Can someone explain the last line of this MatLab expression? I need to convert this to C++ and I do not have any experience in matlab syntax.
LUT = zeros(fix(Max - Min),1);
Bin= 1+LUT(round(Image));
Image is an input image, Min and Max are image minimum and maximum grey levels.
Is Bin going to be an array? What shall it contain? What are the dimensions, same as LUT or Image? What is the '1' stands for (add 1 to each member of array or a shift in array positions? I cannot find any example of this.
Thanks in advance.
LUT is a column vector that has a number of entries that is equal to the difference in maximum and minimum intensities in your image. LUT(round(Image)) retrieves the entries in your vector LUT which are given by the command round(Image). The dimension of Bin will be equal to the size of your matrix Image, and the entries will be equal to the corresponding indices from the LUT vector. So, say you have a 3x3 matrix Image, whose rounded values are as follows:
1 2 3
2 2 4
1 5 1
Then LUT(round(Image)) will return:
LUT(1) LUT(2) LUT(3)
LUT(2) LUT(2) LUT(4)
LUT(1) LUT(5) LUT(1)
And 1+LUT(round(Image)) will return:
1+LUT(1) 1+LUT(2) 1+LUT(3)
1+LUT(2) 1+LUT(2) 1+LUT(4)
1+LUT(1) 1+LUT(5) 1+LUT(1)
Note that this only works if all entries in round(Image) are positive, because you can't use zero/negative indexing in the LUT vector (or any MATLAB matrix/vector, for that matter).
Related
I have a questions which is also mentioed in this answer and this one but I'm using the binary descriptor and need more informations:
I'm using BRAND descriptors as an input of LSH problem. The descriptor's size are 300*32 to 400*32, in which 32 is the length of the descriptor and images have 300 to 400 keypoints. The output of BRAND is a int matrix.
As this answer mentioned the inputs of LSH are vectors of D dimensions, it means every images will insert to the hash table as one vector of D dimension, now here are my question:
A. How can I convert the descriptor matrix of integers to a vector for input? Shall I just copy the matrix rows in sequence as a vector? OR Is it possible to convert each line of the descriptor, convert the items in binary and concatenate them to 256 bit binary and have a vector of D dimension of them?
B. Is it nessesary to convert the integer values of the BRAND descriptor to binary digits, in case of using L1 norm, that as far as I understood is the same Hamming distance for binary vectors?
Thank you very much in advance.
I have observations in vector form and I want to calculate the covariance matrix and the mean from these observations in OpenCV using calcCovarMatrix:
http://docs.opencv.org/modules/core/doc/operations_on_arrays.html
My current function call is:
calcCovarMatrix(descriptors.at(j).descriptor.t(), covar, mean, CV_COVAR_ROWS);
Whereas descriptors.at(j).descriptor.t() is one matrix with 2 columns and 390 rows. So my "random variables" are the rows of this matrix. The covar and mean are empty matrices.
The function calculates covar correctly and retuns a 390x390 matrix. But the mean is just a matrix with 1 row and 2 columns. I do not get this. I am expecting a matrix with 1 columnd and 390 rows (a column vector).
Am I using the wrong variant of the function? If yes, how should I use the correct variant in my case, I am specifically pointing to the value for the nsamples parameter. I don't know two what value to set it.
From time to time I have to port some Matlab Code to OpenCV.
Almost always there is a way to do it and an appropriate function in OpenCV. Nevertheless its not always easy to find.
Therefore I would like to start this summary to find and gather some equivalents between Matlab and OpenCV.
I use the Matlab function as heading and append its description from Matlab help. Afterwards a OpenCV example or links to solutions are appreciated.
Repmat
Replicate and tile an array. B = repmat(A,M,N) creates a large matrix B consisting of an M-by-N tiling of copies of A. The size of B is [size(A,1)*M, size(A,2)*N]. The statement repmat(A,N) creates an N-by-N tiling.
