Flattening SVG matrix transforms in Inkscape - matrix

I have a free clip art SVG file originally created in Inkscape which I'm making modifications to for use in a Windows 8 JavaScript game. It contains numerous instances of a path with a matrix transform applied on a surrounding group, like this:
<g transform="matrix(0.443,0.896,-0.896,0.443,589.739,-373.223)">
<path d="M486,313s27-9,43-29l26,4,1,23-22,5s-25-6-48-3z" />
</g>
I want to flatten that transform by applying it in advance to the path in Inkscape, to reduce browser work during animation. However when I plug the 6 matrix values into the A B C D E F parameters in Inkscape and apply it, it gives the path a completely different rotation and scaling to what the IE10 engine does.
I have checked numerous times that I have the 6 values mapped correctly. What am I doing wrong?
EDIT: OK, here are before and after screenshots from IE10 and Inkscape. For the IE10 case, the SVG resides directly inside the body of an otherwise empty HTML document (the rendering is exactly the same in Firefox). In Inkscape, I simply opened the "before" SVG file which contains only the path element, selected the path, and plugged in the 6 matrix transform values into Object > Transform > Matrix.
I know very little about matrices, I just want to be able to pre-apply these transformations in the same way the browser does, and ideally to understand why there is a difference in Inkscape. Thanks.

Short answer
When typing the transformation matrix params in Inkscape, make sure you have "Edit current matrix" checked, since if you apply a new transformation matrix to an object, you're actually multiplying this new matrix with the existing transformation matrix of the object, so make sure you edit it instead.
Long Answer
How to recalculate everything yourself.
First let us try and understand the transformation matrices a bit.
A transformation matrix is a quick and clever tool for applying affine transformations ( transformation which preserves straight lines) to a vector.
So, if you have a vector (say, 2d coordinates) and a transformation matrix, and multiply the two together, you will end up with transformed coordinates, with the transformations defined in the transformation matrix, applied.
Calculating x' and y' is done like so:
x' = a*x + c*y + e
y' = b*x + d*y + f
Next, we need to understand the svg format a bit.
According to the w3c svg spec the matrix transform takes exactly those 6 parameters (a,b,c,d,e,f) as arguments.
Therefore, from your example,
<g transform="matrix(0.443,0.896,-0.896,0.443,589.739,-373.223)">
we have the following transformation matrix params:
a=0.443
b=0.896
c=-0.896
d=0.443
e=589.739
f=-373.223
Now, if we have the following example coordinate: x=27, y=-9, we can transform it, by using the previously defined transformation matrix like this:
x' = a*x + c*y + e
x' = 0.443*27 + -0.896*-9 + 589.739
x' = 609.764
y' = b*x + d*y + f
y' = 0.896*27 + 0.443*-9 -373.223
y' = −353.018
Neat, huh? You can get more info here
But that is not all. We also need to understand svg path data.
According to the w3c svg path dspecification each letter in the path data represents an instruction. And each of the number pairs that follow an instruction represent a coordinate value.
From your example, we have the following path:
<path d="M486,313s27-9,43-29l26,4,1,23-22,5s-25-6-48-3z" />
Here we see that this path object uses one absolute moveto instruction (uppercase M), a relative smooth curveto cubic Bézier curve (lowercase s), a relative lineto instruction (lowercase l), and another relative smooth curveto cubic Bézier curve instruction, followed by a closepath instruction (lowercase z).
M486,313 is translated to absolute moveto x=486, y=313
s27-9,43-29 is a bit more complicated to read because some comas are omitted because they're not needed if the negative number is negative, so the minus sign acts as a coma - anyways, it translates to relative smooth bezier curveto x=27, y=-9, x=43, y=-29 (one destination point and one control point)
And so on.
So, how do we apply and remove the transformation matrix from your svg group? Like so:
// we read the transformation matrix params
// <g transform="matrix(0.443,0.896,-0.896,0.443,589.739,-373.223)">
a=0.443
b=0.896
c=-0.896
d=0.443
e=589.739
f=-373.223
// we read the path data, and transform each instruction
// <path d="M486,313s27-9,43-29l26,4,1,23-22,5s-25-6-48-3z" />
M486,313 Absolute move to
x' = a*x + c*y + e = a*486 + c*313 + e = 524.589
y' = b*x + d*y + f = b*486 + d*313 + f = 200.892
Move to instruction is now M524.589,200.892
S27-9,43-29 - smooth curveto, repeat the same process for each coordinate, but set the e and f (translation parameters) to 0, since it's a relative instruction not an absolute.
It is now
s20.025,20.205,45.033,25.680999999999997
l26,4,1,23-22,5
will become
l7.934000000000001,25.067999999999998,-20.165,11.085,-14.226,-17.497
s-25-6-48-3
will become
s-5.698999999999999,-25.058000000000003,-18.576,-44.337
And z will remain z
So the resulting transformed path will be:
<path d="M524.589,200.892s20.025,20.205,45.033,25.680999999999997l7.934000000000001,25.067999999999998,-20.165,11.085,-14.226,-17.497s-5.698999999999999,-25.058000000000003,-18.576,-44.337z" />
I hope this makes sense to you.

