So I'm trying to understand convolution and the process on making gradients and I wanted to just see the horizontal gradient of the 1x2 operator on an image named I1. When I tried to use this code I only get a black screen so I'm trying to figure out what went wrong here,sans using conv of course. (I'm also going to try out Sobel too, so I'd like some tips on how to get that going.)
I1 = uint8(round(sum(C1,3)/3));
figure,imshow(I1);
Kern =[-1,1];
Omega = zeros([size(I1,1) size(I1,2)]);
for i=1:ROWS
for j=1:COLS
Work = double(I1(i,j)).*Kern;
Omega(i,j) = sum(Work(:));
end
end
figure,imshow(uint8(Omega));
The problem is that you're only using 1 pixel from I1 to multiply by your kernel. Since you're using one value, the end effect is:
a.*[-1 1]
which gives you
[-a a]
When you sum this, obviously you get zero. To fix this, you'll need to use the same number of pixels from I1 as you have elements in your kernel (in this case, 2). This will also mean that you need to adjust your loop indices:
for i=1:ROWS-1 % avoid accessing outside image
for j=1:COLS-1
Work = double(I1(i,j:j+1)).*Kern; % j:j+1 gives us 2 pixels
Omega(i,j) = sum(Work(:));
end
end
You can also condense the two lines inside the loop into one:
Kern = [-1;1]; % make Kern a column vector
...
for i=1:ROWS-1
for j=1:COLS-1
Omega(i,j) = double(I1(i,j:j+1))*Kern; % vector multiplication, not elementwise
end
end
Another thing you might want to try is use imagesc(Omega) instead of imshow. imagesc will scale the values of the image so it's more visible.
Related
So I am trying to downsample an image using nested for loops. Here, i have a 359x479(widthxheight) image. I am trying to downsample it to a 180x240 image by removing the even rows and columns. Yet, it doesn't seem to be working. I end up getting the same image as the output.
a=imread('w.jpg'); %input image
a=im2double(a); %convert it to double
r=[[1 1 1];[1 1 1];[1 1 1]]/9; % the next 3 steps done to low pass
filter the image
c=imfilter(a,r,'replicate');
imshow(c);
for i=1:359 % for rows
for j=1:479 %for columns
if(mod(i,2)~=0) %to remove even rows
if(mod(j,2)~=0) %to remove odd columns
b(i,j)=c(i,j); %if i and j are odd, the pixel value is assigned to b
end
end
end
end
figure, imshow(b);
should get a 180x240 image but still getting the same image of size 359x479
You also need to assign only one pixel on two ! If you do not, half of yours columns/rows will contain only 0 value.
so you need to use:
b(ceil(i/2),ceil(j/2))=c(i,j);
where 2 correspond to the value of your modulo.
You could also avoid to use some loops by simply writing:
b = c(1:2:259,1:2:194);
I have a problem with the following code. I want to store all the values I am creating in the for loop below so that I can make a plot of it. I have tried several things, but nothing works. Does anyone know a simple method to create a vector of the results and then plot them?
dx=0.1;
t=1;
e=1;
for x=-1:dx:1
lower_bound=-100;
upper_bound=x/(sqrt(4*t*e));
e=1;
u=(1/sqrt(pi))*quad(#integ,lower_bound,upper_bound);
plot(x,u)
hold on
end
hold off
I would like to use as much of this matlab code as possible.
dx=0.1;
t=1;
e=1;
xval=[-1:dx:1].';
upper_bound = zeros(numel(xval),1);
u = zeros(numel(xval),1);
for ii=1:numel(xval)
x = xval(ii)
lower_bound=-100;
upper_bound(ii,1)=x/(sqrt(4*t*e));
u(ii,1)=(1/sqrt(pi))*quad(#integ,lower_bound,upper_bound(ii));
end
figure;
plot(xval,u)
by adding the (ii) behind your statements it saves your variables in an array. I did not use that on your lower_bound since it is a constant.
Note that I first created an array xval and called that with integers in ii, since subscriptindices must be positive integers in MATLAB. I also initialised both upper_bound and u by creating a zero matrix before the loop executes. This is handy since extending an existing vector is very memory and time consuming in MATLAB and since you know how big they will get (same number of elements as xval) you might as well use that.
I also got the plot call outside the loop, to prevent you from plotting 21 blue lines in 1 plot.
I'm currently having some trouble with the 'smooth' command, namely that it seems to have no effect on the generated plot. I have already used the following script to generate a plot
for h=1:4
linespec = {'rx', 'gx', 'bx', 'yx'};
hold on
for b=1:365
y=mean(mean(A(b,6*(h-1)+1:6*h)));
p(h)=plot(b,y,linespec{h});
end
hold off
end
Going row by row in data set A and taking the average of the values in the first six columns, then column 7 through 12, 13 through 18 and 19 through 14; generating four plots in total.
The next step was to smooth the resultant plot by averaging the values over a span of 9. So, I tweaked the script to the following;
for h=1:4
linespec = {'rx', 'gx', 'bx', 'yx'};
hold on
for b=1:365
y=mean(mean(A(b,6*(h-1)+1:6*h)));
w = smooth(y,9,'moving');
p(h)=plot(b,w,linespec{h});
end
hold off
end
Essentially just adding the w variable and replacing y with w in the plot command. Yet this has no effect whatsoever on my plot. Matlab doesn't throw up any errors either, so there doesn't seem to be a problem with the input size. Does anyone have an idea as to what the issue might be?
