求sql和data步程序，每程序50币！

code	date	r
1	1	0.1
1	1	0.2
1	2	0.3
1	2	0.4
2	1	-0.5
2	1	0.5
2	2	0.6
2	2	0.4

求由表格1生成表格2的程序！

r_sum

r_mean

r累乘

分别是累计相加、取均值、累计相乘！
求sql和data步程序，每个程序50币！

向高手致敬先！

code	date	r_sum	r_mean	r累乘
1	1	0.3	0.15	0.02
1	2	0.7	0.35	0.12
2	1	0	0	-0.25
2	2	1	0.5	0.24

SQL是跟soporaeternus学的,不知对不对

1. data raw(index=(code_date=(code date)));
   
2.     input code date r;
   
3. datalines;
   
4. 1 1 0.1
   
5. 1 1 0.2
   
6. 1 2 0.3
   
7. 1 2 0.4
   
8. 2 1 -0.5
   
9. 2 1 0.5
   
10. 2 2 0.6
    
11. 2 2 0.4
    
12. ;
    
13. data out1(drop=r);
    
14.     length code date r_sum r_mean rr 8.;
    
15.     rr=1;
    
16.     do _n_=1 by 1 until(last.date);
    
17.         set raw;
    
18.         by code date;
    
19.         r_sum=sum(r_sum,r);
    
20.         rr=r*rr;
    
21.     end;
    
22.     r_mean=r_sum/_n_;
    
23. run;
    
24. proc sql;
    
25.     create table out2 as
    
26.         select code, date, sum(r) as r_sum, mean(r) as r_mean,
    
27.                min(sign(r))*exp(sum(log(abs(r)))) as rr
    
28.             from raw
    
29.             group by code, date;
    
30. quit;

复制代码

data crackman;
input code date r@;
datalines;
1 1 0.1
1 1 0.2
1 2 0.3
1 2 0.4
2 1 -0.5
2 1 0.5
2 2 0.6
2 2 0.4
;
run;
proc sql;
create table crackman as
select code,date,r ,sum(r) as r_sum,mean(r) as r_mean from crackman
group by 1 ,2;
quit;
data crackman;
set crackman;
codedate=compress(cat(code,date));
run;
data crackman;
set crackman;
by codedate notsorted;
retain multi 1;
if first.codedate then multi=r;
else multi=multi*r;
if last.codedate;
drop codedate;
run;

data crackman;
input code date r@;
datalines;
1 1 0.1
1 1 0.2
1 2 0.3
1 2 0.4
2 1 -0.5
2 1 0.5
2 2 0.6
2 2 0.4
;
run;
proc sql;
create table crackman as
select code,date,r ,sum(r) as r_sum,mean(r) as r_mean from crackman
group by 1 ,2;
quit;
data crackman;
set crackman;
codedate=compress(cat(code,date));
run;
data crackman;
set crackman;
by codedate notsorted;
retain multi 1;
if first.codedate then multi=r;
else multi=multi*r;
if last.codedate;
drop codedate;
run;

2# hopewell
你老有来抢钱了哦

data a;
input code date r;
cards;
1 1 0.1
1 1 0.2
1 2 0.3
1 2 0.4
2 1 -0.5
2 1 0.5
2 2 0.6
2 2 0.4
run;
proc sql;
create table b as
select date, code, sum(r) as r_sum, mean(r) as r_mean
from a
group by code, date;
run;
data c;
set a;
if first.date then do;
r_sum=0;
rn=0;
rprod=1;
end;

r_sum+r;
rn+1;
rprod=rprod*r;
retain rprod;

if  last.date;
by code date;
run;
data d;
set c;
r_mean=r_sum/rn;
drop r rn;
run;

min(sign(r))*exp(sum(log(abs(r))))
这个能否实现有点怀疑！

proc sql;
create table b as
select date, code, sum(r) as r_sum, mean(r) as r_mean ,(sum(sign(r))-1)*exp(sum(log(abs(r)))) as r_prod
from a
group by code, date;
run;

我觉得

1. (-1)**sum(r<0)*exp(sum(log(abs(r))))*((sum(r=0)>0)-1)*(-1)

复制代码

这个严密点

貌似处理正负号和0情况比较麻烦
再完美就是加ifn了，0的时候变成1，靠后面的((sum(r=0)>0)-1)*(-1)来处理为0
这样代码太恶心了
就偷懒利用下sum对缺省值赋0吧