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I'm trying to find a way to generate these pretty correlation plots in MATLAB. These are generated in R using 'corrplot' function, but couldn't find any similar code in MATLAB. Any help would be appreciated.
As a quick description, this function will create a color scale of the correlation values, and create circles in each cell of the correlation matrix/plot with the associated color. The size of the circles is also an indicator of the magnitude of the correlation, with larger circles representing a stronger relationship (positive or negative). More details could be found here.
you can use plot-corrmat (or modify it, depending how articulate you are in matlab), to obtain similar visualizations of correlation matrices (top pic). Or use Correlation circles , that looks somewhat similar as well (bottom pic)...
https://github.com/elayden/plot-corrmat
I could write the below code to generate a similar graph, based on the code provided here
% Produce the input lower triangular matrix data
C = -1 + 2.*rand(12,12);
C = tril(C,-1);
C(logical(eye(size(C)))) = 1;
% Set [min,max] value of C to scale colors
clrLim = [-1,1];
% load('CorrColormap.mat') % Uncomment for custom CorrColormap
% Set the [min,max] of diameter where 1 consumes entire grid square
diamLim = [0.1, 1];
myLabel = {'ICA','Elev','Pr','Rmax','Rmin','Srad','Wspd','Tmin','Tmax','VPD','ET_o','AW'};
% Compute center of each circle
% This assumes the x and y values were not entered in imagesc()
x = 1 : 1 : size(C,2); % x edges
y = 1 : 1 : size(C,1); % y edges
[xAll, yAll] = meshgrid(x,y);
xAll(C==0)=nan; % eliminate cordinates for zero correlations
% Set color of each rectangle
% Set color scale
cmap = jet(256);
% cmap = CorrColormap; % Uncomment for CorrColormap
Cscaled = (C - clrLim(1))/range(clrLim); % always [0:1]
colIdx = discretize(Cscaled,linspace(0,1,size(cmap,1)));
% Set size of each circle
% Scale the size between [0 1]
Cscaled = (abs(C) - 0)/1;
diamSize = Cscaled * range(diamLim) + diamLim(1);
% Create figure
fh = figure();
ax = axes(fh);
hold(ax,'on')
colormap(ax,'jet');
% colormap(CorrColormap) %Uncomment for CorrColormap
tickvalues = 1:length(C);
x = zeros(size(tickvalues));
text(x, tickvalues, myLabel, 'HorizontalAlignment', 'right');
x(:) = length(C)+1;
text(tickvalues, x, myLabel, 'HorizontalAlignment', 'right','Rotation',90);
% Create circles
theta = linspace(0,2*pi,50); % the smaller, the less memory req'd.
h = arrayfun(#(i)fill(diamSize(i)/2 * cos(theta) + xAll(i), ...
diamSize(i)/2 * sin(theta) + yAll(i), cmap(colIdx(i),:),'LineStyle','none'),1:numel(xAll));
axis(ax,'equal')
axis(ax,'tight')
set(ax,'YDir','Reverse')
colorbar()
caxis(clrLim);
axis off
The exact graph is available here:
Fancy Correlation Plots in MATLAB
I would like to convert a image processing program(part of the program below) from Matlab to Simulink and possibly convert the simulink diagram into C code later on. I have 0 experience in Simulink and was wondering if there's any limitations on the types of matlab program/functions that can be converted and how I would go about doing this. Thanks.
clear all
clc
% Read in an image 1
C1 = imread('cloud1.jpg');
Cloud1 = C1(:,:,1); % use only one color
%Cloud1 = Cloud1'; % transpose to get (x,y) instead of (y,x)
Cloud1_xsize = size(Cloud1,2); % get x size of image
Cloud1_ysize = size(Cloud1,1); % get y size of image
%figure(3), imshow(Cloud1) % to plot you need to transpose back to their coordinate system
%hold on
% Read in an image 2
C2 = imread('cloud2.jpg');
Cloud2 = C2(:,:,1); % use only one color
%Cloud2 = Cloud2'; % transpose to get (x,y) instead of (y,x)
Cloud2_xsize = size(Cloud2,2); % get x size of image
Cloud2_ysize = size(Cloud2,1); % get y size of image
%figure(2), imshow(Cloud2)
%hold on
% show the shift in the initial images several times
num = 0;
for k = 1:4
num=num+1;
pause(.5)
figure(1), h1=imshow(C1)
xlabel('FIGURE 1')
F(num) = getframe(gcf);
%image(F.cdata)
%colormap(F.colormap)
pause(0.25)
figure(1), h2=imshow(C2)
xlabel('FIGURE 2')
num=num+1;
F(num) = getframe(gcf);
%image(F.cdata)
%colormap(F.colormap)
end
% Play the movie twenty times
%movie(F,20)
%%%% Set the template size %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% First calc the number of pixels in the shortest direction of the image (usually y direction)
MinSize = min(Cloud1_xsize, Cloud1_ysize); % number of pixels in shortest direction
%%% N is the minimum number of boxes in the shorter direction (usually y direction).
