I have a lot of images in IM{}. I want to calculate mean graylevel of non-black pixels. When I run my code sum has 255 as a maximum value. I don't understand the reason. Why doesn't sum get higher values?
for i=1: length(IM)
[L,W,z]=size( IM{i});
k=1;
sum=0;
for L=1:L
for W=1: W
if IM{i}(L,W)~=0;
sum=IM{i}(L,W)+sum;
k=k+1;
end
end
end
Mean(i)=sum/k
end
That's probably because IM is of type uint8. This data type can't hold values larger than 255. Example:
>> uint8(200) + uint8(200)
ans =
255
To avoid this, you should convert IM to double:
IM = double(IM);
Anyway, your code could be reduced to a single line (including the conversion):
result = mean(double(IM(IM>0)));
With this approach, you could even dispense with double, because mean (actually sum, which is called by mean) converts to double automatically:
result = mean(IM(IM>0));
Related
I have created code that does a Newton's Method approximation. It prints in a table-like format the approximation at each step and the associated error. I want to add a column that shows an integer value that represents the number of correct digits in approximation against the true value.
I am attempting to convert each cell of approximation into a string and counting how many digits are accurate. Example, approx. = 3.14555, true = 3.1555. The number of accurate digits will be 2. Although I have this idea in my head, I am doing it all wrong in my code below. Do you know how to create a proper loop to achieve this? I have less than a year of MATLAB experience; my mental toolbox is limited.
% Program Code of Newton's Method to find root
% This program will not produce a result if initial guess is too far from
% true value
clear;clc;format('long','g')
% Can work for various functions
%FUNCTION: 2*x*log(x)-2*log(x)*x^(3)+2*x^(2)*log(x)-x^(2)+1
%INTIAL GUESS: .01
%ERROR: 1.e-8
a=input('Enter the function in the form of variable x:','s');
x(1)=input('Enter Initial Guess:');
error=input('Enter allowed Error:');
% Passing through the function and calculating the derivative
f=inline(a);
dif=diff(str2sym(a));
d=inline(dif);
% Looping through Newton's Method
for i=1:100
x(i+1)=x(i)-((f(x(i))/d(x(i))));
err(i)=abs(x(i+1)-x(i));
% The loop is broken if acceptable error magnitude is reached
if err(i)<error
break
end
end
root=x(i);
Root = (x(:,1:(end-1)))';
Error = err';
disp('The final approximation is:')
disp(root)
%BELOW IS ALL WRONG, I AM TRYING TO ADD A COLUMN TO 'table'
%THAT SHOWS HOW MANY DIGITS IN APPROXIMATION IS ACCURATE
iter = 0;
y = zeros(1,length(x));
plot(x,y,'+')
zero1 = ('0.327967785331818'); %ACTUAL VALUE
for i = 1:length(Root)
chr = mat2str(Root(i))
for j = 1:length(chr(i))
if chr(i)~=zero1(i)
iter = 0;
return
elseif chr(i)==zero1(i)
iter = iter + 1;
acc(i) = iter
end
end
end
table(Root, Error) %ADD ACCURACY COLUMN HERE
Perhaps something like multiplying both numbers by powers of 10 and then flooring them until the answers are no longer equal:
approx=3.14555;
truth=3.1555;
approx1=0;
truth1=0;
i=0;
while approx1==truth1
approx1=floor(approx*10^i);
truth1=floor(truth*10^i);
i=i+1;
end
acc=i-1;
I do not know what this error means or how to fix it. I am trying to perform an image rotation in a separate space of coordinates. When defining the reference space of the matrix to be at zero, I am getting the error that integers can only be comibined with integers of the same class or scalar doubles. the line is
WZcentered = WZ - [x0;yo]*ones(1,Ncols);
WZ is classified as a 400x299x3 unit 8, in the workspace. It is an image. x0 and y0 are set to 0 when the function is called. How can I fix this issue/what exactly is happening here?
Also, when I do the same thing yet make WZ to be equal to double(WZ) I get the error that 'matrix dimensions must agree.' I am not sure what the double function does however. Here is the whole code.
function [out_flag, WZout, x_final, y_final] = adopted_moveWZ(WZ, x0, y0);
%Initial Test of plot
[Nrows,Ncols]=size(WZ);
if Nrows ~= 2
if Ncols ==2
WZ=transpose(WZ); %take transpose
[Nrows,Ncols]=size(WZ); %reset the number of rows and columns
else
fprintf('ERROR: Input file should have 2-vectors for the input points.\n');
end
end
plot(WZ(1,:),WZ(2,:),'.')
title('These are the original points in the image');
pause(2.0)
%WZorig = WZ;
%centering
WZcentered = WZ - ([x0;y0] * ones(1,Ncols));
FigScale=400;
axis([-FigScale 2*FigScale -FigScale 2*FigScale])
disp('Hit any key to start the animation');
pause;
SceneCenter = zeros(Nrows,Ncols);
WZnew = WZcentered;
for ii=0:20
%rotate
R = [cos(pi/ii) -sin(pi/ii) 0; sin(pi/ii) cos(pi/ii) 0; 0 0 1];
WZnew = R * WZnew;
plot(WZnew(1,:),WZnew(2,:),'.')
