Suppose i have a 256*256 matrix(image pixel values).I want to create that Sliding window 1 to Sliding Window n by shifting exactly one row and one column. All the Sliding Window Matrix Should be in the size of 5*5.
I have tried with this following code. By this i am getting the submatrices but not by the shifting of exactly one row and one column.
I = imread('D:\Study Material\project\Finger Print Database\1_2.png');
J = imresize(I, [128 128]);
C=mat2tiles(J,[5,5]);
Suppose original matrix has 10 row and 10 column.
I'm getting the sub-matrices like (column1-column5 & row1-row5) then (column6-column10 & row6-row10).
But I need like this- (column1-column5 & row1-row5)then (column2-column6 & row1-row5)like this first column will be shifted to 10. After that row will be shifted to 10.
You need im2col with the 'sliding' option, followed by reshape:
blockSize = [5 5];
C = reshape(im2col(J, blockSize, 'sliding'), blockSize(1), blockSize(2), []);
Related
I'm new to MATLAB and its development. I have a image which is 1134 (rows) X 1134 (columns). I want that image to save 3 (columns) X 3 (rows). In order to do that I need 378 cells. For that I used following code, but it gives me an error.
image=imread('C:\Users\ven\Desktop\test\depth.png');
I=reshape(image,1,1134*1134);
chunk_size = [3 3]; % your desired size of the chunks image is broken into
sc = sz ./ chunk_size; % number of chunks in each dimension; must be integer
% split to chunk_size(1) by chunk_size(2) chunks
X = mat2cell(I, chunk_size(1) * ones(sc(1),1), chunk_size(2) *ones(sc(2),1));
Error:
Error using mat2cell (line 97)
Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [1 1285956].'
Unfortunately your code does not work as you think it would.
The ./ operator performs point wise division of two matrices. Short example:
[12, 8] ./ [4, 2] == [12/4, 8/2] == [3, 4]
In order for it to work both matrices must have exactly the same size. In your case you try to perform such an operation on a 1134x1134 matrix (the image) and a 1x2 matrix (chunk_size).
In other words you can not use it to divide matrices into smaller ones.
However, a solution to your problem is to use the mat2cell function to pick out subsets of the matrix. A explanation of how it is done can be found here (including examples): http://se.mathworks.com/matlabcentral/answers/89757-how-to-divide-256x256-matrix-into-sixteen-16x16-blocks.
Hope it helps :)
Behind the C=A./B command is loop over all elements of A(ii,jj,...) and B(ii,jj,..) and each C(ii,jj,..)=A(ii,jj,...)/B(ii,jj,...).
Therefore martices A and B must be of same dimension.
If you want to split matrix into groups you can use
sc=cell(1134/3,1);
kk=0;ll=0;
for ii=2:3:1133
kk=kk+1;
for jj=2:3:1133
ll=ll+1;
sc{kk,ll}=image(ii-1:ii+1,jj-1:jj+1);
end
end
The code allocates cell array sc for resulting submatrices and arbitrary counters kk and ll. Then it loops over ii and jj with step of 3 representing centers of each submatrices.
Edit
Or you can use mat2cell command (type help mat2cell or doc mat2cell in matlab shell)
sc=mat2cell(image,3,3);
In both cases the result is cell array and its iith and jjth elements (matrices) are accessible by sc{ii,jj}. If you want call iith anr jjth number in kkth and llth matrix, do it via sc{kk,ll}(ii,jj).
In short, you divided a 1134 x 1134 by 2 x 1 matrix. That doesn't work.
The error "Matrix dimensions must agree**" is from the dividing a matrix with another matrix that doesn't have the right dimensions.
You used the scalar divide "./" which divided a matrix by another matrix.
You want something like:
n = 1134 / 3 % you should measure the length of the image
I1=image(1:n,1:n); % first row
I2=image(1:n,n:2n);
I3=image(1:n,2n:3n);
I4=image(n:2n,1:n); % second row
I5=image(n:2n,n:2n);
I6=image(n:2n,2n:3n);
I7=image(2n:3n,1:n); % third row
I8=image(2n:3n,n:2n);
I9=image(2n:3n,2n:3n);
from here:
http://au.mathworks.com/matlabcentral/answers/46699-how-to-segment-divide-an-image-into-4-equal-halves
There would be a nice loop you could do it in, but sometimes thinking is hard.
