Plotting for particular values of x-coordinate

5 vues (au cours des 30 derniers jours)
Priya
Priya le 17 Juin 2014
Commenté : Star Strider le 17 Juin 2014
load('wheel_rail_AW.mat')
S = contact_geometry
x_new=S.y;
y_new =S.r_R;
xi = min(x_new) + (max(x_new)-min(x_new)) * rand;
yi = interp1(x_new, y_new, xi, 'linear', 'extrap');
figure(1)
plot(x_new,y_new,'.',xi,yi,'or')
Now, my question is how to execute the above for particular values of xi(given below) instead of random values being generated by xi each time.
I want the xi values to be xi= -0.02, -0.01, 0, 0.01, 0.02
And I want the program to take -0.02 for the first time and -0.01 for the next iteration, likewise for other 3 values
Please do help me.
Thanks

Réponse acceptée

Star Strider
Star Strider le 17 Juin 2014
Hi Priya,
Change the code to:
xi = [-0.02, -0.01, 0, 0.01, 0.02];
xi = xi( (xi >= min(x_new)) & (xi <= max(x_new)) ); % Check for in-range values only
yi = interp1(x_new, y_new, xi, 'linear', 'extrap');
I remember there were problems with xi not being within the bounds of x_new previously, (which is the reason we added 'extrap' that we now don’t need). The added line makes sure you only use your xi values that are within the range of x_new.
  2 commentaires
Priya
Priya le 17 Juin 2014
Modifié(e) : Priya le 17 Juin 2014
Ya, It worked, but I wanted xi to take one point at a time ie., it must take one point out of the 5 for each iteration.
xi=[-0.02 -0.01 0 0.01 0.02];
xi_a=xi(randi(5));
Hope this makes sense. Actually I did this using your previous answer. Thanks for your reply again.
Star Strider
Star Strider le 17 Juin 2014
My pleasure!
It makes sense. I just wasn’t clear on exactly what you wanted to do or how you wanted to do it.
(I added the check for xi in case you change your file but not your xi vector. It keeps it from interpolating out-of-range values.)

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Plus de réponses (1)

dpb
dpb le 17 Juin 2014
You basically answered the question in the question...
xi = [-0.02:0.01:0.02];
Now you'll have to ensure that the values are within the ranges of x_ and y_new or the interp1 call isn't going to work.
  1 commentaire
Priya
Priya le 17 Juin 2014
Sorry, I think there is a lack of clarity in my question. For xi, I have selected particular coordinates but actually its limit is -0.02 to 0.02 where there are a number of other points in between them.
Anyway, I have got it sorted out now. Thanks very much for your reply.

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