%%% In the shorter axis (usually y)there will be N-2 boxes analyzed.
%%% This is because the top and bottom boxes are considered too close to the edge to use.
%%% In the larger direction (usually x) there may be more boxes.
N = 6;
EdgeBoxSize = 1; % the number of edge boxes along each edge
TempWidth = floor(MinSize / N); % the pixel width of each template box
TempHeight = TempWidth; % make the template height and width the same size so corr part works good
%%% Now calculate the exact number of boxes in x and y directions
%%% This depends on the number of x versus y pixels.
Nx = floor(Cloud1_xsize/TempWidth);
Ny = floor(Cloud1_ysize/TempWidth);
I have an array of pixel data frames for use with VideoWriter. I want to overlay lineseries/contour objects into each frame. I don't want to make the movie by iteratively drawing each frame to a figure and capturing it with getframe, because that gives poor resolution and is slow. I tried using getframe on a plot of just the contour, but that returns images scaled to the wrong size with weird margins, especially when using 'axis equal,' which I need.
Updated to accommodate feedback from OP
Getting the contour data as pixel data is not trivial (if possible at all) since using getframe doesn't yield predictable results
What we can do is to extract the contour data and then overlay it on the pixel data frames, forcing them to be to the same scale and then do a getframe on the resultant merged image. This will at least ensure that they two data sets area aligned.
The following code shows the principle though you'd need to modify it for your own needs:
%% Generate some random contours to use
x = linspace(-2*pi,2*pi);
y = linspace(0,4*pi);
[X,Y] = meshgrid(x,y);
Z = sin(X)+cos(Y);
[~,h] = contour(X,Y,Z);
This yields the following contours
Now we get the handles of the children of this image. These will all be 'patch' type objects
patches = get(h,'Children');
Also get the axis limits for the contours
lims = axis;
Next, create a new figure and render the pixel frame data into it
In this example I'm just loading an image but you get the idea.
%% Render frame data
figure
i = imread( some_image_file_png );
This image is actually 194 x 259 x 3. I can display it and rescale the X and Y axes using
%% Set image axes
image(flipdim(i,1),'XData',[lims(1) lims(2)],'YData',[lims(4) lims(3)]);
Note the use of flipdim() to vertically flip the image since the image Y-axis runs in the opposite sense to the contour Y axis. This gives me:
Now I can plot the contours (patches) form the contour plot over the top of the image in the same coordinate space
%% Plot patches
for p =1:length(patches)
xd = get( patches(p), 'XData' );
yd = get( patches(p), 'YData' );
% This causes all contours to be rendered in white. You may
% want to play with this a little
cd = zeros(size(xd));
patch( xd, yd, cd, 'EdgeColor', 'w');
end
This yields
You can now use getframe to extract the frame. If it's important to have coloured contours, you will need to extract colour data from the original contour map and use it to apply an appropriate colouring in the overlaid image.
As a short cut, it's also possible to compile all patch data into a single MxN matrix and render with a single call to patch but I wrote it this way to demonstrate the process.
Well, here's a Bresenham-esque solution based on the ContourMatrix. Not ideal cuz doesn't handle line width, antialiasing, or any more than a single color. But it's pretty efficient (not quite Bham efficient).
