So I want to smooth out Sharkfin signal at the point where it goes down and at the point where it goes up. As shown in the figure below, the Sharkfin waveform has sharp fall and rise at time 2 sec and 4 seconds:
Any idea on how to round of that area so that it is smooth in that section so that it looks something like this:
There are two separate things here - how do you detect a sharp transition, and how do you filter it.
Let's take these in turn.
A sharp transition is characterized by a large curvature - we can easily detect this by taking the diff of the input curve. Using the second parameter = 2 takes diff twice and results in something "like" the second derivative - but it's offset by one. So when we find points where the diff(sharkfin,2) is large, we need to offset by 1 to get the corner points.
Next, the smoothing itself. There are many techniques - I show a simple convolution with a box function. Doing this twice gives a smoothed version of the input. By picking the original "far from the discontinuity" and the filtered version "close to the discontinuity" we get exactly what you were asking for. If you wanted, you could "blend" the points - using a weighted version of filtered and unfiltered depending on how close you were to the corner points. I didn't show that explicitly but it should be easy to see how to expand the code I already wrote:
% generate a "shark fin" function:
fin = exp(-linspace(0,4,60));
shark = [fin (1-fin+fin(end))];
shark = repmat(shark, [1 3]);
D2 = diff(shark, 2);
roundMe = find(abs(D2)>0.1*max(D2))+1; % offset by 1 because second derivative
figure;
subplot(3,1,1)
plot(shark); title 'shark plot'
hold on;
plot(roundMe, shark(roundMe),'r*')
legend('input','corners found')
% take N points on either side of the sharp corners:
N = 3;
% boxplot filtered version of the curve
boxFilt = ones(1, 2*N+1)/(2*N+1);
smoothShark1 = convn(shark, boxFilt, 'same'); % box plot
% second filter - smoother
smoothShark2 = convn(smoothShark1, boxFilt, 'same');
% plot the filtered results:
subplot(3,1,2)
plot(shark)
hold on
plot(smoothShark1);
hold on
plot(smoothShark2);
xlim([114 126])
ylim([0.8,1.1])
legend('original','box','box x2')
title 'smoothed everywhere'
% Now apply filtering only to points near the discontinuity
smoothMe = zeros(size(shark));
smoothMe(roundMe)=1;
smoothMe = convn(smoothMe, boxFilt, 'same');
smoothMe(smoothMe>0)=1; % this finds N points on either side of the corner
subplot(3,1,3)
plot(shark)
finalPlot=shark;
hold on
smoothIndx = find(smoothMe);
finalPlot(smoothIndx)=smoothShark2(smoothIndx);
plot(finalPlot,'g')
plot(smoothIndx, finalPlot(smoothIndx), 'r*')
xlim([114 126])
ylim([0.8,1.1])
legend('original','smoothed','changed')
title 'smoothed only near discontinuity'
Output:
Related
I want to convert an image from Cartesian to Polar and to use it for opengl texture.
So I used matlab referring to the two articles below.
Link 1
Link 2
My code is exactly same with Link 2's anwser
% load image
img = imread('my_image.png');
% convert pixel coordinates from cartesian to polar
[h,w,~] = size(img);
[X,Y] = meshgrid((1:w)-floor(w/2), (1:h)-floor(h/2));
[theta,rho] = cart2pol(X, Y);
Z = zeros(size(theta));
% show pixel locations (subsample to get less dense points)
XX = X(1:8:end,1:4:end);
YY = Y(1:8:end,1:4:end);
tt = theta(1:8:end,1:4:end);
rr = rho(1:8:end,1:4:end);
subplot(121), scatter(XX(:),YY(:),3,'filled'), axis ij image
subplot(122), scatter(tt(:),rr(:),3,'filled'), axis ij square tight
% show images
figure
subplot(121), imshow(img), axis on
subplot(122), warp(theta, rho, Z, img), view(2), axis square
The result was exactly what I wanted, and I was very satisfied except for one thing. It's the area (red circled area) in the picture just below. Considering that the opposite side (blue circled area) is not, I think this part should also be filled. Because of this part is empty, so there is a problem when using it as a texture.
And I wonder how I can fill this part. Thank you.
(little difference from Link 2's answer code like degree<->radian and axis values. but i think it is not important.)
Those issues you show in your question happen because your algorithm is wrong.
What you did (push):
throw a grid on the source image
transform those points
try to plot these colored points and let MATLAB do some magic to make it look like a dense picture
Do it the other way around (pull):
throw a grid on the output
transform that backwards
sample the input at those points
The distinction is called "push" (into output) vs "pull" (from input). Only Pull gives proper results.
