Rebuild a vector defining the sign of the elements

Question

I try to better define my question.

I have a vector q:

q={(2 (3 d^4 + 8 d^3 h - 30 d^2 h^2 + 20 d h^3))/((3 d - 4 h)^2 (d - 2 
h)^2), ( 2 d - 3 h)/(d - 2 h), (-3 d^3 + 5 d^2 h - 2 d h^2)/((3 d - 4 h) (d 
- 2 h)^2), ( 2 d - 3 h)/(d - 2 h), (-3 d^3 + 5 d^2 h - 2 d h^2)/((3 d - 4 h) 
(d - 2 h)^2), 1 + d/(2 (d - 2 h)), ( 2 d - 3 h)/(d - 2 h), 1 + d/( 2 (d - 2 
h)), (2 (d^2 - 2 d h + h^2))/(d - 2 h)^2};
h=1;
qc = Abs[q[[1;;6]]];
qb = -Abs[q[[7;;9]]];
qstar=Join[qc,qb];
Manipulate[ListPlot[qstar]‌,{d,0.01,1}]

It doesn't work...ListPlot doesn't plot...

Hard to judge without the code which does not work but I guess this is the answer: how-are-parameters-evaluated-for-a-plot-in-manipulate. Is that the casE? — Kuba, Jun 26 '17 at 08:37
It is not nice to change the question once answers are given. Now they look off topic and you haven't even given any notice. p.s. see Join. — Kuba, Jun 26 '17 at 14:06
I was wondering what question @Aardvark2012 was answering. :) — MikeY, Jun 26 '17 at 14:17
you are missing a { on q to begin with. Near as I can tell that is the entire problem, voting to close as due to a typo. — george2079, Jun 26 '17 at 14:17

score 0 · Answer 1 · answered Jun 26 '17 at 13:58

0

If you know the dimension of q is always 9 with the first six positive and last three negative, then you can just write:

qstar = { 1, 1, 1, 1, 1, 1,-1,-1,-1} Abs[q]

Otherwise, you can use some trickery to get the multiplying vector.

answered Jun 26 '17 at 13:58

MikeY

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Rebuild a vector defining the sign of the elements

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