So I have written this:
HSRXdistpR = squeeze(comDatape_m1(2,7,1,:,isubj));
HSRXdistpL = squeeze(comDatape_m1(2,4,1,:,isubj));
TocomXdistp = squeeze(comDatape_m1(2,10,1,:,isubj));
for i = 1:2;
HSRXp = NaN(8,3*i);
HSRXp(:,i*3) = [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)];
end
In the first part I am just selecting data from a 5-D matrix, nothing special. All that's important here is that it creates an 8x2 matrix per line (isubj=2). Now I want to add the first column of each matrix into an 8x3 matrix, and then the second column of each matrix into the same matrix (creating an 8x6 matrix). Since the number of my subjects will vary, I want to do this in a for loop. This way, if the isubj increases to 3, it should go on to create an 8x9 matrix.
So I tried to create a matrix that will grow by 3 for each iteration of i, which selects the ith column of each of the 3 matrices and then puts them in there.
However I get the following error:
Subscripted assignment dimension mismatch.
Is it possible to let a matrix grow by more than one in a for loop? Or how should it be done otherwise?
Here is your problem:
HSRXp(:,i*3) = [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)];
You're trying to assign an n x 3 matrix (RHS) into an n x 1 vector (LHS). It would be easier to simply use horizontal concatenation:
HSRXp = [HSRXp, [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)]];
But that would mean reallocation at each step, which might slow your code down if the matrix becomes large.
I have an image of any size that I need to split into non-overlapping regions of 15x15 pixels. Previously I looked this up and used the following code:
newIm = rand(size(im2, 1),size(im2, 2));
subIm = mat2cell(newIm, 15*ones(size(newIm,1)/15,1), 15*ones(size(newIm,2)/15,1));
My problem is that I may not always be able to split the matrix into nice 15x15 regions. How can I fix this? Also, can you explain what exactly happens? I'd like to understand Matlab better for future use!
If you use the above code with a size not perfectly divisible by 15 (say 160), you will get the following error in MATLAB:
Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [160 160].'
So you have to make your second and third input arguments of mat2cell sum to 160. Then you are done.
Code taken from here
blockSize=[15 15];
wholeBlockRows = floor(size(newIm,1)/ blockSize(1));
blockVectorR = [blockSize(1) * ones(1, wholeBlockRows), rem(size(newIm,1), blockSize(1))];
wholeBlockCols = floor(size(newIm,2)/ blockSize(2));
blockVectorC = [blockSize(2) * ones(1, wholeBlockCols), rem(size(newIm,2), blockSize(2))];
% sum of blockVectorR and blockVectorC will be equal to size(newIm,1) and
% size(newIm,2) respectively.
ca = mat2cell(newIm, blockVectorR, blockVectorC, size(newIm,3));
In your output cell array, you will see sub-images in the last row and column where either rows or columns (or both) are equal to: rem(size(newIm,1), blockSize(1)) or (and) rem(size(newIm,2), blockSize(2))
Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
I have a 10 X 10 matrix, A, created in MatLab. All the values in the matrix are between 0 and 100. Say that I want to:
Replace all elements of A < 10 with zeros
Replace all elemtns of A > 90 with infinity
Extract all values between 30 and 50 to a new vector.
Can I do this without writing a script? I can easliy do this through a script with some simple for-loops, but are there any shortcuts available? Any help will be greatly appreciated!
All of these things are really easy to do using logical indexing:
Each of the operations above can be done quite easily using a one or two commands. However each operation must be done independent of the other two. You can't do all 3 in one line.
1.
smallIdx = A<10;
A(smallIdx) = 0;
% One Line Version
A(A<10) = 0;
2.
bigIdx = A>90;
A(bigIdx)=inf;
% One Line Version
A(A>90) = inf;
3.
middleIdx = A>30 & A<50;
newVector = A(middleIdx);
% One Line Version
newVector = A(A>30 & A<50);
new vector is a vector and wont be square like A was
Set up any 3 × 3 matrix a. Write some command-line statements to perform the following
operations on a:
(a) interchange columns 2 and 3;
(b) add a fourth column (of 0s);
(c) insert a row of 1s as the new second row of a (i.e., move the current second and third rows
down);
(d) remove the second column.