I want to reshape pixel intensity by imagesize*1(column vector).
Imvect = reshape(I,imsize,1);
But why these error comes?
Error using reshape
To RESHAPE the number of elements must not change.
Let's start with the syntax used in the documentation:
B = reshape(A,sz1,...,szN)
What reshape does is to take the matrix A, straightens it out, and gives it a new size, that's determined by the 2nd, 3rd to the Nth argument. For this to be possible, you need to have the same number of elements in the input matrix as you have in the output matrix. You can't make a 1x5 vector turn into a 2x3 vector, as one element would be missing. The number of elements in the output matrix will be proportional to the product of sz1, sz2, ..., szN. Now, if you know you want N rows, but don't know exactly how many columns you have, you might use the [] syntax, that tells MATLAB to use as many columns as necessary to make the number of elements be equal.
So reshape(A, 2, [], 3) will become a 2xNx3 matrix, where, for a matrix with 24 elements, N will be 4.
Now, in your case this is not the case. numel(I) ~= imsize. If mod(numel(I), imsize) ~= 0 then your imsize is definitely incorrect. However, if mod(numel(I), imsize) == 0, then your error might be that you want imsize number of rows, and a number of columns that makes this possible. If it's the latter, then this should work:
Imvect = reshape(I,imsize, []);
If you simply want to make you matrix I a vector of size (numel(I), 1), then you should use the colon operator :, as such:
Imvect = I(:);
An alternative, if you really want to use reshape, is to specify that you want a single column, and let MATLAB select the number of rows, as such:
Imvect = reshape(I, [], 1);
Related
I have an n x n matrix in MATLAB. In every row, if the value in each element is higher than a certain threshold, replace that element with a 1. Else, with a 0.
NOTICE: In every row, we compare the value of element with different threshold.
For element-wise comparison of two matrices of the same size, use the ">" operator e.g. result = data > threshold (this will return ones and zeroes depending on whether the condition is satisfied or not).
Suppose you have your data in a matrix called data and your thresholds in a column vector called thresholds (i.e. length(thresholds) == size(data, 1)). You can create an array the same size as the data matrix using repmat: thresholdsMatrix = repmat(thresholds, 1, size(data, 2)).
You can then compare this to your data:
result = data > repmat(thresholds, 1, size(data, 2)).
This should give you the result you want.
[Note that you can also directly compare the vector to the matrix without using repmat i.e. result = data > thresholds, but IMO this can be unclear and may lead to unexpected behaviour]
I have a column vector A with dimensions (35064x1) that I want to reshape into a matrix with 720 lines and as many columns as it needs.
In MATLAB, it'd be something like this:
B = reshape(A,720,[])
in which B is my new matrix.
However, if I divide 35604 by 720, there'll be a remainder.
Ideally, MATLAB would go about filling every column with 720 values until the last column, which wouldn't have 720 values; rather, 504 values (48x720+504 = 35064).
Is there any function, as reshape, that would perform this task?
Since I am not good at coding, I'd resort to built-in functions first before going into programming.
reshape preserves the number of elements but you achieve the same in two steps
b=zeros(720*ceil(35604/720),1); b(1:35604)=a;
reshape(b,720,[])
A = rand(35064,1);
NoCols = 720;
tmp = mod(numel(A),NoCols ); % get the remainder
tmp2 = NoCols -tmp;
B = reshape([A; nan(tmp2,1)],720,[]); % reshape the extended column
This first gets the remainder after division, and then subtract that from the number of columns to find the amount of missing values. Then create an array with nan (or zeros, whichever suits your purpose best) to pad the original and then reshape. One liner:
A = rand(35064,1);
NoCols = 720;
B = reshape([A; nan(NoCols-mod(numel(A),NoCols);,1)],720,[]);
karakfa got the right idea, but some error in his code.
Fixing the errors and slightly simplifying it, you end up with:
B=nan(720,ceil(numel(a)/720));
B(1:numel(A))=A;
Create a matrix where A fits in and assingn the elemnent of A to the first numel(A) elements of the matrix.
An alternative implementation which is probably a bit faster but manipulates your variable b
%pads zeros at the end
A(720*ceil(numel(A)/720))=0;
%reshape
B=reshape(A,720,[]);
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.
