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I have 2 greyscale images that i am trying to align using scalar scaling 1 , rotation matrix [2,2] and translation vector [2,1]. I can calculate image1's transformed coordinates as
y = s*R*x + t;
Below the resulting images are shown.
The first image is image1 before transformation,
the second image is image1 (red) with attempted interpolation using interp2 shown on top of image2 (green)
The third image is when i manually insert the pixel values from image1 into an empty array (that has the same size as image2) using the transformed coordinates.
From this we can see that the coordinate transformation must have been successful, as the images are aligned although not perfectly (which is to be expected since only 2 coordinates were used in calculating s, R and t) .
How come interp2 is not producing a result more similar to when i manually insert pixel values?
Below the code for doing this is included:
Interpolation code
function [transformed_image] = interpolate_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
% doesn't help if i use get_grid that the other function is using here
[~, grid_xr, grid_yr] = get_ipgrid(im_r);
[x_t, grid_xt, grid_yt] = get_ipgrid(im_t);
y = s*R*x_t + t;
yx = reshape(y(1,:), m,n);
yy = reshape(y(2,:), m,n);
transformed_image = interp2(grid_xr, grid_yr, im_r, yx, yy, 'nearest');
end
function [x, grid_x, grid_y] = get_ipgrid(image)
[m,n] = size(image);
[grid_x,grid_y] = meshgrid(1:n,1:m);
x = [reshape(grid_x, 1, []); reshape(grid_y, 1, [])]; % X is [2xM*N] coordinate pairs
end
The manual code
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(round(yx(i,j)),round(yy(i,j))) = im_r(i,j);
end
end
end
end
function [x, grid_x, grid_y] = get_grid(image)
[m,n] = size(image);
[grid_y,grid_x] = meshgrid(1:n,1:m);
x = [grid_x(:) grid_y(:)]'; % X is [2xM*N] coordinate pairs
end
Can anyone see what i'm doing wrong with interp2? I feel like i have tried everything
Turns out i got interpolation all wrong.
In my question i calculate the coordinates of im1 in im2.
However the way interpolation works is that i need to calculate the coordinates of im2 in im1 such that i can map the image as shown below.
This means that i also calculated the wrong s,R and t since they were used to transform im1 -> im2, where as i needed im2 -> im1. (this is also called the inverse transform). Below is the manual code, that is basically the same as interp2 with nearest neighbour interpolation
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(i,j) = im_r(round(yx(i,j)),round(yy(i,j)));
end
end
end
end
I am able to plot a point using the x and y coordinates using the following code.
figure(1);
plot(x(1),y(1),'o');
h_compound = gce();
h_compound.children.mark_size = 20;
h_compound.children.mark_background = 2;
h_axes = gca();
h_axes.data_bounds = [0,0;100,100];
My program contains a loop which keeps on refreshing the coordinate values. Every time the loop is executed the point is plotted in the same graphic, such that the new points overlap the older ones. How do I make the old points disappear as the new points are plotted so that an animation-like sequence is generated?
scf(1);clf;
x=linspace(0,10,100);
y=sin(x);
plot(x(1),y(1),"o")
h_compound = gce();
h_point=h_compound.children
h_point.mark_size = 20;
h_point.mark_background = 2;
h_axes = gca();
h_axes.data_bounds = [0,-1;10,1];
realtimeinit(0.1);
for i=1:100
realtime(i);//wait 0.1 second before drawing the new position
h_point.data=[x(i),y(i)];
end
I want to find the size of an image (height, width, and area) and save the size value in txt file. I tried this command:
img = imread('TN4.jpg');
abu=rgb2gray(img);
imwrite(abu,'abu.jpg');
cb=imclearborder(abu);
imwrite(cb,'cb.jpg')
thresh=graythresh(cb);
b=im2bw(cb,thresh);
imwrite(b,'b.jpg');
bw=bwareaopen(b,100);
bwfill=imfill(bw,'holes');
imwrite(bwfill,'bw.jpg');
s = regionprops(bwfill,'BoundingBox');
out = bsxfun(#times,img,uint8(bwfill));
objects=cell(numel(s),1);
for idx = 1:numel(s)
bb=floor(s(idx).BoundingBox);
objects{idx} = out(bb(2):bb(2)+bb(4), bb(1):bb(1)+bb(3),:);
end
k1=objects{1};
c1=rgb2gray(k1);
t1=graythresh(c1);
biner1=im2bw(c1,t1);
[height, width] = size(biner1);
a1 = bwarea(biner1);
h1=height(biner1);
w1=width(biner1);
X = [(h1);(w1);(a1)];
save datagrading.txt X -ascii
but i got warning massage
??? Index exceeds matrix dimensions.