B = repeat(A, M, N)
OpenCV Docs
Find
Find indices of nonzero elements. I = find(X) returns the linear indices corresponding to the nonzero entries of the array X. X may be a logical expression. Use IND2SUB(SIZE(X),I) to calculate multiple subscripts from the linear indices I.
Similar to Matlab's find
Conv2
Two dimensional convolution. C = conv2(A, B) performs the 2-D convolution of matrices A and B. If [ma,na] = size(A), [mb,nb] = size(B), and [mc,nc] = size(C), then mc = max([ma+mb-1,ma,mb]) and nc = max([na+nb-1,na,nb]).
Similar to Conv2
Imagesc
Scale data and display as image. imagesc(...) is the same as IMAGE(...) except the data is scaled to use the full colormap.
SO Imagesc
Imfilter
N-D filtering of multidimensional images. B = imfilter(A,H) filters the multidimensional array A with the multidimensional filter H. A can be logical or it can be a nonsparse numeric array of any class and dimension. The result, B, has the same size and class as A.
SO Imfilter
Imregionalmax
Regional maxima. BW = imregionalmax(I) computes the regional maxima of I. imregionalmax returns a binary image, BW, the same size as I, that identifies the locations of the regional maxima in I. In BW, pixels that are set to 1 identify regional maxima; all other pixels are set to 0.
SO Imregionalmax
Ordfilt2
2-D order-statistic filtering. B=ordfilt2(A,ORDER,DOMAIN) replaces each element in A by the ORDER-th element in the sorted set of neighbors specified by the nonzero elements in DOMAIN.
SO Ordfilt2
Roipoly
Select polygonal region of interest. Use roipoly to select a polygonal region of interest within an image. roipoly returns a binary image that you can use as a mask for masked filtering.
SO Roipoly
Gradient
Approximate gradient. [FX,FY] = gradient(F) returns the numerical gradient of the matrix F. FX corresponds to dF/dx, the differences in x (horizontal) direction. FY corresponds to dF/dy, the differences in y (vertical) direction. The spacing between points in each direction is assumed to be one. When F is a vector, DF = gradient(F)is the 1-D gradient.
SO Gradient
Sub2Ind
Linear index from multiple subscripts. sub2ind is used to determine the equivalent single index corresponding to a given set of subscript values.
SO sub2ind
backslash operator or mldivide
solves the system of linear equations A*x = B. The matrices A and B must have the same number of rows.
cv::solve
Some details about my problem:
I'm trying to realize corner detector in openCV (another algorithm, that are built-in: Canny, Harris, etc).
I've got a matrix filled with the response values. The biggest response value is - the biggest probability of corner detected is.
I have a problem, that in neighborhood of a point there are few corners detected (but there is only one). I need to reduce number of false-detected corners.
Exact problem:
I need to walk through the matrix with a kernel, calculate maximum value of every kernel, leave max value, but others values in kernel make equal zero.
Are there build-in openCV functions to do this?
This is how I would do it:
Create a kernel, it defines a pixels neighbourhood.
Create a new image by dilating your image using this kernel. This dilated image contains the maximum neighbourhood value for every point.
Do an equality comparison between these two arrays. Wherever they are equal is a valid neighbourhood maximum, and is set to 255 in the comparison array.
Multiply the comparison array, and the original array together (scaling appropriately).
This is your final array, containing only neighbourhood maxima.
This is illustrated by these zoomed in images:
9 pixel by 9 pixel original image:
After processing with a 5 by 5 pixel kernel, only the local neighbourhood maxima remain (ie. maxima seperated by more than 2 pixels from a pixel with a greater value):
There is one caveat. If two nearby maxima have the same value then they will both be present in the final image.