Very little MATLAB code is necessary. You just need pol2cart and interp2, and a meshgrid.
With interp2 you get to choose the interpolation (linear, cubic, ...). Nearest-neighbor interpolation leaves visible artefacts.
im = im2single(imread("PQFax.jpg"));
% center of polar map, manually picked
cx = 10 + 409/2;
cy = 7 + 413/2;
% output parameters
radius = 212;
dRho = 1;
dTheta = 2*pi / (2*pi * radius);
Thetas = pi/2 - (0:dTheta:2*pi);
Rhos = (0:dRho:radius);
% polar mesh
[Theta, Rho] = meshgrid(Thetas, Rhos);
% transform...
[Xq,Yq] = pol2cart(Theta, Rho);
% translate to sit on the circle's center
Xq = Xq + cx;
Yq = Yq + cy;
% sample image at those points
Ro = interp2(im(:,:,1), Xq,Yq, "cubic");
Go = interp2(im(:,:,2), Xq,Yq, "cubic");
Bo = interp2(im(:,:,3), Xq,Yq, "cubic");
Vo = cat(3, Ro, Go, Bo);
Vo = imrotate(Vo, 180);
imshow(Vo)
The other way around (get a "torus" from a "ribbon") is quite similar. Throw a meshgrid on the torus space, subtract center, transform from cartesian to polar, and use those to sample from the "ribbon" image into the "torus" image.
I'm more familiar with OpenCV than with MATLAB. Perhaps MATLAB has something like OpenCV's warpPolar(), or a generic remap(). In any case, the operation is trivial to do entirely "by hand" but there are enough supporting functions to take the heavy lifting off your hands (interp2, pol2cart, meshgrid).
1.- The white arcs tell that the used translation pol-cart introduces significant errors.
2.- Reversing the following script solves your question.
It's a script that goes from cart-pol without introducing errors or ignoring input data, which is what happens when such wide white arcs show up upon translation apparently correct.
clear all;clc;close all
clc,cla;
format long;
A=imread('shaffen dass.jpg');
[sz1 sz2 sz3]=size(A);
szx=sz2;szy=sz1;
A1=A(:,:,1);A2=A(:,:,2);A3=A(:,:,3); % working with binary maps or grey scale images this wouldn't be necessary
figure(1);imshow(A);
hold all;
Cx=floor(szx/2);Cy=floor(szy/2);
plot(Cx,Cy,'co'); % because observe image centre not centered
Rmin=80;Rmax=400; % radius search range for imfindcircles
[centers, radii]=imfindcircles(A,[Rmin Rmax],... % outer circle
'ObjectPolarity','dark','Sensitivity',0.9);
h=viscircles(centers,radii);
hold all; % inner circle
[centers2, radii2]=imfindcircles(A,[Rmin Rmax],...
'ObjectPolarity','bright');
h=viscircles(centers2,radii2);
% L=floor(.5*(radii+radii2)); % this is NOT the length X that should have the resulting XY morphed graph
L=floor(2*pi*radii); % expected length of the morphed graph
cx=floor(.5*(centers(1)+centers2(1))); % coordinates of averaged circle centres
cy=floor(.5*(centers(2)+centers2(2)));
plot(cx,cy,'r*'); % check avg centre circle is not aligned to figure centre
plot([cx 1],[cy 1],'r-.');
t=[45:360/L:404+1-360/L]; % if step=1 then we only get 360 points but need an amount of L points
% if angle step 1/L over minute waiting for for loop to finish
R=radii+5;x=R*sind(t)+cx;y=R*cosd(t)+cy; % build outer perimeter
hL1=plot(x,y,'m'); % axis equal;grid on;
% hold all;
% plot(hL1.XData,hL1.YData,'ro');
x_ref=hL1.XData;y_ref=hL1.YData;
% Sx=zeros(ceil(R),1);Sy=zeros(ceil(R),1);
Sx={};Sy={};
for k=1:1:numel(hL1.XData)
Lx=floor(linspace(x_ref(k),cx,ceil(R)));
Ly=floor(linspace(y_ref(k),cy,ceil(R)));
% plot(Lx,Ly,'go'); % check
% plot([cx x(k)],[cy y(k)],'r');
% L1=unique([Lx;Ly]','rows');
Sx=[Sx Lx'];Sy=[Sy Ly'];
end
sx=cell2mat(Sx);sy=cell2mat(Sy);
[s1 s2]=size(sx);
B1=uint8(zeros(s1,s2));
B2=uint8(zeros(s1,s2));
B3=uint8(zeros(s1,s2));
for n=1:1:s2
for k=1:1:s1
B1(k,n)=A1(sx(k,n),sy(k,n));
B2(k,n)=A2(sx(k,n),sy(k,n));
B3(k,n)=A3(sx(k,n),sy(k,n));
end
end
C=uint8(zeros(s1,s2,3));
C(:,:,1)=B1;
C(:,:,2)=B2;
C(:,:,3)=B3;
figure(2);imshow(C);
the resulting
3.- let me know if you'd like some assistance writing pol-cart from this script.
