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I want to pad an object based on edge pixel to replace the zeros.
I am not sure if padarray is applicable to this, I am showing a sample code below to replicate my need. I am able to do it but I think this is not an efficient way as I am scanning each row at a time to find and pad the zeros.
%% Example code to recreate my need
image = imread('moon.tif');
[~, ncols] = size(image);
image(image <50) = 0;
image = fliplr(image(:,1:round(ncols/2)));
% figure, imshow(image,[])
BW = bwareafilt(logical(image),1);
% create bounding box
boxProps=regionprops(BW,'BoundingBox');
cords_BoundingBox = boxProps(1).BoundingBox;
% Extract sub_Image
sub_Image = imcrop(image, cords_BoundingBox);
% figure, imshow(sub_Image,[])
%% This is the part I want to use better or existing function for padding
duplicate_sub_Image = sub_Image;
[nrows, ~] = size(duplicate_sub_Image);
for nrow = 1:nrows
% current_row_inverted = fliplr(sub_Image(nrow,:));
[~,col,pad_value] = find(duplicate_sub_Image(nrow,:),1,'last');
duplicate_sub_Image(nrow,col+1:end) = pad_value;
end
figure,
subplot(131),imshow(image,[]), title('original image');
subplot(132),imshow(sub_Image,[]), title('bounding box image');
subplot(133),imshow(duplicate_sub_Image,[]), title('row padded image');
Any suggestions to improve this code or use of existing functions to address this issue?
Thanks
Here is a way without using loops:
[~,imin] = min(sub_Image, [], 2);
col = max(1, imin-1);
ind = sub2ind(size(sub_Image), (1:numel(col)).', col);
duplicate_sub_Image = sub_Image(ind) .* ~sub_Image + sub_Image;
Based on Shai and Biguri's codes and comments, I have finished a color picture like this:
A problem arises, how to remove the white edge and make it smooth? One solution may be to build 3x3 matrix or bigger and average. But the calculations should be large for every white-edge points. Or there may be some useful functions in Matlab to deal well with this problems?
If you have a license for the image processing toolbox, you can try using for example medfilt2 to apply a median filter on the image. A 11 by 11 median filter should do the trick. It is not very difficult to reimplement the filter yourself if you don't have the toolbox.
This is just one of the possibilities, you can use many different filters that will have different impacts on sharpness ang edge removal.
Edit:
Here is a quick median filter implementation (it may contain errors and could be optimized):
function ret = imageMedianFilter(im, np)
if(size(np,2) == 1)
npx = np;
npy = np;
else
npx = np(1);
npy = np(2);
end
ret = zeros(size(im,1),size(im,2));
for xpos = 1:size(im,1)
for ypos = 1:size(im,2)
curval = double(0);
if(xpos + npx - 1) > size(im,1)
npixels_x = size(im,1) - xpos + 1;
else
npixels_x = npx;
end
if(ypos + npy - 1) > size(im,2)
npixels_y = size(im,2) - ypos + 1;
else
npixels_y = npy;
end
a = im(xpos:xpos+npixels_x-1 , ypos:ypos+npixels_y-1);
a = reshape(a,1,size(a,1)*size(a,2));
curval = median(a);
ret(xpos , ypos) = curval;
end
end
ret = uint8(ret);
end
You can use it on R,G and B components as shown by Rotem below:
RGB = cat(3, imageMedianFilter(RGB(:,:,1), [11,11]), imageMedianFilter(RGB(:,:,2), [11,11]), imageMedianFilter(RGB(:,:,3), [11,11]));
(assuming your image is named RGB).
Here is my solution. I take n*n patch to average the near RGB. But there is a problem arising. Why the right down side of processed picture showing black lines?
clc;clf;close all;clear all;
img = imread('sample2color_t1.bmp'); %// read image
bw = img(:,:,1) > 128; %// convert to binary mask
[lb,lab] = bwlabel(bw,4); %// extract distinct regions
[a,b,c]=size(img);
R=ones(a,b);
G=ones(a,b);
B=ones(a,b);
%I have omitted other colors process codes. Below it is the white edges code.
r=[];c=[];
[r,c] = find(lb ==0);
for i=1:length(r)
R(r(i),c(i))=1;
G(r(i),c(i))=1;
B(r(i),c(i))=1;
end
scale=5;%步长1,8连通
for i=1:length(r)
sumR=0;sumG=0;sumB=0;
for j=0:2*scale
for k=0:2*scale
sumR=sumR+R(r(i)-scale+j,c(i)-scale+k);
sumG=sumG+G(r(i)-scale+j,c(i)-scale+k);
sumB=sumB+B(r(i)-scale+j,c(i)-scale+k);
end
end
R(r(i),c(i))=sumR/(2*scale+1)^2;
G(r(i),c(i))=sumG/(2*scale+1)^2;
B(r(i),c(i))=sumB/(2*scale+1)^2;
end
imPaint=cat(3,R,G,B);
figure;
imshow(imPaint);
I have created a big heat map using matlab's imagesc command. It plots the error output for each combination of the values in x and y axes. As can be seen in the figure there are too many axes labels. This might become even denser as I plan to increase the number of points in both x and y axes - which means I will get more outputs on a finer grid.