You can bake the coords selecting the path then using Path -> Union (CTRL++).
Hope this helps

Paste In Place can help you:
Double click the group in Inkscape, to enter it.
Select all the contents of the group by pressing Ctrl+A, and copy them with Ctrl+C.
Double click outside the group to leave the group.
Edit > Paste In Place (Ctrl+Alt+V) – at this point, group transformations are applied to the obects you paste.
Group the objects again (Ctrl+G)
Move the new group to the same depth as the original, and delete the original group. (This is probably easier with the XML editor, Ctrl+Shift+X.)

Following the answer of #andraaspar, you can also try ungrouping (Ctrl-U) and grouping again (Ctrl-G). It worked for me.

Thanks ArtBIT for all the info ! I had some issues about this on a PHP app, and wrote a library that manipulates font data (from SVG file) and does any sort of transformation on it. Anyone else interested may give it a try from the GitHub :
https://github.com/kartsims/easysvg
Usage example :
require 'easySVG.php';
$svg = new EasySVG();
$svg->setFont("paris-bold-webfont.svg", 100, "#000000");
$svg->addText("Simple text display");
$svg->addAttribute("width", "800px");
$svg->addAttribute("height", "100px");
echo $svg->asXML();
SVG data manipulation example :
$def = 'YOUR SVG DEFINITION HERE';
$easySVG = new EasySVG();
// rotate by 40°
$rotated_def = $easySVG->defRotate($def, 40)
// rotate by 40° with center at (200,100)
$rotated_def2 = $easySVG->defRotate($def, 40, 200, 100)
// scale transform : width*4
$scaled_def = $easySVG->defScale($def, 4)

Related

Fitting of a sphere using SVD/LMS

I would like to fit a MR binary data of 281*398*104 matrix which is not a perfect sphere, and find out the center and radius of sphere and error also. I know LMS or SVD is a good choice to fit for sphere.
I have tried sphereFit from matlab file exchange but got an error,
>> sphereFit(data)
Warning: Matrix is singular to working precision.
> In sphereFit at 33
ans =
NaN NaN NaN
Would you let me know where is the problem, or any others solution?
If you want to use sphere fitting algorithm you should first extract the boundary points of the object you assume to be a sphere. The result should be represented by a N-by-3 array containing coordinates of the points. Then you can apply sphereFit function.
In order to obtain boundary point of a binary object, there are several methods. One method is to apply morphological erosion (you need the "imerode" function from the image processing toolbox) with small structuring element, then compute set difference between the two images, and finally use the "find" function to transform binary image into a coordinate array.
the idea is as follow:
dataIn = imerode(data, ones([3 3 3]));
bnd = data & ~data2;
inds = find(bnd);
[y, x, z] = ind2sub(size(data), inds); % be careful about x y order
points = [x y z];
sphere = sphereFitting(points);
By the way, the link you gave refers to circle fitting, I suppose you wanted to point to a sphere fitting submission?
regards,

