i am asking for help.. I want to animate the Kaczmarz method on Matlab. It's method allows to find solution of system of equations by the serial projecting solution vector on hyperplanes, which which is given by the eqations of system.
And i want make animation of this vector moving (like the point is going on the projected vectors).
%% System of equations
% 2x + 3y = 4;
% x - y = 2;
% 6x + y = 15;
%%
A = [2 3;1 -1; 6 1];
f = [4; 2; 15];
resh = pinv(A)*f
x = -10:0.1:10;
e1 = (1 - 2*x)/3;
e2 = (x - 2);
e3 = 15 - 6*x;
plot(x,e1)
grid on
%
axis([0 4 -2 2])
hold on
plot(x,e2)
hold on
plot(x,e3)
hold on
precision = 0.001; % точность
iteration = 100; % количество итераций
lambda = 0.75; % лямбда
[m,n] = size(A);
x = zeros(n,1);
%count of norms
for i = 1:m
nrm(i) = norm(A(i,:));
end
for i = 1:1:iteration
j = mod(i-1,m) + 1;
if (nrm(j) <= 0), continue, end;
predx = x;
x = x + ((f(j) - A(j,:)*x)*A(j,:)')/(nrm(j))^2;
p = plot(x);
set(p)
%pause 0.04;
hold on;
if(norm(predx - x) <= precision), break, end
end
I wrote the code for this method, by don't imagine how make the animation, how I can use the set function.
In your code there are a lot of redundant and random pieces. Do not call hold on more than once, it does nothing. Also set(p) does nothing, you want to set some ps properties to something, then you use set.
Also, you are plotting the result, but not the "change". The change is a line between the previous and current, and that is the only reason you'd want to have a variable such as predx, to plot. SO USE IT!
Anyway, this following code plots your algorithm. I added a repeated line to plot in green and then delete, so you can see what the last step does. I also changed the plots in the begging to just plot in red so its more clear what is each of the things.
Change your loop for:
for i = 1:1:iteration
j = mod(i-1,m) + 1;
if (nrm(j) <= 0), continue, end;
predx = x;
x = x + ((f(j) - A(j,:)*x)*A(j,:)')/(nrm(j))^2;
plot([predx(1) x(1)],[predx(2) x(2)],'b'); %plot line
c=plot([predx(1) x(1)],[predx(2) x(2)],'g'); %plot it in green
pause(0.1)
children = get(gca, 'children'); %delete the green line
delete(children(1));
drawnow
% hold on;
if(norm(predx - x) <= precision), break, end
end
This will show:
Related
While using Matlab for image processing (exactly improving img by Fuzzy Logic) I found a really strange thing. My fuzzy function is correct, I tested it on random values and they are basically simple linear functions.
function f = Udark(z)
if z < 50
f = 1;
elseif z > 125
f = 0;
elseif (z >= 50) && (z <= 125)
f = -z/75 + 125/75;
end
end
where z is a value of a pixel (in grayscale). Now there is a really strange thing going on.
f = -z/75 + 125/75;, where a is an image. However, it is giving really different results if used as an input. I.e. if I use a variable p = 99, the output of the function is 0.3467 as it should be, when if I use A(i,j) it is giving me result f=2. Since it is clearly impossible, I do not know where is the problem. I thought that maybe there is a case with the type of the variable but if I change it to uint8 it stays the same... If you know what's going on, please, let me know :)
1.Changed line:
f = (125/75) - (z/75);
After editing the third condition the resultant/transformed image has no pixel values of 2. Not sure if you intend to work with decimals. If decimals are necessary using the im2double() function to convert the image and scaling it up by a factor of 255 might suffice your needs. See heading 3 for rounding details.
2.Reading in Image and Testing:
%Reading in the image and applying the function%
Image = imread("RGB_Image.png");
Greyscale_Image = rgb2gray(Image);
[Image_Height,Image_Width] = size(Greyscale_Image);
Transformed_Image = zeros(Image_Height,Image_Width);
for Row = 1: +1: Image_Height
for Column = 1: +1: Image_Width
Pixel_Value = Greyscale_Image(Row,Column);
[Transformed_Pixel_Value] = Udark(Pixel_Value);
Transformed_Image(Row,Column) = Transformed_Pixel_Value;
end
end
subplot(1,2,1); imshow(Greyscale_Image);
subplot(1,2,2); imshow(Transformed_Image);
%Checking that no transformed pixels falls in this impossible range%
Check = (Transformed_Image > (125/75)) & (Transformed_Image ~= 1);
Check_Flag = any(Check,'all');
%Function to transform pixel values%
function f = Udark(z)
if z < 50
f = 1;
elseif z > 125
f = 0;
elseif (z >= 50) && (z <= 125)
f = (125/75) - (z/75);
end
end
3.Evaluating the Specifics of the Third Condition
Working with integers (uint8) will force the values to be rounded to the nearest integer. Any number that falls between the range (50,125] will evaluate to 1 or 0.
