I am having trouble understanding the code about the Matlab
a = imread('Untitled2.png');
faceDetector = vision.CascadeObjectDetector;
bbox=step(faceDetector,a);
for j=1:size(bbox)
xbox=bbox(j,:);
subImage = imcrop(a, xbox);
H = fspecial('disk',10);
blurred = imfilter(subImage,H);
a(xbox(2):xbox(2)+xbox(4),xbox(1):xbox(1)+xbox(3),1:end) = blurred;
end
imshow(a);
Can anyone please kindly explain me that what is the for loop doing? I tried to use my own method to blur the face that I detected, but I just manage to crop out the face and blur the cropped image but I don't know how to put it back to original image. Then I tried to use the source code above I get from internet and the internet source code is able to blur the face and I can't understand the for loop logic. Kindly please explain to me, T^T.
Thanks.
As you can see there :
BBOX = step(detector,I) returns BBOX, an M-by-4 matrix defining M bounding boxes containing the detected objects. This method performs multiscale object detection on the input image, I. Each row of the output matrix, BBOX, contains a four-element vector, [x y width height], that specifies in pixels, the upper-left corner and size of a bounding box. The input image I, must be a grayscale or truecolor (RGB) image.
Are you sure that it's j=1:size(bbox) and not j=1:size(bbox,1) in the code?
Basically, The definition of BBox speaks by itself. The loop just iterates over all the boxes detected..
You then extract the informations about the jth box.
Then you extract the subImage given the position and size of xbox (xbox is a vector containing [x y width height]).
Then you define you filter.
Then you blur your subImage.
Then you override you image a with the blurred subimage using the informations in xbox.
EDIT : If you've already suceeded in blurring the cropped image, you just need to override you input image with your blurred image!
Related
I have this image (8 bit, pseudo-colored, gray-scale):
And I want to create an intensity band of a specific measure around it's border.
I tried erosion and other mathematical operations, including filtering to achieve the desired band but the actual image intensity changes as soon as I use erosion to cut part of the border.
My code so far looks like:
clear all
clc
x=imread('8-BIT COPY OF EGFP001.tif');
imshow(x);
y = imerode(x,strel('disk',2));
y1=imerode(y,strel('disk',7));
z=y-y1;
figure
z(z<30)=0
imshow(z)
The main problem I am encountering using this is that it somewhat changes the intensity of the original images as follows:
So my question is, how do I create such a band across image border without changing any other attribute of the original image?
Going with what beaker was talking about and what you would like done, I would personally convert your image into binary where false represents the background and true represents the foreground. When you're done, you then erode this image using a good structuring element that preserves the roundness of the contours of your objects (disk in your example).
The output of this would be the interior of the large object that is in the image. What you can do is use this mask and set these locations in the image to black so that you can preserve the outer band. As such, try doing something like this:
%// Read in image (directly from StackOverflow) and pseudo-colour the image
[im,map] = imread('http://i.stack.imgur.com/OxFwB.png');
out = ind2rgb(im, map);
%// Threshold the grayscale version
im_b = im > 10;
%// Create structuring element that removes border
se = strel('disk',7);
%// Erode thresholded image to get final mask
erode_b = imerode(im_b, se);
%// Duplicate mask in 3D
mask_3D = cat(3, erode_b, erode_b, erode_b);
%// Find indices that are true and black out result
final = out;
final(mask_3D) = 0;
figure;
imshow(final);
Let's go through the code slowly. The first two lines take your PNG image, which contains a grayscale image and a colour map and we read both of these into MATLAB. Next, we use ind2rgb to convert the image into its pseudo-coloured version. Once we do this, we use the grayscale image and threshold the image so that we capture all of the object pixels. I threshold the image with a value of 10 to escape some quantization noise that is seen in the image. This binary image is what we will operate on to determine those pixels we want to set to 0 to get the outer border.
Next, we declare a structuring element that is a disk of a radius of 7, then erode the mask. Once I'm done, I duplicate this mask in 3D so that it has the same number of channels as the pseudo-coloured image, then use the locations of the mask to set the values that are internal to the object to 0. The result would be the original image, but having the outer contours of all of the objects remain.
The result I get is:
How to identify boundaries of a binary image to crop in matlab?
ie. the input binary image has no noises. only has one black object in white background.
You can use the edge command in MATLAB.
E = edge(I);
I would be an input grayscale or binary image. This will return a binary image with only the edges.
This can provide further assistance:
http://www.mathworks.com/help/images/ref/edge.html
If your image is just black-and-white and has a single object, you can likely make use of the Flood fill algorithm, for which Matlab has built-in support!
Try the imfill function (ref).
This should give you the extents of the object, which would allow you to crop at will.
You can also invert the image, then do regionprops to extract all of the properties for separate objects. You need to invert the image as regionprops assumes that the objects are white while the background is black. A good thing about this approach is that it generalizes for multiple objects and you only need about a few lines of code to do it.