Need to automatically eliminate noise in image and outer boundary of a object

i am mechanical engineering student working on a project to automatically detect the weld seam (The seam is a edge that is to be welded) present in a workshop. This gives a basic terminology involved in welding (http://i.imgur.com/Hfwjq0w.jpg).
To separate the weldment from the other objects, i have taken the background image and subtracted the foreground image having the weldment to obatin only the weldment(http://i.imgur.com/v7yBWs1.jpg). After image subtraction,there are the shadow ,glare and remnant noises of subtracted background are still present.
As i want to automatically identify only the weld seam without the outer boundary of weldment, i have tried to detect the edges in the weldment image using canny algorithm and tried to eliminate the isolated noises using the function bwareopen.I have somehow obtained the approximate boundary of weldment and weld seam. The threshold i have used are purely on trial and error approach as dont know a way to automatically set a threshold to detect them.
The problem now i am facing is that i cant specify an definite threshold as this algorithm should be able to identify the seam of any material regardless of its surface texture,glare and shadow present there. I need some assistance to remove the glare,shadow and isolated points from the background subtracted image.
Also i need help to get rid of the outer boundary and obtain only smooth weld seam from starting point to end point.
i have tried to use the following code:
a=imread('imageofworkpiece.jpg'); %http://i.imgur.com/3ngu235.jpg
b=imread('background.jpg'); %http://i.imgur.com/DrF6wC2.jpg
Ip = imsubtract(b,a);
imshow(Ip) % weldment separated %http://i.imgur.com/v7yBWs1.jpg
BW = rgb2gray(Ip);
c=edge(BW,'canny',0.05); % by trial and error
figure;imshow(c) % %http://i.imgur.com/1UQ8E3D.jpg
bw = bwareaopen(c, 100); % by trial and error
figure;imshow(bw) %http://i.imgur.com/Gnjy2aS.jpg
Can anybody please suggest me a adaptive way to set a threhold and remove the outer boundary to detect only the seam? Thank you
Well this doesn't solve your problem of finding an automatic thresholding algorithm. but I can help with isolation the seam. The seam is along the y axis (will this always be the case?) so I used hough transform to isolate only near vertical lines. Normally it finds all lines but I restricted the theta search parameter. The code I'm using now happens to highlight the longest line segment (I got it directly from the matlab website) and it is coincidentally the weld seam. This was purely coincidental. But using your bwareaopened image as input the hough line detector is able to find the seam. Of course it required a bit of playing around to work, so you are stuck at your original problem of finding optimal settings somehow
Maybe this can be a springboard for someone else
a=imread('weldment.jpg'); %http://i.imgur.com/3ngu235.jpg
b=imread('weld_bg.jpg'); %http://i.imgur.com/DrF6wC2.jpg
Ip = imsubtract(b,a);
imshow(Ip) % weldment separated %http://i.imgur.com/v7yBWs1.jpg
BW = rgb2gray(Ip);
c=edge(BW,'canny',0.05); % by trial and error
bw = bwareaopen(c, 100); % by trial and error
figure(1);imshow(c) ;title('canny') % %http://i.imgur.com/1UQ8E3D.jpg
figure(2);imshow(bw);title('bw area open') %http://i.imgur.com/Gnjy2aS.jpg
[H,T,R] = hough(bw,'RhoResolution',1,'Theta',-15:5:15);
figure(3)
imshow(H,[],'XData',T,'YData',R,...
'InitialMagnification','fit');
xlabel('\theta'), ylabel('\rho');
axis on, axis normal, hold on;
P = houghpeaks(H,5,'threshold',ceil(0.5*max(H(:))));
x = T(P(:,2)); y = R(P(:,1));
plot(x,y,'s','color','white');
% Find lines and plot them
lines = houghlines(BW,T,R,P,'FillGap',2,'MinLength',30);
figure(4), imshow(BW), hold on
max_len = 0;
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','green');
% Plot beginnings and ends of lines
plot(xy(1,1),xy(1,2),'x','LineWidth',2,'Color','yellow');
plot(xy(2,1),xy(2,2),'x','LineWidth',2,'Color','red');
% Determine the endpoints of the longest line segment
len = norm(lines(k).point1 - lines(k).point2);
if ( len > max_len)
max_len = len;
xy_long = xy;
end
end
% highlight the longest line segment
plot(xy_long(:,1),xy_long(:,2),'LineWidth',2,'Color','blue');
