I have a string with the length of 480000 , it represents a 800x600 frame.
meaning 600 rows and 800 columns. i need to find a square of 3X3 were all the 9 values are in a specific range. i have the range itself. and return the index of the middle.
this is what i am doing right now:
# find 3x3 minValue < square < maxVlaue
for row in range(598):
for col in range(798):
if checkRow(frame,row,col,minValue,maxValue):
if checkRow(frame,row+1,col,minValue,maxValue):
if checkRow(frame,row+2,col,minValue,maxValue):
string = str(row+1),str(col+1)
print string
return
def checkRow(Frame,row,col,minValue,maxValue):
if (Frame[row*800+col])>minValue) and (Frame[row*800+col])<maxValue):
if (Frame[row*800+col+1])>minValue) and (Frame[row*800+col+1])<maxValue):
if (Frame[row*800+col+2])>minValue) and (Frame[row*800+col+2])<maxValue):
return True
return False
this way i check each "pixel" until the square is found.
is there some kind of special function in python for this job?
or maybe a faster and more efficient algorithm?
i thought about checking a pixel every 2 columns and skipping 2 rows. this way i can make sure i wont miss the 3x3 square, but this way when i hit a pixel , i need to check 9 different options for the square itself. so i am not sure how much faster it will be.
thanks
Related
I would like to calculate the median of each pixel in a set of images or "video". However, when MATLAB starts calculating this, it takes a very long time and finishes randomly with an index error. Why?
This is the code:
V = VideoReader('hall_monitor.avi');
info = get(V);
M = info.Width;
N = info.Height;
nb_frames_bk = 5;
v_pixel = zeros([nb_frames_bk 3]);
IB=zeros([M N 3],'double');
for i=1:M
for j=1:N
for k=1:nb_frames_bk
frm=read(V,k);
v_pixel(k,:)=frm(i,j,:);
end
IB(i,j,:)=median(v_pixel(:,:));
end
end
IB=uint8(IB);
imshow(IB);
This code can benefit from a lot of refactoring. For one thing, you are re-reading frames when you can just read them once, store them and use them after you're done.
Secondly, iterating over all pixels to compute your median is going to be very slow. From what it looks like in your code, for each spatial position over the first nb_frames_bk frames, you collect all of the RGB values within these frames and calculate the median RGB value.
Also as a minor note, you are getting a dimension exceeds error because you defined the output matrix wrong. You defined it as M x N with M being the width and N being the height. This needs to be swapped. Remember that matrices are defined as height first, width second. However, this is unnecessary with what I'm going to suggest for implementing this properly.
Instead of reading the frames one at a time, specify a range of frames. This way, you will get a 4D matrix where the first three dimensions references an image, with the fourth dimension representing the frame number. You can then take the median in the fourth dimension to find the median RGB value over all frames.
In other words, simply do this:
V = VideoReader('hall_monitor.avi');
nb_frames_bk = 5;
frms = read(V, [1 nb_frames_bk]);
IB = median(frms, 4);
imshow(IB);
This is much better, to the point and guaranteed to be faster. You also don't need to obtain the width and height of each frame as it is no longer needed as we are no longer looping over each pixel.
I would like to resize a 512X512 image into 363X762 image which will be larger than the original image(of size 512X512). Those extra pixel values must be different values in the range of 0-255.
I tried the following code:
I=imread('photo.jpg'); %photo.jpg is a 512X512 image
B=zeros(363,726);
sizeOfMatrixB=size(B);
display(sizeOfMatrixB);
B(1:262144)=I(1:262144);
imshow(B);
B(262155:263538)=0;
But I think this is a lengthy one and the output is also not as desired. Could anyone suggest me with a better piece of code to perform this. Thank you in advance.
I think that the code you have is actually pretty close to ideal except that you have a lot of hard-coded values in there. Those should really be computed on the fly. We can do that using numel to count the number of elements in B.
