I have obtained the latitude, longitude and height information from a TIFF image into Matlab and plotting it using the surfc function as surfc(X,Y,Z).
surfc output:
Image of the tree:
But how do I get Z to display as a contour (something like a cone) corresponding to the height of an object in the image?
Thanks for any answers
You can change the color mapping of the surf. Example:
clear;clc;close all
tif4 = imread('RzwK3.tif');
THeight = rgb2gray(tif4(:,:,1:3));
imshow(THeight)
x = 1:size(THeight,1); % can be changed to coordinates
y = 1:size(THeight,2);
[X,Y] = ndgrid(x,y);
% make contour color on the surface
M = 4; % # color in contour
Cvec = parula(M); % any Mx3 RGB triplet
hs = surf(X,Y,THeight,'EdgeAlpha',.1);
colormap(Cvec)
colorbar
If parula or other color generation function is not available in your Matlab version, you can assign Cvec manually. Each row of the matrix is a RGB color triplet with values between 0 and 1 (you can divide a web RGB color by 256) and there should be M rows in the matrix. Example: the following is the output of parula(4), which can be manually input by replacing the line of code.
Cvec = [
0.2081 0.1663 0.5292; % R1 G1 B1
0.0265 0.6137 0.8135; % R2 G2 B2
0.6473 0.7456 0.4188; % R3 G3 B3
0.9763 0.9831 0.0538]; %R4 G4 B4
If you have the pionts as cartesian coordinates use plot3
plot3(X,Y,Z,'.');
If X and Y are the independent variables and Z the dependent one (The corresponding height for a given X and Y) then use surf or mesh
surf(X,Y,Z);
mesh(X,Y,Z);
Otherwise put some sample of your data and results so we can better understand your problem.
EDIT: You are actually obtaining a contour plot, but surfc should also plot the surface.
Take a look at the documentation of surfc and you will see that it combines two plots in one figure (the surface and the contour)
[X,Y,Z] = peaks(30);
figure
surfc(X,Y,Z)
Related
Assume I have a 2x2 matrix filled with values which will represent a plane. Now I want to rotate the plane around itself in a 3-D way, in the "z-Direction". For a better understanding, see the following image:
I wondered if this is possible by a simple affine matrix, thus I created the following simple script:
%Create a random value matrix
A = rand*ones(200,200);
%Make a box in the image
A(50:200-50,50:200-50) = 1;
Now I can apply transformations in the 2-D room simply by a rotation matrix like this:
R = affine2d([1 0 0; .5 1 0; 0 0 1])
tform = affine3d(R);
transformed = imwarp(A,tform);
However, this will not produce the desired output above, and I am not quite sure how to create the 2-D affine matrix to create such behavior.
I guess that a 3-D affine matrix can do the trick. However, if I define a 3-D affine matrix I cannot work with the 2-D representation of the matrix anymore, since MATLAB will throw the error:
The number of dimensions of the input image A must be 3 when the
specified geometric transformation is 3-D.
So how can I code the desired output with an affine matrix?
The answer from m3tho correctly addresses how you would apply the transformation you want: using fitgeotrans with a 'projective' transform, thus requiring that you specify 4 control points (i.e. 4 pairs of corresponding points in the input and output image). You can then apply this transform using imwarp.
The issue, then, is how you select these pairs of points to create your desired transformation, which in this case is to create a perspective projection. As shown below, a perspective projection takes into account that a viewing position (i.e. "camera") will have a given view angle defining a conic field of view. The scene is rendered by taking all 3-D points within this cone and projecting them onto the viewing plane, which is the plane located at the camera target which is perpendicular to the line joining the camera and its target.
Let's first assume that your image is lying in the viewing plane and that the corners are described by a normalized reference frame such that they span [-1 1] in each direction. We need to first select the degree of perspective we want by choosing a view angle and then computing the distance between the camera and the viewing plane. A view angle of around 45 degrees can mimic the sense of perspective of normal human sight, so using the corners of the viewing plane to define the edge of the conic field of view, we can compute the camera distance as follows:
camDist = sqrt(2)./tand(viewAngle./2);
Now we can use this to generate a set of control points for the transformation. We first apply a 3-D rotation to the corner points of the viewing plane, rotating around the y axis by an amount theta. This rotates them out of plane, so we now project the corner points back onto the viewing plane by defining a line from the camera through each rotated corner point and finding the point where it intersects the plane. I'm going to spare you the mathematical derivations (you can implement them yourself from the formulas in the above links), but in this case everything simplifies down to the following set of calculations:
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
And outP now contains your normalized set of control points in the output image. Given an image of size s, we can create a set of input and output control points as follows:
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
And you can apply the transformation like so:
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
outputView = imref2d(s);
newImage = imwarp(oldImage, tform, 'OutputView', outputView);
The only issue you may come across is that a rotation of 90 degrees (i.e. looking end-on at the image plane) would create a set of collinear points that would cause fitgeotrans to error out. In such a case, you would technically just want a blank image, because you can't see a 2-D object when looking at it edge-on.