from your image it looks like the weld seam will be usually very dark with sharp intensity edge so why don't you use that ?
do not use background
create derivation image
dx[y][x]=pixel[y][x]-pixel[y][x-1]
do this for whole image (if on place then x must decrease in loop!!!)
filter out all derivations lower then thresholds
if (|dx[y][x]|<threshold) dx[y][x]=0; else pixel[y][x]=255;` // or what ever values you use
how to obtain threshold value ?
compute min and max intensity and set threshold as (max-min)*scale where scale is value lower then 1.0 (start with 0.02 or 0.1 for example ...
do this also for y axis
so compute dy[][]... and combine dx[][] and dy[][] together. Either with OR or by AND logical functions
filter out artifacts
you can use morphologic filters or smooth threshold for this. After all this you will have mask of pixels of weld seam
if you need boundig box then just loop through all pixels and remember min,max x,y coords ...
[Notes]
if your images will have good lighting then you can ignore the derivation and threshold the intensity directly with something like:
threshold = 0.5*(average_intensity+lowest_intensity)
if you want really fully automate this then you have to use adaptive thresholds. So try more thresholds in a loop and remember result closest to desired output based on geometry size,position etc ...
[edit1] finally have some time/mood for this so
Intensity image threshold
you provided just single image which is far from enough to make reliable algorithm. This is the result
as you can see without further processing this is not good approach
Derivation image threshold
threshold derivation by x (10%)
threshold derivation by y (5%)
AND combination of both 10% di/dx and 1.5% di/dy
The code in C++ looks like this (sorry do not use Matlab):
int x,y,i,i0,i1,tr2,tr3;
pic1=pic0; // copy input image pic0 to pic1
pic2=pic0; // copy input image pic0 to pic2 (just to resize to desired size for derivation)
pic3=pic0; // copy input image pic0 to pic3 (just to resize to desired size for derivation)
pic1.rgb2i(); // RGB -> grayscale
// abs derivate by x
for (y=pic1.ys-1;y>0;y--)
for (x=pic1.xs-1;x>0;x--)
{
i0=pic1.p[y][x ].dd;
i1=pic1.p[y][x-1].dd;
i=i0-i1; if (i<0) i=-i;
pic2.p[y][x].dd=i;
}
// compute min,max derivation
i0=pic2.p[1][1].dd; i1=i0;
for (y=1;y<pic1.ys;y++)
for (x=1;x<pic1.xs;x++)
{
i=pic2.p[y][x].dd;
if (i0>i) i0=i;
if (i1<i) i1=i;
}
tr2=i0+((i1-i0)*100/1000);
// abs derivate by y
for (y=pic1.ys-1;y>0;y--)
for (x=pic1.xs-1;x>0;x--)
{
i0=pic1.p[y ][x].dd;
i1=pic1.p[y-1][x].dd;
i=i0-i1; if (i<0) i=-i;
pic3.p[y][x].dd=i;
}
// compute min,max derivation
i0=pic3.p[1][1].dd; i1=i0;
for (y=1;y<pic1.ys;y++)
for (x=1;x<pic1.xs;x++)
{
i=pic3.p[y][x].dd;
if (i0>i) i0=i;
if (i1<i) i1=i;
}
tr3=i0+((i1-i0)*15/1000);
// threshold the derivation images and combine them
for (y=1;y<pic1.ys;y++)
for (x=1;x<pic1.xs;x++)
{
// copy original (pic0) pixel for non thresholded areas the rest fill with green color
if ((pic2.p[y][x].dd>=tr2)&&(pic3.p[y][x].dd>=tr3)) i=0x00FF00;
else i=pic0.p[y][x].dd;
pic1.p[y][x].dd=i;
}
pic0 is input image
pic1 is output image
pic2,pic3 are just temporary storage for derivations
pic?.xy,pic?.ys is the size of pic?
pic.p[y][x].dd is pixel axes (dd means access pixel as DWORD ...)
as you can see there is a lot of stuff around (nod visible in the first image you provided) so you need to process this further
segmentate and separate...,
use hough transform ...
filter out small artifacts ...
identify object by expected geometry properties (aspect ratio,position,size)
Adaptive thresholds:
you need for this to know the desired output image properties (not possible to reliably deduce from single image input) then create function that do the above processing with variable tr2,tr3. Try in loop more options of tr2,tr3 (loop through all values or iterate to better results and remember the best output (so you also need some function that detects the quality of output) for example:
quality=0.0; param=0.0;
for (a=0.2;a<=0.8;a+=0.1)
{
pic1=process_image(pic0,a);
q=detect_quality(pic1);
if (q>quality) { quality=q; param=a; pico=pic1; }
}
after this the pic1 should hold the relatively best threshold image ... You should process like this all threshold separately inside the process_image the targeted threshold must be scaled by a for example tr2=i0+((i1-i0)*a);