B = zeros(363, 726);
%// Assign the first 262144 elements of B to the values in I
%// all of the rest will remain as 0
B(1:numel(I)) = I;
This flexibility is important and the importance is actually demonstrated via the typo in your last line:
B(262155:263538)=0;
%// Should be
B(262144:263538)=0;
Also, you don't need these extra assignments to zero at the end because you initialize the matrix to be all zeros in the first place.
A Side Note
It looks like you are spreading the original image data for each column across multiple columns. I'm guessing this isn't what you want. You probably only want to grab the first 363 rows of I to be placed into B. You can do that this way:
B = zeros(363, 726);
B(1:size(B, 1), 1:size(I, 2)) = I(1:size(B, 1), :);
Update
If you want the other values to be something other than zero, you can initialize your matrix to be that value instead.
value = 2;
B = zeros(363, 726) + value;
B(1:numel(I)) = I;
If you want them to be random integers between 0 and 255, use randi to initialize the matrix.
B = randi([0 255], 363, 726);
B(1:numel(I)) = I;
I have two black and white images and I need to calculate the mutual information.
Image 1 = X
Image 2 = Y
I know that the mutual information can be defined as:
MI = entropy(X) + entropy(Y) - JointEntropy(X,Y)
MATLAB already has built-in functions to calculate the entropy but not to calculate the joint entropy. I guess the true question is: How do I calculate the joint entropy of two images?
Here is an example of the images I'd like to find the joint entropy of:
X =
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
Y =
0 0 0 0 0 0
0 0 0.38 0.82 0.38 0.04
0 0 0.32 0.82 0.68 0.17
0 0 0.04 0.14 0.11 0
0 0 0 0 0 0
To calculate the joint entropy, you need to calculate the joint histogram between two images. The joint histogram is essentially the same as a normal 1D histogram but the first dimension logs intensities for the first image and the second dimension logs intensities for the second image. This is very similar to what is commonly referred to as a co-occurrence matrix. At location (i,j) in the joint histogram, it tells you how many intensity values we have encountered that have intensity i in the first image and intensity j in the second image.
What is important is that this logs how many times we have seen this pair of intensities at the same corresponding locations. For example, if we have a joint histogram count of (7,3) = 2, this means that when we were scanning both images, when we encountered the intensity of 7, at the same corresponding location in the second image, we encountered the intensity of 3 for a total of 2 times.
Constructing a joint histogram is very simple to do.
First, create a 256 x 256 matrix (assuming your image is unsigned 8-bit integer) and initialize them to all zeroes. Also, you need to make sure that both of your images are the same size (width and height).
Once you do that, take a look at the first pixel of each image, which we will denote as the top left corner. Specifically, take a look at the intensities for the first and second image at this location. The intensity of the first image will serve as the row while the intensity of the second image will serve as the column.
Find this location in the matrix and increment this spot in the matrix by 1.
Repeat this for the rest of the locations in your image.
After you're done, divide all entries by the total number of elements in either image (remember they should be the same size). This will give us the joint probability distribution between both images.
One would be inclined to do this with for loops, but as it is commonly known, for loops are notoriously slow and should be avoided if at all possible. However, you can easily do this in MATLAB in the following way without loops. Let's assume that im1 and im2 are the first and second images you want to compare to. What we can do is convert im1 and im2 into vectors. We can then use accumarray to help us compute the joint histogram. accumarray is one of the most powerful functions in MATLAB. You can think of it as a miniature MapReduce paradigm. Simply put, each data input has a key and an associated value. The goal of accumarray is to bin all of the values that belong to the same key and do some operation on all of these values. In our case, the "key" would be the intensity values, and the values themselves are the value of 1 for every intensity value. We would then want to add up all of the values of 1 that map to the same bin, which is exactly how we'd compute a histogram. The default behaviour for accumarray is to add all of these values. Specifically, the output of accumarray would be an array where each position computes the sum of all values that mapped to that key. For example, the first position would be the summation of all values that mapped to the key of 1, the second position would be the summation of all values that mapped to the key of 2 and so on.
