On a shape from a logical image, I am trying to extract the field of view from any point inside the shape on matlab :
I tried something involving to test each line going through the point but it is really really long.(I hope to do it for each points of the shape or at least each point of it's contour wich is quite a few times)
I think a faster method would be working iteratively by the expansion of a disk from the considered point but I am not sure how to do it.
How can I find this field of view in an efficient way?
Any ideas or solution would be appreciated, thanks.
Here is a possible approach (the principle behind the function I wrote, available on Matlab Central):
I created this test image and an arbitrary point of view:
testscene=zeros(500);
testscene(80:120,80:120)=1;
testscene(200:250,400:450)=1;
testscene(380:450,200:270)=1;
viewpoint=[250, 300];
imsize=size(testscene); % checks the size of the image
It looks like this (the circle marks the view point I chose):
The next line computes the longest distance to the edge of the image from the viewpoint:
maxdist=max([norm(viewpoint), norm(viewpoint-[1 imsize(2)]), norm(viewpoint-[imsize(1) 1]), norm(viewpoint-imsize)]);
angles=1:360; % use smaller increment to increase resolution
Then generate a set of points uniformly distributed around the viewpoint.:
endpoints=bsxfun(#plus, maxdist*[cosd(angles)' sind(angles)'], viewpoint);
for k=1:numel(angles)
[CX,CY,C] = improfile(testscene,[viewpoint(1), endpoints(k,1)],[viewpoint(2), endpoints(k,2)]);
idx=find(C);
intersec(k,:)=[CX(idx(1)), CY(idx(1))];
end
What this does is drawing lines from the view point to each directions specified in the array angles and look for the position of the intersection with an obstacle or the edge of the image.
This should help visualizing the process:
Finally, let's use the built-in roipoly function to create a binary mask from a set of coordinates:
FieldofView = roipoly(testscene,intersec(:,1),intersec(:,2));
Here is how it looks like (obstacles in white, visible field in gray, viewpoint in red):
Related
I have a segmented image as shown here
i want to fit a curve along the top pixels of the segmented image(show as red curve) and i want to find the top point along the curve show in blue. I have already worked on basic idea like traversing through the top to bottom and collecting the top point along each column. i want to know is there any easy solution for this problem like directly taking out the boundary pixels and find the top point.I am using MATLAB for this problem
%download the image
img = logical(imread('http://i.stack.imgur.com/or2iX.png'));
%for some reason it appeared RGB with big solid borders.
%to monochrome
img = img(:,:,1);
%remove borders
img = img(~all(img,2), ~all(img,1));
%split into columns
cimg = num2cell(img,1);
%find first nonzero element per column
ridx = cellfun(#(x) find(x,1,'first'), cimg);
figure, imshow(img)
hold on
%image dim1 is Y, dim2 is X
plot(1:size(img,2),ridx-1,'r','linewidth',2)
%find top point
[yval, xval] = min(ridx);
If you want a smoother curve, try polyfit/polyval
#EDIT
If we want the line to have break at gaps between connected components, we should change the code to something like
bord_idx = sub2ind(size(img), ridx, 1:size(img,2));
regs=regionprops(bwlabel(img),'pixelidxlist');
regs_idx = struct2cell(regs);
split_step = cellfun(#(x) sum(ismember(bord_idx,x)), regs_idx);
split_step = split_step(split_step>0);
split_yvals = mat2cell(ridx',split_val);
split_xvals = mat2cell([1:size(img,2)]',split_val);
figure, imshow(img)
hold on
for k = 1:length(split_step),
plot(split_xvals{k}, split_yvals{k}, 'r', 'linewidth', 2),
end
However, the result is not ideal if one region is positioned over the other. If the "shadowed" points are needed, you should try bwtraceboundary or convexhull and find where the border turns down
As far as "simplest matlab solution" by which I think you mean built in matlab functions: imclose()->edge()->bwboundaries()->findpeaks()'on each boundary'->'filter results based on width and magnitude of peaks'. *you will need to tune all the parameters in these functions, I am just listing what would get you there if appropriately applied.
As far as processing speed is concerned, I think I would have done exactly what you did, basically collecting the top edge from a top down column search and then looking for the point of highest inflection. As soon as you start doing processing of any type, you start doing several operations per pixel which will quickly become more expensive than your initial search (just requires that your image and target are simple enough)
That being said, here are some ideas that may help:
1:If you run a sufficiently heavy closing (dilate->erode), that should fill in all that garbage at the bottom.
2: If you know that your point of interest is not at left or right of picture (boundaries), you could take the right and left edge points and calculate a slope to be applied as an offset to flatten the whole image.
3: If your image always has the large dark linear region below the peak as seen here, you could locate those edges with houghlines looking for verticals and then search only the columns between them.
4: If speed is a concern, you could do a more sophisticated search pattern than left to right, as your peak has a pretty good distribution around it which could help with faster localization of maxima.