Resize an image with bilinear interpolation without imresize

I've found some methods to enlarge an image but there is no solution to shrink an image. I'm currently using the nearest neighbor method. How could I do this with bilinear interpolation without using the imresize function in MATLAB?
In your comments, you mentioned you wanted to resize an image using bilinear interpolation. Bear in mind that the bilinear interpolation algorithm is size independent. You can very well use the same algorithm for enlarging an image as well as shrinking an image. The right scale factors to sample the pixel locations are dependent on the output dimensions you specify. This doesn't change the core algorithm by the way.
Before I start with any code, I'm going to refer you to Richard Alan Peters' II digital image processing slides on interpolation, specifically slide #59. It has a great illustration as well as pseudocode on how to do bilinear interpolation that is MATLAB friendly. To be self-contained, I'm going to include his slide here so we can follow along and code it:
Please be advised that this only resamples the image. If you actually want to match MATLAB's output, you need to disable anti-aliasing.
MATLAB by default will perform anti-aliasing on the images to ensure the output looks visually pleasing. If you'd like to compare apples with apples, make sure you disable anti-aliasing when comparing between this implementation and MATLAB's imresize function.
Let's write a function that will do this for us. This function will take in an image (that is read in through imread) which can be either colour or grayscale, as well as an array of two elements - The image you want to resize and the output dimensions in a two-element array of the final resized image you want. The first element of this array will be the rows and the second element of this array will be the columns. We will simply go through this algorithm and calculate the output pixel colours / grayscale values using this pseudocode:
function [out] = bilinearInterpolation(im, out_dims)
%// Get some necessary variables first
in_rows = size(im,1);
in_cols = size(im,2);
out_rows = out_dims(1);
out_cols = out_dims(2);
%// Let S_R = R / R'
S_R = in_rows / out_rows;
%// Let S_C = C / C'
S_C = in_cols / out_cols;
%// Define grid of co-ordinates in our image
%// Generate (x,y) pairs for each point in our image
[cf, rf] = meshgrid(1 : out_cols, 1 : out_rows);
%// Let r_f = r'*S_R for r = 1,...,R'
%// Let c_f = c'*S_C for c = 1,...,C'
rf = rf * S_R;
cf = cf * S_C;
%// Let r = floor(rf) and c = floor(cf)
r = floor(rf);
c = floor(cf);
%// Any values out of range, cap
r(r < 1) = 1;
c(c < 1) = 1;
r(r > in_rows - 1) = in_rows - 1;
c(c > in_cols - 1) = in_cols - 1;
%// Let delta_R = rf - r and delta_C = cf - c
delta_R = rf - r;
delta_C = cf - c;
%// Final line of algorithm
%// Get column major indices for each point we wish
%// to access
in1_ind = sub2ind([in_rows, in_cols], r, c);
in2_ind = sub2ind([in_rows, in_cols], r+1,c);
in3_ind = sub2ind([in_rows, in_cols], r, c+1);
in4_ind = sub2ind([in_rows, in_cols], r+1, c+1);
%// Now interpolate
%// Go through each channel for the case of colour
%// Create output image that is the same class as input
out = zeros(out_rows, out_cols, size(im, 3));
out = cast(out, class(im));
for idx = 1 : size(im, 3)
chan = double(im(:,:,idx)); %// Get i'th channel
%// Interpolate the channel
tmp = chan(in1_ind).*(1 - delta_R).*(1 - delta_C) + ...
chan(in2_ind).*(delta_R).*(1 - delta_C) + ...
chan(in3_ind).*(1 - delta_R).*(delta_C) + ...
chan(in4_ind).*(delta_R).*(delta_C);
out(:,:,idx) = cast(tmp, class(im));
end
Take the above code, copy and paste it into a file called bilinearInterpolation.m and save it. Make sure you change your working directory where you've saved this file.
Except for sub2ind and perhaps meshgrid, everything seems to be in accordance with the algorithm. meshgrid is very easy to explain. All you're doing is specifying a 2D grid of (x,y) co-ordinates, where each location in your image has a pair of (x,y) or column and row co-ordinates. Creating a grid through meshgrid avoids any for loops as we will have generated all of the right pixel locations from the algorithm that we need before we continue.
