I have an image with dark blue spots on a black background. I want to convert this to inverse gray scale. By inverse, I mean, I want the black ground to be white.
When I convert it to gray scale, it makes everything look black and it makes it very hard to differentiate.
Is there a way to do an inverse gray scale where the black background takes the lighter shades?
Or, another preferable option is to represent the blue as white and the black as black.
I am using img = rgb2gray(img); in MATLAB for now.
From mathworks site:
IM2 = imcomplement(IM)
Is there a way to do an inverse gray scale where the black
background takes the lighter shades?
Based on your image description I created an image sample.png:
img1 = imread('sample.png'); % Read rgb image from graphics file.
imshow(img1); % Display image.
Then, I used the imcomplement function to obtain the complement of the original image (as suggested in this answer).
img2 = imcomplement(img1); % Complement image.
imshow(img2); % Display image.
This is the result:
Or, another preferable option is to represent the blue as white and
the black as black.
In this case, the simplest option is to work with the blue channel. Now, depending on your needs, there are two approaches you can use:
Approach 1: Convert the blue channel to a binary image (B&W)
This comment suggests using the logical operation img(:,:,3) > 0, which will return a binary array of the blue channel, where every non-zero valued pixel will be mapped to 1 (white), and the rest of pixels will have a value of 0 (black).
While this approach is simple and valid, binary images have the big disadvantage of loosing intensity information. This can alter the perceptual properties of your image. Have a look at the code:
img3 = img1(:, :, 3) > 0; % Convert blue channel to binary image.
imshow(img3); % Display image.
This is the result:
Notice that the round shaped spots in the original image have become octagon shaped in the binary image, due to the loss of intensity information.
Approach 2: Convert the blue channel to grayscale image
A better approach is to use a grayscale image, because the intensity information is preserved.
The imshow function offers the imshow(I,[low high]) overload, which adjusts the color axis scaling of the grayscale image through the DisplayRange parameter.
One very cool feature of this overload, is that we can let imshow do the work for us.
From the documentation:
If you specify an empty matrix ([]), imshow uses [min(I(:)) max(I(:))]. In other words, use the minimum value in I as black, and the maximum value as white.
Have a look at the code:
img4 = img1(:, :, 3); % Extract blue channel.
imshow(img4, []); % Display image.
This is the result:
Notice that the round shape of the spots is preserved exactly as in the original image.
Related
clear all;
clc;
imag = imread('286502.png');
image_binary = im2bw(imag,0.85); %converte image to binary
image_binary = not(image_binary);
figure(1);clf
imagesc(image_binary);colormap(gray)
I am using this code to genarate a binary image of an ellipse having its inside as white while outside as black.But , the problem i am facing is that whenever the foreground and background of my input ellipse is either light or dark then the binary image becomes all black or all white.
if you want to keep the gray-scale nature of our image, you may remove the line im2bw command because that flattens the gray-scale values into binary (0-1 or white/black).
if you just want to show a binary image in gray color, you can try customizing the colormap
colormap([0 0 0;0.5 0.5 0.5]);
im2bw(imag,0.85) thresholds at a fixed value, which will work for some images but not for others. I recommend that you binarize with a method such as Otsu, which determines an optimal threshold for each image individually.
image_binary = imbinarize(imag);
I'm learning about statistical feature of an image.A quote that I'm reading is
For the first method which is statistical features of texture, after
the image is loaded, it is converted to gray scale image. Then the
background is subtracted from the original image. This is done by
subtract the any blue intensity pixels for the image. Finally, the ROI
is obtained by finding the pixels which are not zero value.
The implementation :
% PREPROCESSING segments the Region of Interest (ROI) for
% statistical features extraction.
% Convert RGB image to grayscale image
g=rgb2gray(I);
% Obtain blue layer from original image
b=I(:,:,3);
% Subtract blue background from grayscale image
r=g-b;
% Find the ROI by finding non-zero pixels.
x=find(r~=0);
f=g(x);
My interpretation :
The purpose of substracting the blue channel here is related to the fact that the ROI is non blue background? Like :
But in the real world imaging like for example an object but surrounded with more than one colors? What is the best way to extract ROI in that case?
like for example (assuming only 2 colors on all parts of the bird which are green and black, & geometri shaped is ignored):
what would I do in that case? Also the picture will be transformed to gray scale right? while there's a black part of the ROI (bird) itself.
