I am trying to take a gradient of an image using the Prewitt filter. Can you tell me if this approach is correct?
I = imread('image.jpg')
Gx = [-1 0 1; -1 0 1; -1 0 1];
Gy = [1 1 1; 0 0 0; 1 1 1];
D = conv2(conv2(I, Gx), Gy)
imshow(D)
Is that correct? Is there a cleaner way to do it (no need to call conv2 twice)? Is conv2(I, Gx) the same as conv2(Gx, I)? (i.e. commutative?)
Thanks.
Judging by my wikipedia-ing it looks like what you should do is:
I = imread('image.jpg')
Gx = [-1 0 1; -1 0 1; -1 0 1];
Gy = [1 1 1; 0 0 0; -1 -1 -1];
A = sqrt( conv2(I,Gx).^2 + conv2(I,Gy).^2 );
imshow(A);
Link to Wikipedia Article
Related
I tried to make a 3dof(3 degrees of freedom mechanism) in matlab but i get this error and i don't know why.
this is for a school project and i need to simulate a human finger.
the code is running normal but after i enter the values for the angles it says that A,B,C are
undefined and i don't know why
a1 = input('valuarea lui q1(grade):');
a2 = input('valuarea lui q2(grade):');
a3 = input('valuarea lui q3(grade):');
L1=35;
L2=45;
L3=30;
z = [-10 10];
plot(z,10);
grid ON;
O=[0;0;0;1];
m= linspace(pi/2,pi/2+a1*pi/180,100);
n = linspace(-pi/2,a2*pi/180,100);
k=linspace(-pi/2,a3*pi/180,100);
for a=1:100
[A1,B1,C1] = Transform(m(a),n(a),k(a),L1,L2,L3);
x = [O(1) A(1) B(1) C(1)];
y = [O(2) A(2) B(2) C(2)];
Cx(i)= C1(1);
Cy(i) = C1(2);
i=i+1;
Plot = plot(x,y,'r',...1
'LineWidth',1);
title('Sumularea unui deget');
plot(Cx,Cy,'--g',...
'LineWidth',1);
pause(0.075);
delete(Plot);
end
plot(x,y,'r',...
'LineWidth',3);
function [A,B,C ] = Transform( m,n,p,l1,l2,l3 )
P = [0;0;0;1];
T1 = [cos(m) -sin(m) 0 0;sin(m) cos(m) 0 0;0 0 1 0; 0 0 0 1];
T2 = [cos(n) -sin(n) 0 11;sin(n) cos(n) 0 0;0 0 1 0; 0 0 0 1];
T3 = [cos(p) -sin(p) 0 12;sin(p) cos(p) 0 0;0 0 1 0; 0 0 0 1];
T4 = [1 0 0 13;0 1 0 0; 0 0 1 0; 0 0 0 1];
A = T1*T2*P;
B = T1*T2*T3*P;
C = T1*T2*T3*T3*P;
end
In your main function, this is the first use of the variables:
x = [O(1) A(1) B(1) C(1)];
They are never written previously. Instead A1 is written, which is a different variable. I guess you mixed the two up.
I have a code for a 1D heat equation. Im trying to format a for loop so that the A matrix will follow a certain pattern of 1 -2 1 down the entire diagonal of a matrix that could be infinite. The pattern starts to take shape when I mess around with the initialized count at the beginning of the for loop but this changes the size of the matrix which fails the rest of the code.
My current code is below. The commented A matrix edits are what it should be.
N = 5;
%A(2,1:3) = [1 -2 1];
%A(3,2:4) = [1 -2 1];
%A(4,3:5) = [1 -2 1];
%A(5,4:6) = [1 -2 1];
A = zeros(N+1,N+1);
A(1,1) = 1;
for count=N:N+1
A(count+1,count:count+2) = [1 -2 1];
end
A(N+1,N+1) = 1;
In Matlab you can often avoid loops. In this case you can get the desired result with 2D convolution:
>> N = 6;
>> A = [1 zeros(1,N-1); conv2(eye(N-2), [1 -2 1]); zeros(1,N-1) 1]
A =
1 0 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 0 1
Or, depending on what you want,
>> A = conv2(eye(N), [1 -2 1], 'same')
A =
-2 1 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 1 -2
There are many simple ways of creating this matrix.
Your loop can be amended as follows:
N = 5;
A = zeros(N+1,N+1);
A(1,1) = 1;
for row = 2:N
A(row, row-1:row+1) = [1 -2 1];
end
A(N+1,N+1) = 1;
I've renamed count to row, we're indexing each row (from 2 to N, skipping the first and last rows), then finding with row-1:row+1 the three indices for that row that you want to address.
Directly indexing the diagonal and off-diagonal elements. Diagonal elements for an NxN matrix are 1:N+1:end. This is obviously more complex, I'd prefer the loop:
N = 6;
A = zeros(N,N);
A(1:N+1:end) = -2;
A(2:N+1:end-2*N) = 1; % skip last row
A(2*N+2:N+1:end) = 1; % skip first row
A(1,1) = 1;
A(N,N) = 1;
Using diag. We need to special-case the first and last rows:
N = 6;
A = diag(-2*ones(N,1),0) + diag(ones(N-1,1),1) + diag(ones(N-1,1),-1);
A(1,1:2) = [1,0];
A(end,end-1:end) = [0,1];
I was searching for the implementation of Perona & Malik filter in Matlab, when i found this link "the link has an implementation for Perona and Malik filter using Matlab"
but there is the matrices that i didn't understand what is the use of them:
% 2D convolution masks - finite differences.
hN = [0 1 0; 0 -1 0; 0 0 0];
hS = [0 0 0; 0 -1 0; 0 1 0];
hE = [0 0 0; 0 -1 1; 0 0 0];
hW = [0 0 0; 1 -1 0; 0 0 0];
hNE = [0 0 1; 0 -1 0; 0 0 0];
hSE = [0 0 0; 0 -1 0; 0 0 1];
hSW = [0 0 0; 0 -1 0; 1 0 0];
hNW = [1 0 0; 0 -1 0; 0 0 0];
any one has an idea what are these matrices, or what do they mean?
These are the kernels for getting the gradients in North, East, South and West directions, and also for dialgonals (NE = North East etc.).
hN = [0 1 0; 0 -1 0; 0 0 0];
or nicely formatted:
0 1 0
0 -1 0
0 0 0
Convolution of this kernel with an image yields the gradient in "North direction". Basically, this is a mask of factors that you apply to all your 3x3 areas of your image and since all except two entries are 0, what you get is the pixel at the top minus that in the center, thus the gradient in y-direction.
I am using Matlab and Euler Angles in order to reorient a 3axes coordinate system. Specifically,
Rz = [cos(ψ) sin(ψ) 0;-sin(ψ) cos(ψ) 0;0 0 1];
Ry = [cos(φ) 0 -sin(φ);0 1 0;sin(φ) 0 cos(φ)];
Rx = [1 0 0;0 cos(θ) -sin(θ);0 sin(θ) cos(θ)];
Rtotal = Rz*Ry*Rz
Then I loop through my old system coordinates (x,y,z) and make a vector coord_old. Then I get the reoriented system with (xn,yn,zn)
for i=1:size(num,1)
coord_old = [x(i,1);y(i,1);z(i,1)];
coord_new = Rtotal*coord_old;
xn(i,1) = coord_new(1,1);
yn(i,1) = coord_new(2,1);
zn(i,1) = coord_new(3,1);
end
My issue is that when θ,φ,ψ≃0 then x->-y and y->x and when θ,φ≃0 and ψ=90 then x and y will not rotate! That means that when x,y should rotate they don't and when they shouldn't rotate they stay as they were!
--EDIT--
For example, when ψ=20.0871, φ=0.0580 and θ=0.0088 I get these results
See that x->-y and y->x while z doesn't change at all!
Any thoughts?
Ok, I see two main problems here:
Rtotal = Rz*Ry*Rz is probably not what you want since Rz is multiplied twice. I think you mean Rtotal = Rz*Ry*Rx.
Your rotation matrix seems to be incorrect. Check this Wikipedia artice to get the correct signs.
Here a corrected rotation matrix:
Rz = [cos(psi) -sin(psi) 0; sin(psi) cos(psi) 0; 0 0 1];
Ry = [cos(phi) 0 sin(phi); 0 1 0; -sin(phi) 0 cos(phi)];
Rx = [1 0 0; 0 cos(theta) -sin(theta); 0 sin(theta) cos(theta)];
Rtotal = Rz*Ry*Rx;
With this matrix I get the correct results:
x=1; y=2; z=3;
psi=0; phi=0; theta=0;
[xn,yn,zn] >> 1 2 3
x=1; y=2; z=3;
psi=90/180*pi; phi=0; theta=0;
[xn,yn,zn] >> -2 1 3
And here a full graphical example of a cube in 3d-space:
% Create cube (not in origin)
DVert = [0 0 0; 0 1 0; 1 1 0; 1 0 0 ; ...
0 0 1; 0 1 1; 1 1 1; 1 0 1];
DSide = [1 2 3 4; 2 6 7 3; 4 3 7 8; ...
1 5 8 4; 1 2 6 5; 5 6 7 8];
DCol = [0 0 1; 0 0.33 1; 0 0.66 1; ...
0 1 0.33; 0 1 0.66; 0 1 1];
% Rotation angles
psi = 20 /180*pi; % Z
phi = 45 /180*pi; % Y
theta = 0 /180*pi; % X
% Rotation matrix
Rz = [cos(psi) -sin(psi) 0; sin(psi) cos(psi) 0; 0 0 1];
Ry = [cos(phi) 0 sin(phi); 0 1 0; -sin(phi) 0 cos(phi)];
Rx = [1 0 0; 0 cos(theta) -sin(theta); 0 sin(theta) cos(theta)];
Rtotal = Rz*Ry*Rz;
% Apply rotation
DVertNew = Rtotal * DVert';
% Plot cubes
figure;
patch('Faces',DSide,'Vertices',DVert,'FaceColor','flat','FaceVertexCData',DCol);
patch('Faces',DSide,'Vertices',DVertNew','FaceColor','flat','FaceVertexCData',DCol);
% Customize view
grid on;
axis equal;
view(30,30);
When I use your code and insert 0 for all angles, I get Rtotal:
Rtotal =
1 0 0
0 1 0
0 0 1
This is the identity matrix and will not change your values.
You have an error in your matrix multiplication. I think you should multiply: Rtotal*coord_old. I think you are missing the _old. depending on what is in you coordvariable, this may be the bug.
When I run:
for i=1:size(1,1)
coord_old = [1;2;3];
coord_new = Rtotal*coord_old;
xn(i,1) = coord_new(1,1);
yn(i,1) = coord_new(2,1);
zn(i,1) = coord_new(3,1);
end
I get the correct result:
coord_new =
1
2
3
Thank you both #Steffen and #Matt. Unfortunately, my reputation is not high enough to vote Up your answers!
The problem was not with Rtotal as #Matt correctly stated. It should be as it was Rz*Ry*Rx. However, both your ideas helped me test my code with simple examples (5 sets of coordinates and right hand rule), and realize where my (amateur) mistake was.
I had forgotten I had erased parts of codes where I was expressing my angles to degrees... I should be using sind & cosd instead of sin and cos.
I want to generate a matrix that is "stairsteppy" from a vector.
Example input vector: [8 12 17]
Example output matrix:
[1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
Is there an easier (or built-in) way to do this than the following?:
function M = stairstep(v)
M = zeros(length(v),max(v));
v2 = [0 v];
for i = 1:length(v)
M(i,(v2(i)+1):v2(i+1)) = 1;
end
You can do this via indexing.
A = eye(3);
B = A(:,[zeros(1,8)+1, zeros(1,4)+2, zeros(1,5)+3])
Here's a solution without explicit loops:
function M = stairstep(v)
L = length(v); % M will be
V = max(v); % an L x V matrix
M = zeros(L, V);
% create indices to set to one
idx = zeros(1, V);
idx(v + 1) = 1;
idx = cumsum(idx) + 1;
idx = sub2ind(size(M), idx(1:V), 1:V);
% update the output matrix
M(idx) = 1;
EDIT: fixed bug :p
There's no built-in function I know of to do this, but here's one vectorized solution:
v = [8 12 17];
N = numel(v);
M = zeros(N,max(v));
M([0 v(1:N-1)]*N+(1:N)) = 1;
M(v(1:N-1)*N+(1:N-1)) = -1;
M = cumsum(M,2);
EDIT: I like the idea that Jonas had to use BLKDIAG. I couldn't help playing with the idea a bit until I shortened it further (using MAT2CELL instead of ARRAYFUN):
C = mat2cell(ones(1,max(v)),1,diff([0 v]));
M = blkdiag(C{:});
A very short version of a vectorized solution
function out = stairstep(v)
% create lists of ones
oneCell = arrayfun(#(x)ones(1,x),diff([0,v]),'UniformOutput',false);
% create output
out = blkdiag(oneCell{:});
You can use ones to define the places where you have 1's:
http://www.mathworks.com/help/techdoc/ref/ones.html