function renderContour
clc
close all
x = randn(100,70);
[c,h] = contour(x,[0 0],'LineColor','r');
axis equal
if ~isnumeric(h.LineColor)
error('not handled')
end
cs = nan(size(c,2),4);
k = 0;
ci = 1;
for i = 1:size(c,2)
if k <= 0
k = c(2,i);
else
if k > 1
cs(ci,:) = reshape(c(:,i+[0 1]),[1 4]);
ci = ci + 1;
end
k = k - 1;
end
end
pix = renderLines(cs(1:ci-1,:),[1 1;fliplr(size(x))],10,h.LineColor);
figure
image(pix)
axis equal
end
function out = renderLines(cs,rect,res,color)
% cs = [x1(:) y1(:) x2(:) y2(:)]
% rect = [x(1) y(1);x(2) y(2)]
% doesnt handle line width, antialiasing, etc
% could do those with imdilate, imfilter, etc.
test = false;
if test
if false
cs = [0 0 5 5; 0 5 2.5 2.5];
rect = [0 0; 10 10];
else
cs = 100 * randn(1000,4);
rect = 200 * randn(2);
end
res = 10;
color = [1 .5 0];
end
out = nan(abs(res * round(diff(fliplr(rect)))));
cs = cs - repmat(min(rect),[size(cs,1) 2]);
d = [cs(:,1) - cs(:,3) cs(:,2) - cs(:,4)];
lens = sqrt(sum(d.^2,2));
for i = 1:size(cs,1)
n = ceil(sqrt(2) * res * lens(i));
if false % equivalent but probably less efficient
pts = linspace(0,1,n);
pts = round(res * (repmat(cs(i,1:2),[length(pts) 1]) - pts' * d(i,:)));
else
pts = round(res * [linspace(cs(i,1),cs(i,3),n);linspace(cs(i,2),cs(i,4),n)]');
end
pts = pts(all(pts > 0 & pts <= repmat(fliplr(size(out)),[size(pts,1) 1]),2),:);
out(sub2ind(size(out),pts(:,2),pts(:,1))) = 1;
end
out = repmat(flipud(out),[1 1 3]) .* repmat(permute(color,[3 1 2]),size(out));
if test
image(out)
axis equal
end
end
I am trying to rotate an image with Matlab without using imrotate function. I actually made it by using transformation matrix.But it is not good enough.The problem is, the rotated image is "sliding".Let me tell you with pictures.
This is my image which I want to rotate:
But when I rotate it ,for example 45 degrees, it becomes this:
I am asking why this is happening.Here is my code,is there any mathematical or programming mistakes about it?
image=torso;
%image padding
[Rows, Cols] = size(image);
Diagonal = sqrt(Rows^2 + Cols^2);
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;
degree=45;
%midpoints
midx=ceil((size(imagepad,1)+1)/2);
midy=ceil((size(imagepad,2)+1)/2);
imagerot=zeros(size(imagepad));
%rotation
for i=1:size(imagepad,1)
for j=1:size(imagepad,2)
x=(i-midx)*cos(degree)-(j-midy)*sin(degree);
y=(i-midx)*sin(degree)+(j-midy)*cos(degree);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1)
imagerot(x,y)=imagepad(i,j); % k degrees rotated image
end
end
end
figure,imagesc(imagerot);
colormap(gray(256));
The reason you have holes in your image is because you are computing the location in imagerot of each pixel in imagepad. You need to do the computation the other way around. That is, for each pixel in imagerot interpolate in imagepad. To do this, you just need to apply the inverse transform, which in the case of a rotation matrix is just the transpose of the matrix (just change the sign on each sin and translate the other way).
Loop over pixels in imagerot:
imagerot=zeros(size(imagepad)); % midx and midy same for both
for i=1:size(imagerot,1)
for j=1:size(imagerot,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y=-(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
imagerot(i,j)=imagepad(x,y); % k degrees rotated image
end
end
end
Also note that your midx and midy need to be calculated with size(imagepad,2) and size(imagepad,1) respectively, since the first dimension refers to the number of rows (height) and the second to width.
NOTE: The same approach applies when you decide to adopt an interpolation scheme other than nearest neighbor, as in Rody's example with linear interpolation.
