I'm trying to write a "weighted moving window" without nested loops for speed improvement.
I already tried using arrayfun without getting exciting results, but maybe I did it in a wrong way.
The window has a different weight in each position (stored in B) and should be superimposed on a matrix A returning the values of the matrix A that lie inside the window, times the weight of the window in that position (read from B).
Also, the windows can overlap one on the other and in this case the maximum value should be kept.
Finally window's dimension and shift should be parameters of the function.
It looks more difficult that it actually is, so I show you the code that I would like to improve:
A = reshape([1:35],7,5)'; % values matrix
B = [1:3;4:6]; % window s weight matrix
% matrices size
[m n] = size(A);
[a b] = size(B);
% window s parameters
shift = 2; % window s movement at each iteration
zone = 3; % window s size (zone x zone)
% preallocation
C = ones(m,n); % to store the right weight to be applied in each position
% loop through positions and find the best weight when they overlap
for i=1:m
for j=1:n
C(i,j) = max(max(B( max(round((i-zone)/shift)+1,1) : min(ceil(i/shift),a) , max(round((j-zone)/shift)+1,1) : min(ceil(j/shift),b))));
end
end
% find the output of the windows
result = C.*A;
I hope that I made myself clear, but if you need more details please ask.
Thank you in advance for your help!
If you have access to the Image Processing Toolbox, you'll want to check out how to perform sliding neighborhood operations. In particular, I think the function NLFILTER can be used to achieve the result you want:
A = reshape([1:35],7,5)'; %'# Matrix to be filtered
B = [1:3;4:6]; %# Window weights
result = nlfilter(A,[2 3],#(M) max(M(:).*B(:)));
I would use im2col. Assuming your image is j x k, and your window is m x n, you'll get a matrix that is mn x (j-m+1)*(k-n+1). Then, you can just take every other column.
Example Code:
%A = your_image
B = im2col(A, [m n],'sliding');
C = B(:,1:2:end);
And there's your sliding window with "shift 2".
Try filter.
For example, to do a windowed average, over 5 elements:
outdata = filter([ 0.2 0.2 0.2 0.2 0.2 ], 1, indata);
Related
I need to plot population growth model with different values of b
where b>d
and b=d
using for loops. Both results should be visible on the same graph with different colors.
Here is my function and intial value, but I am not getting how to get the plot .
[T,Y]=ode45(H,[0:1:time],[N0]);
Intitial value
b= 0.1 0.001
d=0.001
N0=400````
I need two line for each b on the same plot.
```***
Intitial value
b= 0.1 0.001;
d=0.001;
N0=400;
H=#(t,N)[b*N-d*N];
[T,Y]=ode45(H,[0:1:time],[N0]);
plot(T,Y)
***```
First, you need to put the code in a proper format, now you copied the initial value block twice (I guess you had some issue putting the code format. I hope you can focus on that when asking new questions.
The current code gives an error on the definition of b as you did not define it as a vector. To have two lines, it is best to store the output Y as a matrix of two columns (each column being one vector). MATLAB then outputs each column as a line in the plot.
The following code solves your issue:
% Initial value
b = [0.1, 0.001];
d = 0.001;
N0= 400;
simTime = 0:1:300;
% Preallocate output for the loop
Y = zeros(length(simTime), length(b));
% Run for loop
for i = 1:length(b)
b_i = b(i);
H=#(t,N)[b_i*N-d*N];
[T,y]=ode45(H, simTime, [N0]);
Y(:,i) = y;
end
% Plot output
plot(T,Y)
I was asked to do circular convolution between two functions by sampling them, using the functions cconv. A known result of this sort of convolution is: CCONV( sin(x), sin(x) ) == -pi*cos(x)
To test the above I did:
w = linspace(0,2*pi,1000);
l = linspace(0,2*pi,1999);
stem(l,cconv(sin(w),sin(w))
but the result I got was:
which is absolutely not -pi*cos(x).
Can anybody please explain what is wrong with my code and how to fix it?
In the documentation of cconv it says that:
c = cconv(a,b,n) circularly convolves vectors a and b. n is the length of the resulting vector. If you omit n, it defaults to length(a)+length(b)-1. When n = length(a)+length(b)-1, the circular convolution is equivalent to the linear convolution computed with conv.
I believe that the reason for your problem is that you do not specify the 3rd input to cconv, which then selects the default value, which is not the right one for you. I have made an animation showing what happens when different values of n are chosen.
If you compare my result for n=200 to your plot you will see that the amplitude of your data is 10 times larger whereas the length of your linspace is 10 times bigger. This means that some normalization is needed, likely a multiplication by the linspace step.