f = -z/75 + 125/75;
If z = 50.1,
-50.1/75 + 125/75 = 74.9/75 ≈ 0.9987 → rounds to 1
Using MATLAB version: R2019b
I'm using the next code to plot in a pie chart the percentage of values in a matrix that are greater/smaller than 1. The thing is that when I want to put the title above the graph, it overlaps with the label of one of the groups.
I tried replacing it with text() but it didn't worked, and Documentation on pie say nothing to this. How can I avoid this overlap?
eigen = []; % Modes array
c2 = 170; % Sound speed divided by 2
%% Room dimensions
lx = 5.74;
ly = 8.1;
lz = 4.66;
i = 1; % Index for modes array
for nz = 0:50
for ny = 0:50
for nx = 0:50
aux = c2 * sqrt((nx/lx)^2+(ny/ly)^2+(nz/lz)^2);
if aux < 400 %% If value is into our range of interest
eigen(i) = aux;
i=i+1;
end
end
end
end
eigen = round(sort(eigen'),1);
eigen
% dif = eigen(2:end)-eigen(1:end-1); % Distance between modes
x = 0; %% dif >= 1
y = 0; %% dif <= 1
dif = [];
for i=2:length(eigen)
if eigen(i)-eigen(i-1) >= 1
x = x+1;
else
y = y+1;
end
end
figure
dif = [x,y];
explode = [1 1];
graf = pie(dif,explode);
hText = findobj(graf,'Type','text');
percentValues = get(hText,'String');
txt = {'Smaller than 1 Hz: ';'Greater than 1 Hz: '};
combinedtxt = strcat(txt,percentValues);
oldExtents_cell = get(hText,'Extent');
oldExtents = cell2mat(oldExtents_cell);
hText(1).String = combinedtxt(1);
hText(2).String = combinedtxt(2);
title('Distance between modes')
You can rotate the pie chart so that the figure look better. Further, you can use position to allocate your text as follows,
figure
dif = [x,y];
explode = [1 1];
graf = pie(dif,explode);
hText = findobj(graf,'Type','text');
percentValues = get(hText,'String');
txt = {'Smaller than 1 Hz: ';'Greater than 1 Hz: '};
combinedtxt = strcat(txt,percentValues);
oldExtents_cell = get(hText,'Extent');
oldExtents = cell2mat(oldExtents_cell);
hText(1).String = combinedtxt(1);
hText(2).String = combinedtxt(2);
view([90 90]) % this is to rotate the chart
textPositions_cell = get(hText,{'Position'});
textPositions = cell2mat(textPositions_cell);
textPositions(:,1) = textPositions(:,1) + 0.2; % replace 0.2 with any offset value you want
hText(1).Position = textPositions(1,:);
hText(2).Position = textPositions(2,:);
title('Distance between modes')
You can change only the text position (without rotation) by deleting view command.
Actually i have two intersecting circles as specified in the figure
i want to find the area of each part separately using Monte carlo method in Matlab .
The code doesn't draw the rectangle or the circles correctly so
i guess what is wrong is my calculation for the x and y and i am not much aware about the geometry equations for solving it so i need help about the equations.
this is my code so far :
n=1000;
%supposing that a rectangle will contain both circles so :
% the mid point of the distance between 2 circles will be (0,6)
% then by adding the radius of the left and right circles the total distance
% will be 27 , 11 from the left and 16 from the right
% width of rectangle = 24
x=27.*rand(n-1)-11;
y=24.*rand(n-1)+2;
count=0;
for i=1:n
if((x(i))^2+(y(i))^2<=25 && (x(i))^2+(y(i)-12)^2<=100)
count=count+1;
figure(2);
plot(x(i),y(i),'b+')
hold on
elseif(~(x(i))^2+(y(i))^2<=25 &&(x(i))^2+(y(i)-12)^2<=100)
figure(2);
plot(x(i),y(i),'y+')
hold on
else
figure(2);
plot(x(i),y(i),'r+')
end
end
Here are the errors I found:
x = 27*rand(n,1)-5
y = 24*rand(n,1)-12
The rectangle extents were incorrect, and if you use rand(n-1) will give you a (n-1) by (n-1) matrix.
and
first If:
(x(i))^2+(y(i))^2<=25 && (x(i)-12)^2+(y(i))^2<=100
the center of the large circle is at x=12 not y=12
Second If:
~(x(i))^2+(y(i))^2<=25 &&(x(i)-12)^2+(y(i))^2<=100
This code can be improved by using logical indexing.
