Bob Jagow suggests changing "date(j)" to "date(j+1)" if you want the date to be the day when the slope moved up (not the day before). I'm not channeling Bob or reading his mind -- he's been answering some of my probing e-mail inquiries (such as "Huh?" or "Duh?"). Here's the whole scan again (for positively the last time, as that indefatigable troupe in "Nicholas Nickleby" said):
//Slope formula by Bob Jagow
//Back-testing version by Roy Yorgensen output = "linregslope.lst";
input="portfoli.lst";
//issuetype common;
integer first, i, j, S, Sx;
float b,b1, Sxx, Sxy, Sy, pctgain;
first:= -10;
Daystoload = 50 -first;
DaysRequired = 22 - first;
S := 21; // set here
b1 := 0; // only initialized first time
for j = 0 to first step -1 do //why 0 to -10?
// reset all variables at start of 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 := .000001+(S*Sxy - Sx*Sy)/(S*Sxx - Sx*Sx);
if b = 0 then print symbol,"b was zero with j: ",j; endif;
pctgain := abs((b1/b) -1); //gives div by 0 -- strange?
if j < 0 // eliminate first pass thru loop -- clumsy
and b < b1 then
//pctgain := abs((b1/b)-1); // not % and constrained to be > 0 ;)
if pctgain > 1 then
println symbol,",", date(j+1),",",close(0):6:3,","," B: ",
b:4:3,","," B1: ", b1:4:3,","," Percent gain: ", pctgain:4:2;
endif; endif;
b1:=b; // must alway set to last day's val
next j; |