Regards
John BG
I'm working on images with overlapping line shapes (left plot). Ultimately I want to segment single objects. I'm working with a Hough transform to achieve this and it works well in finding lines of (significantly) different orientation - e.g. represented by the two maxima in the hough space below (middle plot).
the green and yellow lines (left plot) and crosses (right plot) stem from an approach to do something with the thickness of the line. I couldn't figure out how to extract a broad line though, so I didn't follow up.
I'm aware of the ambiguity of assigning the "overlapping pixels". I will address that later.
Since I don't know, how many line objects one connected region may contain, my idea is to iteratively extract the object corresponding to the hough line with the highest activation (here painted in blue), i.e. remove the line shaped object from the image, so that the next iteration will find only the other line.
But how do I detect, which pixels belong to the line shaped object?
The function hough_bin_pixels(img, theta, rho, P) (from here - shown in the right plot) gives pixels corresponding to the particular line. But that obviously is too thin of a line to represent the object.
Is there a way to segment/detect the whole object that is orientied along the strongest houghline?
The key is knowing that thick lines in the original image translate to wider peaks on the Hough Transform. This image shows the peaks of a thin and a thick line.
You can use any strategy you like to group all the pixels/accumulator bins of each peak together. I would recommend using multithresh and imquantize to convert it to a BW image, and then bwlabel to label the connected components. You could also use any number of other clustering/segmentation strategies. The only potentially tricky part is figuring out the appropriate thresholding levels. If you can't get anything suitable for your application, err on the side of including too much because you can always get rid of erroneous pixels later.
Here are the peaks of the Hough Transform after thresholding (left) and labeling (right)
Once you have the peak regions, you can find out which pixels in the original image contributed to each accumulator bin using hough_bin_pixels. Then, for each peak region, combine the results of hough_bin_pixels for every bin that is part of the region.
Here is the code I threw together to create the sample images. I'm just getting back into matlab after not using it for a while, so please forgive the sloppy code.
% Create an image
image = zeros(100,100);
for i = 10:90
image(100-i,i)=1;
end;
image(10:90, 30:35) = 1;
figure, imshow(image); % Fig. 1 -- Original Image
% Hough Transform
[H, theta_vals, rho_vals] = hough(image);
figure, imshow(mat2gray(H)); % Fig. 2 -- Hough Transform
% Thresholding
thresh = multithresh(H,4);
q_image = imquantize(H, thresh);
q_image(q_image < 4) = 0;
q_image(q_image > 0) = 1;
figure, imshow(q_image) % Fig. 3 -- Thresholded Peaks
% Label connected components
L = bwlabel(q_image);
figure, imshow(label2rgb(L, prism)) % Fig. 4 -- Labeled peaks
% Reconstruct the lines
[r, c] = find(L(:,:)==1);
segmented_im = hough_bin_pixels(image, theta_vals, rho_vals, [r(1) c(1)]);
for i = 1:size(r(:))
seg_part = hough_bin_pixels(image, theta_vals, rho_vals, [r(i) c(i)]);
segmented_im(seg_part==1) = 1;
end
region1 = segmented_im;
[r, c] = find(L(:,:)==2);
segmented_im = hough_bin_pixels(image, theta_vals, rho_vals, [r(1) c(1)]);
for i = 1:size(r(:))
seg_part = hough_bin_pixels(image, theta_vals, rho_vals, [r(i) c(i)]);
segmented_im(seg_part==1) = 1;
end
region2 = segmented_im;
figure, imshow([region1 ones(100, 1) region2]) % Fig. 5 -- Segmented lines
% Overlay and display
out = cat(3, image, region1, region2);
figure, imshow(out); % Fig. 6 -- For fun, both regions overlaid on original image
I am struggling with template matching in the Fourier domain in Matlab. Here are my images (the artist is RamalamaCreatures on DeviantArt):
My aim is to place a bounding box around the ear of the possum, like this example (where I performed template matching using normxcorr2):
Here is the Matlab code I am using:
clear all; close all;
template = rgb2gray(imread('possum_ear.jpg'));
background = rgb2gray(imread('possum.jpg'));
%% calculate padding
bx = size(background, 2);
by = size(background, 1);
tx = size(template, 2); % used for bbox placement
ty = size(template, 1);
%% fft
c = real(ifft2(fft2(background) .* fft2(template, by, bx)));
%% find peak correlation
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = find(c == max(c(:)));
figure; surf(c), shading flat; % plot correlation
%% display best match
hFig = figure;
hAx = axes;
position = [xpeak(1)-tx, ypeak(1)-ty, tx, ty];
imshow(background, 'Parent', hAx);
imrect(hAx, position);
The code is not functioning as intended - it is not identifying the correct region. This is the failed result - the wrong area is boxed:
This is the surface plot of the correlations for the failed match:
Hope you can help! Thanks.