EDIT: I'm assuming you are using a loop for demonstrative purposes, but in practice there is no need for loops. Here's an example of nearest neighbor interpolation (what you are using), keeping the same size image, but you can modify this to produce a larger image that includes the whole source image:
imagepad = imread('peppers.png');
[nrows ncols nslices] = size(imagepad);
midx=ceil((ncols+1)/2);
midy=ceil((nrows+1)/2);
Mr = [cos(pi/4) sin(pi/4); -sin(pi/4) cos(pi/4)]; % e.g. 45 degree rotation
% rotate about center
[X Y] = meshgrid(1:ncols,1:nrows);
XYt = [X(:)-midx Y(:)-midy]*Mr;
XYt = bsxfun(#plus,XYt,[midx midy]);
xout = round(XYt(:,1)); yout = round(XYt(:,2)); % nearest neighbor!
outbound = yout<1 | yout>nrows | xout<1 | xout>ncols;
zout=repmat(cat(3,1,2,3),nrows,ncols,1); zout=zout(:);
xout(xout<1) = 1; xout(xout>ncols) = ncols;
yout(yout<1) = 1; yout(yout>nrows) = nrows;
xout = repmat(xout,[3 1]); yout = repmat(yout,[3 1]);
imagerot = imagepad(sub2ind(size(imagepad),yout,xout,zout(:))); % lookup
imagerot = reshape(imagerot,size(imagepad));
imagerot(repmat(outbound,[1 1 3])) = 0; % set background value to [0 0 0] (black)
To modify the above to linear interpolation, compute the 4 neighboring pixels to each coordinate in XYt and perform a weighted sum using the fractional components product as the weights. I'll leave that as an exercise, since it would only serve to bloat my answer further beyond the scope of your question. :)
The method you are using (rotate by sampling) is the fastest and simplest, but also the least accurate.
Rotation by area mapping, as given below (this is a good reference), is much better at preserving color.
But: note that this will only work on greyscale/RGB images, but NOT on colormapped images like the one you seem to be using.
image = imread('peppers.png');
figure(1), clf, hold on
subplot(1,2,1)
imshow(image);
degree = 45;
switch mod(degree, 360)
% Special cases
case 0
imagerot = image;
case 90
imagerot = rot90(image);
case 180
imagerot = image(end:-1:1, end:-1:1);
case 270
imagerot = rot90(image(end:-1:1, end:-1:1));
% General rotations
otherwise
% Convert to radians and create transformation matrix
a = degree*pi/180;
R = [+cos(a) +sin(a); -sin(a) +cos(a)];
% Figure out the size of the transformed image
[m,n,p] = size(image);
dest = round( [1 1; 1 n; m 1; m n]*R );
dest = bsxfun(#minus, dest, min(dest)) + 1;
imagerot = zeros([max(dest) p],class(image));
% Map all pixels of the transformed image to the original image
for ii = 1:size(imagerot,1)
for jj = 1:size(imagerot,2)
source = ([ii jj]-dest(1,:))*R.';
if all(source >= 1) && all(source <= [m n])
% Get all 4 surrounding pixels
C = ceil(source);
F = floor(source);
% Compute the relative areas
A = [...
((C(2)-source(2))*(C(1)-source(1))),...
((source(2)-F(2))*(source(1)-F(1)));
((C(2)-source(2))*(source(1)-F(1))),...
((source(2)-F(2))*(C(1)-source(1)))];
% Extract colors and re-scale them relative to area
cols = bsxfun(#times, A, double(image(F(1):C(1),F(2):C(2),:)));
% Assign
imagerot(ii,jj,:) = sum(sum(cols),2);
end
end
end
end
subplot(1,2,2)
imshow(imagerot);
Output:
Rotates colored image according to angle given by user without any cropping of image in matlab.
Output of this program is similar to output of inbuilt command "imrotate" .This program dynamically creates background according to angle input given by user.By using rotation matrix and origin shifting, we get relation between coordinates of initial and final image.Using relation between coordinates of initial and final image, we now map the intensity values for each pixel.
img=imread('img.jpg');
[rowsi,colsi,z]= size(img);
angle=45;
rads=2*pi*angle/360;
%calculating array dimesions such that rotated image gets fit in it exactly.
% we are using absolute so that we get positve value in any case ie.,any quadrant.
rowsf=ceil(rowsi*abs(cos(rads))+colsi*abs(sin(rads)));
colsf=ceil(rowsi*abs(sin(rads))+colsi*abs(cos(rads)));
% define an array withcalculated dimensionsand fill the array with zeros ie.,black
C=uint8(zeros([rowsf colsf 3 ]));
%calculating center of original and final image
xo=ceil(rowsi/2);
yo=ceil(colsi/2);
midx=ceil((size(C,1))/2);
midy=ceil((size(C,2))/2);
% in this loop we calculate corresponding coordinates of pixel of A
% for each pixel of C, and its intensity will be assigned after checking
% weather it lie in the bound of A (original image)
for i=1:size(C,1)
for j=1:size(C,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y= -(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+xo;
y=round(y)+yo;
if (x>=1 && y>=1 && x<=size(img,1) && y<=size(img,2) )
C(i,j,:)=img(x,y,:);
end
end
end
imshow(C);
Check this out.
