CALCUATE AVERAGE FOR EACH GROUP OF ITERATIONS ***;
proc summary data=simulation_interpolation  nway;
    class member_count;
    var px;
    output out=stats  mean=mean;
run;

我不确定这是你想要的，我不理解你关于插值的描述，但希望这可以帮助你在你的问题上取得进展。

如何在SAS中通过数组进行线性插值？

1 个答案: