I have 4 point locations in source and destination images. I wrote it on a homography matrix. I need a help to find H matrix on MATLAB.
sx1 = 285; sx2 = 576; sx3 = 583; sx4 = 295; %source image x locations
sy1 = 505; sy2 = 454; sy3 = 868; sy4 = 858; %source image y locations
dx1 = 439; dx2 = 786; dx3 = 450; dx4 = 789; %destination image x locations
dy1 = 539; dy2 = 528; dy3 = 878; dy4 = 882; %destination image y locations
A = [sx1 sy1 1 0 0 0 -sx1*dx1 -sy1*dx1; sx2 sy2 1 0 0 0 -sx2*dx2 -sy2*dy2; sx3 sy3 1 0 0 0 -sx3*dx3 -sy3*dy3; sx4 sy4 1 0 0 0 -sx4*dx4 -sy4*dy4;
0 0 0 sx1 sy1 1 -sx1*dy1 -sy1*dy1; 0 0 0 sx2 sy2 1 -sx2*dy2 -sy2*dy2; 0 0 0 sx3 sy3 1 -sx3*dy3 -sy3*dy3; 0 0 0 sx4 sy4 1 -sx4*dy4 -sy4*dy4];
B = [dx1; dx2; dx3; dx4; dy1; dy2; dy3; dy4];
Related
I need to make an array of zeros and ones in this particular fractal pattern:
0 0 0 0 0 1 0 0 0 0 0 0
0 0 1 1 1 1 1 1 1 0 0 0
0 1 0 0 0 1 0 0 0 1 0 0
0 1 0 0 1 1 1 0 0 1 0 0
0 1 0 1 0 1 0 1 0 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 1 0 1 0 1 0 1 0 0
0 1 0 0 1 1 1 0 0 1 0 0
0 1 0 0 0 1 0 0 0 1 0 0
0 0 1 1 1 1 1 1 1 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0
The actual array should be 100 x 100. The pattern should start from the middle (x,y) coordinate, and expand to look like the pattern (using loops).
So far, I have only managed to make a pattern that looks like a '+' sign. I am not sure how to continue it.
This is my code so far:
n = zeros(16); % using 16x16 array for practice
x = length(n)/2;
y = length(n)/2;
z = length(n) - 1;
xso = length(n)/2; % x axis south movement
xno = length(n)/2; % x axis north movement
yea = length(n)/2; % y axis east movement
ywe = length(n)/2; % y axis west movement
for i = 1:1:z
newyea = move('east', x, yea);
n(x, newyea) = 1;
yea = newyea;
newywe = move('west', x, ywe);
n(x, newywe) = 1;
ywe = newywe;
newxso = move('south', xso, y);
n(newxso, y) = 1;
xso = newxso;
newxno = move('north', xno, y);
n(newxno, y) = 1;
xno = newxno;
end
I also have a user defined function:
function newval = move(dir, x, y)
switch dir
case 'east'
newval = y + 1;
case 'west'
newval = y - 1;
case 'south'
newval = x + 1;
case 'north'
newval = x - 1;
end
Since there were no restrictions given on the appearance of the loop, I would propose the below solution. But, before, let's have a look at your given example:
Either you should restrict the desired dimension d to be odd, i.e. d = 11, 13, 15, ... or you should specify, how the pattern should be continued in case of an even dimension d, like here d = 12. For my solution, I decided to rely on the dimension d to be odd.
Here's the code:
d = 15; % Dimension
A = zeros(d); % Initialize output array A
c = (d + 1) / 2; % Calculate center index (row, column)
A(:, c) = 1; % Add: Cross
A(c, :) = 1;
J = 0; % Auxiliary index
for I = (c+2):2:d % For every second row (or column) from center to border
J = J + 1;
l = 4 * J - 1; % Calculate length of line to draw
s = c - (l-1)/2; % Calculate start point of line
e = c + (l-1)/2; % Calculate end point of line
A(I, s:e) = 1; % Add: "South" line
A(s:e, I) = 1; % Add: "East" line
A(c - 2*J, s:e) = 1; % Add: "North" line
A(s:e, c - 2*J) = 1; % Add: "West" line
end
figure(1); % Show image
imagesc(A);
Output for d = 15 (to compare to given example):
Output for d = 99:
Hope that helps!
If you have some more stricter limitations on the for loop, let me/us know. Then, I will try to modify my code accordingly.
I am working on a MATLAB code that involves deep learning using neural networks.
The image or the data is being fed in the form of matrices.
But I am getting an error "Matrix dimensions must agree".
Can Someone please help me with this problem?
I tried to solve this problem by using .* in stead of matrix multiplication * but the method didn't work.
Function Deeplearningoriginal:
function [w1,w2,w3,w4] = Deeplearningoriginal(w1,w2,w3,w4,input_Image,correct_output)
alpha=0.01;
N=5;
for k = 1:N
input_Image = reshape( input_Image( :, :,k ),25 ,1);
input_of_hidden_layer1 = w1* input_Image;
output_of_hidden_layer1 = ReLU(input_of_hidden_layer1);
input_of_hidden_layer2 = w2* output_of_hidden_layer1;
output_of_hidden_layer2 = ReLU( input_of_hidden_layer2);
input_of_hidden_layer3 = w3* output_of_hidden_layer2;
output_of_hidden_layer3 = ReLU(input_of_hidden_layer3);
input_of_output_node = w4* output_of_hidden_layer3;
final_output = Softmax(input_of_output_node);
correct_output_transpose = correct_output(k,:);
error = correct_output_transpose - final_output;
delta4 = error;
error_of_hidden_layer3 = w4'* delta4;
delta3 = (input_of_hidden_layer3>0).*error_of_hidden_layer3;
error_of_hidden_layer2 = w3'* delta3;
delta2 = (input_of_hidden_layer2>0).* error_of_hidden_layer2;
error_of_hidden_layer1 = w2'*delta2;
delta1 = (input_of_hidden_layer1>0).* error_of_hidden_layer1;
adjustment_of_w4 = alpha*delta4*output_of_hidden_layer3';
adjustment_of_w3 = alpha*delta3*output_of_hidden_layer2';
adjustment_of_w2 = alpha*delta2*output_of_hidden_layer1';
adjustment_of_w1 = alpha*delta1*reshaped_input_image';
w1 = w1 + adjustment_of_w1;
w2 = w2 + adjustment_of_w2;
w3 = w3 + adjustment_of_w3;
w4 = w4 + adjustment_of_w4;
end
end
Training network:
input_Image = zeros (5,5,5);
input_Image(:,:,1) = [ 1 0 0 1 1;
1 1 0 1 1;
1 1 0 1 1;
1 1 0 1 1;
1 0 0 0 1;
];
input_Image(:,:,2) = [ 0 0 0 0 1;
1 1 1 1 0;
1 0 0 0 1;
0 1 1 1 1;
0 0 0 0 0;
];
input_Image(:,:,3) = [ 0 0 0 0 1;
1 1 0 0 1;
1 0 1 0 1;
0 0 0 0 0;
1 1 1 0 1;
];
input_Image(:,:,4) = [ 1 1 1 0 1;
1 1 0 0 1;
1 0 1 0 1;
0 0 0 0 0;
1 1 1 0 1;
];
input_Image(:,:,5) = [ 0 0 0 0 0;
0 1 1 1 1;
0 0 0 0 1;
1 1 1 1 0;
0 0 0 0 1;
];
correct_output = [ 1 0 0 0 0;
0 1 0 0 0;
0 0 1 0 0;
0 0 0 1 0;
0 0 0 0 1;
];
w1 = 2* rand(20,25) -1;
w2 = 2* rand(20,20) -1;
w3 = 2* rand(20,20) -1;
w4 = 2* rand(5,20) -1;
for epoch = 1:100
[w1,w2,w3,w4] = Deeplearningoriginal(w1,w2,w3,w4,input_Image,correct_output);
end
I expected this code to run but because of the error I am not able to proceed.
The problem is the reshape (actually, two problems). After the
input_image = reshape(input_image(:,:,k), 25,1);
input_image is an array with 25 rows and 1 column, whereas w2, w3, and w4 have only 20 columns. To do the matrix multiplication A*B, A must have as many columns as B has rows.
The other problem with the reshape as written is that after the first pass through the loop, input_image is no longer a 5x5x5 array, it is a 25x1 array that contains only the elements of input_image(:,:,1). It is necessary to use a different name on the left-hand-side of the assignment (and throughout the rest of the loop) to avoid loosing the content of input_image.
Hope this helps,
JAC
I am writing a function in Matlab to model the length of stay in hospital of stroke patients. I am having difficulty in storing my output values.
Here is my function:
function [] = losdf(age, strokeType, dest)
% function to mdetermine length of stay in hospitaal of stroke patients
% t = time since admission (days);
% age = age of patient;
% strokeType = 1. Haemorhagic, 2. Cerebral Infarction, 3. TIA;
% dest = 5.Death 6.Nursing Home 7. Usual Residence;
alpha1 = 6.63570;
beta1 = -0.03652;
alpha2 = -3.06931;
beta2 = 0.07153;
theta0 = -8.66118;
theta1 = 0.08801;
mu1 = 22.10156;
mu2 = 2.48820;
mu3 = 1.56162;
mu4 = 0;
nu1 = 0;
nu2 = 0;
nu3 = 1.27849;
nu4 = 0;
rho1 = 0;
rho2 = 11.76860;
rho3 = 3.41989;
rho4 = 63.92514;
for t = 1:1:365
p = (exp(-exp(theta0 + (theta1.*age))));
if strokeType == 1
initialstatevec = [1 0 0 0 0 0 0];
elseif strokeType == 2
initialstatevec = [0 1 0 0 0 0 0];
else
initialstatevec = [0 0 (1-p) p 0 0 0];
end
lambda1 = exp(alpha1 + (beta1.*age));
lambda2 = exp(alpha2 + (beta2.*age));
Q = [ -(lambda1+mu1+nu1+rho1) lambda1 0 0 mu1 nu1 rho1;
0 -(lambda2+mu2+nu2+rho2) lambda2 0 mu2 nu2 rho2;
0 0 -(mu3+nu3+rho3) 0 mu3 nu3 rho3;
0 0 0 -(mu4+nu4+rho4) mu4 nu4 rho4;
0 0 0 0 0 0 0;
0 0 0 0 0 0 0;
0 0 0 0 0 0 0];
Pt = expm(t./365.*Q);
Pt = Pt(strokeType, dest);
Ft = sum(initialstatevec.*Pt);
Ft
end
end
Then to run my function I use:
losdf(75,3,7)
I want to plot my values of Ft in a graph from from 0 to 365 days. What is the best way to do this?
Do I need to store the values in an array first and if so what is the best way to do this?
Many ways to do this, one straightforward way is to save each data point to a vector while in the loop and plot that vector after you exit your loop.
...
Ft = zeros(365,1); % Preallocate Ft as a vector of 365 zeros
for t = 1:365
...
Ft(t) = sum(initialstatevec.*Pt); % At index "t", store your output
...
end
plot(1:365,Ft);
I know the code below :
N = 5;
assert(N>1 && mod(N,2)==1);
A = zeros(N);
% diamond mask
N2 = fix(N/2);
[I,J] = meshgrid(-N2:N2);
mask = (abs(I) + abs(J)) == N2;
% fill with zeros
A(mask) = 1;
which transforms matrix A to this:
A=
0 0 1 0 0
0 1 0 1 0
1 0 0 0 1
0 1 0 1 0
0 0 1 0 0
But I want the diamond to be filled with 1.
What should I do?
Here's a vectorized approach using bsxfun -
Nh = (N+1)/2;
range_vec = [1:Nh Nh-1:-1:1];
out = bsxfun(#plus,range_vec(:),range_vec) > Nh
Sample runs -
1) N = 5 :
out =
0 0 1 0 0
0 1 1 1 0
1 1 1 1 1
0 1 1 1 0
0 0 1 0 0
2) N = 9 :
out =
0 0 0 0 1 0 0 0 0
0 0 0 1 1 1 0 0 0
0 0 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 0
1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 0
0 0 1 1 1 1 1 0 0
0 0 0 1 1 1 0 0 0
0 0 0 0 1 0 0 0 0
You can use tril and flip functions:
mat = tril(ones(N), round((N-1)/2)) - tril(ones(N), round((-N-1)/2));
out = mat & flip(mat)
Odd values of N:
% N = 5;
out =
0 0 1 0 0
0 1 1 1 0
1 1 1 1 1
0 1 1 1 0
0 0 1 0 0
Even values of N:
% N = 4;
out =
0 1 1 0
1 1 1 1
1 1 1 1
0 1 1 0
What you need is to return a 1 or a 0 based on the Manhattan distance from each array location to the center of your diamond
N = 5;
assert(N>1 && mod(N,2)==1);
A = false(N);
[m, n] = size(A); %dimensions of A
X = floor([m, n]/2); %floored division gives integer indices of center of array
x = X(1); y = X(2);
radius = m/2; %half the height gives the radius
for a = 1 : m
for b = 1 : n
A(a,b) = abs(a-x)+abs(b-y) <= radius; %test if manhatten distance <= radius
end
end
This naturally will need editing to suit your particular case... In particular, the center of your diamond can realistically be placed anywhere by modifying x, y, and the radius can be either smaller or larger than half the width of the array if you so choose.
Just add a for loop and fill all diagonals:
N = 5;
assert(N>1 && mod(N,2)==1);
A = zeros(N);
% diamond mask
N2 = fix(N/2);
[I,J] = meshgrid(-N2:N2);
for id = 0:N2
A((abs(I) + abs(J)) == id) = 1;
end
I need to assign boundary values to a 3d box.
Assuming I have z = rand(L,M,N), is there a better way to address all the faces of this box than processing all 6 faces individually, z(:,:,1), z(:,:,end), z(:,1,:), z(:,end,:), z(1,:,:), z(end,:,:)?
This is what I have right now:
L=3;M=4;N=5;
z = rand(L,M,N);
bv = 0;
z([1,end],:,:) = bv;
z(:,[1,end],:) = bv;
z(:,:,[1,end]) = bv;
I would like to be able to do something like z(indices) = bv.
If you have the Image Processing toolbox, using padarray would work:
z = padarray(z(2:end-1,2:end-1,2:end-1), [1 1 1], bv);
This just takes the inner block of the cube and adds 1 copy of bv on all sides.
Not sure if this is any better than your code, but here it goes:
%// Data
L = 3;
M = 4;
N = 5;
z = rand(L,M,N)
newValue = 0;
%// Let's go
indL = false(L, 1, 1);
indM = false(1, M, 1);
indN = false(1, 1, N);
indL([1 end]) = true;
indM([1 end]) = true;
indN([1 end]) = true;
ind = bsxfun(#or, bsxfun(#or, indL, indM), indN); %// linear index of all faces
z(ind) = newValue
Before:
z(:,:,1) =
0.2653 0.7302 0.1078 0.8178
0.8244 0.3439 0.9063 0.2607
0.9827 0.5841 0.8797 0.5944
z(:,:,2) =
0.0225 0.1615 0.0942 0.6959
0.4253 0.1788 0.5985 0.6999
0.3127 0.4229 0.4709 0.6385
z(:,:,3) =
0.0336 0.5309 0.8200 0.5313
0.0688 0.6544 0.7184 0.3251
0.3196 0.4076 0.9686 0.1056
z(:,:,4) =
0.6110 0.0908 0.2810 0.4574
0.7788 0.2665 0.4401 0.8754
0.4235 0.1537 0.5271 0.5181
z(:,:,5) =
0.9436 0.2407 0.6718 0.2548
0.6377 0.6761 0.6951 0.2240
0.9577 0.2891 0.0680 0.6678
After:
z(:,:,1) =
0 0 0 0
0 0 0 0
0 0 0 0
z(:,:,2) =
0 0 0 0
0 0.1788 0.5985 0
0 0 0 0
z(:,:,3) =
0 0 0 0
0 0.6544 0.7184 0
0 0 0 0
z(:,:,4) =
0 0 0 0
0 0.2665 0.4401 0
0 0 0 0
z(:,:,5) =
0 0 0 0
0 0 0 0
0 0 0 0