For example, using R, you could do (Matlab code is left as an excercise):
n = 10000
x = 27*runif(n)-5
y = 24*runif(n)-12
plot(x,y)
r = (x^2 + y^2)<=25 & ((x-12)^2 + y^2)<=100
g = (x^2 + y^2)<=25
b = ((x-12)^2 + y^2)<=100
points(x[g],y[g],col="green")
points(x[b],y[b],col="blue")
points(x[r],y[r],col="red")
which gives:
Here is my generic solution for any two circles (without any hardcoded value):
function [ P ] = circles_intersection_area( k1, k2, N )
%CIRCLES_INTERSECTION_AREA Summary...
% Adnan A.
x1 = k1(1);
y1 = k1(2);
r1 = k1(3);
x2 = k2(1);
y2 = k2(2);
r2 = k2(3);
if sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)) >= (r1 + r2)
% no intersection
P = 0;
return
end
% Wrapper rectangle config
a_min = x1 - r1 - 2*r2;
a_max = x1 + r1 + 2*r2;
b_min = y1 - r1 - 2*r2;
b_max = y1 + r1 + 2*r2;
% Monte Carlo algorithm
n = 0;
for i = 1:N
rand_x = unifrnd(a_min, a_max);
rand_y = unifrnd(b_min, b_max);
if sqrt((rand_x - x1)^2 + (rand_y - y1)^2) < r1 && sqrt((rand_x - x2)^2 + (rand_y - y2)^2) < r2
% is a point in the both of circles
n = n + 1;
plot(rand_x,rand_y, 'go-');
hold on;
else
plot(rand_x,rand_y, 'ko-');
hold on;
end
end
P = (a_max - a_min) * (b_max - b_min) * n / N;
end
Call it like: circles_intersection_area([-0.4,0,1], [0.4,0,1], 10000) where the first param is the first circle (x,y,r) and the second param is the second circle.
Without using For loop.
n = 100000;
data = rand(2,n);
data = data*2*30 - 30;
x = data(1,:);
y = data(2,:);
plot(x,y,'ro');
inside5 = find(x.^2 + y.^2 <=25);
hold on
plot (x(inside5),y(inside5),'bo');
hold on
inside12 = find(x.^2 + (y-12).^2<=144);
plot (x(inside12),y(inside12),'g');
hold on
insidefinal1 = find(x.^2 + y.^2 <=25 & x.^2 + (y-12).^2>=144);
insidefinal2 = find(x.^2 + y.^2 >=25 & x.^2 + (y-12).^2<=144);
% plot(x(insidefinal1),y(insidefinal1),'bo');
hold on
% plot(x(insidefinal2),y(insidefinal2),'ro');
insidefinal3 = find(x.^2 + y.^2 <=25 & x.^2 + (y-12).^2<=144);
% plot(x(insidefinal3),y(insidefinal3),'ro');
area1=(60^2)*(length(insidefinal1)/n);
area3=(60^2)*(length(insidefinal2)/n);
area2= (60^2)*(length(insidefinal3)/n);
I'm writing a "Peak finder" in Matlab. I've never used Matlab or anything similar before this project, so I'm new to "vectorizing" my code. Essentially, the program needs to take a video of molecules and plot circles on the molecules present in each frame of the video. If a molecule is crowded then it gets a red circle, but if it is not crowded it gets a green circle.
My problem is that some of these videos have 2000 frames and my program takes up to ~25 seconds to process a single frame, which is not practical.