How sub2ind works is that it takes in a row and column location in a 2D matrix (well... it can really be any amount of dimensions you want), and it outputs a single linear index. If you're not aware of how MATLAB indexes into matrices, there are two ways you can access an element in a matrix. You can use the row and column to get what you want, or you can use a column-major index. Take a look at this matrix example I have below:
A =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
If we want to access the number 9, we can do A(2,4) which is what most people tend to default to. There is another way to access the number 9 using a single number, which is A(11)... now how is that the case? MATLAB lays out the memory of its matrices in column-major format. This means that if you were to take this matrix and stack all of its columns together in a single array, it would look like this:
A =
1
6
11
2
7
12
3
8
13
4
9
14
5
10
15
Now, if you want to access element number 9, you would need to access the 11th element of this array. Going back to the interpolation bit, sub2ind is crucial if you want to vectorize accessing the elements in your image to do the interpolation without doing any for loops. As such, if you look at the last line of the pseudocode, we want to access elements at r, c, r+1 and c+1. Note that all of these are 2D arrays, where each element in each of the matching locations in all of these arrays tell us the four pixels we need to sample from in order to produce the final output pixel. The output of sub2ind will also be 2D arrays of the same size as the output image. The key here is that each element of the 2D arrays of r, c, r+1, and c+1 will give us the column-major indices into the image that we want to access, and by throwing this as input into the image for indexing, we will exactly get the pixel locations that we want.
There are some important subtleties I'd like to add when implementing the algorithm:
You need to make sure that any indices to access the image when interpolating outside of the image are either set to 1 or the number of rows or columns to ensure you don't go out of bounds. Actually, if you extend to the right or below the image, you need to set this to one below the maximum as the interpolation requires that you are accessing pixels to one over to the right or below. This will make sure that you're still within bounds.
You also need to make sure that the output image is cast to the same class as the input image.
I ran through a for loop to interpolate each channel on its own. You could do this intelligently using bsxfun, but I decided to use a for loop for simplicity, and so that you are able to follow along with the algorithm.
As an example to show this works, let's use the onion.png image that is part of MATLAB's system path. The original dimensions of this image are 135 x 198. Let's interpolate this image by making it larger, going to 270 x 396 which is twice the size of the original image:
im = imread('onion.png');
out = bilinearInterpolation(im, [270 396]);
figure;
imshow(im);
figure;
imshow(out);
The above code will interpolate the image by increasing each dimension by twice as much, then show a figure with the original image and another figure with the scaled up image. This is what I get for both:
Similarly, let's shrink the image down by half as much:
im = imread('onion.png');
out = bilinearInterpolation(im, [68 99]);
figure;
imshow(im);
figure;
imshow(out);
Note that half of 135 is 67.5 for the rows, but I rounded up to 68. This is what I get:
One thing I've noticed in practice is that upsampling with bilinear has decent performance in comparison to other schemes like bicubic... or even Lanczos. However, when you're shrinking an image, because you're removing detail, nearest neighbour is very much sufficient. I find bilinear or bicubic to be overkill. I'm not sure about what your application is, but play around with the different interpolation algorithms and see what you like out of the results. Bicubic is another story, and I'll leave that to you as an exercise. Those slides I referred you to does have material on bicubic interpolation if you're interested.
Good luck!