I want to be flexible with the labels, and skip some of them. I want to do this for both X and Y. I also want to be flexible with the "ticks" and draw either all of them or maybe skip some of them. Keep in mind that both the X and Y values are not increasing in order, at first the increment is 0.01 for 9 points, then 0.1, then 1 or 3 or whatever. I will change these increments too.
I tried to show what I want the graph look like in the second image. I want roughly the labels shown in red boxes only. As I said these are not set values, and I will make the increments smaller which will lead to denser plot.
Thank you for your help.
OS: Windows 7, 8 (64 bit)
Matlab version: Matlab 2014 a
You can manipulate the ticks and labels like this:
ticksarray=[1 33 41 100 ...] % edit these to whatever you want
tickslabels={'1', '33', '41', '100'; ...} % match the size of both arrays
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
The same thing applies to the y-axis.
Small working example:
x=1:100;
y=2*x.^2-3*x+2;
plot(x,y)
ticksarray=[1 33 41 100];
tickslabels={'1', '33', '41', '100'};
set(gca,'XTick',ticksarray)
set(gca,'XTickLabel',tickslabels)
Example:
figure(1)
load clown
subplot(211)
imagesc(X);
subplot(212)
imagesc(X);
h = gca;
Now you can either set a maximum number of labels per axis:
%// define maximum number of labels
maxLabel = 3;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxLabel);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxLabel);
or define how many labels should be skipped:
%// define number of labels to skip
skipLabel = 2;
h.XTick = h.XTick(1:skipLabel:end);
h.YTick = h.YTick(1:skipLabel:end)
You can also get a different number of ticks and labels, more complicated though:
maxLabel = 3;
maxTicks = 6;
h.XTick = linspace(h.xlim(1),h.xlim(2),maxTicks);
h.YTick = linspace(h.ylim(1),h.ylim(2),maxTicks);
h.XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
h.YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ) ) = repmat({''},1,maxTicks-maxLabel);
If you use a prior version of Matlab 2014b, then you will need the set command to set all properties:
%// define maximum number of labels
maxLabel = 3;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxLabel));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxLabel));
%// or define number of labels to skip
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
%// or combined
maxLabel = 3;
maxTicks = 6;
Xlim = get(h,'Xlim');
Ylim = get(h,'Ylim');
set(h,'XTick', linspace(Xlim(1),Xlim(2),maxTicks));
set(h,'YTick', linspace(Ylim(1),Ylim(2),maxTicks));
XTickLabel = cellstr(get(h,'XTickLabel'));
YTickLabel = cellstr(get(h,'YTickLabel'));
XTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
YTickLabel( setdiff( 1:maxTicks, 1:maxTicks/maxLabel:maxTicks ),: ) = repmat({''},1,maxTicks-maxLabel);
set(h,'XTickLabel',XTickLabel);
set(h,'YTickLabel',YTickLabel);
After applying the second method proposed by #thewaywewalk I got the second figure below. Apparently the labels need to be structured as well, because they only take the first so many labels.
Then I tried to manipulate the labels as shown below, and got the third image.
skipLabel = 2;
XTick = get(h,'XTick');
YTick = get(h,'YTick');
set(h,'XTick', XTick(1:skipLabel:end));
set(h,'YTick', YTick(1:skipLabel:end));
XTickLabel = get(h,'XTickLabel');
labelsX = cell( length(1: skipLabel:length(XTick)) , 1);
j = 1;
for i = 1: skipLabel:length(XTick)
labelsX{j} = XTickLabel(i, :);
j = j + 1;
end
set(h,'XTickLabel', labelsX);
YTickLabel = get(h,'YTickLabel');
labelsY = cell( length(1: skipLabel:length(YTick)) , 1);
j = 1;
for i = 1: skipLabel:length(YTick)
labelsY{j} = YTickLabel(i, :);
j = j + 1;
end
set(h,'YTickLabel', labelsY);
The Y axis labels seem to be in place as before (right next to tick), however the X axis labels seem to be shifted to the left a little. How can I correct this?
