I've got a 250 x 250 image, I want to have a scatter plot of the intensity of every pixel and its nearest neighborhood. This is my code:
I = imread(image);
i = [1,249];
j = [1,250];
X = I(i,j);
Y = I(i+1,j);
scatter(X,Y);
why do I get the " X and Y vectors must be the same length" error? They are the same length !
Because scatter(X, Y) is only used for vectors, not matrix. In your example, both X and Y are 2x2 matrices, not vectors.
From its documentation:
scatter(X,Y) displays circles at the locations specified by the vectors X and Y. This type of graph is also known as a bubble plot.
Edit: if you want to plot matrix, use plotmatrix() instead:
plotmatrix(X,Y)
Scatter(X,Y) is used only for vectors as herohuyongtao correctly mentioned. You could try to do the following:
m = 250;
X = I(m+1:end);
Y = I(1:end-m);
scatter(X,Y);
You convert your image matrix I into a vector X while ignoring the first column and in a vector Y while ignoring the last column. X(n) is thus the neighbour of Y(n) on the right side.
I hope this helps!
Related
I have a discrete dataset, X=[x1,x2,..,x12] & Y=[y1,y2,...,y12]. X ranges from [-25, 0] and Y ranges from [1e-6, 1e0]. X does not increase uniformly - as x approaches a value of 0, data sampling density increases from increments of 2.5 to increments of 1. Each x value is units of cm. I cannot get a good fit to the data from fitting a function (I've tried quite a few). I'm left with the discrete data. My need is to sweep the X, Y data completly around the Z axis and put the resulting swept data values into a matrix Z of size (51, 51). I've tried using the cylinder function, [u,v,w] = cylinder(Y) thinking I could extract the data or create a matrix Z from [u, v, w]. I can't seem to sort that out. surf(u,v,w) plots almost correctly - the scaling on the (u, v) axes ranges from [-1, 1] instead of [-25, 25]. This is, I assume, because I'm using cylinder(Y). When I try [u,v,w] = cylinder(X,Y) I get, error: linspace: N must be a scalar. It seems like there should be a better way then my approach of using cylinder to take the X & Y data, interpolate between points (to fill Z where data isn't), rotate it, and put the result into a matrix Z. Any suggestions are welcome. I'm using Octave 6.3.0. Thank you in advance.
Create a matrix R containing distance from origin values.
Use for loops and single value interpolation to cover the R space.
Place the interpolated values into the matrix Z.
% Original data X = [-25,-22.5,...,0]; size(X) = 1 12
% Original data Y = [1e-6, 1.3e-6,...,1] size(Y) = 1 12
u = [-range(X):1:range(X)]; v = [-range(X):1:range(X)]';
R = -sqrt(u.^2.+v.^2);
Z = zeros( 2 .* range(X) + 1);
for i = 1:size(R,1)
for j = 1:size(R,2)
if R(i,j) < min(X); Z(i,j) = 0; endif
if R(i,j) >= min(X); Z(i,j) = interp1(X,Y,R(i,j)); endif
endfor
endfor
I have two arrays
x = [0 9.8312 77.1256 117.9810 99.9979];
y = [0 2.7545 4.0433 5.3763 5.0504];
figure; plot(x, y)
I want to make more samples of x and y then I interpolated both arrays. I tried this code
xi =min(x):.1:max(x);
yi = interp1(x,y,xi);
figure; plot(xi, yi)
but the trajectory is not same as previous plot. Its because the xi is not fluctuating same as x. How should I interpolate both arrays with same trajectory as original one?
This is an issue because when interpolating, MATLAB is going to ignore the order that you feed in the points and instead just sort them based upon their x location.
Rather than interpolating in x/y coordinates, you can instead use a parameter which represents the cumulative arc length of the line segments and use that to interpolate both the x and y coordinates. This will provide you with an interpolant that respects the order and guarantees monotonicity even for multiple values at the same x coordinate.
% Compute the distance between all points.
distances = sqrt(diff(x).^2 + diff(y).^2);
% Compute the cumulative arclength
t = cumsum([0 distances]);
% Determine the arclengths to interpolate at
tt = linspace(t(1), t(end), 1000);
% Now interpolate x and y at these locations
xi = interp1(t, x, tt);
yi = interp1(t, y, tt);
I am trying to graph a surface with a diagonal matrix, the equation I am trying graph is f = x^TDx, x is a 2 by 1 vector and D is a 2 by 2 matrix.
Here is what have so far, but I keep getting error.
x = linspace(-10,10);
y = linspace(-10,10);
[X,Y] = meshgrid(x,y);
D = [1 0; 0 1];
f = #(x,y) [x,y]*D*[x,y].'; % [x,y] is 1 by 2
contour (X,Y,f(X,Y))
Can someone tell me how to get rid of the error? Thanks
Since x and y have the same length, your diagonal matrix D must be a square matrix of size n x n, with n equal to two times the length of your x or y vectors. The reason why you need to multiply the length by two is because the operation [x,y] concatenates the arrays horizontally thus duplicating one of the dimensions.