As an example, let's artificially create a circle in the centre of an image that is black on a white background as you have suggested. Let's assume this is also a binary image.
im = true(200, 200);
[X,Y] = meshgrid(1:200, 1:200);
ind = (X-100).^2 + (Y-100).^2 <= 1000;
im(ind) = false;
imshow(im);
This is what your circle will look like:
Now let's go ahead and invert this so that it's a white circle on black background:
imInvert = ~im;
imshow(imInvert);
This is what your inverted circle will look like:
Now, invoke regionprops to find properties of all of the objects in our image. In this case, there should only be one.
s = regionProps(imInvert, 'BoundingBox');
As such, s contains a structure that is 1 element long, and has a single field called BoundingBox. This field is a 4 element array that is structured in the following way:
[x y w h]
x denotes the column/vertical co-ordinate while y denotes the row/horizontal co-ordinate of the top-left corner of the bounding box. w,h are the width and height of the rectangle. Our output of the above code is:
s =
BoundingBox: [68.5000 68.5000 63 63]
This means that the top-left corner of our bounding box is located at (x,y) = (68.5,68.5), and has a width and height of 63 each. Therefore, the span of our bounding box goes from rows (68.5,131.5) and columns (68.5,131.5). To make sure that we have the right bounding box, you can draw a rectangle around our shape by using the rectangle command.
imshow(im);
rectangle('Position', s.BoundingBox);
This is what your image will look like with a rectangle drawn around the object. As you can see, the bounding box given from regionprops is the minimum spanning bounding box required to fully encapsulate the object.
If you wish to crop the object, you can do the following:
imCrop = imcrop(imInvert, s.BoundingBox);
This should give you the cropped image that is defined by the bounding box that we talked about earlier.
Hope this is what you're looking for. Good luck!
I'm trying to detect the screen border from the image (In need the 4 corners).
This is the Image:
I used HOUGH transform to detect lines and intersection points (the black circles) and this is the result:
Now I need to find the 4 corners or the 4 lines.. everything that will help me to crop the image, What can I do?
Maybe use the screen aspect ratio? but how?
I'm using Matlab.
Thanks.
A naive first approach that would do the trick if and only if you have same image conditions (background and laptop).
Convert your image to HSV (examine that in HSV the image inside the
screen is the only portion of the image with high Saturation, Value
values)
Create a mask by hard thresholding the Saturation and Value channels
Dilate the mask to connect disconnected regions
Compute the convex hull to get the mask boundaries
See a quick result:
Here is the same mask with the original image portion that makes it through the mask:
Here is the code to do so:
img = imread( 'imagename.jpg'); % change the image name
hsv = rgb2hsv( img);
mask = hsv(:,:,2)>0.25 & hsv(:,:,3)>0.5;
strel_size = round(0.025*max(size(mask)));
dilated_mask=imdilate(mask,strel('square',strel_size));
s=regionprops(dilated_mask,'BoundingBox','ConvexHull');
% here Bounding box produces a box with the minimum-maximum white pixel positions but the image is not actually rectangular due to perspective...
imshow(uint8(img.*repmat(dilated_mask,[1 1 3])));
line(s.ConvexHull(:,1),s.ConvexHull(:,2),'Color','red','LineWidth',3);
You may, of course, apply some more sophisticated processing to be a lot more accurate and to correct the convex hull to be just a rectangular shape with perspective, but this is just a 5 minutes attempt just to showcase the approach...
I would like to crop an image but I want to retain the part of image that is outside of the rectangle. How can this can be done?
It seems that with imcrop only the part within the rectangle can be retained.
An image in Matlab is represented by a matrix, just like any other matrix, you can read more about representation forms here.
It seems that what you want to do is to take the area that you don't want and change the values of the corresponding cells in the matrix to the color that you want to put instead (each cell in the matrix is a pixel in the image). That is if you know the place where your unwanted data is.
If you don't know where it is, and want to use the tool given by imcrop to manually choose the "cropped" area, you can take the resulting matrix, and find the part of the original image which is an exact match with the cropped part, and to color it as you wish.
The code for doing this:
I=imread('img_9.tif');
I2=imcrop(I,[60,50,85,85]);
n_big=size(I);
n_small=size(I2);
for j1=1:(n_big(1)-n_small(1))
for j2=1:(n_big(2)-n_small(2))
Itest=I(j1:j1+n_small(1)-1,j2:j2+n_small(2)-1,:);
if ( Itest == I2)
I(j1:j1+n_small(1)-1,j2:j2+n_small(2)-1,:) = zeros(n_small(1),n_small(2),3);
end
end
end
figure(1);
imshow(I);
figure(2);
imshow(I2);
The results of my test were:
original:
cropped:
resulting image:
maybe what you want to do is first a mask with the inverse area of what you want to crop and save this result.
There is an image, surrounded by black region, I would like to find the exact locaton of this image (which is surrounded by black region). I mean the coordinates of four corners of this image. Thanks.
You could apply a second derivative mask on your image which will then accurately pick up the points at which the colour goes from black to the content of the picture. You can then extract the first and last row and column each and you've got your coordinates.
If you are in matlab:
Mask out the black area. Either try:
Image = logical(Image)
Or find the intensity of a black voxel (probably zero) and say:
Image = ind2sub(size(Image), find(Image ~= blackPixelIntensity))
Once you have the binary non-black part of the image (so just the object and not the background), you just want to find the min, max corners of each voxel. Say:
[x y] = ind2sub(size(Image), find(Image ~= 0))
c1 = [min(x) min(y)]
c2 = [max(x) min(y)]
c3 = [max(x) max(y)]
c4 = [min(x) min(y)]
Where c1,...,c4 are your corners :)
Lemme know about any syntax error as I don't have access to matlab atm.
tylerthemiler
Edit: if you simply want the entire perimeter of the non-black part of the image, just do whichever of the first two lines of code above works, and then say:
Imperim = bwperim(Image)
Edit2: Note that Image is the 2D array, you can use whatever you want to load in the jpg :P