I have the follow code in matlab which is supposed to draw a polygon on a image (has to be a 2d image, be just a patch).
numCorners=8;
dotPos=[];
for rr=1:numCorners
dotPos(end+1)=(cos(rr/numCorners*2*pi))*100;
dotPos(end+1)=(sin(rr/numCorners*2*pi))*100;
end
BaseIm=zeros(1000,1000);
dotpos=[500,500];
imageMatrix =drawpolygon(BaseIm, dotPos, 1); or how else do draw a white polygon here?
imshow(imageMatrix);
This doesn't work as drawpolygon does not appear to exist in this way any idea how to do this?
Note that the resulting data must be an image of equal size of baseIM and must be an array of doubles (ints can be converted) as this is test data for another algorithm.
I have since found the inpolygon(xi,yi,xv,yv); function which I could combine with a for loop if I knew how to properly call it.
If you just need to plot two polygons, you can use the fill function.
t=0:2*pi;
x=cos(t)*2;
y=sin(t)*2
fill(x,y,'r')
hold on
fill(x/2,y/2,'g')
As an alternative, you can use the patch function:
figure
t=0:2*pi;
x=cos(t)*2;
y=sin(t)*2
patch(x,y,'c')
hold on
patch(x/2,y/2,'k')
Edit
The fill and patch functions allow to add polygons also over an actual image too.
% Load an image on the axes
imshow('Jupiter_New_Horizons.jpg')
hold on
% Get the axis limits (just to center the polygons
x_lim=get(gca,'xlim')
y_lim=get(gca,'ylim')
% Create the polygon's coords
t=0:2*pi;
x=cos(t)*50+x_lim(2)/2;
y=sin(t)*50+y_lim(2)/2
% Add the two polygons to the image
f1_h=fill(x,y,'r')
hold on
f1_h=fill(x/2,y/2,'g')
Hope this helps.
I am currently doing some seismic modelling and processing in MATLAB, and would like to come up with an easy way of muting parts of various datasets. If I plot the frequency-wavenumber spectrum of some of my data, for instance, I obtain the following result:
Now, say that I want to mute some of the data present here. I could of course attempt to run through the entire matrix represented here and specify a threshold value where everything above said value should be set equal to zero, but this will be very difficult and time-consuming when I later will work with more complicated fk-spectra. I recently learned that MATLAB has an inbuilt function called impoly which allows me to interactively draw a polygon in plots. So say I, for instance, draw the following polygon in my plot with the impoly-function:
Is there anything I can do now to set all points within this polygon equal to zero? After defining the polygon as illustrated above I haven't found out how to proceed in order to mute the information contained in the polygon, so if anybody can give me some help here, then i would greatly appreciate it!
Yes, you can use the createMask function that's part of the impoly interface once you delineate the polygon in your figure. Once you use create this mask, you can use the mask to index into your data and set the right regions to zero.
Here's a quick example using the pout.tif image in MATLAB:
im = imread('pout.tif');
figure; imshow(im);
h = impoly;
I get this figure and I draw a polygon inside this image:
Now, use the createMask function with the handle to the impoly call to create a binary mask that encapsulates this polygon:
mask = createMask(h);
I get this mask:
imshow(mask);
You can then use this mask to index into your data and set the right regions to 0. First make a copy of the original data then set the data accordingly.
im_zero = im;
im_zero(mask) = 0;
I now get this:
imshow(im_zero);
Note that this only applies to single channel (2D) data. If you want to apply this to multi-channel (3D) data, then perhaps a multiplication channel-wise with the opposite of the mask may be prudent.
Something like this:
im_zero = bsxfun(#times, im, cast(~mask, class(im)));
The above code takes the opposite of the polygon mask, converts it into the same class as the original input im, then performs an element-wise multiplication of this mask with each channel of the input separately. The result will zero each spatial location that's defined in the mask over all channels.
I'm searching for an certain object in my photograph:
Object: Outline of a rectangle with an X in the middle. It looks like a rectangular checkbox. That's all. So, no fill, just lines. The rectangle will have the same ratios of length to width but it could be any size or any rotation in the photograph.
I've looked a whole bunch of image recognition approaches. But I'm trying to determine the best for this specific task. Most importantly, the object is made of lines and is not a filled shape. Also, there is no perspective distortion, so the rectangular object will always have right angles in the photograph.
Any ideas? I'm hoping for something that I can implement fairly easily.
Thanks all.
You could try using a corner detector (e.g. Harris) to find the corners of the box, the ends and the intersection of the X. That simplifies the problem to finding points in the right configuration.
Edit (response to comment):
I'm assuming you can find the corner points in your image, the 4 corners of the rectangle, the 4 line endings of the X and the center of the X, plus a few other corners in the image due to noise or objects in the background. That simplifies the problem to finding a set of 9 points in the right configuration, out of a given set of points.
