How can i plot a contour plot with the information below?

8 visualizaciones (últimos 30 días)
Alexander Taylor
Alexander Taylor el 28 de Mzo. de 2018
Comentada: Walter Roberson el 28 de Mzo. de 2018
I have a 57 x 3 vector where the first 2 columns represent the x and y coordinates respectively and I'd like to join the coordinates with a curve based on the value in the 3rd column.
i.e. all coordinates with a "1" in the third column will be joined by a single curve, all coordinates with a "2" in the third column will be joined by a separate curve and so on, until all 57 points have been plotted.
The vector is called "new" shown below:
0.814723686393179 0.959291425205444 1
0.905791937075619 0.547215529963803 1
0.126986816293506 0.138624442828679 1
0.913375856139019 0.149294005559057 1
0.632359246225410 0.257508254123736 1
0.0975404049994095 0.840717255983663 1
0.278498218867048 0.254282178971531 1
0.546881519204984 0.814284826068816 1
0.957506835434298 0.243524968724989 1
0.964888535199277 0.929263623187228 1
0.157613081677548 0.349983765984809 1
0.970592781760616 0.196595250431208 1
0.957166948242946 0.251083857976031 1
0.485375648722841 0.616044676146639 1
0.800280468888800 0.473288848902729 1
0.141886338627215 0.351659507062997 1
0.421761282626275 0.830828627896291 1
0.915735525189067 0.585264091152724 1
0.792207329559554 0.549723608291140 1
0.959492426392903 0.917193663829810 1
0.655740699156587 0.285839018820374 1
0.0357116785741896 0.757200229110721 1
0.849129305868777 0.753729094278495 1
0.933993247757551 0.380445846975357 1
0.678735154857774 0.567821640725221 1
0.757740130578333 0.0758542895630636 1
0.743132468124916 0.0539501186666072 1
0.392227019534168 0.530797553008973 1
0.655477890177557 0.779167230102011 2
0.171186687811562 0.934010684229183 2
0.706046088019609 0.129906208473730 2
0.0318328463774207 0.568823660872193 2
0.276922984960890 0.469390641058206 2
0.0461713906311539 0.0119020695012414 2
0.0971317812358475 0.337122644398882 2
0.823457828327293 0.162182308193243 2
0.694828622975817 0.794284540683907 2
0.317099480060861 0.311215042044805 2
0.950222048838355 0.528533135506213 2
0.0344460805029088 0.165648729499781 3
0.438744359656398 0.601981941401637 3
0.381558457093008 0.262971284540144 3
0.765516788149002 0.654079098476782 3
0.795199901137063 0.689214503140008 3
0.186872604554379 0.748151592823710 3
0.489764395788231 0.450541598502498 3
0.445586200710900 0.0838213779969326 3
0.646313010111265 0.228976968716819 4
0.709364830858073 0.913337361501670 4
0.754686681982361 0.152378018969223 4
0.276025076998578 0.825816977489547 4
0.679702676853675 0.538342435260057 5
0.655098003973841 0.996134716626886 5
0.162611735194631 0.0781755287531837 5
0.118997681558377 0.442678269775446 5
0.498364051982143 0.106652770180584 5
0.959743958516081 0.961898080855054 5

Respuestas (2)

Walter Roberson
Walter Roberson el 28 de Mzo. de 2018
new = [ ...];
[unique_labs, lab_idx] = unique(new(:,3)); %I do not rely upon them being integer or consecutive
num_unique = length(unique_lab);
for K = length(unique_lab) : -1 : 1
this_label = unique_labs(K);
this_xy = new(lab_idx == K, 1:2);
line_handles(K) = plot(this_xy(:,1), this_xy(:,2), 'DisplayName', sprintf('group %g', this_label) );
end
legend( line_handles, 'show')
  2 comentarios
Alexander Taylor
Alexander Taylor el 28 de Mzo. de 2018
Editada: Alexander Taylor el 28 de Mzo. de 2018

Hi Walter,

What is the purpose of:

this_xy = new(lab_idx == K, 1:2);

I tried to run the code and got a Subscripted assignment dimension mismatch error on the following line.

Walter Roberson
Walter Roberson el 28 de Mzo. de 2018
If we knew for sure that the labels were consecutive integers starting from 1, then the code could be shortened to
for K = 1 : 5
this_xy = new(new(:,3) == K, 1:2);
line_handles(K) = plot(this_xy(:,1), this_xy(:,2), 'DisplayName', sprintf('group %g', K) );
hold on
end
hold off
legend(line_handles, 'show')
and if we knew that there were no other items plotted we could shorten further to
for K = 1 : 5
this_xy = new(new(:,3) == K, 1:2);
plot(this_xy(:,1), this_xy(:,2));
hold on
end
hold off
legend('group 1', 'group 2', 'group 3', 'group 4', 'group 5');

Iniciar sesión para comentar.


Birdman
Birdman el 28 de Mzo. de 2018
Editada: Birdman el 28 de Mzo. de 2018
[~,ia,~]=unique(new(:,3));
ia=[ia;size(new,1)];
for i=1:numel(ia)-1
figure(i);
contour(new(ia(i):ia(i+1),1:2))
end

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