Is it possible to extract data from a Mathematica plot and plot it with the Origin software?

Question

I have combined three plots with "Show[p1,p2,p3]. Can I extract the data points from my plot and want to plot with origin?? `

F = 0.025;
kv = 0.0043;
kw = 0.046;
A = (F + kv);
B = (F + kw);
Du = 290;
Dv = 100.443;
Dw = 7.3;
G = (F - Sqrt[F^2 - 4*F*(A^2 + B^2)])^2;
j11 = (-((4*F^2*(A^2 + B^2))*G^(-1)) - F);
j12 = -(2*A);
j13 = -(2*B);
j21 = (4*A^2*F^2)/(G);
j22 = (A);
j31 = (4*B^2*F^2)/G;
j33 = B;

C1 = (k^2 (Du + Dv + Dw) - j11 - j22 - j33);

C2 = ((-j11*(Dv + Dw) - j22*(Du + Dw) - j33*(Du + Dv))*
     k^2 + (Du*Dv + Du*Dw + Dv*Dw)*k^4 - j12*j21 - j13*j31 + j11*j22 +
     j11*j33 + j22*j33);

C3 = ((Du*Dv*Dw)*k^6 + (-j11*Dv*Dw - j22*Du*Dw - j33*Du*Dv)*
     k^4 + (Dw*j11*j22 + Dv*j11*j33 + Du*j22*j33 - Dw*j12*j21 - 
       Dv*j13*j31)*k^2 - j11*j22*j33 + j12*j21*j33 + j13*j31*j22);
p1 = ContourPlot[
  x^3 + (C1) *x^2 + (C2)*x + C3 == 0, {k, 0.005, 0.020}, {x, -0.01, 
   0.008}, ContourStyle -> Magenta]

` Then repeated the same code with Different Dv=149 and Dv=174, combine the three plots with Show[p1,p2, p3]

Strongly related: https://mathematica.stackexchange.com/q/14222/1871 https://mathematica.stackexchange.com/q/31858/1871 — xzczd, Jan 14 '19 at 07:39
This question might be a duplicate, but also might not be, depending how the mentioned combined plot looks like. Can you provide an image of your combined plot? — Anton Antonov, Jan 14 '19 at 14:39
Anton AntonovThank you so much for your reply I have added the image file here. — Albert, Jan 17 '19 at 01:33
Do you have the code to generate this plot, Albert, or do you only have an image of the plot? — Carl Lange, Jan 17 '19 at 01:47
@CarlLange Yes I have but I am new here, so i don't know ho to add code in my comments — Albert, Jan 17 '19 at 01:48
Click "edit" under your question and paste the code into your question. Then select your code and click the icon in the textbox toolbar that looks like {}. That will format your code. Then save your question. — Carl Lange, Jan 17 '19 at 01:53
@Albert I still don't get it - the point $(0,0)$ is visible in the plot - do you want to center it around that point? Also, why can you not change the original plot command? Why do you want to extract the data from an existing plot? (which is way messier) — Lukas Lang, Jan 17 '19 at 14:38
@AntonAntonov from "origin" I mean to Plot in "origin" software. I need text file of the Datapoints, so that I use an other software to plot it — Albert, Jan 17 '19 at 15:03
All this fuss and to-do over the lack of an uppercase O on origin in the original post ! — High Performance Mark, Jan 17 '19 at 15:11
@AntonAntonov Thank you dude, I really appreciate your effort. sorry for taking your time, I think I didn't convey my question properly.
sorry again for taking your time — Albert, Jan 18 '19 at 02:02

Anton Antonov · Answer 1 · 2019-01-17T15:18:14.800

@AntonAntonov I want to plot with origin. – Albert

I think OP wants this (it is not entirely clear):

F = 0.025;
kv = 0.0043;
kw = 0.046;
A = (F + kv);
B = (F + kw);
Du = 290;
(*Dv=100.443;*)
Dw = 7.3;
G = (F - Sqrt[F^2 - 4*F*(A^2 + B^2)])^2;
j11 = (-((4*F^2*(A^2 + B^2))*G^(-1)) - F);
j12 = -(2*A);
j13 = -(2*B);
j21 = (4*A^2*F^2)/(G);
j22 = (A);
j31 = (4*B^2*F^2)/G;
j33 = B;

{kmin, kmax} = {0.005, 0.020};

grs =
  MapThread[
   Function[{Dv, color},
    C1 = (k^2 (Du + Dv + Dw) - j11 - j22 - j33);
    C2 = ((-j11*(Dv + Dw) - j22*(Du + Dw) - j33*(Du + Dv))*
        k^2 + (Du*Dv + Du*Dw + Dv*Dw)*k^4 - j12*j21 - j13*j31 + 
       j11*j22 + j11*j33 + j22*j33);
    C3 = ((Du*Dv*Dw)*k^6 + (-j11*Dv*Dw - j22*Du*Dw - j33*Du*Dv)*
        k^4 + (Dw*j11*j22 + Dv*j11*j33 + Du*j22*j33 - Dw*j12*j21 - 
          Dv*j13*j31)*k^2 - j11*j22*j33 + j12*j21*j33 + j13*j31*j22);
    ContourPlot[
     x^3 + (C1)*x^2 + (C2)*x + C3 == 0, {k, kmin, kmax}, {x, -0.01, 
      0.008}, ContourStyle -> color, PlotRange -> All]
    ], {{100.443, 149, 174}, {Red, Blue, Green}}];

Show[grs, PlotRange -> {{-kmin, kmax}, All}, GridLines -> {{0}, {0}}, 
 Epilog -> {Red, PointSize[0.03], Point[{0, 0}], Black, 
   Text[Style["Origin", Bold], {0, 0}, {-1, -1}]}]

@QuantumPenguin Thanks, I spent too much time coding in order to get OP to say from "origin" I mean to Plot in "origin" software. — Anton Antonov, Jan 17 '19 at 15:35

Is it possible to extract data from a Mathematica plot and plot it with the Origin software?

1 Answers1