%place WZnew at a different place in the scene
SceneCenter = (ii*[30;40])*ones(1,Ncols);
plot(SceneCenter(1,:) + WZnew(1,:), SceneCenter(2,:) + WZnew(2,:),'.')
axis([-FigScale 2*FigScale -FigScale 2*FigScale])
pause(1.0);
end
%Set final values for output at end of program
x_final = SceneCenter(1,1);
y_final = SceneCenter(2,1);
PPout = PPnew + SceneCenter;
This happens due to WZ and ([x0;y0] * ones(1,Ncols)) being of different data types. You might think MATLAB is loosely typed, and hence should do the right thing when you have a floating point type operated with an integer type, but this rule breaks every once in a while. A simpler example to demonstrate this is here:
X = uint8(magic(5))
Y = zeros(5)
X - Y
This breaks with the same error that you are reporting. One way to fix this is to force cast one of the operands to the other, typically up-casted to make sure the math works. When you do this, both the numbers you are working on are floating point (double precision), and so they are represented in the same byte formatting sequence in memory. This way, the '-' sign is valid, in the same way that you can say 3 apples + 4 apples = 7 apples, but 3 oranges (uint8) + 4 apples (double) = ?. The double(X) makes it clear that you really mean to use double precision arithmetic, and hence fixes the error. This is how it looks now:
double(X) - Y
After having identified this, the new error is 'matrix dimensions do not match'. This means exactly what it says. WZ is a 400x299x3 matrix, and the right hand side matrix is 2xnCols. Now can you subtract a 2D matrix from a 3D matrix of different sizes meaningfully?
Depending on what your code is really intending to do, you can pad the RHS matrix, or find out other ways to make the sizes equal.
All of this is why MATLAB includes routines to do image rotation, namely http://www.mathworks.com/help/images/ref/imrotate.html . This is part of the Image Processing Toolbox, though.
I'm trying to quantize a set of double type samples with 128 level uniform quantizer and I want my output to be double type aswell. When I try to use "quantize" matlab gives an error: Inputs of class 'double' are not supported. I tried "uencode" as well but its answer was nonsense. I'm quite new to matlab and I've been working on this for hours. Any help appriciated. Thanks
uencode is supposed to give integer results. Thats the point of it. but the key point is that it assumes a symmetric range. going from -x to +x where x is the largest or smallest value in your data set. So if your data is from 0-10 your result looks like nonsense because it quantizes the values on the range -10 to 10.
In any event, you actually want the encoded value and the quantized value. I wrote a simple function to do this. It even has little help instructions (really just type "help ValueQuantizer"). I also made it very flexible so it should work with any data size (assuming you have enough memory) it can be a vector, 2d array, 3d, 4d....etc
here is an example to see how it works. Our number is a Uniform distribution from -0.5 to 3.5 this shows that unlike uencode, my function works with nonsymmetric data, and that it works with negative values
a = 4*rand(2,4,2) - .5
[encoded_vals, quant_values] = ValueQuantizer(a, 3)
produces
a(:,:,1) =
0.6041 2.1204 -0.0240 3.3390
2.2188 0.1504 1.4935 0.8615
a(:,:,2) =
1.8411 2.5051 1.5238 3.0636
0.3952 0.5204 2.2963 3.3372
encoded_vals(:,:,1) =
1 4 0 7
5 0 3 2
encoded_vals(:,:,2) =
4 5 3 6
1 1 5 7
quant_values(:,:,1) =
0.4564 1.8977 -0.0240 3.3390
2.3781 -0.0240 1.4173 0.9368
quant_values(:,:,2) =
1.8977 2.3781 1.4173 2.8585
0.4564 0.4564 2.3781 3.3390
so you can see it returns the encoded values as integers (just like uencode but without the weird symmetric assumption). Unlike uencode, this just returns everything as doubles rather than converting to uint8/16/32. The important part is it also returns the quantized values, which is what you wanted
here is the function
function [encoded_vals, quant_values] = ValueQuantizer(U, N)
% ValueQuantizer uniformly quantizes and encodes the input into N-bits
% it then returns the unsigned integer encoded values and the actual
% quantized values
%
% encoded_vals = ValueQuantizer(U,N) uniformly quantizes and encodes data
% in U. The output range is integer values in the range [0 2^N-1]
%
% [encoded_vals, quant_values] = ValueQuantizer(U, N) uniformly quantizes
% and encodes data in U. encoded_vals range is integer values [0 2^N-1]
% quant_values shows the original data U converted to the quantized level
% representing the number
if (N<2)
disp('N is out of range. N must be > 2')
return;
end
quant_values = double(U(:));
max_val = max(quant_values);
min_val = min(quant_values);
%quantizes the data
quanta_size = (max_val-min_val) / (2^N -1);
quant_values = (quant_values-min_val) ./ quanta_size;
%reshapes the data
quant_values = reshape(quant_values, size(U));
encoded_vals = round(quant_values);
%returns the original numbers in their new quantized form
quant_values = (encoded_vals .* quanta_size) + min_val;
end
As far as I can tell this should always work, but I haven't done extensive testing, good luck
There's a peculiar bug in my code that I can't seem to figure out. For context, I is an image, a matrix, consisting of scaled values from 0 to 255. GetSpatAvg is a function that is not included here. The problem I'm facing is that the elements in ngtdm always max out at 255. When this function finishes, I get the matrix ngtdm back consisting of many many values that are 255. The code is below exactly as it shows on my computer.
