Finding y-intercept for loop

Hi,

In the m.uvals file I 24 x 1 x 31 matrix, 24 are the index of xivals and 31 are the values of u and m.array file is the y array.

Please how can find the y-intercept for each value of xivals? I have tried code below for one of xivals, which seems to work. Also why is the value of the intercept not at the point where y exactly intercept x ?

xivals =[0:1:21 60 100];

y=yarray;

V=squeeze(u(1, :,:))';

loc=diff(sign(V)) & abs(y(1:end-1))<0.001;

yintercept=y(loc);

figure(1)

plot(y,V,yintercept,0,'rx');

yline(0);

xlabel y-coordinate;

ylabel Velocity

Answer 1

[Star Strider]

I made some minor changes to your code, and added a loop to return reasonably precise values of ‘y’ for each intersection. Your idea was appropriate, however the indices themselves will not be accurate enough for most purposes. It is necessary to interpolate to get reasonably precise values.

Try this —

LD1 = load('uval.mat');
u = squeeze(LD1.ux);
Sizeu = size(u)

Sizeu = 1×2

24 31

LD2 = load('yarray.mat');
yarray = LD2.yarray;

xivals =[0:1:21 60 100];

y=yarray;
V=u(1,:); 
loc=diff(sign(V)) & abs(y(1:end-1))<0.001;
% yintercept=y(loc)

intsx = find(diff(sign(V)));                                    % Approximate Indices Of Intersections

for k = 1:numel(intsx)
    idxrng = max(1,intsx(k)-1) : min(numel(V),intsx(k)+1);      % Index Range
    yintercept(k) = interp1(V(idxrng), y(idxrng), 0);           % Interpolated Values Of Intersections
end

yintercept

yintercept = 1×2

1.0e-06 * -0.6311 1.0000

figure(1)
plot(y,V,yintercept,0,'rx'); 
yline(0);
xlabel y-coordinate; 
ylabel Velocity

.

[University]

Thank you so much for your assisstance. I'm finding it difficult to plot the intercept against xivals. Is there any way I can do that? I was expected the length of yintercept to be the same as the length of xivals. From the V=u(1,:); I guess you just pick the first index of xivals. I want to loop over all the xivals so that I can plot plot(xivals,yintercept).

[Star Strider]

As always, my pleasure!

The problem is that they are not the same lengths, so plotting them against each other is not possible. They share the row size of‘u’ (‘u’ is 24x31) not the column size, and that defines ‘V’. The only way to make them the same lengtrhs is to interpolate ‘u’ to be the same size as ‘xivals’ (or the reverse of that) or to plot the columns of ‘u’ against ‘xivals’, however that does not work because of how ‘u’ is defined. It does not appear to have any intercepts in that direction. You could possibly take the intercepts of all the rows of ‘V’ and plot that against ‘xivals’.

LD1 = load('uval.mat');
u = squeeze(LD1.ux);
Sizeu = size(u)

Sizeu = 1×2

24 31

LD2 = load('yarray.mat');
yarray = LD2.yarray;

xivals =[0:1:21 60 100];

y=yarray;
V=u(1,:); 
loc=diff(sign(V)) & abs(y(1:end-1))<0.001;
% yintercept=y(loc)


for k1 = 1:size(u,1)
    V = u(k1,:);
    intsx = find(diff(sign(V)));                                    % Approximate Indices Of Intersections
    for k2 = 1:numel(intsx)
        idxrng = max(1,intsx(k2)-1) : min(numel(V),intsx(k2)+1);      % Index Range
        yintercept(k1,k2) = interp1(V(idxrng), y(idxrng), 0);           % Interpolated Values Of Intersections
    end
end

yintercept

yintercept = 24×2

1.0e-06 * -0.6311 1.0000 -0.0016 1.0000 -0.0845 1.0000 -0.4855 1.0000 -0.0900 1.0000 -0.0956 1.0000 -0.1011 1.0000 -0.1067 1.0000 -0.1124 1.0000 -0.1181 1.0000

v1 = ones(size(u(1,:)));

figure(1)
hold on
for k = 1:size(u,1)
    plot3(y,u(k,:),v1*xivals(k))
    plot3(yintercept(k,:),[0 0], [1 1]*xivals(k),'rx'); 
end
hold off
zl = zlim;
patch([xlim flip(xlim)], zeros(1,4), [[1 1]*zl(1) [1 1]*zl(2)], [1 1 1]*0.5, 'FaceAlpha',0.25, 'EdgeColor','none')
xlabel y-coordinate; 
ylabel Velocity
zlabel xivals
grid on
view(-27.5,30)

Fortunately, there are only two values of ‘yintercept’ for each row of ‘u’ so regular matrices (not cell arrays) will work to store them. The gray patch replaces the yline call here.

.

