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,[]);
Related
I am trying to remove rows from a matrix based on a condition. I have a 371000x5 double matrix, and a 371000x1 vector of dummies (1 and 0). I want to remove each row from the original matrix, where the value of the vector of dummies is 1.
I have tried the following, but it is taking very long:
for i = 1:size(matrix_num,1)
if missing_matrix(i,1) >=0
matrix_num(i,:) = [];
end
end
My Matlab has been busy for over 30 minutes now, so I am not even sure if the code is right. Is there a more efficient way to do this?
Additionally, I have to do the same action on a cell matrix (categorical data). Should I expect any huge difference from the numerical matrix?
The programmatic way of doing this is:
new_matrix = old_matrix(missing_vector==1,:)
for keeping lines with missing_vector 1
new_matrix = old_matrix(missing_vector==0,:)
for removing lines with missing_vector 1
For educational values, if you want the loop to work, don't do that row by row. Your solution causes the matrix to be copied and re-allocated on every row removed.
So, you would be better off if you calculate the resulting matrix size in advance:
new_matrix = zeros(sum(missing_vector), 5)
and then your iteration would work:
index_new=1
for index_old = 1:size(old_matrix,1)
if missing_vector(index_old) ==0
new_matrix(index_new,:) = old_matrix(index_old,:);
end
end
Try compact MATLAB code
matrix_num(missing_matrix>=0,:)=[]
Note : You must make a vector for missing_matrix variable. If this variable is matrix, you need other form of code .
As I know, you can use it in cell array too.
I need to know if there is any efficient way of doing the following in MATLAB.
I have several big sparse matrices, the size of each one is roughly 9000000x9000000.
I need to access multiple element of such matrix and assign to each selected element a different value stored in another array. I'll give an example:
What I have:
SPARSE MATRIX of size 9000000x9000000
Matrix with the list of indexes and values I want to access, this is a matrix like this:
[row1, col1, value1;
row2, col2, value2;
...
rowN, colN, valueN]
Where N is the length of such matrix.
What I need:
Assign to the SPARSE MATRIX the corresponding value to the corresponding index, this is:
SPARSE_MATRIX(row1, col1) = value1
SPARSE_MATRIX(row2, col2) = value2
...
SPARSE_MATRIX(rowN, colN) = valueN
Thanks in advance!
EDIT:
Thank you to both for answering, I think I did not explain myself well, I'll try again.
I already have a large SPARSE MATRIX of about 9000000 rows x 9000000 columns, it is a SPARSE MATRIX filled with zeros.
Then I have another array or matrix, let's call it M with N number of rows, where N could take values from 0 to 9000000; and 3 columns. The first two columns are used to index an element of my SPARSE MATRIX, and the third column stores the value I want to transfer to the SPARSE MATRIX, this is, given a random row of M, i:
SPARSE_MATRIX(M(i, 1), M(i, 2)) = M(i, 3)
The idea is to do that for all the rows, I have tried it with common indexing:
SPARSE_MATRIX(M(:, 1), M(:, 2)) = M(:, 3)
Now I would like to do this assignation for all the rows in M as fast as possible, because if I use a loop or common indexing it takes ages (I am using a 7th Gen i7 processor with 16 GB of RAM). And I also need to keep the zeros in the SPARSE_MATRIX.
EDIT 2: SOLVED! Thank you Metahominid, I was not thinking through, but yes the sparse function does solve my problem, I just think my brain circuits were shortcircuited yesterday and was unable to see through it hahaha. Thank you to both anyway!
Regards!
You can construct a sparse matrix like this.
A = sparse(i,j,v)
S = sparse(i,j,v) generates a sparse matrix S from the triplets i, j,
and v such that S(i(k),j(k)) = v(k). The max(i)-by-max(j) output
matrix has space allotted for length(v) nonzero elements. sparse adds
together elements in v that have duplicate subscripts in i and j.
So you can simply construct the row vector, column vector and value vector.
I am answering in part because I cannot comment. You question seems a little confusing to me. The sparse() function in MATLAB does just this.
You can enter your arrays of indices and values directly into the interface, or declare a sparse matrix of zeros and set each individually.
Given your data format make three vectors, ROWS = [row1; ...; rown], COLS = [col1; ...; coln], and DATA = [val1; ... valn]. I am assuming that your size is the overall size of the full matrix and not the sparse portion.
Then
A = sparse(ROWS, COLS, DATA) will do just what you want. You can even specify the original matrix size.
A = sparse(ROWS, COLS, DATA, 90...., 90....).
I have a complex matrix of size 3x2x372 complex double. I would like to work with only one specific of these three dimensions. Therefore, I used the following code to make the table easier to read:
new_output = abs(output);
In fact, the new matrix is of size 3x2x372 double. I guess it makes the further computation simpler. So I obtain the following output:
I would now like to create a matrix that only refers to the highlighted values. So it should ideally be of size 2x372 double.
Make a for-loop and assign the last row to a new matrix.
mat = zeroes(372, 2)
for k = 1:372
a = val(:, :, k)
mat(k, :) = a(1, :)
end
Edit: above gives you a 372x2 matrix. Use below to get a 2x372 matrix
mat = zeroes(2, 372)
And
mat(:, k) = a(1,:).'
in the loop
Actually, you need the last row from each "slice", so you can get it by:
new_output=data(size(data,1),:,:);
But that will give you the same dimensions as the original matrix, with 3D. To directly get it as 2D matrix, use squeeze:
new_output=squeeze(data(size(data,1),:,:));
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);
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.