x=imread('test.jpg');
imshow(x);
total = 0;
for i=1:2
for j=1:2
for k=1:2
total = total + abs(x(i,j,k));
end
end
end
total
The above code prints total as 255 no matter what are the max values for i,j,k. Please explain
It prints out 255 because matlab doesn't overflow integers, and the datatype is uint8
a = repmat(uint8(100),5, 1)
a(1)+a(2)
a(1)+a(2)+a(3)
The outputs will be 200 and 255, because Matlab clamps the output at the maximum value, rather than wrapping around. If you use the sum function as given by Dennis, then you get the correct value as Matlab converts to double first
sum(a)
should give 500 as the output.
Not sure what your code fragment is, but if you want to sum of the absolute values of the array it is really easy:
sum(abs(x(:)))
If you just want the submatrix containing the first 2 values from the corner:
subM= x(1:2,1:2,1:2)
sum(abs(subM(:)))
Related
I have a 64 X 64 matrix that I need to find the column-wise mean values for.
However, instead of dividing by the total number of elements in each column (i.e. 64), I need to divide by the total number of non-zeros in the matrix.
I managed to get it to work for a single column as shown below. For reference, the function that generates my matrix is titled fmu2(i,j).
q = 0;
for i = 1:64
if fmu2(i,1) ~= 0;
q = q + 1;
end
end
for i = 1:64
mv = (1/q).*sum(fmu2(i,1));
end
This works for generating the "mean" value of the first column. However, I'm having trouble looping this procedure so that I will get the mean for each column. I tried doing a nested for loop, but it just calculated the mean for the entire 64 X 64 matrix instead of one column at a time. Here's what I tried:
q = 0;
for i = 1:64
for j = 1:64
if fmu2(i,j) ~= 0;
q = q +1;
end
end
end
for i = 1:64
for j = 1:64
mv = (1/q).*sum(fmu2(i,j));
end
end
Like I said, this just gave me one value for the entire matrix instead of 64 individual "means" for each column. Any help would be appreciated.
For one thing, do not call the function that generates your matrix in each iteration of a loop. This is extremely inefficient and will cause major problems if your function is complex enough to have side effects. Store the return value in a variable once, and refer to that variable from then on.
Secondly, you do not need any loops here at all. The total number of nonzeros is given by the nnz function (short for number of non-zeros). The sum function accepts an optional dimension argument, so you can just tell it to sum along the columns instead of along the rows or the whole matrix.
m = fmu2(i,1)
averages = sum(m, 1) / nnz(m)
averages will be a 64-element array with an average for each column, since sum(m, 1) is a 64 element sum along each column and nnz(m) is a scalar.
One of the great things about MATLAB is that it provides vectorized implementations of just about everything. If you do it right, you should almost never have to use an explicit loop to do any mathematical operations at all.
If you want the column-wise mean of non-zero elements you can do the following
m = randi([0,5], 5, 5); % some data
avg = sum(m,1) ./ sum(m~=0,1);
This is a column-wise sum of values, divided by the column-wise number of elements not equal to 0. The result is a row vector where each element is the average of the corresponding column in m.
Note this is very flexible, you could use any condition in place of ~=0.
i want to know the coding to find the maximum value, over a certain range.
i already coded like below.
f=f'
ac_yyyy_f=ac_yyyy_f'
[row,col] = ind2sub(size(ac_yyyy_f),find(ac_yyyy_f==max(ac_yyyy_f)))
but the problem is, sometimes the maximum value of Y axis choosen by my code is not what i want.
the X axis has the range of 0 to 100000 and i want the maximum between 20000 to 100000. the problem is sometimes the max value show up at the range of 0 to 20000.
How can i figure this out?
Use the max() function:
%let R be your range of values
R = [2 1 7 4];
[value, index] = max(R);
In the above example, value will be 7 and index will be 3
For more info: http://fr.mathworks.com/help/matlab/ref/max.html
I'm using a vector of random integers to stand in for your function output.
a = floor(rand(1,100000)*100);
[val, idx] = max(a(20000:100000));
You want to use the max function here to find the maximum value rather than find.
Now, the other part of the task is getting the max value from a certain part of your matrix. You can pass just a subset of a vector or matrix to a function by indexing it with a range of values. Note that idx gives you the position of val within a(20000:100000). If you need the position within a, you need to use idx+19999.
Also, you should take a look at the matrix indexing reference—there are many different and fun ways to index a matrix—because indexing is one of the most important features of matlab.