[University]

Thank you so much Star.... So if can make u size the same as that of the xivals size, I can then plot (xivals, intercept) ? So here i have attached uxl.m file which is the same size as xivals

xivals =[0 0.001 0.01 0.1 0.2 0.5 0.6 0.7 0.8 1:1:20 60 100];

I would appreciate it if you could use this to plot (xivals, y-intercept). xivals is 31 and uxl is 31 now.

[Star Strider]

As always, my pleasure!

The row size of ‘u’ (24) ias already the same size as ‘xivals’. The only way to plot ‘yintercept’ as a function of ‘xivals’ is the way I did it using a 3D representation. Interpolating ‘xivals’ to be (1x31) could be possible, however I doubt that it would be correct.

If you want to do it, this works —

xivals =[0:1:21 60 100]

xivals = 1×24

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 60 100

xq = linspace(1, numel(xivals), 31);                            % Query Vector
xivals31 = interp1((1:numel(xivals)), xivals, xq)

xivals31 = 1×31

0 0.7667 1.5333 2.3000 3.0667 3.8333 4.6000 5.3667 6.1333 6.9000 7.6667 8.4333 9.2000 9.9667 10.7333 11.5000 12.2667 13.0333 13.8000 14.5667 15.3333 16.1000 16.8667 17.6333 18.4000 19.1667 19.9333 20.7000 39.2000 69.3333

.

[University]

Sorry, I had to upload the u which has the same size as xivals now.

xivals =[0 0.001 0.01 0.1 0.2 0.5 0.6 0.7 0.8 1:1:20 60 100];

and uxl.m file has 31 x 1x 31 now.

[Star Strider]

The ‘u’ originally provided was (24x31). A (31x31) version should work with the original code, however ‘xivals’ is still (1x24) so that problem persists, unless there is a new version of ‘xivals’.

[University]

So sorry , the new xivals is 31 also Strider.

size(xivals)

ans = 1 31

That's new xivals.

[Star Strider]

What do you want to do with the new ‘xivals’ vector?

If you want my help with it, I need the new ‘u’ matrix as well, and to know what you want to do.

-----

(My apologies for the delay. I just wasted about 30 minutes recovering my desktop machine from yet another Win 11 crash caused by some sort of bloatware call ‘autopilot’ that then completely corrupted my desktop layout as well. I hate Windows. My next computer is going to be Linux. If that also means that I can’t use any other Micro$oft apps, being completely free from Micro$oft works for me.)

[University]

Sorry about that Strider. Attached the new u values and corresponding new xivals is

xivals =[0 0.001 0.01 0.1 0.2 0.5 0.6 0.7 0.8 1:1:20 60 100];

I just want a measure of plot(xivals, yintercept(1:, 2)). Something like that.

[Star Strider]

I am a bit confused.

Do you want to do the same thing with ‘y’ or should ‘xivals’ replace ‘y’ in the code?

Try this (previous version with new ‘u’ and ‘xivals’) —

LD1 = load('uval31.mat');
u = squeeze(LD1.ux);
Sizeu = size(u)

Sizeu = 1×2

31 31

LD2 = load('yarray.mat');
yarray = LD2.yarray;

xivals =[0 0.001 0.01 0.1 0.2 0.5 0.6 0.7 0.8 1:1:20 60 100];

y=yarray;
V=u(1,:); 
loc=diff(sign(V)) & abs(y(1:end-1))<0.001;
% yintercept=y(loc)


for k1 = 1:size(u,1)
    V = u(k1,:);
    intsx = find(diff(sign(V)));                                    % Approximate Indices Of Intersections
    for k2 = 1:numel(intsx)
        idxrng = max(1,intsx(k2)-1) : min(numel(V),intsx(k2)+1);      % Index Range
        yintercept(k1,k2) = interp1(V(idxrng), xivals(idxrng), 0);           % Interpolated Values Of Intersections
    end
end

yintercept

yintercept = 31×2

0.5533 100.0000 0.5428 100.0000 0.2978 100.0000 13.5207 100.0000 9.9913 100.0000 7.7154 100.0000 7.4728 100.0000 7.2979 100.0000 7.1652 100.0000 6.9767 100.0000

v1 = ones(size(u(1,:)));

figure(1)
hold on
for k = 1:size(u,1)
    plot3(y,u(k,:),v1*xivals(k))
    plot3(yintercept(k,:),[0 0], [1 1]*xivals(k),'rx'); 
end
hold off
zl = zlim;
patch([xlim flip(xlim)], zeros(1,4), [[1 1]*zl(1) [1 1]*zl(2)], [1 1 1]*0.5, 'FaceAlpha',0.25, 'EdgeColor','none')
xlabel y-coordinate; 
ylabel Velocity
zlabel xivals
grid on
view(-27.5,30)

.

[University]

Thank you so much for your help. I have been able to get through it. I just did this:

plot(xivals, yintercept(:, 1), xivals, yintercept(:, 2)).

[Star Strider]

As always, my pleasure!

I am glad it now works as you want it to!