However, for the joint histogram, you want to figure out which values map to the same intensity pair of (i,j), and so the keys here would be a pair of 2D coordinates. As such, any intensities that have an intensity of i in the first image and j in the second image in the same spatial location shared between the two images go to the same key. Therefore in the 2D case, the output of accumarray would be a 2D matrix where each element (i,j) contains the summation of all values that mapped to key (i,j), similar to the 1D case that was mentioned previously which is exactly what we are after.
In other words:
indrow = double(im1(:)) + 1;
indcol = double(im2(:)) + 1; %// Should be the same size as indrow
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
With accumarray, the first input are the keys and the second input are the values. A note with accumarray is that if each key has the same value, you can simply assign a constant to the second input, which is what I've done and it's 1. In general, this is an array with the same number of rows as the first input. Also, take special note of the first two lines. There will inevitably be an intensity of 0 in your image, but because MATLAB starts indexing at 1, we need to offset both arrays by 1.
Now that we have the joint histogram, it's really simple to calculate the joint entropy. It is similar to the entropy in 1D, except now we are just summing over the entire joint probability matrix. Bear in mind that it will be very likely that your joint histogram will have many 0 entries. We need to make sure that we skip those or the log2 operation will be undefined. Let's get rid of any zero entries now:
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
Take notice that I searched the joint histogram instead of the joint probability matrix. This is because the joint histogram consists of whole numbers while the joint probability matrix will lie between 0 and 1. Because of the division, I want to avoid comparing any entries in this matrix with 0 due to numerical roundoff and instability. The above will also convert our joint probability matrix into a stacked 1D vector, which is fine.
As such, the joint entropy can be calculated as:
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
If my understanding of calculating entropy for an image in MATLAB is correct, it should calculate the histogram / probability distribution over 256 bins, so you can certainly use that function here with the joint entropy that was just calculated.
What if we have floating-point data instead?
So far, we have assumed that the images that you have dealt with have intensities that are integer-valued. What if we have floating point data? accumarray assumes that you are trying to index into the output array using integers, but we can still certainly accomplish what we want with this small bump in the road. What you would do is simply assign each floating point value in both images to have a unique ID. You would thus use accumarray with these IDs instead. To facilitate this ID assigning, use unique - specifically the third output from the function. You would take each of the images, put them into unique and make these the indices to be input into accumarray. In other words, do this instead:
[~,~,indrow] = unique(im1(:)); %// Change here
[~,~,indcol] = unique(im2(:)); %// Change here
%// Same code
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
Note that with indrow and indcol, we are directly assigning the third output of unique to these variables and then using the same joint entropy code that we computed earlier. We also don't have to offset the variables by 1 as we did previously because unique will assign IDs starting at 1.
Aside
You can actually calculate the histograms or probability distributions for each image individually using the joint probability matrix. If you wanted to calculate the histograms / probability distributions for the first image, you would simply accumulate all of the columns for each row. To do it for the second image, you would simply accumulate all of the rows for each column. As such, you can do:
histogramImage1 = sum(jointHistogram, 1);
histogramImage2 = sum(jointHistogram, 2);
After, you can calculate the entropy of both of these by yourself. To double check, make sure you turn both of these into PDFs, then compute the entropy using the standard equation (like above).
How do I finally compute Mutual Information?
To finally compute Mutual Information, you're going to need the entropy of the two images. You can use MATLAB's built-in entropy function, but this assumes that there are 256 unique levels. You probably want to apply this for the case of there being N distinct levels instead of 256, and so you can use what we did above with the joint histogram, then computing the histograms for each image in the aside code above, and then computing the entropy for each image. You would simply repeat the entropy calculation that was used jointly, but apply it to each image individually:
%// Find non-zero elements for first image's histogram
indNoZero = histogramImage1 ~= 0;
%// Extract them out and get the probabilities
prob1NoZero = histogramImage1(indNoZero);
prob1NoZero = prob1NoZero / sum(prob1NoZero);
%// Compute the entropy
entropy1 = -sum(prob1NoZero.*log2(prob1NoZero));
%// Repeat for the second image
indNoZero = histogramImage2 ~= 0;
prob2NoZero = histogramImage2(indNoZero);
prob2NoZero = prob2NoZero / sum(prob2NoZero);
entropy2 = -sum(prob2NoZero.*log2(prob2NoZero));
%// Now compute mutual information
mutualInformation = entropy1 + entropy2 - jointEntropy;
Hope this helps!