Indeed, after proper scaling and choice of n we get the right result:
res = 100; % resolution
w = linspace(0,2*pi,res);
dx = diff(w(1:2)); % grid step
stem( linspace(0,2*pi,res), dx * cconv(sin(w),sin(w),res) );
This is the code I used for the animation:
hF = figure();
subplot(1,2,1); hS(1) = stem(1,cconv(1,1,1)); title('Autoscaling');
subplot(1,2,2); hS(2) = stem(1,cconv(1,1,1)); xlim([0,7]); ylim(50*[-1,1]); title('Constant limits');
w = linspace(0,2*pi,100);
for ind1 = 1:200
set(hS,'XData',linspace(0,2*pi,ind1));
set(hS,'YData',cconv(sin(w),sin(w),ind1));
suptitle("n = " + ind1);
drawnow
% export_fig(char("D:\BLABLA\F" + ind1 + ".png"),'-nocrop');
end
How to reduce for loops in a moving window based operation? I'm using a 15x15 window across two images and performing multiplication to get average value per pixel.
[ma,na]=size(g);
z= (win1 -1)/2;%centre of window
ini=z+1;
for i= ini :(ma-z)
for j= ini:(na-z)
for a= (i-z):(i+z)
for b=(j-z):(j+z)
W(pp,qq)= g(a, b);%window on image
Es(pp,qq)=edg(a,b);%window on edge
qq=qq+1;
end
qq=1;
pp=pp+1;
end
pp=1;
E(i,j)=sum(sum(W.*Es))/sum(sum(Es));
end
end
I might have gotten lost in your loops and i can't exactly read the formula (it's a bit fuzzy) but i think this is what you want:
g = rand(5); %sample img1
edg = rand(5); %sample img2
windowsize = 3; %set this to 15 for real images
A = g.*edg; % multiply each element beforehand, corresponds to mu*sigma in formula
B = movsum(movsum(A,windowsize,2),windowsize,1); % get moving window sum of A, corresponds to numerator in formula
C = movsum(movsum(edg,windowsize,2),windowsize,1); % get moving window sum of edg, corresponds to denominator in formula
E = B./C; %hopefully what you wanted
Ps: You need 2016a or newer to run this
I have found several questions/answers for vectorizing and speeding up routines for multiplying a matrix and a vector in a single loop, but I am trying to do something a little more general, namely multiplying an arbitrary number of matrices together, and then performing that operation an arbitrary number of times.
I am writing a general routine for calculating thin-film reflection from an arbitrary number of layers vs optical frequency. For each optical frequency W each layer has an index of refraction N and an associated 2x2 transfer matrix L and 2x2 interface matrix I which depends on the index of refraction and the thickness of the layer. If n is the number of layers, and m is the number of frequencies, then I can vectorize the index into an n x m matrix, but then in order to calculate the reflection at each frequency, I have to do nested loops. Since I am ultimately using this as part of a fitting routine, anything I can do to speed it up would be greatly appreciated.
This should provide a minimum working example:
W = 1260:0.1:1400; %frequency in cm^-1
N = rand(4,numel(W))+1i*rand(4,numel(W)); %dummy complex index of refraction
D = [0 0.1 0.2 0]/1e4; %thicknesses in cm
[n,m] = size(N);
r = zeros(size(W));
for x = 1:m %loop over frequencies
C = eye(2); % first medium is air
for y = 2:n %loop over layers
na = N(y-1,x);
nb = N(y,x);
%I = InterfaceMatrix(na,nb); % calculate the 2x2 interface matrix
I = [1 na*nb;na*nb 1]; % dummy matrix
%L = TransferMatrix(nb) % calculate the 2x2 transfer matrix
L = [exp(-1i*nb*W(x)*D(y)) 0; 0 exp(+1i*nb*W(x)*D(y))]; % dummy matrix
C = C*I*L;
end
a = C(1,1);
c = C(2,1);
r(x) = c/a; % reflectivity, the answer I want.
end
Running this twice for two different polarizations for a three layer (air/stuff/substrate) problem with 2562 frequencies takes 0.952 seconds while solving the exact same problem with the explicit formula (vectorized) for a three layer system takes 0.0265 seconds. The problem is that beyond 3 layers, the explicit formula rapidly becomes intractable and I would have to have a different subroutine for each number of layers while the above is completely general.
Is there hope for vectorizing this code or otherwise speeding it up?
(edited to add that I've left several things out of the code to shorten it, so please don't try to use this to actually calculate reflectivity)
Edit: In order to clarify, I and L are different for each layer and for each frequency, so they change in each loop. Simply taking the exponent will not work. For a real world example, take the simplest case of a soap bubble in air. There are three layers (air/soap/air) and two interfaces. For a given frequency, the full transfer matrix C is:
C = L_air * I_air2soap * L_soap * I_soap2air * L_air;
and I_air2soap ~= I_soap2air. Thus, I start with L_air = eye(2) and then go down successive layers, computing I_(y-1,y) and L_y, multiplying them with the result from the previous loop, and going on until I get to the bottom of the stack. Then I grab the first and third values, take the ratio, and that is the reflectivity at that frequency. Then I move on to the next frequency and do it all again.
I suspect that the answer is going to somehow involve a block-diagonal matrix for each layer as mentioned below.