Another note: How can I change the scientific values into normal numbers? Also, probably there is a better approach at manipulating the labels.
I've an image over which I would like to compute a local histogram within a circular neighborhood. The size of the neighborhood is given by a radius. Although the code below does the job, it's computationally expensive. I run the profiler and the way I'm accessing to the pixels within the circular neighborhoods is already expensive.
Is there any sort of improvement/optimization based maybe on vectorization? Or for instance, storing the neighborhoods as columns?
I found a similar question in this post and the proposed solution is quite in the spirit of the code below, however the solution is still not appropriate to my case. Any ideas are really welcomed :-) Imagine for the moment, the image is binary, but the method should also ideally work with gray-level images :-)
[rows,cols] = size(img);
hist_img = zeros(rows, cols, 2);
[XX, YY] = meshgrid(1:cols, 1:rows);
for rr=1:rows
for cc=1:cols
distance = sqrt( (YY-rr).^2 + (XX-cc).^2 );
mask_radii = (distance <= radius);
bwresponses = img(mask_radii);
[nelems, ~] = histc(double(bwresponses),0:255);
% do some processing over the histogram
...
end
end
EDIT 1 Given the received feedback, I tried to update the solution. However, it's not yet correct
radius = sqrt(2.0);
disk = diskfilter(radius);
fun = #(x) histc( x(disk>0), min(x(:)):max(x(:)) );
output = im2col(im, size(disk), fun);
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
circfilter = (dist <= radius);
end
Following on the technique I described in my answer to a similar question you could try to do the following:
compute the index offsets from a particular voxel that get you to all the neighbors within a radius
Determine which voxels have all neighbors at least radius away from the edge
Compute the neighbors for all these voxels
Generate your histograms for each neighborhood
It is not hard to vectorize this, but note that
It will be slow when the neighborhood is large
It involves generating an intermediate matrix that is NxM (N = voxels in image, M = voxels in neighborhood) which could get very large
Here is the code:
% generate histograms for neighborhood within radius r
A = rand(200,200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy zz] = meshgrid(1:sz(2), 1:sz(1), 1:sz(3));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 + (zz - center(3)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2), center(3));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius & ...
zz > radius + 1 & zz < sz(3) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
A 2D version of the same code (which might be all you need, and is considerably faster):
% generate histograms for neighborhood within radius r
A = rand(200,200);
radius = 2.5;
tic
sz=size(A);
[xx yy] = meshgrid(1:sz(2), 1:sz(1));
center = round(sz/2);
centerPoints = find((xx - center(1)).^2 + (yy - center(2)).^2 < radius.^2);
centerIndex = sub2ind(sz, center(1), center(2));
% limit to just the points that are "far enough on the inside":
inside = find(xx > radius+1 & xx < sz(2) - radius & ...
yy > radius + 1 & yy < sz(1) - radius);
offsets = centerPoints - centerIndex;
allPoints = 1:prod(sz);
insidePoints = allPoints(inside);
indices = bsxfun(#plus, offsets, insidePoints);
hh = histc(A(indices), 0:0.1:1); % <<<< modify to give you the histogram you want
toc
You're right, I don't think that colfilt can be used as you're not applying a filter. You'll have to check the correctness, but here's my attempt using im2col and your diskfilter function (I did remove the conversion to double so it now output logicals):
function circhist
% Example data
im = randi(256,20)-1;
% Ranges - I do this globally for the whole image rather than for each neighborhood
mini = min(im(:));
maxi = max(im(:));
edges = linspace(mini,maxi,20);
% Disk filter
radius = sqrt(2.0);
disk = diskfilter(radius); % Returns logical matrix
% Pad array with -1
im_pad = padarray(im, (size(disk)-1)/2, -1);
% Convert sliding neighborhoods to columns
B = im2col(im_pad, size(disk), 'sliding');
% Get elements from each column that correspond to disk (logical indexing)
C = B(disk(:), :);
% Apply histogram across columns to count number of elements
out = histc(C, edges)
% Display output
figure
imagesc(out)
h = colorbar;
ylabel(h,'Counts');
xlabel('Neighborhood #')
ylabel('Bins')
axis xy
function disk = diskfilter(radius)
height = 2*ceil(radius)+1;
width = 2*ceil(radius)+1;
[XX,YY] = meshgrid(1:width,1:height);
dist = sqrt((XX-ceil(width/2)).^2+(YY-ceil(height/2)).^2);
disk = (dist <= radius);
If you want to set your ranges (edges) based on each neighborhood then you'll need to make sure that the vector is always the same length if you want to build a big matrix (and then the rows of that matrix won't correspond to each other).