Using tic and toc I've found the trouble maker: A for-loop which calls a function that contains a for-loop.
function getPeaks( orgnl_img, processed_img, cut_off, radius )
% find large peaks peaks by thresholding, i.e. you accept a peak only
% if its more than 'threshold' higher than its neighbors
threshold = 2*std2(orgnl_img);
peaks = (orgnl_img - processed_img) > threshold;
% m and n are dimensions of the frame, named peaks
[m, n] = size(peaks);
cc_centroids = regionprops(peaks, 'centroid');
% Pre-allocate arrays
x_centroid = zeros(1, length(cc_centroids));
y_centroid = zeros(1, length(cc_centroids));
for i = 1:length(cc_centroids)
% Extract the x and y components from cc_centroids struct
x_centroid(i) = cc_centroids(i).Centroid(1);
y_centroid(i) = cc_centroids(i).Centroid(2);
row = int64(x_centroid(i));
col = int64(y_centroid(i));
% Assure that script doesnt attempt to exceed frame
if col-2 > 0 && row-2 > 0 && col+2 < m && row+2 < n
region_of_interest = orgnl_img(col-2:col+2,row-2:row+2);
local_intensity = max(region_of_interest(:));
% Do not plot circle when intensity is 'cut off' or lower
if local_intensity > cut_off
dist_bool = findDistance(cc_centroids, x_centroid(i), y_centroid(i), radius);
if dist_bool == 1
color = 'g';
else
color = 'r';
end
plotCircle(color, x_centroid(i), y_centroid(i), radius)
end
end
end
end
And here is the findDistance function which contains another for-loop and determines if the circles overlap:
function dist_bool = findDistance( all_centroids, x_crrnt, y_crrnt, radius )
x_nearby = zeros(1, length(all_centroids));
y_nearby = zeros(1, length(all_centroids));
for i = 1:length(all_centroids)
if all_centroids(i).Centroid(1) < (x_crrnt+2*radius) &&...
all_centroids(i).Centroid(1) > (x_crrnt-2*radius) &&...
all_centroids(i).Centroid(2) < (y_crrnt+2*radius) &&...
all_centroids(i).Centroid(2) > (y_crrnt-2*radius)
x_nearby(i) = all_centroids(i).Centroid(1);
y_nearby(i) = all_centroids(i).Centroid(2);
pts_of_interest = [x_nearby(i),y_nearby(i);x_crrnt,y_crrnt];
dist = pdist(pts_of_interest);
if dist == 0 || dist > 2*radius
dist_bool = 1;
else
dist_bool = 0;
break
end
end
end
end
I think that there must be much improvement to be done here. I would appreciate any advice.
Thanks! :)
UPDATE: Here are the profiler results. The first section of code is the "getPeaks" function
http://i.stack.imgur.com/VaLdH.png
The hold comes from the plot circle function:
function plotCircle( color, x_centroid, y_centroid, radius )
hold on;
th = 0:pi/50:2*pi;
xunit = radius * cos(th) + x_centroid;
yunit = radius * sin(th) + y_centroid;
plot(xunit, yunit, color);
hold off;
end
:) I'm trying to code a Least Squares algorithm and I've come up with this:
function [y] = ex1_Least_Squares(xValues,yValues,x) % a + b*x + c*x^2 = y
points = size(xValues,1);
A = ones(points,3);
b = zeros(points,1);
for i=1:points
A(i,1) = 1;
A(i,2) = xValues(i);
A(i,3) = xValues(i)^2;
b(i) = yValues(i);
end
constants = (A'*A)\(A'*b);
y = constants(1) + constants(2)*x + constants(3)*x^2;
When I use this matlab script for linear functions, it works fine I think. However, when I'm passing 12 points of the sin(x) function I get really bad results.
These are the points I pass to the function:
xValues = [ -180; -144; -108; -72; -36; 0; 36; 72; 108; 144; 160; 180];
yValues = [sind(-180); sind(-144); sind(-108); sind(-72); sind(-36); sind(0); sind(36); sind(72); sind(108); sind(144); sind(160); sind(180) ];
And the result is sin(165°) = 0.559935259380508, when it should be sin(165°) = 0.258819
There is no reason why fitting a parabola to a full period of a sinusoid should give good results. These two curves are unrelated.
MATLAB already contains a least square polynomial fitting function, polyfit and a complementary function, polyval. Although you are probably supposed to write your own, trying out something like the following will be educational:
xValues = [ -180; -144; -108; -72; -36; 0; 36; 72; 108; 144; 160; 180];
% you may want to experiment with different ranges of xValues
yValues = sind(xValues);
% try this with different values of n, say 2, 3, and 4
p = polyfit(xValues,yValues,n);
x = -180:36:180;
y = polyval(p,x);
plot(xValues,yValues);
hold on
plot(x,y,'r');
Also, more generically, you should avoid using loops as you have in your code. This should be equivalent:
points = size(xValues,1);
A = ones(points,3);
A(:,2) = xValues;
A(:,3) = xValues.^2; % .^ and ^ are different
The part of the loop involving b is equivalent to doing b = yValues; either name the incoming variable b or just use the variable yValues, there's no need to make a copy of it.