To segment above image i want to use line height. I don't know even any algorithm for this. I want a useful link pls help..
just i want an efficient algorithm link for this.....
I shall be very thankful to you for this..
// i ahve tried till now
function [avmax avmin avgwidth]=firstsvg(Imag)
%Imag=imread('D:\THAPAR\poj\images\ndpj.jpg');
imtool(Imag);
G=Imag;
%xlswrite('G.xlsx',Imag(:,:,1));
[y,x]=size(G); % y dentoes rows and x denotes columns
T=160;%sum(sum(I))/(y*x)%T dentoes threshhold value - i.e avg
HYT=zeros(1,1);
GY=zeros(y,1); %single column of zeros
for j=1:y
for i=1:x
if (G(j,i)<T)
GY(j,1)=GY(j,1)+1; % count of no. of black pixel
end
end
end
for c=1:y
if (GY(c,1)> min)
min = GY(c,1);
gt(ce,1)=GY(c,1);
ce=ce+1;
end
end
dgt=zeros(ce,1);
for b=1:(ce-2)
dgt(b,1)= gt(b+1,1)- gt(b,1);
end
mdgt= mean(dgt);
avgwidth= thyt;
avmax =thyt;
avmin = gt(1,1);
just i want an algorithm link to calculate line height ???
Lets say you have image of M x N dimension where M is row and N is column. Perform logical OR for pixels in row. If you get 1, then you have atleast one black pixel in particular row.
Just see the number of consecutive 1's you are getting, it will be the approximate height of each sentence.
I have one image in bmp format, with size of 512*512. I want to count the number of pixels with values more than 11 and then find the average of these pixels. Here is my code. I don't know what is the problem but the sum of pixel values is wrong and it is always 255. I tried with different images.
Could you please help me to figure it out?
A=imread('....bmp');
sum=0; count=0;
for i=1:512
for j=1:512
if (A(i,j)>=11)
sum=sum+A(i,j);
count=count+1;
end
end
end
disp('Number of pixels grater than or equal to 11')
disp(count)
disp('sum')
disp(sum)
disp('Average')
Avrg=sum/count;
disp(Avrg)
Why doesn't your code work
Difficult to tell, could you display a portion of your matrix and the size using something like
disp(A(1:10,1:10))
disp(size(A))
% possibly also the min and max...
disp(min(A(:))
disp(max(A(:))
just to be sure the format of A is as you expect - imread could have given you a 512x512x3 matrix if the image was read in color, or the image may be in the interval [0,1].
Better approach
Once you're sure that the matrix is indeed 512x512, and has values above 11, you're best off by generating a mask, i.e.
mask = A > 11;
numabove11 = sum(mask(:));
avabove11 = mean(A(mask));
Also in your code you use >= i.e. greater than or equal to, but you say 'greater than' - pick which you want and be consistent.
Explanation
So what do these 3 lines do?
Generate a logical matrix, same size as A that is true wherever A > 11, else false.
Sum the logical matrix, which means sum values that are 1 everywhere that A > 11, else 0 (boolean values are converted to floats for this summation).
Index in to matrix A using logical indexing, and take the mean of those values.
Avoid shadowing builtins
In your code you use the variable sum - this is bad practice as there is a builtin matlab function with the same name, which becomes unusable if you use a variable of the same name.
I also faced a similar problem and actually the solution lies in the fact that matlab stores A(i,j) in uint8 format whose maximum value is 255, so, just change the statement:
sum=sum+A(i,j);
to
sum=sum+double(A(i,j));
I hope this helps.