My first try would be to look at each corner point A. Then I'd iterate over the points B close to A. Now if I assume that (e.g.) A is the upper left corner of the rectangle and B is the lower right corner, I can easily calculate, where I would expect the other corner points to be in the image. I'd use some nearest-neighbor search (or a library like FLANN) to see if there are corners where I'd expect them. If I can find a set of points that matches these expected positions, I know where the symbol would be, if it is present in the image.
You have to try if that is good enough for your application. If you have too many false positives (sets of corners of other objects that accidentially form a rectangle + X), you could check if there are lines (i.e. high contrast in the right direction) where you would expect them. And you could check if there is low contrast where there are no lines in the pattern. This should be relatively straightforward once you know the points in the image that correspond to the corners/line endings in the object you're looking for.
I'd suggest the Generalized Hough Transform. It seems you have a fairly simple, fixed shape. The generalized Hough transform should be able to detect that shape at any rotation or scale in the image. You many need to threshold the original image, or pre-process it in some way for this method to be useful though.
You can use local features to identify the object in image. Feature detection wiki
For example, you can calculate features on some referent image which contains only the object you're looking for and save the results, let's say, to a plain text file. After that you can search for the object just by comparing newly calculated features (on images with some complex scenes containing the object) with the referent ones.
Here's some good resource on local features:
Local Invariant Feature Detectors: A Survey
Greetings,
I'm working on a game project that uses a 3D variant of hexagonal tile maps. Tiles are actually cubes, not hexes, but are laid out just like hexes (because a square can be turned to a cube to extrapolate from 2D to 3D, but there is no 3D version of a hex). Rather than a verbose description, here goes an example of a 4x4x4 map:
(I have highlighted an arbitrary tile (green) and its adjacent tiles (yellow) to help describe how the whole thing is supposed to work; but the adjacency functions are not the issue, that's already solved.)
I have a struct type to represent tiles, and maps are represented as a 3D array of tiles (wrapped in a Map class to add some utility methods, but that's not very relevant).
Each tile is supposed to represent a perfectly cubic space, and they are all exactly the same size. Also, the offset between adjacent "rows" is exactly half the size of a tile.
That's enough context; my question is:
Given the coordinates of two points A and B, how can I generate a list of the tiles (or, rather, their coordinates) that a straight line between A and B would cross?
That would later be used for a variety of purposes, such as determining Line-of-sight, charge path legality, and so on.
BTW, this may be useful: my maps use the (0,0,0) as a reference position. The 'jagging' of the map can be defined as offsetting each tile ((y+z) mod 2) * tileSize/2.0 to the right from the position it'd have on a "sane" cartesian system. For the non-jagged rows, that yields 0; for rows where (y+z) mod 2 is 1, it yields 0.5 tiles.
I'm working on C#4 targeting the .Net Framework 4.0; but I don't really need specific code, just the algorithm to solve the weird geometric/mathematical problem. I have been trying for several days to solve this at no avail; and trying to draw the whole thing on paper to "visualize" it didn't help either :( .
Thanks in advance for any answer
Until one of the clever SOers turns up, here's my dumb solution. I'll explain it in 2D 'cos that makes it easier to explain, but it will generalise to 3D easily enough. I think any attempt to try to work this entirely in cell index space is doomed to failure (though I'll admit it's just what I think and I look forward to being proved wrong).
So you need to define a function to map from cartesian coordinates to cell indices. This is straightforward, if a little tricky. First, decide whether point(0,0) is the bottom left corner of cell(0,0) or the centre, or some other point. Since it makes the explanations easier, I'll go with bottom-left corner. Observe that any point(x,floor(y)==0) maps to cell(floor(x),0). Indeed, any point(x,even(floor(y))) maps to cell(floor(x),floor(y)).
Here, I invent the boolean function even which returns True if its argument is an even integer. I'll use odd next: any point point(x,odd(floor(y)) maps to cell(floor(x-0.5),floor(y)).
Now you have the basics of the recipe for determining lines-of-sight.
You will also need a function to map from cell(m,n) back to a point in cartesian space. That should be straightforward once you have decided where the origin lies.
Now, unless I've misplaced some brackets, I think you are on your way. You'll need to:
decide where in cell(0,0) you position point(0,0); and adjust the function accordingly;
decide where points along the cell boundaries fall; and
generalise this into 3 dimensions.
Depending on the size of the playing field you could store the cartesian coordinates of the cell boundaries in a lookup table (or other data structure), which would probably speed things up.
Perhaps you can avoid all the complex math if you look at your problem in another way:
I see that you only shift your blocks (alternating) along the first axis by half the blocksize. If you split up your blocks along this axis the above example will become (with shifts) an (9x4x4) simple cartesian coordinate system with regular stacked blocks. Now doing the raytracing becomes much more simple and less error prone.