I mean in the bird case how can I extract only green & black parts? and remove the rest colors (which are considered as background ) of it?
Background removal in an image is a large and potentielly complicated subject in a general case but what I understand is that you want to take advantage of a color information that you already have about your background (correct me if I'm wrong).
If you know the colour to remove, you can for instance:
switch from RGB to Lab color space (Wiki link).
after converting your image, compute the Euclidean from the background color (say orange), to all the pixels in your image
define a threshold under which the pixels are background
In other words, if coordinates of a pixel in Lab are close to orange coordinates in Lab, this pixel is background. The advantage of using Lab is that Euclidean distance between points relates to human perception of colours.
I think this should work, please give it a shot or let me know if I misunderstood the question.
This question already has an answer here:
Access RGB channels in an image in MATLAB
(1 answer)
Closed 6 years ago.
I'm working with images similar to this: a cell image, and I want to extract only the red-pink sections. As of now I'm using img(:,:,1) to pull out the red values but this produces a binary image. I wanted to know if there was a way to extract the "red" values and produce a grayscale image based on their degree of "redness" or intensity. Any help would be awesome.
You are likely visualizing the result using imshow which will automatically set the color limits of the axes to be between 0 and 1. Your image is RGB and the values of the red channel are going to range from 0 to 255. Because of this, if you only specify one input to imshow, you will get an image where all values > 1 will appear as white and all zero-values will be black. So your image isn't really binary, it just appears that way.
You want to either display your image with imagesc which will automatically scale the color limits to match your data:
imagesc(img(:,:,1));
colormap gray
Or you can specify the second input to imshow to cause it to also scale to fit your data range:
imshow(img(:,:,1), [])
The reason that this isn't an issue when you are visualizing all channels is that if you specify red, green, and blue channels, this is considered a true color image and all axes color limits are ignored.
The data you capture will be correct (and is grayscale), but the visualization may be incorrect. When trying to visualize a 2D matrix (same as your result img(:,:,1)), matlab applies the default colormap and the result is:
[x,y]=meshgrid(1:200, 1:200);
z=x.^2.*sin(y/max(y(:))*pi);
figure;imagesc(z);
If you want to avoid the applied jet colormap, either change the colormap:
colormap('gray')
or change your 2D matrix into a 3D one, explicitely specifying the colors to display (must be values between 0 and 1):
z3d = z(:,:,[1 1 1]); % more efficient than repmat
z3d = (z3d - min(z(:)))./range(z(:)); % make sure values in range [0; 1]
You see banding in the colormap version, because by default a colormap contains 64 different colors; the 3d matrix version doesn't have this problem as it directly displays the colors.
If I may add to your question, it seems to me you're simply trying to isolate and visualise the red, green, and blue fluorofores separately (or in combination). I specifically think this because you mention 'pink'.
It may be nicer to just isolate the channels:
>> F_red = F; F_red(:,:,[2,3]) = 0;
>> F_green = F; F_green(:,:,[1,3]) = 0;
>> F_blue = F; F_blue(:,:,[1,2]) = 0;
>> F_pink = F; F_pink(:,:,2) = 0;
Here's a subplot of the result:
Furthermore, you should know that the 'naive' way of producing a grayscale image does not preserve the 'luminosity' of colours as perceived by the human eye, since 'green' at the same intensity as 'red' and 'blue' will actually be perceived as brighter by the human eye, and similarly 'red' is brighter than 'blue'. Matlab provides an rgb2gray function which converts an rgb image to a grayscale image that correctly preserves luminance. This is irrelevant for your pure red, green, and blue conversions, but it may be something to think about with respect to a 'pink-to-grayscale' image. For instance, compare the two images below, you will see subtle contrast differences.
>> F_pinktogray_naive = mean(F(:,:,[1,3]), 3);
>> F_pinktogray_luminance = rgb2gray(F_pink);
A subplot of the two:
In a sense, you probably care more about the left (naive) one, because you don't care about converting the pink one to a gray one "visually", but you care more about the red and blue fluorofores being "comparable" in terms of their intensity on the grayscale image instead (since they represent measurements rather than a visual scene). But it's an important distinction to keep in mind when converting rgb images to grayscale.
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:
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...