Here's the reference for the max function:
http://www.mathworks.com/help/matlab/ref/max.html
And the reference for indexing:
http://www.mathworks.com/help/matlab/math/matrix-indexing.html
You can use the max function with a subset of your array. It will return the maximum value, as well as the index where it is located. Be sure to correct the index it returns you based on your desired range. Like this:
%//create an array of 100,000 values to play with.
f=floor(rand(100000,1).*100);
%//find the max between f(20000) and f(100000)
[myMax, I] = max( f(20000:100000) );
%//correct the index based on where we started looking
%//for the max. Subtract 1 because it's MATLAB!
myIndex = I+20000-1;
This results in:
>> myMax
myMax =
99
>> myIndex
myIndex =
20045
>> f(myIndex)
ans =
99
How to increment a variable by a infinite set of numbers, in Matlab. I'v a variable which I want to increment till the loop ends by 0.1 every time but through set of range.
I'm currently doing this by: K=K*0.1; %K = 2 initially but I want this same by Matlab's trick of ranged values like [0.1:0.1:9] where 9 is the loop condination.
My Code:
K=2;
for ii=1:9
K=K*0.1;
end
If I understand correctly:
for K = 2 * 0.1.^(1:9)
%// do something with K
end
You can try using the cumprod command which returns the cumulative product of the elements in a matrix or vector. For your example, something like:
K=cumprod([2 repmat(0.1,1,9)]); % returns a row vector of 9 elements
repmat just creates a row vector of nine elements each set to the value 0.1. The last element in the vector, K(end), will be the product returned by your example. i.e.K = 2*0.1^9;
I am new to Matlab and am trying to implement code to perform the same function as histeq without actual use of the function. In my code the image colour I get changes drastically when it should not change that much. The average intensity in the image (ranging between 0 and 255) is 105.3196. The image is of an open source pollen particle.
Any help would be much appreciated. The sooner the better! Please could any help be simplified as my Matlab understanding is limited. Thanks.
clc;
clear all;
close all;
pollenJpg = imread ('pollen.jpg', 'jpg');
greyscalePollen = rgb2gray (pollenJpg);
histEqPollen = histeq(greyscalePollen);
averagePollen = mean2 (greyscalePollen)
sizeGreyScalePollen = size(greyscalePollen);
rowsGreyScalePollen = sizeGreyScalePollen(1,1);
columnsGreyScalePollen = sizeGreyScalePollen(1,2);
for i = (1:rowsGreyScalePollen)
for j = (1:columnsGreyScalePollen)
if (greyscalePollen(i,j) > averagePollen)
greyscalePollen(i,j) = greyscalePollen(i,j) + (0.1 * averagePollen);
if (greyscalePollen(i,j) > 255)
greyscalePollen(i,j) = 255;
end
elseif (greyscalePollen(i,j) < averagePollen)
greyscalePollen(i,j) = greyscalePollen(i,j) - (0.1 * averagePollen);
if (greyscalePollen(i,j) > 0)
greyscalePollen(i,j) = 0;
end
end
end
end
figure;
imshow (pollenJpg);
title ('Original Image');
figure;
imshow (greyscalePollen);
title ('Attempted Histogram Equalization of Image');
figure;
imshow (histEqPollen);
title ('True Histogram Equalization of Image');
To implement the equalisation algorithm described on the Wikipedia page, follow these these steps:
Decide on a binSize to group greyscale values. (This is a tweakable, the larger the bin, the less accurate the result from the ideal case, but I think it can cause problems if chosen too small on real images).
Then, calculate the probability of a pixel being a shade of grey:
pixelCount = imageWidth * imageHeight
histogram = all zero
for each pixel in image at coordinates i, j
histogram[floor(pixel / 255 / 10) + 1] += 1 / pixelCount // 1-based arrays, not 0-based
// Note a technicality here: you may need to
// write special code to handle pixels of 255,
// because they will fall in their own bin. Or instead use rounding with an offset.
The histogram in this calculation is scaled (divided by the pixel count) so that the values make sense as probabilities. You can of course factor the division out of the for loop.
Now you need to calculate the accumulative sum of this:
histogramSum = all zero // length of histogramSum must be one bigger than histogram
for i == 1 .. length(histogram)
histogramSum[i + 1] = histogramSum[i] + histogram[i]
Now you have to invert this function and this is the tricky part. The best is to not calculate an explicit inverse, but calculate it on the spot, and apply it on the image. The basic idea is to search for the pixel value in the histogramSum (find the closest index below), and then do a linear interpolation between the index and the next index.
foreach pixel in image at coordinates i, j
hIndex = findIndex(pixel, histogramSum) // You have to write findIndex, it should be simple
equilisationFactor = (pixel - histogramSum[hIndex])/(histogramSum[hIndex + 1] - histogramSum[hIndex]) * binSize
// This above is the linear interpolation step.