In either version of the loop, you appear to be plotting individual values of y against individual values of b. I presume, then, that y is a single value. You can't smooth a point, so the smooth operation is having no effect.
From the start, you don't need to make a loop to calculate the various means; mean can take a 2D matrix and return a vector. Calculate y in one go, then smooth that vector (should have length 365, I presume - depends on the size of input A). e.g.:
b = 1:365;
y=mean(A(:,6*(h-1)+1:6*h),2);
w = smooth(y,9,'moving');
plot(b,y,'rx');
hold on
plot(b,w,'gx');
I have recently been tasked with using a first derivative filter on an image of myself. The instructor said that I should first fix the value of y and preform f(x+1) - f(x) on the rows and then fix the new "X" values and preform f(y+1)-f(y) on the columns.
Note: I have been asked to do this task manually, not using filter2() or any other programmed function, so please do not suggest that I use filter2() or similar. Thanks!
I tried calling up all the pixels and subtracting each successive one by doing
fid = fopen('image.raw')
myimage = fread(fid,[512 683], '*int8')
fclose(fid)
imsz = size(myimage)
x = imsz(1)
for I = 1:512
for J = 1:683
X(I) - X(I-1) = XX
But it doesnt seem to work, and I dont quite understand why. If you could help me, or point me in the right direction, I would be very appreciative.
First of all, your code is syntatically incorrect:
There is no end statement to any of your loops, and besides, you don't even need loops here.
You seem to read your image into the variable myimage, but you're using an undefined variable X when attempting to calculate the derivative.
The order of your assignment statements is reversed. The variable you wish to assign to should be written in the left hand part of the assignement.
I strongly suggest that you read online tutorials and get yourself familiar with MATLAB basics before taking on more complicated tasks.
As to your specific problem:
MATLAB encourages vectorized operations, i.e operations on entire arrays (vectors or matrices) at once. To subtract adjacent values in an array, what you're basically doing is subtracting two arrays, shifted by one element with respect to each other. For one dimensional arrays, that would translate in MATLAB to:
a(2:end) - a(1:end-1)
where a is your array. The end keyword specifies the last index in the array.
To compute the derivative of an image (a 2-D matrix), you need to decide along which axis you want to perform that operation. To approximate the derivate along the y-axis, do this:
X(2:end, :) - X(1:end-1, :)
You can verify that this gives you the same result as diff(X, 1) (or simply diff(X)). To compute the approximate derivative along the x-axis, which is equivalent to diff(X, 2), do this:
X(:, 2:end) - X(:, 1:end-1)
The colon (:) is the same as writing 1:end as the array subscript for the corresponding dimension.
If your filtered image is div then
for Y = 1:682
for X = 1:511
div(X, Y) = myimage(X + 1, Y + 1) - myimage(X,Y);
end
end
Remember the last row and the last column are not filtered!
I'm attempting to run this simple diffusion case (I understand that it isn't ideal generally), and I'm doing fine with getting the inside of the solid, but need some help with the outer edges.
global M
size=100
M=zeros(size,size);
M(25,25)=50;
for diffusive_steps=1:500
oldM=M;
newM=zeros(size,size);
for i=2:size-1;
for j=2:size-1;
%we're considering the ij-th pixel
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
It's a pretty simple piece of code, and I know that. I'm not very good at using Octave yet (chemist by trade), so I'd appreciate any help!
If you have concerns about the border of your simulation you could pad your matrix with NaN values, and then remove the border after the simulation has completed. NaN stands for not a number and is often used to denote blank data. There are many MATLAB functions work in a useful way with these values.
e.g. finding the mean of an array which has blanks:
nanmean([0 nan 5 nan 10])
ans =
5
In your case, I would start by adding a border of NaNs to your M matrix. I'm using 'n' instead of 'size', since size is an important function in MATLAB, and using it as a variable can lead to confusing errors.
n=100;
blankM=zeros(n+2,n+2);
blankM([1,end],:) = nan;
blankM(:, [1,end]) = nan;
Now we can define 'M'. N.B that the first column and row will be NaNs so we need to add an offset (25+1):
M = blankM;
M(26,26)=50;
Run the simulation through,
m = size(blankM, 1);
n = size(blankM, 2);
for diffusive_steps=1:500
oldM = M;
newM = blankM;
for i=2:m-1;
for j=2:n-1;
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
and then extract the area of interest
finalResult = M(2:end-1, 2:end-1);
One simple change you might make is to add a boundary of ghost cells, or halo, around the domain of interest. Rather than mis-use the name size I've used a variable called sz. Replace:
M=zeros(sz,sz)
with
M=zeros(sz+2,sz+2)
and then compute your diffusion over the interior of this augmented matrix, ie over cells (2:sz+1,2:sz+1). When it comes to considering the results, discard or just ignore the halo.
Even simpler would be to simply take what you already have and ignore the cells in your existing matrix which are on the N,S,E,W edges.
This technique is widely used in problems such as, and similar to, yours and avoids the need to write code which deals with the computations on cells which don't have a full complement of neighbours. Setting the appropriate value for the contents of the halo cells is a problem-dependent matter, 0 isn't always the right value.