Much, much better, Brooke ;)
Making pctgain divbyzero-proof also eliminates the bothersome
b:= .0001+(num/den); kludge.
As I e-mailed you, the logic is simpler if j is stepped
forward; assuming Roy will abide that, here is my version
with the pctgain test and some minor cosmetic changes. -Bob
--------------------
// Brooke's latest with j advancing, pctgain test, and close(j)
//Back-testing version by Roy Yorgensen
output = "linregslope.lst";
//input="3kVol.lst";
issuetype common;
integer i,j, S, Sx;
float b,lastB, Sxx, Sxy, Sy, pctgain;
S := 21;
daystoload = 50 + S ;
daysrequired = 2 + 10 + S; //may need for newish issues
lastB:= 9999; // large first time flag won't require testing
for j = -10 to 0 do // reset all variables at start of i loop
Sx := 0; Sxx := 0; Sxy := 0; Sy := 0;
for i = j+(1 - S) to j do
Sx := Sx + i;
Sy := Sy + close(i);
Sxx := Sxx + i*i;
Sxy := Sxy + i*close(i);
next i;
b := (S*Sxy - Sx*Sy)/(S*Sxx - Sx*Sx);
pctgain:= (b -lastb)/abs(lastb+.0000001)*100;
if pctgain > 25 then
println symbol,",", date(j),",",close(j):6:3,","," B: ",
b:4:3,","," LastB: ", lastb:4:3,","," Pctgain: ", pctgain:4:2;
endif;
lastb:=b; // save current for pctgain
next j; |