What you're doing in your code is actually not correlation at all. You are using the template and performing convolution with the input image. If you recall from the Fourier Transform, the multiplication of the spectra of two signals is equivalent to the convolution of the two signals in time/spatial domain.
Basically, what you are doing is that you are using the template as a kernel and using that to filter the image. You are then finding the maximum response of this output and that's what is deemed to be where the template is. Where the response is being boxed makes sense because that region is entirely white, and using the template as the kernel with a region that is entirely white will give you a very large response, which is why it most likely identified that area to be the maximum response. Specifically, the region will have a lot of high values (~255 or so), and naturally performing convolution with the template patch and this region will give you a very large output due to the operation being a weighted sum. As such, if you used the template in a dark area of the image, the output would be small - which is false because the template is also consisting of dark pixels.
However, you can certainly use the Fourier Transform to locate where the template is, but I would recommend you use Phase Correlation instead. Basically, instead of computing the multiplication of the two spectra, you compute the cross power spectrum instead. The cross power spectrum R between two signals in the frequency domain is defined as:
Source: Wikipedia
Ga and Gb are the original image and the template in frequency domain, and the * is the conjugate. The o is what is known as the Hadamard product or element-wise product. I'd also like to point out that the division of the numerator and denominator of this fraction is also element-wise. Using the cross power spectrum, if you find the (x,y) location here that produces the absolute maximum response, this is where the template should be located in the background image.
As such, you simply need to change the line of code that computes the "correlation" so that it computes the cross power spectrum instead. However, I'd like to point out something very important. When you perform normxcorr2, the correlation starts right at the top-left corner of the image. The template matching starts at this location and it gets compared with a window that is the size of the template where the top-left corner is the origin. When finding the location of the template match, the location is with respect to the top-left corner of the matched window. Once you compute normxcorr2, you traditionally add the half of the rows and half of the columns of the maximum response to find the centre location.
Because we are more or less doing the same operations for template matching (sliding windows, correlation, etc.) with the FFT / frequency domain, when you finish finding the peak in this correlation array, you must also take this into account. However, your call to imrect to draw a rectangle around where the template matches takes in the top left corner of a bounding box anyway, so there's no need to do the offset here. As such, we're going to modify that code slightly but keep the offset logic in mind when using this code for later if want to find the centre location of the match.
I've modified your code as well to read in the images directly from StackOverflow so that it's reproducible:
clear all; close all;
template = rgb2gray(imread('http://i.stack.imgur.com/6bTzT.jpg'));
background = rgb2gray(imread('http://i.stack.imgur.com/FXEy7.jpg'));
%% calculate padding
bx = size(background, 2);
by = size(background, 1);
tx = size(template, 2); % used for bbox placement
ty = size(template, 1);
%% fft
%c = real(ifft2(fft2(background) .* fft2(template, by, bx)));
%// Change - Compute the cross power spectrum
Ga = fft2(background);
Gb = fft2(template, by, bx);
c = real(ifft2((Ga.*conj(Gb))./abs(Ga.*conj(Gb))));
%% find peak correlation
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = find(c == max(c(:)));
figure; surf(c), shading flat; % plot correlation
%% display best match
hFig = figure;
hAx = axes;
%// New - no need to offset the coordinates anymore
%// xpeak and ypeak are already the top left corner of the matched window
position = [xpeak(1), ypeak(1), tx, ty];
imshow(background, 'Parent', hAx);
imrect(hAx, position);
With that, I get the following image:
I also get the following when showing a surface plot of the cross power spectrum:
There is a clear defined peak where the rest of the output has a very small response. That's actually a property of Phase Correlation and so obviously, the location of the maximum value is clearly defined and this is where the template is located.
Hope this helps!
