Finding y-intercept for loop

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University 2023-10-27

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https://ww2.mathworks.cn/matlabcentral/answers/2039291-finding-y-intercept-for-loop

评论： Star Strider 2023-10-27

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

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Answer 1

Star Strider 2023-10-27

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在 MATLAB Online 中打开

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

.

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Star Strider 2023-10-27

在 MATLAB Online 中打开

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

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.

.

Star Strider 2023-10-27

在 MATLAB Online 中打开

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

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 2023-10-27

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 2023-10-27

As always, my pleasure!

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

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Finding y-intercept for loop

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