Error in ==> exercise at 35
h1=height(biner1);
can anyone help me?
Thank you
height and width are scalars yet (it looks like to me...) you are assuming that they are functions.... no they're not functions. Just use height and width on its own and place it into your X matrix:
[height, width] = size(biner1);
a1 = bwarea(biner1);
%h1=height(biner1); %// Remove
%w1=width(biner1); %// Remove
X = [height;width;a1]; %// Change
save datagrading.txt X -ascii
Since you have multiple objects and you want to save the height, width and area of these objects, just loop over each one, find the quantities you desire, place them into an overall matrix and save it:
X = zeros(3, numel(objects));
for k = 1 : numel(objects)
k1=objects{k}; %// Change
c1=rgb2gray(k1);
t1=graythresh(c1);
biner1=im2bw(c1,t1);
[height, width] = size(biner1);
a1 = bwarea(biner1);
X(:,k) = [height;width;a1];
end
save datagrading.txt X -ascii
X is now a 3 x s matrix where s is the total number of objects you have. Therefore, each column contains the height, width and area of each object.
You declared
[height, width] = size(biner1);
So height(biner1) will refer to the biner1-th element of height, which is not existent.
Use X = [(height);(width);(a1)]; instead.
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 am new to d3, learning a lot. I have an issue I cannot find an example for:
I have two y axes with positive and negative values with vastly different domains, one being large dollar amounts the other being percentages.
The resulting graph from cobbling together examples looks really awesome with one slight detail, the zero line for each y axis is in a slightly different position. Does anyone know of a way in d3 to get the zero line to be at the same x position?
I would like these two yScales/axes to share the same zero line
// define yScale
var yScale = d3.scale.linear()
.range([height, 0])
.domain(d3.extent(dataset, function(d) { return d.value_di1; }))
;
// define y2 scale
var yScale2 = d3.scale.linear()
.range([height, 0])
.domain(d3.extent(dataset, function(d) { return d.calc_di1_di2_percent; }))
;
Here is a link to a jsfiddle with sample data:
http://jsfiddle.net/jglover/XvBs3/1/
(the x-axis ticks look horrible in the jsfiddle example)
In general, there's unfortunately no way to do this neatly. D3 doesn't really have a concept of several things lining up and therefore no means of accomplishing it.
In your particular case however, you can fix it quite easily by tweaking the domain of the second y axis:
.domain([d3.min(dataset, function(d) { return d.calc_di1_di2_percent; }), 0.7])
Complete example here.
To make the 0 level the same position, a strategy is to equalize the length/proportion of the y axes.
Here are the concepts to the solution below:
The alignment of baseline depends on the length of the y axes.
To let all value shown in the bar, we need to extend the shorter side of the dimension, which compares to the other, to make the proportion of the two axes equal.
example:
// dummy data
const y1List = [-1000, 120, -130, 1400],
y2List = [-0.1, 0.2, 0.3, -0.4];
// get proportion of the two y axes
const totalY1Length = Math.abs(d3.min(y1List)) + Math.abs(d3.max(y1List)),
totalY2Length = Math.abs(d3.min(y2List)) + Math.abs(d3.max(y2List)),
maxY1ToY2 = totalY2Length * d3.max(y1List) / totalY1Length,
minY1ToY2 = totalY2Length * d3.min(y1List) / totalY1Length,
maxY2ToY1 = totalY1Length * d3.max(y2List) / totalY2Length,
minY2ToY1 = totalY1Length * d3.min(y2List) / totalY2Length;
// extend the shorter side of the upper dimension with corresponding value
let maxY1Domain = d3.max(y1List),
maxY2Domain = d3.max(y2List);
if (maxY1ToY2 > d3.max(y2List)) {
maxY2Domain = d3.max(y2List) + maxY1ToY2 - d3.max(y2List);
} else {
maxY1Domain = d3.max(y1List) + maxY2ToY1 - d3.max(y1List);
}
// extend the shorter side of the lower dimension with corresponding value
let minY1Domain = d3.min(y1List),
minY2Domain = d3.min(y2List);
if (minY1ToY2 < d3.min(y2List)) {
minY2Domain = d3.min(y2List) + minY1ToY2 - d3.min(y2List);
} else {
minY1Domain = d3.min(y1List) + minY2ToY1 - d3.min(y1List);
}
// finally, we get the domains for our two y axes
const y1Domain = [minY1Domain, maxY1Domain],
y2Domain = [minY2Domain, maxY2Domain];