You should note that the shape of the disk returned by fspecial is not as circular as what you were using. It's meant to be used a smoothing/averaging filter so the edges are fuzzy (anti-aliased). Thus when you use ~=0 it will grab more pixels. It'd stick with your own function, which is faster anyways.
You could try processing with an opposite logic (as briefly explained in the comment)
hist = zeros(W+2*R, H+2*R, Q);
for i = 1:R+1;
for j = 1:R+1;
if ((i-R-1)^2+(j-R-1)^2 < R*R)
for q = 0:1:Q-1;
hist(i:i+W-1,j:j+H-1,q+1) += (image == q);
end
end
end
end
I have read in an image file to MATLAB and I am trying to stretch it in one direction, but a variable amount (sinusoidal). This would create an accordion affect on the image. I have toyed around with imresize, however that only resizes the image linearly. I would like the amount of "stretch" to vary for each image line. I tried to convey this with the following code:
periods = 10; % Number of "stretch" cycles
sz = size(original_image,2)/periods;
s = 0;
x = 0;
for index = 1:periods
B = original_image(:,round(s+1:s+sz));
if mod(index,2) == 0
amp = 1.5;
else
amp = 0.75;
end
xi = size(B,2)*amp;
new_image(:,x+1:x+xi) = imresize(B, [size(B,1) size(B,2)*amp]);
s = s + sz;
x = x+xi;
end
You can see that segments of the image are stretched, then compressed, then stretched, etc, like an accordion. However, each segment has a uniform amount of stretch, whereas I'd like it to be increasing then decreasing as you move along the image.
I have also looked at MATLAB's example of Applying a Sinusoidal Transformation to a Checkerboard which seems very applicable to my problem, however I have been trying and I cannot get this to produce the desired result for my image.
Any help is much appreciated.
UPDATE:
Thank you for Answer #1. I was unable to get it to work for me, but also realized it would resulted in loss of data, as the code only called for certian lines in the original image, and other lines would have been ignored.
After experimenting further, I developed the code below. I used a checkerboard as an example. While combersome, it does get the job done. However, upon trying the script with an actual high-resolution image, it was extremely slow and ended up failing due to running out of memory. I believe this is because of the excessive number of "imresize" commands that are used in loop.
I = checkerboard(10,50);
I = imrotate(I,90);
[X Y] = size(I);
k = 4; % Number of "cycles"
k = k*2;
x = 1;
y = 2;
z = 2;
F = [];
i = 1;
t = 0;
s = 0;
for j = 1:k/2
t = t + 1;
for inc = round(s+1):round(Y/k*t)
Yi = i + 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z + 1;
i = i + 1;
end
y = y - 2;
z = z - 4;
for inc = round(Y/k*t+1):round(Y/k*(t+1))
Yi = i - 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z - 1;
i = i - 1;
end
y = y + 2;
z = z + 4;
s = Y/k*(t+1);
t = t + 1;
end
Fn = imresize(F, [X Y]);
imshow(Fn);
Does anyone know of a simpler way to achieve this? If you run the code above, you can see the effect I am trying to achieve. Unfortunately, my method above does not allow me to adjust the amplitude of the "stretch" either, only the number of "cycles," or frequency. Help on this would also be appreciated. Much thanks!
Here is how I would approach it:
Determine how the coordinate of each point in your Final image F maps into your Initial image I of size (M,N)
Since you want to stretch horizontally only, given a point (xF,yF) in your final image, that point would be (xI,yI) in your initial image where xI and yI can be obtained as follows:
yI = yF;
xI = xF + Lsin(xFK);
Notes:
these equations do not guarantee that xI remains within the range [1:N] so cropping needs to be added
K controls the how many wrinkles you want to have in your accordion effect. For example, if you only want one wrinkle, K would be 2*pi/N
L controls how much stretching you want to apply
Then simply express your image F from image I with the transforms you have in 1.
Putting it all together, the code below creates a sample image I and generates the image F as follows:
% Generate a sample input image
N=500;
xF=1:N;
I=(1:4)'*xF/N*50;
% Set the parameters for your accordion transform
K=2*pi/N;
L=100;
% Apply the transform
F=I(:, round(min(N*ones(1,N), max(ones(1,N), (xF + L*sin(xF*K))))) );
% Display the input and output images side by side
image(I);
figure;
image(F);
If you run this exact code you get:
As you can see, the final image on the right stretches the center part of the image on the left, giving you an accordion effect with one wrinkle.
You can fiddle with K and L and adjust the formula to get the exact effect you want, but note how by expressing the transform in a matrix form MATLAB executes the code in a fraction of second. If there is one take away for you is that you should stay away from for loops and complex processing whenever you can.
Have fun!