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I already have variables "i, discont, a", and I want to creat a variable of b as below.

Thanks,


i   discount   a       b;
1   .97          2.7     =(0.97+0.94) + 0.7*0.93
2   .94          3.5     =(.97+.94+.93)+0.5*0.90
3   .93          4.1     =(0.97+0.94+0.93+0.90)+0.1*0.85
4   .90
5   .85

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关键词：NEEDED Help Need elp Variables Help NEEDED

sdu0632 发表于 2010-3-22 00:22
I already have variables "i, discont, a", and I want to creat a variable of b as below.

Thanks,


i   discount   a       b;
1   .97          2.7     =(0.97+0.94) + 0.7*0.93
2   .94          3.5     =(.97+.94+.93)+0.5*0.90
3   .93          4.1     =(0.97+0.94+0.93+0.90)+0.1*0.85
4   .90
5   .85

为什么i=4和5没有a和b值呢？

sdu0632 发表于 2010-3-22 00:22
I already have variables "i, discont, a", and I want to creat a variable of b as below.

Thanks,


i   discount   a       b;
1   .97          2.7     =(0.97+0.94) + 0.7*0.93
2   .94          3.5     =(.97+.94+.93)+0.5*0.90
3   .93          4.1     =(0.97+0.94+0.93+0.90)+0.1*0.85
4   .90
5   .85

1. 
   
2. DATA AA;
   
3. MISSING Z;
   
4. INPUT I DISCOUNT A;
   
5. DATALINES;
   
6. 1   .97   2.7     
   
7. 2   .94   3.5     
   
8. 3   .93   4.1     
   
9. 4   .90  Z
   
10. 5   .85  Z
    
11. ;
    
12. DATA BB;
    
13. MERGE AA
    
14.      AA(KEEP = DISCOUNT FIRSTOBS=2 RENAME = (DISCOUNT = DISCOUNT2))
    
15.      AA(KEEP = DISCOUNT FIRSTOBS=3 RENAME = (DISCOUNT = DISCOUNT3));
    
16. IF _N_ = 1 THEN S = DISCOUNT;
    
17. S+DISCOUNT2;
    
18. B = S+(A-INT(A))*DISCOUNT3;
    
19. KEEP I DISCOUNT A B;
    
20. RUN;

复制代码

3# jingju11

Well done. Done well

It helps a lot. However, my actual data is a lit bit more complicated. Could you please complete the code for me?

Appreciate it.



data test;
do i=1 to 70;
dis=1/(1.03**(i-1));
output;
end;
run;
data test1;
input agegroup $ ageofdeath;
datalines;
<1 0.3
1-4 3.0
5-9 7.5
10-14 12.5
15-29 17.5
20-24 22.5
25-29 27.5
30-34 32.5
35-39 37.5
40-44 42.5
45-49 47.5
50-54 52.5
55-59 57.5
60-64 62.5
65-69 67.5
;
run;
data test2;
merge test test1;
a=70-ageofdeath;
run;

I am still waiting...

1# sdu0632

你要算的是不是可以表达成这样：

[sum(r_i) over (i=1 to n)] + r_(n+1) + [a-truncation(a)]*r_(n+2)

用SAS里的RETAIN，你可以得到第一部分；后两部分用LAG/LEAD函数应该就可以了。

I do not understand. Could you please give me the codes. Thanks,

8# sdu0632

这两天太忙，只能等到周末看看。你手上有SAS的MANUAL吗？可以查一查RETAIN、LEAD/LAD的SYNTAX，应该不是太困难吧？

版主删了我的悬赏帖， 我只有自己顶了。