Here is some Python code that does it, it should be very easy to convert to c++:
import cv
im = cv.LoadImage('fish2.png',cv.CV_LOAD_IMAGE_GRAYSCALE)
maxed = cv.CreateImage((im.width, im.height), cv.IPL_DEPTH_8U, 1)
comp = cv.CreateImage((im.width, im.height), cv.IPL_DEPTH_8U, 1)
#Create a 5*5 kernel anchored at 2,2
kernel = cv.CreateStructuringElementEx(5, 5, 2, 2, cv.CV_SHAPE_RECT)
cv.Dilate(im, maxed, element=kernel, iterations=1)
cv.Cmp(im, maxed, comp, cv.CV_CMP_EQ)
cv.Mul(im, comp, im, 1/255.0)
cv.ShowImage("local max only", im)
cv.WaitKey(0)
I didn't realise until now, but this is what #sansuiso suggested in his/her answer.
This is possibly better illustrated with this image, before:
after processing with a 5 by 5 kernel:
solid regions are due to the shared local maxima values.
I would suggest an original 2-step procedure (there may exist more efficient approaches), that uses opencv built-in functions :
Step 1 : morphological dilation with a square kernel (corresponding to your neighborhood). This step gives you another image, after replacing each pixel value by the maximum value inside the kernel.
Step 2 : test if the cornerness value of each pixel of the original response image is equal to the max value given by the dilation step. If not, then obviously there exists a better corner in the neighborhood.
If you are looking for some built-in functionality, FilterEngine will help you make a custom filter (kernel).
http://docs.opencv.org/modules/imgproc/doc/filtering.html#filterengine
Also, I would recommend some kind of noise reduction, usually blur, before all processing. That is unless you really want the image raw.
I am using letter_regcog example from OpenCV, it used dataset from UCI which have structure like this:
Attribute Information:
1. lettr capital letter (26 values from A to Z)
2. x-box horizontal position of box (integer)
3. y-box vertical position of box (integer)
4. width width of box (integer)
5. high height of box (integer)
6. onpix total # on pixels (integer)
7. x-bar mean x of on pixels in box (integer)
8. y-bar mean y of on pixels in box (integer)
9. x2bar mean x variance (integer)
10. y2bar mean y variance (integer)
11. xybar mean x y correlation (integer)
12. x2ybr mean of x * x * y (integer)
13. xy2br mean of x * y * y (integer)
14. x-ege mean edge count left to right (integer)
15. xegvy correlation of x-ege with y (integer)
16. y-ege mean edge count bottom to top (integer)
17. yegvx correlation of y-ege with x (integer)
example:
T,2,8,3,5,1,8,13,0,6,6,10,8,0,8,0,8
I,5,12,3,7,2,10,5,5,4,13,3,9,2,8,4,10
now I have segmented image of letter and want to transform it into data like this to put recognize it but I don't understand the mean of all value like "6. onpix total # on pixels" what is it mean ? Can you please explain the mean of these value. thanks.
I am not familiar with OpenCV's letter_recog example, but this appears to be a feature vector, or set of statistics about the image of a letter that is used to classify the future occurrences of the letter. The results of your segmentation should leave you with a binary mask with 1's on the letter and 0's everywhere else. onpix is simply the total count of pixels that fall on the letter, or in other words, the sum of your binary mask.
Most of the rest values in the list need to be calculated based on the set of pixels with a value of 1 in your binary mask. x and y are just the position of the pixel. For instance, x-bar is just the sample mean of all of the x positions of all pixels that have a 1 in the mask. You should be able to easily find references on the web for mathematical definitions of mean, variance, covariance and correlation.
14-17 are a little different since they are based on edge pixels, but the calculations should be similar, just over a different set of pixels.
My name is Antonio Bernal.
In page 3 of this article you will find a good description for each value.
Letter Recognition Using Holland-Style Adaptive Classifiers.
If you have any doubt let me know.
I am trying to make this algorithm work, but my problem is that I do not know how to scale the values to fit them to the range 0-15.
Do you have any idea how to do this?
Another Link from Google scholar -> Letter Recognition Using Holland-Style Adaptive Classifiers