Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I am calculating eigenvlaues. Some are negative, some are positive. I define the index nn and mm (as well as np and mp) of each matrix element by some functions that assign positive, zero and negative integer values for nn (np) and strictly positive integers for mm (mp). My code is much more complex, but a simplified version of it looks like this:

nn = 2; mm = 3; som = 2 (nn + 1) mm; RX = 1; PO = 5;
n[a_] := Floor[(a - 1)/mm] - Floor[(a - 1)/((nn + 1) mm)] - nn; 
m[a_] := Mod[a - 1, mm] + 1;
np[b_] := Floor[(b - 1)/mm] - Floor[(b - 1)/((nn + 1) mm)] - nn; 
mp[b_] := Mod[b - 1, mm] + 1;
Vnozer[nn1_, np1_, mm1_, 
   mp1_] := (2 nn1 + 3 np1 + 4 mm1 + 5 mp1) Exp[kx];
Vzer[nn1_, np1_, mm1_, 
   mp1_] := (-4 nn1 - 3 np1 - 2 mm1 - 1 mp1) Exp[kx];
V[nn1_, np1_, mm1_, mp1_] := 
  If[Abs[nn1] > 0 && Abs[np1] > 0, Vnozer[nn1, np1, mm1, mp1], 
   Vzer[nn1, np1, mm1, mp1]];
EN[nn1_] := Sign[nn1] Sqrt[Abs[nn1]];
c[a_, b_] := 
  EN[n[a]] KroneckerDelta[Abs[n[a]], Abs[np[b]]] KroneckerDelta[m[a], 
     mp[b]] KroneckerDelta[Sign[n[a]], Sign[n[b]]] + 
   V[n[a], np[b], m[a], mp[b]];
Q = Array[c, {som, som}];
MF = Table[{kx, Eigenvalues[Q]}, {kx, -RX, RX, (2 RX)/(PO - 1)}];
Array[ou, som]; Array[pl, som];
Table[ou[y] = Table[{MF[[x, 1]], MF[[x, 2]][[y]]}, {x, 1, PO}], {y, 1,
    som}];
Table[pl[y] = ListPlot[ou[y], PlotRange -> All], {y, 1, som}];
FPL = Table[pl[z], {z, 1, som}];
Show[FPL]

Eigenvalue plot looks like this.

I would like to get Red PlotMarkers when my eigenvalues are positive and, say, Blue when my eigenvalues are negative. I played around with an If statement, but I was not able to figure out the right syntax. Something along the lines

PlotMarkers->{If[x]>0, {Red, dotsize}, {Blue, dotsize}}
share|improve this question
    
your code doesn't work on my computer (hanging) –  andre Feb 24 '13 at 23:19
1  
pts = RandomInteger[{-10, 10}, 100]; ListPlot@GatherBy[pts, Positive] –  ssch Feb 24 '13 at 23:22
add comment

2 Answers 2

up vote 1 down vote accepted
 lp = ListPlot[Table[ou[y], {y, 1, som}],
 PlotRange -> All, PlotMarkers -> {Automatic, Medium},
 Joined -> True, ImageSize -> 500]

enter image description here

Post-process to change the colors of markers:

 Normal[lp] /. Inset[a_, {x_, y_}] :> Sequence[If[y >= 0, Red, Blue], Inset[a, {x, y}]]

enter image description here

share|improve this answer
add comment

Perhaps (I saw later that ssch commented a solution around the same lines above):

ListPlot[GatherBy[Flatten[Table[ou[y], {y, 1, som}], 1], Positive[#[[2]]] &], 
   PlotStyle -> {Red, Blue}]

Mathematica graphics

share|improve this answer
    
This looks good in terms of positive eigenvalues being colored red, negatives being colored blue. What I really want to accomplish is to plot in red all eigenvalues that come from matrix elements labeled by nn (np) values that are positive and plot blue the eigenvalues that are labeled by nn (np) values that are negative (I will then have to deal with nn=0, but that is the next step). The eigenvalue itself may be positive OR negative. This is why I thought an If statement may be the way to go. Perhaps I should have suggested PlotMarkers->{If[nn]>0&&[np]>0, {Red, dotsize}, {Blue, dotsize}} –  Rainforest Frog Feb 24 '13 at 23:45
4  
@RainforestFrog But that is not what you said in the question... –  rm -rf Feb 25 '13 at 0:00
    
@rm-rf I quit .. –  belisarius Feb 25 '13 at 0:47
2  
@RainforestFrog i hope you appreciate that it is frustrating for people who contribute time and expertise to offer an answer to a specific question only to be confronted with details or "what ifs" not previously mentioned in the question. It creates frustration and ill will and makes it less likely that you will get positive replies to your questions in the future. –  Mike Honeychurch Feb 25 '13 at 2:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.