I want to create a 4 by 4 sparse matrix A. I want assign values (e.g. 1) to following entries:
A(2,1), A(3,1), A(4,1)
A(2,2), A(3,2), A(4,2)
A(2,3), A(3,3), A(4,3)
A(2,4), A(3,4), A(4,4)
According to the manual page, I know that I should store the indices by row and column respectively. That is, for row indices,
r=[2,2,2,2,3,3,3,3,4,4,4,4]
Also, for column indices
c=[1,2,3,4,1,2,3,4,1,2,3,4]
Since I want to assign 1 to each of the entries, so I use
value = ones(1,length(r))
Then, my sparse matrix will be
Matrix = sparse(r,c,value,4,4)
My problem is this:
Indeed, I want to construct a square matrix of arbitrary dimension. Says, if it is a 10 by 10 matrix, then my column vector will be
[1,2,..., 10, 1,2, ..., 10, 1,...,10, 1,...10]
For row vector, it will be
[2,2,...,2,3,3,...,3,...,10, 10, ...,10]
I would like to ask if there is a quick way to build these column and row vector in an efficient manner? Thanks in advance.
I think the question aims to create vectors c,r in an easy way.
n = 4;
c = repmat(1:n,1,n-1);
r = reshape(repmat(2:n,n,1),1,[]);
Matrix = sparse(r,c,value,n,n);
This will create your specified vectors in general.
However as pointed out by others full sparse matrixes are not very efficient due to overhead. If I recall correctly a sparse matrix offers advantages if the density is lower than 25%. Having everything except the first row will result in slower performance.
You can sparse a matrix after creating its full version.
A = (10,10);
A(1,:) = 0;
B = sparse(A);
I am currently trying to code up a function to assign probabilities to a collection of vectors using a histogram count. This is essentially a counting exercise, but requires some finesse to be able to achieve efficiently. I will illustrate with an example:
Say that I have a matrix X = [x1, x2....xM] with N rows and M columns. Here, X represents a collection of M, N-dimensional vectors. IN other words, each of the columns of X is an N-dimensional vector.
As an example, we can generate such an X for M = 10000 vectors and N = 5 dimensions using:
X = randint(5,10000)
This will produce a 5 x 10000 matrix of 0s and 1s, where each column is represents a 5 dimensional vector of 1s and 0s.
I would like to assign a probability to each of these vectors through a basic histogram count. The steps are simple: first find the unique columns of X; second, count the number of times each unique column occurs. The probability of a particular occurrence is then the #of times this column was in X / total number of columns in X.
Returning to the example above, I can do the first step using the unique function in MATLAB as follows:
UniqueXs = unique(X','rows')'
The code above will return UniqueXs, a matrix with N rows that only contains the unique columns of X. Note that the transposes are due to weird MATLAB input requirements.
However, I am unable to find a good way to count the number of times each of the columns in UniqueX is in X. So I'm wondering if anyone has any suggestions?
Broadly speaking, I can think of two ways of achieving the counting step. The first way would be to use the find function, though I think this may be slow since find is an elementwise operation. The second way would be to call unique recursively as it can also provide the index of one of the unique columns in X. This should allow us to remove that column from X and redo unique on the resulting X and keep counting.
Ideally, I think that unique might already be doing some counting so the most efficient way would probably be to work without the built-in functions.
Here are two solutions, one assumes all values are either 0's or 1's (just like the example in your description), the other does not. Both codes should be very fast (more so the one with binary values), even on large data.
1) only zeros and ones
%# random vectors of 0's and 1's
x = randi([0 1], [5 10000]); %# RANDINT is deprecated, use RANDI instead
%# convert each column to a binary string
str = num2str(x', repmat('%d',[1 size(x,1)])); %'
%# convert binary representation to decimal number
num = (str-'0') * (2.^(size(s,2)-1:-1:0))'; %'# num = bin2dec(str);
%# count frequency of how many each number occurs
count = accumarray(num+1,1); %# num+1 since it starts at zero
%# assign probability based on count
prob = count(num+1)./sum(count);
2) any positive integer
%# random vectors with values 0:MAX_NUM
x = randi([0 999], [5 10000]);
%# format vectors as strings (zero-filled to a constant length)
nDigits = ceil(log10( max(x(:)) ));
frmt = repmat(['%0' num2str(nDigits) 'd'], [1 size(x,1)]);
str = cellstr(num2str(x',frmt)); %'
%# find unique strings, and convert them to group indices
[G,GN] = grp2idx(str);
%# count frequency of occurrence
count = accumarray(G,1);
%# assign probability based on count
prob = count(G)./sum(count);
Now we can see for example how many times each "unique vector" occurred:
>> table = sortrows([GN num2cell(count)])
table =
'000064850843749' [1] # original vector is: [0 64 850 843 749]
'000130170550598' [1] # and so on..
'000181606710020' [1]
'000220492735249' [1]
'000275871573376' [1]
'000525617682120' [1]
'000572482660558' [1]
'000601910301952' [1]
...
Note that in my example with random data, the vector space becomes very sparse (as you increase the maximum possible value), thus I wouldn't be surprised if all counts were equal to 1...