Here's some code illustrating the above transformations by animating a spinning image:
img = imread('peppers.png');
s = size(img);
outputView = imref2d(s);
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
viewAngle = 45;
camDist = sqrt(2)./tand(viewAngle./2);
for theta = linspace(0, 360, 360)
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
spinImage = imwarp(img, tform, 'OutputView', outputView);
if (theta == 0)
hImage = image(spinImage);
set(gca, 'Visible', 'off');
else
set(hImage, 'CData', spinImage);
end
drawnow;
end
And here's the animation:
You can perform a projective transformation that can be estimated using the position of the corners in the first and second image.
originalP='peppers.png';
original = imread(originalP);
imshow(original);
s = size(original);
matchedPoints1 = [1 1;1 s(1);s(2) s(1);s(2) 1];
matchedPoints2 = [1 1;1 s(1);s(2) s(1)-100;s(2) 100];
transformType = 'projective';
tform = fitgeotrans(matchedPoints1,matchedPoints2,'projective');
outputView = imref2d(size(original));
Ir = imwarp(original,tform,'OutputView',outputView);
figure; imshow(Ir);
This is the result of the code above:
Original image:
Transformed image:
I have the 2D coordinates of the edge points [x,y] of the objects in gery image (img). Where x and y are nx1 vectors. The objects have irregular shapes.
My question is how can we crop (or extract the image of the object from the original image (img) using matlab.
The following sample of my code:
Input_Grey = rgb2gray(Input_RGB);
[B,L] = bwboundaries(CC); % where CC is the binary image with the edges of the objects.
boundary = B{1}; % the edge points of the first object.
x = boundary(:, 2);
y = boundary(:, 1);
I m trying to compute an efficient way to transform an image in cartesian coordinates into a polar representation. I know some functions such as ImToPolar are doing it and it works perfectly but takes a considerable much time for big images, especially when they require to be processed back and forth.
Here´s my input image:
and then I generate a polar mesh using a cartesian mesh centered at 0 and the function cart2pol(). Finally, I plot my image using mesh(theta, r, Input).
And here´s what I obtain:
Its exactly the image I need and it´s the same as ImToPolar or maybe better.
Since MATLAB knows how to compute it, does anybody know how to extract a matrix in polar representation from this output? Or maybe a fast (like in fast fourier transform) way to compute a Polar transform (and inverse) on MATLAB?
pol2cart and meshgrid and interp2 are sufficient to create the result:
I=imread('http://i.stack.imgur.com/HYSyb.png');
[r, c,~] = size(I);
%rgb image can be converted to indexed image to prevent excessive copmutation for each color
[idx, mp] = rgb2ind(I,32);
% add offset to image coordinates
x = (1:c)-(c/2);
y = (1:r)-(r/2);
% create distination coordinates in polar form so value of image can be interpolated in those coordinates
% angle ranges from 0 to 2 * pi and radius assumed that ranges from 0 to 400
% linspace(0,2*pi, 200) leads to a stretched image try it!
[xp yp] = meshgrid(linspace(0,2*pi), linspace(0,400));
%translate coordinate from polar to image coordinates
[xx , yy] = pol2cart(xp,yp);
% interpolate pixel values for unknwon coordinates
out = interp2(x, y, idx, xx, yy);
% save the result to a file
imwrite(out, mp, 'result.png')
I'd like to take in an RGB image, find the points in the image that are white, and get the cartesian coordinates of those points in the image. I've gotten most of the way there, but when I try to plot the cartesian coordinates, I get a vertically tiled image (i.e. 5 overlapped copies of what I should see). Anyone know what could be causing this?
,
Code: (JPG comes in as 2448 x x3264 x 3 uint8)
I = imread('IMG_0245.JPG');
imshow(I); % display unaltered image
% Convert image to grayscale
I = rgb2gray(I);
% Convert image to binary (black/white)
I = im2bw(I, 0.9);
% Generate cartesian coordinates of image
imageSize = size(I);
[x, y] = meshgrid( 1:imageSize(1), 1:imageSize(2) );
PerspectiveImage = [x(:), y(:), I(:)];
% Get indices of white points only
whiteIndices = find(PerspectiveImage(:,3));
figure; plot( PerspectiveImage(whiteIndices, 1), PerspectiveImage(whiteIndices, 2),'.');
% Flip vertically to correct indexing vs. plotting issue
axis ij
Very simple. You're declaring your meshgrid wrong. It should be:
[x, y] = meshgrid( 1:imageSize(2), 1:imageSize(1) );
The first parameter denotes the horizontal extents of the 2D grid, and so you want to make this vary for as many columns as you have. Similarly, the second parameter denotes the vertical extents of the 2D grid, and so you want to make this for as many rows as you have.
I had to pre-process some of your image to get some good results because your original image had a large white border surrounding the image. I had to remove this border by removing all pure white pixels. I also read in the image directly from StackOverflow:
I = imread('http://s7.postimg.org/ovb53w4ff/Track_example.jpg');
mask = all(I == 255, 3);
I = bsxfun(#times, I, uint8(~mask));
This was the image I get after doing my pre-processing:
Once I do this and change your meshgrid call, I get this:
I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)