Just ended up implementing the same with python with similar ideas as #rayryeng's using scipy.fftpack.fftn() / ifftn() functions with the following result on the same target and template images:
import numpy as np
import scipy.fftpack as fp
from skimage.io import imread
from skimage.color import rgb2gray, gray2rgb
import matplotlib.pylab as plt
from skimage.draw import rectangle_perimeter
im = 255*rgb2gray(imread('http://i.stack.imgur.com/FXEy7.jpg')) # target
im_tm = 255*rgb2gray(imread('http://i.stack.imgur.com/6bTzT.jpg')) # template
# FFT
F = fp.fftn(im)
F_tm = fp.fftn(im_tm, shape=im.shape)
# compute the best match location
F_cc = F * np.conj(F_tm)
c = (fp.ifftn(F_cc/np.abs(F_cc))).real
i, j = np.unravel_index(c.argmax(), c.shape)
print(i, j)
# 214 317
# draw rectangle around the best match location
im2 = (gray2rgb(im)).astype(np.uint8)
rr, cc = rectangle_perimeter((i,j), end=(i + im_tm.shape[0], j + im_tm.shape[1]), shape=im.shape)
for x in range(-2,2):
for y in range(-2,2):
im2[rr + x, cc + y] = (255,0,0)
# show the output image
plt.figure(figsize=(10,10))
plt.imshow(im2)
plt.axis('off')
plt.show()
Also, the below animation shows the result obtained while locating a bird's template image inside a set of (target) frames extracted from a video with a flock of birds.
One thing to note: the output is very much dependent on the similarity of the size and shape of the object that is to be matched with the template, if it's quite different from that of the template image, the template may not be matched at all.
I have a cube in which I have something as a anomaly, consider it as a cube of zeros and I put a smaller cube of numbers in it using the following code:
anomaly = zeros(100,100,100);
for j = 40:60
for jj = 40:60
for jjj=40:60
anomaly(j,jj,jjj)=10;
end
end
end
I want to somehow view it in a 3D figure, for which I can see the anomaly it it. How can I do it?
Go for isosurface. With isosurface you will be able to extract the inclusion. If you also want to show the whole box, I recommend you to do the following: Create a small box (so it has few faces when converted to isosurface) and then scale it to the real size of the box. You can see this in the following code:
anomaly = zeros(100,100,100); % The anomaly
an2=zeros(100/10+2,100/10+2,100/10+2); % For plotting pourposes
an2(2:10,2:10,2:10)=1; % Create a 10x10x10 box
for j = 40:60
for jj = 40:60
for jjj=40:60
anomaly(j,jj,jjj)=10;
end
end
end
iso=isosurface(anomaly,5); % Extract surface of inclusion
iso2=isosurface(an2,0.5); % create a box
iso2.vertices=(iso2.vertices-1)*10; % offset( so its in 1-10 range)
% and scaling to be 100x100x100
patch(iso,'facecolor','r','edgecolor','none')
patch(iso2,'facecolor','none');
% this last ones are for having a nicer plot
camlight
axis off
axis equal
The function called DicePlot simulates rolling 10 dice 5000 times.
The function calculates the sum of values of the 10 dice of each roll, which will be a 1 ⇥ 5000 vector, and plot relative frequency histogram with edges of bins being selected in where each bin in the histogram represents a possible value of for the sum of the dice.
The mean and standard deviation of the 1 ⇥ 5000 sums of dice values will be computed, and the probability density function of normal distribution (with the mean and standard deviation computed) on top of the relative frequency histogram will be plotted.
Below is my code so far - What am I doing wrong? The graph shows up but not the extra red line on top? I looked at answers like this, and I don't think I'll be plotting anything like the Gaussian function.
% function[]= DicePlot()
for roll=1:5000
diceValues = randi(6,[1, 10]);
SumDice(roll) = sum(diceValues);
end
distr=zeros(1,6*10);
for i = 10:60
distr(i)=histc(SumDice,i);
end
bar(distr,1)
Y = normpdf(X)
xlabel('sum of dice values')
ylabel('relative frequency')
title(['NumDice = ',num2str(NumDice),' , NumRolls = ',num2str(NumRolls)]);
end
It is supposed to look like
But it looks like
The red line is not there because you aren't plotting it. Look at the documentation for normpdf. It computes the pdf, it doesn't plot it. So you problem is how do you add this line to the plot. The answer to that problem is to google "matlab hold on".
Here's some code to get you going in the right direction:
% Normalize your distribution
normalizedDist = distr/sum(distr);
bar(normalizedDist ,1);
hold on
% Setup your density function using the mean and std of your sample data
mu = mean(SumDice);
stdv = std(SumDice);
yy = normpdf(xx,mu,stdv);
xx = linspace(0,60);
% Plot pdf
h = plot(xx,yy,'r'); set(h,'linewidth',1.5);