In this example D is the Identity matrix. See eye for more information.
x = linspace(-10,10); % x is 1x100
y = linspace(-10,10); % y is 1x100
[X,Y] = meshgrid(x,y); % X is 100x100 and Y is 100x100
D = eye(2*numel(x)); % D is 2*100x2*100 = 200x200
f = #(x,y) [x,y]*D*[x,y].'; % [X,Y] is 100x200 and [X,Y].' is 200x100
contour (X,Y,f(X,Y))
If you want D to be a random diagonal matrix, you can accomplish this combining diag with one of the Random Number Generation functions available, like for example randn.
On the previous example, replace D with the following instruction:
D = diag(randn(1,2*numel(x)));
You can also give the coefficients you choose to the diagonal matrix. To do so, you will need to create the vector of coefficients manually, making sure that it has the adequate length, so that it satisfies the conditions explained at the beginning of this post.
Try now replacing D with the following instructions:
v = 1:2*numel(x); % vector of coefficients: v = [1 2 ... 200]
D = diag(v);
Let's say I have a vector of x coordinates, a matrix of y coordinates and corresponding z values:
xcoordinates = [1 2 3 4 5];
ycoordinates = repmat(xcoordinates,5,1)+rand(5,5);
z = zeros(5,5);
for x=xcoordinates
for y=1:5
z(x,y) = sqrt(x^2+ycoordinates(x,y)^2);
end
end
How do I plot a surface determined by the z values at the points given by the corresponding x and y values? The first x value defines the x value for all y values in the first row of the matrix, the second x value to all values in the second row and so on.
(If the answer is griddata I would like some additional pointers. How can I get my data into the correct format?)
mesh(repmat(xcoordinates,5,1), ycoordinates, z)
By the way, you could easily vectorize this computation:
x = repmat(1:5, 5, 1);
y = x + rand(5,5);
z = sqrt(x.^2+y.^2);
mesh(x', y, z)
I have a function which takes a voxel representation of a 3D landscape and can plot a X-Y section to show the middle of the landscape. The voxel representation is stored in a 3 dimensional matrix with a number that represents something important. Obviously the matrix is
1,1,1
2,2,2
in terms of accessing the elements but the actual 3D locations are found in the following method:
(index-1)*resolution+0.5*resolution+minPos;
where resolution is the grid size :
resolution
<-->
__ __ __
|__|__|__|
<- Min pos
and minPos is where the grid starts.
Now in terms of the actual question, i would like to extract a single X-Y section of this voxel representation and display it as a surf. This can be done by just doing this:
surf(voxel(:, :, section))
however then you get this:
The obvious problem is that the grid will start at 0 because that is how the matrix representation is. How can i set the minimum and cell size for surf, ie so that the grid will start at the minimum (shown above) and will have the grid spacing of resolution (shown above).
Read the documentation of surf, you can also provide x and y coordinates corresponding to your data points.
surf(X,Y,Z)
X and Y can be either vectors or matrices:
surf(X,Y,Z) uses Z for the color data and surface height. X and Y are vectors or matrices defining the x and y components of a surface. If X and Y are vectors, length(X) = n and length(Y) = m, where [m,n] = size(Z). In this case, the vertices of the surface faces are (X(j), Y(i), Z(i,j)) triples. To create X and Y matrices for arbitrary domains, use the meshgrid function
Example
Z=[ 0 1 2 3;
7 6 5 4;
8 9 10 11];
x=[-1 0 1 2];
y=[-2 0 2];
surf(x,y,Z);
Of course you have to match Z, x and y matrices/vectors as clearly described in the doc^^
Just remember that elements in columns of Z are surf'ed as values along the y-axis, elements in rows of Z are surf'ed as values along the x-axis. This is clearly to be seen in the example picture.
Solution
I think you switched the x and y-axis around, which you can fix by just transposing z:
s = size(voxel);
xi = (minPosX:resolution:(minPosX+resolution*s(1)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(2)-1));
z = (voxel(:,:,section));
surf(xi, yi, z');
or that you're picking the wrong numbers for constructing xi and yi and it should be this instead:
xi = (minPosX:resolution:(minPosX+resolution*s(2)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(1)-1));
z = (voxel(:,:,section));
surf(xi, yi, z);
So it was easy enough to do:
lets say we have a 3D matrix "voxel";
s = size(voxel);
xi = (minPosX:resolution:(minPosX+resolution*s(1)-1));
yi = (minPosY:resolution:(minPosY+resolution*s(2)-1));
z = (voxel(:,:,section));
[x y] = meshgrid(xi, yi);
x = x';
y = y';
surf(x, y, z);
Provides the following plot:
This is rotated which is annoying, I cant seem to get it to rotate back (I could just visualise around the other way but that's ok)