this is fastest way that you can do.
img = imread('Koala.jpg');
theta = pi/10;
rmat = [
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1];
mx = size(img,2);
my = size(img,1);
corners = [
0 0 1
mx 0 1
0 my 1
mx my 1];
new_c = corners*rmat;
T = maketform('affine', rmat); %# represents translation
img2 = imtransform(img, T, ...
'XData',[min(new_c(:,1)) max(new_c(:,1))],...
'YData',[min(new_c(:,2)) max(new_c(:,2))]);
subplot(121), imshow(img);
subplot(122), imshow(img2);
I am trying to rotate an image with Matlab without using imrotate function. I actually made it by using transformation matrix.But it is not good enough.The problem is, the rotated image is "sliding".Let me tell you with pictures.
This is my image which I want to rotate:
But when I rotate it ,for example 45 degrees, it becomes this:
I am asking why this is happening.Here is my code,is there any mathematical or programming mistakes about it?
image=torso;
%image padding
[Rows, Cols] = size(image);
Diagonal = sqrt(Rows^2 + Cols^2);
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;
degree=45;
%midpoints
midx=ceil((size(imagepad,1)+1)/2);
midy=ceil((size(imagepad,2)+1)/2);
imagerot=zeros(size(imagepad));
%rotation
for i=1:size(imagepad,1)
for j=1:size(imagepad,2)
x=(i-midx)*cos(degree)-(j-midy)*sin(degree);
y=(i-midx)*sin(degree)+(j-midy)*cos(degree);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1)
imagerot(x,y)=imagepad(i,j); % k degrees rotated image
end
end
end
figure,imagesc(imagerot);
colormap(gray(256));
The reason you have holes in your image is because you are computing the location in imagerot of each pixel in imagepad. You need to do the computation the other way around. That is, for each pixel in imagerot interpolate in imagepad. To do this, you just need to apply the inverse transform, which in the case of a rotation matrix is just the transpose of the matrix (just change the sign on each sin and translate the other way).
Loop over pixels in imagerot:
imagerot=zeros(size(imagepad)); % midx and midy same for both
for i=1:size(imagerot,1)
for j=1:size(imagerot,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y=-(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
imagerot(i,j)=imagepad(x,y); % k degrees rotated image
end
end
end
Also note that your midx and midy need to be calculated with size(imagepad,2) and size(imagepad,1) respectively, since the first dimension refers to the number of rows (height) and the second to width.
NOTE: The same approach applies when you decide to adopt an interpolation scheme other than nearest neighbor, as in Rody's example with linear interpolation.
EDIT: I'm assuming you are using a loop for demonstrative purposes, but in practice there is no need for loops. Here's an example of nearest neighbor interpolation (what you are using), keeping the same size image, but you can modify this to produce a larger image that includes the whole source image:
imagepad = imread('peppers.png');
[nrows ncols nslices] = size(imagepad);
midx=ceil((ncols+1)/2);
midy=ceil((nrows+1)/2);
Mr = [cos(pi/4) sin(pi/4); -sin(pi/4) cos(pi/4)]; % e.g. 45 degree rotation
% rotate about center
[X Y] = meshgrid(1:ncols,1:nrows);
XYt = [X(:)-midx Y(:)-midy]*Mr;
XYt = bsxfun(#plus,XYt,[midx midy]);
xout = round(XYt(:,1)); yout = round(XYt(:,2)); % nearest neighbor!
outbound = yout<1 | yout>nrows | xout<1 | xout>ncols;
zout=repmat(cat(3,1,2,3),nrows,ncols,1); zout=zout(:);
xout(xout<1) = 1; xout(xout>ncols) = ncols;
yout(yout<1) = 1; yout(yout>nrows) = nrows;
xout = repmat(xout,[3 1]); yout = repmat(yout,[3 1]);
imagerot = imagepad(sub2ind(size(imagepad),yout,xout,zout(:))); % lookup
imagerot = reshape(imagerot,size(imagepad));
imagerot(repmat(outbound,[1 1 3])) = 0; % set background value to [0 0 0] (black)
To modify the above to linear interpolation, compute the 4 neighboring pixels to each coordinate in XYt and perform a weighted sum using the fractional components product as the weights. I'll leave that as an exercise, since it would only serve to bloat my answer further beyond the scope of your question. :)
The method you are using (rotate by sampling) is the fastest and simplest, but also the least accurate.