Not next to a matlab, so that's only a starter,
Instead of the double loop you can write na*nb as Nab=N(1:end-1,:).*N(2:end,:);
The term in the exponent nb*W(x)*D(y) can be written as e=N(2:end,:)*W'*D;
The result of I*L is a 2x2 block matrix that has this form:
M = [1, Nab; Nab, 1]*[e-, 0;0, e+] = [e- , Nab*e+ ; Nab*e- , e+]
with e- as exp(-1i*e), and e+ as exp(1i*e)'
see kron on how to get the block matrix form, to vectorize the propagation C=C*I*L just take M^n
#Lama put me on the right path by suggesting block matrices, but the ultimate answer ended up being more complicated, and so I put it here for posterity. Since the transfer and interface matrix is different for each layer, I leave in the loop over the layers, but construct a large sparse block matrix where each block represents a frequency.
W = 1260:0.1:1400; %frequency in cm^-1
N = rand(4,numel(W))+1i*rand(4,numel(W)); %dummy complex index of refraction
D = [0 0.1 0.2 0]/1e4; %thicknesses in cm
[n,m] = size(N);
r = zeros(size(W));
C = speye(2*m); % first medium is air
even = 2:2:2*m;
odd = 1:2:2*m-1;
for y = 2:n %loop over layers
na = N(y-1,:);
nb = N(y,:);
% get the reflection and transmission coefficients from subroutines as a vector
% of length m, one value for each frequency
%t = Tab(na, nb);
%r = Rab(na, nb);
t = rand(size(W)); % dummy vector for MWE
r = rand(size(W)); % dummy vector for MWE
% create diagonal and off-diagonal elements. each block is [1 r;r 1]/t
Id(even) = 1./t;
Id(odd) = Id(even);
Io(even) = 0;
Io(odd) = r./t;
It = [Io;Id/2].';
I = spdiags(It,[-1 0],2*m,2*m);
I = I + I.';
b = 1i.*(2*pi*D(n).*nb).*W;
B(even) = -b;
B(odd) = b;
L = spdiags(exp(B).',0,2*m,2*m);
C = C*I*L;
end
a = spdiags(C,0);
a = a(odd).';
c = spdiags(C,-1);
c = c(odd).';
r = c./a; % reflectivity, the answer I want.
With the 3 layer system mentioned above, it isn't quite as fast as the explicit formula, but it's close and probably can get a little faster after some profiling. The full version of the original code clocks at 0.97 seconds, the formula at 0.012 seconds and the sparse diagonal version here at 0.065 seconds.
I am working towards comparing multiple images. I have these image data as column vectors of a matrix called "images." I want to assess the similarity of images by first computing their Eucledian distance. I then want to create a matrix over which I can execute multiple random walks. Right now, my code is as follows:
% clear
% clc
% close all
%
% load tea.mat;
images = Input.X;
M = zeros(size(images, 2), size (images, 2));
for i = 1:size(images, 2)
for j = 1:size(images, 2)
normImageTemp = sqrt((sum((images(:, i) - images(:, j))./256).^2));
%Need to accurately select the value of gamma_i
gamma_i = 1/10;
M(i, j) = exp(-gamma_i.*normImageTemp);
end
end
My matrix M however, ends up having a value of 1 along its main diagonal and zeros elsewhere. I'm expecting "large" values for the first few elements of each row and "small" values for elements with column index > 4. Could someone please explain what is wrong? Any advice is appreciated.
Since you're trying to compute a Euclidean distance, it looks like you have an error in where your parentheses are placed when you compute normImageTemp. You have this:
normImageTemp = sqrt((sum((...)./256).^2));
%# ^--- Note that this parenthesis...
But you actually want to do this:
normImageTemp = sqrt(sum(((...)./256).^2));
%# ^--- ...should be here
In other words, you need to perform the element-wise squaring, then the summation, then the square root. What you are doing now is summing elements first, then squaring and taking the square root of the summation, which essentially cancel each other out (or are actually the equivalent of just taking the absolute value).
Incidentally, you can actually use the function NORM to perform this operation for you, like so:
normImageTemp = norm((images(:, i) - images(:, j))./256);
The results you're getting seem reasonable. Recall the behavior of the exp(-x). When x is zero, exp(-x) is 1. When x is large exp(-x) is zero.
Perhaps if you make M(i,j) = normImageTemp; you'd see what you expect to see.
Consider this solution:
I = Input.X;
D = squareform( pdist(I') ); %'# euclidean distance between columns of I
M = exp(-(1/10) * D); %# similarity matrix between columns of I
PDIST and SQUAREFORM are functions from the Statistics Toolbox.
Otherwise consider this equivalent vectorized code (using only built-in functions):
%# we know that: ||u-v||^2 = ||u||^2 + ||v||^2 - 2*u.v
X = sum(I.^2,1);
D = real( sqrt(bsxfun(#plus,X,X')-2*(I'*I)) );
M = exp(-(1/10) * D);
As was explained in the other answers, D is the distance matrix, while exp(-D) is the similarity matrix (which is why you get ones on the diagonal)
there is an already implemented function pdist, if you have a matrix A, you can directly do
Sim= squareform(pdist(A))