Vector decomposition in matlab

this is my situation: I have a 30x30 image and I want to calculate the radial and tangent component of the gradient of each point (pixel) along the straight line passing through the centre of the image (15,15) and the same (i,j) point.
[dx, dy] = gradient(img);
for i=1:30
for j=1:30
pt = [dx(i, j), dy(i,j)];
line = [i-15, j-15];
costh = dot(line, pt)/(norm(line)*norm(pt));
par(i,j) = norm(costh*line);
tang(i,j) = norm(sin(acos(costh))*line);
end
end
is this code correct?
I think there is a conceptual error in your code, I tried to get your results with a different approach, see how it compares to yours.
[dy, dx] = gradient(img);
I inverted x and y because the usual convention in matlab is to have the first dimension along the rows of a matrix while gradient does the opposite.
I created an array of the same size as img but with each pixel containing the angle of the vector from the center of the image to this point:
[I,J] = ind2sub(size(img), 1:numel(img));
theta=reshape(atan2d(I-ceil(size(img,1)/2), J-ceil(size(img,2)/2)), size(img))+180;
The function atan2d ensures that the 4 quadrants give distinct angle values.
Now the projection of the x and y components can be obtained with trigonometry:
par=dx.*sind(theta)+dy.*cosd(theta);
tang=dx.*cosd(theta)+dy.*sind(theta);
Note the use of the .* to achieve point-by-point multiplication, this is a big advantage of Matlab's matrix computations which saves you a loop.
Here's an example with a well-defined input image (no gradient along the rows and a constant gradient along the columns):
img=repmat(1:30, [30 1]);
The results:
subplot(1,2,1)
imagesc(par)
subplot(1,2,2)
imagesc(tang)
colorbar

How to use Haar wavelet to detect LINES on an image?

So I have an image like this:
I want to get something like this (I haven't drawn all lines I want but I hope you can get my idea):
I want to use SURF ( (Speeded Up Robust Features) is a robust image descriptor, first presented by Herbert Bay et al. in 2006 ) or something that is based on sums of 2D Haar wavelet responses and makes an efficient use of integral images for finding all straight lines on image. I want to get relative to picture pixel coords start and end points of lines.
So on this picture to find all lines between tiles and those 2 black lines on top.
Is there any such Code Example (with lines search capability) to start from?
I love C and C++ but any other readable code will probably work for me=)
The following is a complete example of applying Hough Transform to detect lines. I am using MATLAB for the job..
The trick is to divide the image into regions and process each differently; this is because you have different "textures" in your scene (tiles on the upper region of the wall are quite different from the darker ones on the bottom, and processing the image all at once wont be optimal).
As a working example, consider this one:
%# load image, blur it, then find edges
I0 = rgb2gray( imread('http://www.de-viz.ru/catalog/new2/Holm/hvannaya.jpg') );
I = imcrop(I0, [577 156 220 292]); %# select a region of interest
I = imfilter(I, fspecial('gaussian', [7 7], 1), 'symmetric');
BW = edge(I, 'canny');
%# Hough Transform and show accumulated matrix
[H T R] = hough(BW, 'RhoResolution',2, 'Theta',-90:0.5:89.5);
imshow(imadjust(mat2gray(H)), [], 'XData',T, 'YData',R, ...
'InitialMagnification','fit')
xlabel('\theta (degrees)'), ylabel('\rho')
axis on, axis normal, colormap(hot), colorbar, hold on
%# detect peaks
P = houghpeaks(H, 20, 'threshold',ceil(0.5*max(H(:))));
plot(T(P(:,2)), R(P(:,1)), 'gs', 'LineWidth',2);
%# detect lines and overlay on top of image
lines = houghlines(BW, T, R, P, 'FillGap',50, 'MinLength',5);
figure, imshow(I), hold on
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1), xy(:,2), 'g.-', 'LineWidth',2);
end
hold off
You could try the same procedure for other regions while tuning the parameters to get good results..
Have you tried a simpler approach such as the Hough transform for finding lines? A function to perform this and example are included in OpenCV called cvHoughLines2.
Two-dimensional wavelet transforms are implemented in R using the package waveslim. Specifically, the function dwt2D() uses a C "backend" for speed. You can then apply thresholding to find the lines.

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