Hi,
I'm a new struggler with IML, and have a noobish question.
What I have here, or would have, is an inner function that for each "timevalue" x makes constant vector c=x, a running time vector t from moment x till the end of data t(x:m), deducts this t from constant x and multiplies with "price" vector p from moment x onwards (e=p[x:m]). Finally s(x) is sum of these rows.
Simple thing expained in complex way but simply but I would like to construct g(x)=SUM operator from x to m [(x-t(x))*p(x)]. Inner function is valid and for example x=3 creates:
t c e d s
3 3 2 0 -104
4 3 4 -4
5 3 7 -14
6 3 2 -6
7 3 4 -16
8 3 8 -40
9 3 4 -24
But how can I insert x there one by one from 1 to m and get a result vector of s(x:m)?
p={5,6,2,4,7,2,4,8,4};
m=nrow(p);
Start uber(p,m);
w=0;
do i=1 to nrow(p);
x=w+i;
q=(x:m); /*m is sample size, a constant from outside of this loop*/
t=q`; /*just transposing to get column vector into row vector*/
c=j(nrow(t),1,x); /*creates row vector of constant x*/
e=p[x:m];
d=(c-t)#e; /*distance times price*/
s=sum(d);
end;
return (s); /*given series of prices p, inserting x into this loop gives s(x)=cumulative sum of weight times price for (x:T), OR SHOULD DO SO*/
finish;
I'm not 100% sure that I follow your example, but it sounds like you want ti compute the sum s for each value of x in the loop. To do that, just create s as a vector outside of the loop and assign to s inside the loop:
s = j(m,1);
do i = 1 to nrow(p);
...
s = sum(d);
end;
return( s);
I'll mention that you can accomplish the same computation with many fewer variables. You don't need x (which equals i) and you don't need w, q, e, or c. Here's a simpler implementation:
s = j(m,1);
do i=1 to nrow(p);
t=T(i:m); /*m is sample size, a constant */
d=(i-t)#p
s=sum(d);
end;
I'm not 100% sure that I follow your example, but it sounds like you want ti compute the sum s for each value of x in the loop. To do that, just create s as a vector outside of the loop and assign to s inside the loop:
s = j(m,1);
do i = 1 to nrow(p);
...
s = sum(d);
end;
return( s);
I'll mention that you can accomplish the same computation with many fewer variables. You don't need x (which equals i) and you don't need w, q, e, or c. Here's a simpler implementation:
s = j(m,1);
do i=1 to nrow(p);
t=T(i:m); /*m is sample size, a constant */
d=(i-t)#p
s=sum(d);
end;
That indeed does the trick, and also enlightens how to use a loop, thanks!
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