It is building a number which contains the 6th, 7th, 14th, 15th, 22nd and 23rd bits from the original image’s colour. i.e. it is producing a crude 6-bit colour from a 24-bit colour. e.g. 000000000rrrrrrrrrggggggggbbbbbbbb becomes the top bits of rrbbgg solved java-image processing [closed]
You’ve basically outlined the right approach which I’ll summarize here: 1) Do the 2D FFTs. 2) Multiply them by a weighting factor. If you’re FFTs are in (Real, Imaginary) form, multiply both components by the weight, and if (Magnitude, phase) just multiple the magnitude. If you’re interested in just high and low frequencies, this will … Read more
Following is a naive method to achieve the desired functionality: Generate a mask of pixel indices with dummy values for pixel with zero value. Count the number of non-zero pixels Create an output vector with length equal to non-zero count. Copy the non-zero pixel indices from the generated mask to the output vector (a process … Read more
Why don’t you simply use regionprops with ‘Image’ property? img = imread(‘http://i.stack.imgur.com/zpYa5.png’); %// read the image bw = img(:,:,1) > 128; %// conver to mask Use some minor morphological operations to handle spurious pixels dbw = imdilate(bw, ones(3)); lb = bwlabel(dbw).*bw; %// label each character as a connected component Now you can use regionprops to … Read more
This seems to work for the files I was able to find to test it with. #include <iostream> #include <fstream> #include <string> #include <vector> struct PBM { unsigned int width; unsigned int height; std::string comment; std::vector<std::vector<char>> data; bool ReadAscii(std::ifstream& f) { std::string id; if(!std::getline(f, id) || id != “P1”) { return false; } if(f.peek() == … Read more
In case you don’t like the command-line approach, I have recast my ImageMagick answer in Python. I am quite a beginner in Python so there may be inefficiencies and poor practices in my code, but it works. The technique is exactly the same as the other answer: convert to HSV colourspace, and locate highly saturated … Read more
Keep adjusting the compression in this example until the image size in bytes is below the limit. Screen shot Typical outputs Fractional Metrics: true Nonantialiased text mode PNG size: 7390 bytes JPG size: 7036 bytes quality: 35 Fractional Metrics: true Antialiased text mode PNG size: 8741 bytes JPG size: 8663 bytes quality: 55 Fractional Metrics: … Read more
I tend to agree with nikie that you should be working from a book if you are at this basic level, but here is an answer anyway. I = imread(‘your_image’); # convert I to grayscale double as appropriate using rgb2gray(), double(), etc. # calculate the gradients and the formula you provided [dIx, dIy] = gradient(I); … Read more
If you get the same values for every pixel there are several possible reasons. a) your image has the same values in every pixel b) you do not change x and y between your calls to getRGB c) you read something else but the return value of getRGB solved getRGB(x,y) in Java awt returns for … Read more
You can draw homogeneity criteria from the histogram of gray-levels or the histogram of gradient intensities. A low dispersion (small variance) reveals a quasi-constant lightness, but is sensitive to smooth variations. A low gradient average reveals a lack of detail, but is sensitive to texture/noise. 4 solved how homogeneity is calculated? [closed]
The upper bounds of your for-loops are the cause for your troubles. You specify: imh=size(ab,2); imw=size(ab,1); However, your for-loops have these conditions: j=1:imh+wh-1 and i=1:imw+ww-1 So you move past both the ‘height’ and the ‘width’ dimension. Try this instead: for j=1:imh-wh for i=1:imw-ww w1=ab(j:j+wh,i:i+wh,:); w2=salt(j:j+wh,i:i+wh,:); w3=w1-w2; end k=k+1; end 1 solved How to apply sliding … Read more
You have to project your image coordinates to a 2d plane. For a cone you could proceed like this: For a simple approach you could ignore the fact that it’s a truncated cone and treat it like a cylinder. For each point (alpha,r,z) on the cone calculate the angle alpha, distance r and height z. … Read more
If you have an RGB colour, you can convert it to HSV using suitable Color methods. Then adjust the saturation value as appropriate, and convert it back. 2 solved How to adjust saturation value in android [closed]
You can use a CNN: The input is then not 3 * w * h but (3*number of images) * w * h – so you can just concatenate the stuff in depth The output is just an image instead of a class. So no flattening in between… or a reshape has to be added. … Read more
maybe is the best to load all .mat files which contain parameters you want to use for training of neural network and put all your input parameters into one variable, and than to use that variable as the input for neural network. solved Training neural network in matlab