Rotation by area mapping, as given below (this is a good reference), is much better at preserving color.
But: note that this will only work on greyscale/RGB images, but NOT on colormapped images like the one you seem to be using.
image = imread('peppers.png');
figure(1), clf, hold on
subplot(1,2,1)
imshow(image);
degree = 45;
switch mod(degree, 360)
% Special cases
case 0
imagerot = image;
case 90
imagerot = rot90(image);
case 180
imagerot = image(end:-1:1, end:-1:1);
case 270
imagerot = rot90(image(end:-1:1, end:-1:1));
% General rotations
otherwise
% Convert to radians and create transformation matrix
a = degree*pi/180;
R = [+cos(a) +sin(a); -sin(a) +cos(a)];
% Figure out the size of the transformed image
[m,n,p] = size(image);
dest = round( [1 1; 1 n; m 1; m n]*R );
dest = bsxfun(#minus, dest, min(dest)) + 1;
imagerot = zeros([max(dest) p],class(image));
% Map all pixels of the transformed image to the original image
for ii = 1:size(imagerot,1)
for jj = 1:size(imagerot,2)
source = ([ii jj]-dest(1,:))*R.';
if all(source >= 1) && all(source <= [m n])
% Get all 4 surrounding pixels
C = ceil(source);
F = floor(source);
% Compute the relative areas
A = [...
((C(2)-source(2))*(C(1)-source(1))),...
((source(2)-F(2))*(source(1)-F(1)));
((C(2)-source(2))*(source(1)-F(1))),...
((source(2)-F(2))*(C(1)-source(1)))];
% Extract colors and re-scale them relative to area
cols = bsxfun(#times, A, double(image(F(1):C(1),F(2):C(2),:)));
% Assign
imagerot(ii,jj,:) = sum(sum(cols),2);
end
end
end
end
subplot(1,2,2)
imshow(imagerot);
Output:
Rotates colored image according to angle given by user without any cropping of image in matlab.
Output of this program is similar to output of inbuilt command "imrotate" .This program dynamically creates background according to angle input given by user.By using rotation matrix and origin shifting, we get relation between coordinates of initial and final image.Using relation between coordinates of initial and final image, we now map the intensity values for each pixel.
img=imread('img.jpg');
[rowsi,colsi,z]= size(img);
angle=45;
rads=2*pi*angle/360;
%calculating array dimesions such that rotated image gets fit in it exactly.
% we are using absolute so that we get positve value in any case ie.,any quadrant.
rowsf=ceil(rowsi*abs(cos(rads))+colsi*abs(sin(rads)));
colsf=ceil(rowsi*abs(sin(rads))+colsi*abs(cos(rads)));
% define an array withcalculated dimensionsand fill the array with zeros ie.,black
C=uint8(zeros([rowsf colsf 3 ]));
%calculating center of original and final image
xo=ceil(rowsi/2);
yo=ceil(colsi/2);
midx=ceil((size(C,1))/2);
midy=ceil((size(C,2))/2);
% in this loop we calculate corresponding coordinates of pixel of A
% for each pixel of C, and its intensity will be assigned after checking
% weather it lie in the bound of A (original image)
for i=1:size(C,1)
for j=1:size(C,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y= -(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+xo;
y=round(y)+yo;
if (x>=1 && y>=1 && x<=size(img,1) && y<=size(img,2) )
C(i,j,:)=img(x,y,:);
end
end
end
imshow(C);
Check this out.
this is fastest way that you can do.
img = imread('Koala.jpg');
theta = pi/10;
rmat = [
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1];
mx = size(img,2);
my = size(img,1);
corners = [
0 0 1
mx 0 1
0 my 1
mx my 1];
new_c = corners*rmat;
T = maketform('affine', rmat); %# represents translation
img2 = imtransform(img, T, ...
'XData',[min(new_c(:,1)) max(new_c(:,1))],...
'YData',[min(new_c(:,2)) max(new_c(:,2))]);
subplot(121), imshow(img);
subplot(122), imshow(img2);