// Notice the technicality that you need to handle:
// histogramSum[hIndex + 1] may be out of bounds
equalisedImage[i, j] = pixel * equilisationFactor
Edit: without drilling into the maths, I can't be 100% sure, but I think that division by 0 errors are possible. These can occur if one bin is empty, so consecutive sums are equal. So you need special code to handle this case too. The best you can do is take the value for the factor as halfway between hIndex, hIndex + n, where n is the highest value for which histogramSum[hIndex + n] == histogramSum[hIndex].
And that should be it, once you have dealt with all the technicalities.
The above algorithm is slow (especially in the findIndex step). You may be able to optimize this with a special lookup datastructure. But only do that when it's working, and only if necessary.
One more thing about your Matlab code: the rows and columns are inverted. Because of the symmetry in the algorithm, the result is the same, but it can cause puzzling bugs in other algorithms, and be very confusing if you examine pixel values during debugging. In the pseudocode above I used them the same as you, though.
Relatively few (5) lines of code can do this. I used a low contrast file called 'pollen.jpg' that I found at http://commons.wikimedia.org/wiki/File%3ALepismium_lorentzianum_pollen.jpg
I read it in using your code, run all the above, then do the following:
% find out the index of pixels sorted by intensity:
[gv gi] = sort(greyscalePollen(:));
% create a table of "approximately equal" intensity values:
N = numel(gv);
newVals = repmat(0:255, [ceil(N/256) 1]);
% perform lookup:
% the pixels in sorted order need new values from "equal bins" table:
newImg = zeros(size(greyscalePollen));
newImg(gi) = newVals(1:N);
% if the size of the image doesn't divide into 256, the last bin will have
% slightly fewer pixels in it than the others
When I run this algorithm, and then create a composite of the four images (original, your attempt, my attempt, and histeq), you get the following:
I think it's convincing. The images are not exactly identical - I believe that is because the matlab histeq routine ignores all pixels with value 0. Since it is fully vectorized it is also pretty fast (although not nearly as fast as histeq by about a factor 15 on my image.
EDIT: a bit of explanation might be in order. The repmat command I use to create the newVals matrix creates a matrix that looks like this:
0 1 2 3 4 ... 255
0 1 2 3 4 ... 255
0 1 2 3 4 ... 255
...
0 1 2 3 4 ... 255
Since matlab stores matrices in "first index first" order, if you read this matrix with a single index (as I do in the line newVals(1:N)), you access first all the zeros, then all the ones, etc:
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 ...
So - when I know the indices of the pixels in the order of their intensity (as returned by the second argument of the sort command, which I called gi), then I can easily assign the value 0 to the first N/256 pixels, the value 1 to the next N/256 etc, with the command I used:
newImg(gi) = newVals(1:N);
I hope this makes the code a little easier to understand.
hi i was doing the following in matlab
values = [0;1;0;0;1;0;1;0]; % can contain only 0s and 1s
h = modem.oqpskmod;
y = modulate(h, values);
g = modem.oqpskdemod(h);
z = demodulate(g,y);
BER = sum(logical(values(:)-z(:)))/numel(values); % thanks to gnovice!
before calculating the BER, do you think that i have to convert the array z into array of 1s and 0s only... for example, my z may consist of 3s and 0s, then do i have to convert all numbers bigger than 0 to 1s and all 0s unchanged. so that at the end i get the correct comparison for the BER?
This depends on how you want to do the comparison of the input and output. The code I gave you for computing the BER will count any difference between the input and output as an error. So, a value of 3 for an output and a corresponding 1 for an input are counted as different.
If you want to compute the BER in another way, here are a couple more options you could use. Each of these assumes values contains only zeroes and ones:
Count any non-zero output value as a 1: In this case, you would use the function LOGICAL to convert any non-zero value of z (i.e. the values 3, 0.5, -1, etc.) to a 1, leaving the values that are exactly zero unchanged:
BER = sum(values(:)-logical(z(:)))/numel(values);
Count positive values as 1 and negative values as 0: This solution just uses a comparison operator instead of LOGICAL:
BER = sum(values(:)-(z(:) > 0))/numel(values);
You may also want to try setting the 'OutputType' property of the modem.oqpskdemod object to 'bit' to ensure a binary output. Then you shouldn't have to do anything to z:
...
g = modem.oqpskdemod(h,'OutputType','bit');
z = demodulate(g,y);
BER = sum(values(:)-z(:))/numel(values);