function ngtdm = getNGTDM(I,d)
[rowI, colI] = size(I);
ngtdm = zeros(256, 1);
for r=1+d:rowI-d
for c=1+d:colI-d
term = I(r,c)-getSpatAvg(r,c);
ngtdm(I(r,c)+1)=ngtdm(I(r,c)+1)+term;
end
end
end
I isolated a specific value in I, 254, in the code below.
function ngtdm = getNGTDM(I,d)
[rowI, colI] = size(I);
ngtdm = zeros(256, 1);
for r=1+d:rowI-d
for c=1+d:colI-d
if(I(r,c)==254)
term = I(r,c)-getSpatAvg(r,c);
disp(term);
ngtdm(I(r,c)+1)=ngtdm(I(r,c)+1)+term;
end
end
end
end
The variable 'term', in this instance was always 222. There were 369 instances of 222. Therefore the value at ngtdm(255) (254+1=255) should be larger than 255, yet the element still maxed out at 255. When I replace term with 222 as shown below:
ngtdm(I(r,c)+1)=ngtdm(I(r,c)+1)+222;
I get the correct value, a number larger than 255.
I can't seem to figure out why my element is always maxing out at 255. Could it be something to do with the fact that the values of I are scaled between 0 and 255. I'm pretty positive that getSpatAvg is not the issue because the correct value is being returned.
Thank You
It's not a bug - it sounds like you are using a uint8 datatype. If you convert to a datatype with more bits - e.g. uint16, uint32, single, double, etc - you will not run into this problem. I guess you are working with images, as images read using imread are read in as uint8 by default to save memory. Quickest fix: use I=double(I); either at the start of your function, or on the variable I before you put it into the functions.
I am trying to understand why the original image is not coming with this code. The resulting image receive is yellowish in color, instead of being similar to the image Img_new.
Img=imread(‘lena_color.tif’);
Img_new=rgb2gray(img);
Send=zeroes(size(Img_new);
Receive= zeroes(size(Img_new);
Mask= rand(size(Img_new);
for i=1 :256
for j=1:256
Send(i,j)=xor( Img_new(i,j),mask(i,j));
End
End
image(send);
imshow(send);
for i=1 :256
for j=1:256
receive(i,j)=xor( send(i,j),mask(i,j));
End
End
image(receive);
imshow(receive);
What am I doing wrong?
There are several problems in your code.
MATLAB is case sensitive so end and End is not the same. Same goes for receive, and send.
MATLAB has a lot of matrix-based operations so please use for loops as a last resort since most of these operations can be executed by MATLAB's optimized routines for matrices.
MATLAB's xor returns the logical xor so when it sees two values (or matrices of values) it doesn't matter if it's 234 xor 123 or 12 xor 23 since it is equivalent to 1 xor 1 and 1 xor 1. You're looking for bitxor which does the bitwise xor on each element of the matrix and I've used it in my code below. This is the only way you can retrieve the information with the pixel == xor(xor(pixel,key),key) operation (assuming that's what you want to do).
rand returns a real value from 0 - 1 ; therefore, to do a successful bitwise xor, you need numbers from 0 - 255. Hence, in my code, you'll see that mask has random values from 0-255.
Note: I've used peppers.png since it's available in MATLAB. Replace it with lena_color.tif.
%%# Load and convert the image to gray
img = imread('peppers.png');
img_new = rgb2gray(img);
%%# Get the mask matrix
mask = uint8(rand(size(img_new))*256);
%%# Get the send and receive matrix
send = bitxor(img_new,mask);
receive = bitxor(send,mask);
%%# Check and display
figure;imshow(send,[0 255]);
figure;imshow(receive,[0 255]);
Update:
%%# Get mask and img somehow (imread, etc.)
img = double(img);
mask_rgb = double(repmat(mask,[1 1 3]));
bitxor(img,mask);
If instead, you choose to make everything uint8 instead of double, then I urge you to check if you are losing data anywhere. img is uint8 so there is no loss, but if any of the values of mask is greater than 255 then making it double will lead to a loss in data.