I have matrix A = 50x2
How to convert the data into cell array.
Should be I have 10 cell which each cell contain data [5x2].
Thank you for help.
That is what mat2cell does:
A = rand(50,2); % example matrix
N = 10; % number of cells in which to split the first dimension
result = mat2cell(A, repmat(size(A,1)/N, 1, N), size(A,2));
One can use num2cell:
N_ROWS = 5; N_COLS = 2;
A = rand(50,2);
B = num2cell(reshape(A,N_ROWS,N_COLS,[]),[1,2]); % B is 1x1x10 cell array
What this does is turn your input array into 5x2 "slices" stacked along the 3rd dimension.
You can add a squeeze(B), B = B(:) or a reshape(B,[],1) at the end if you need the output as a column vector.
Related
I have:
A matrix 3D: A = (m, n, k).
An array of choices for the third dimension corresponding to each index of the first dimension. idn = (m, 1) (wherein the value of any idn is a random integer in [1,k].
I need to capture the 2D matrix B (m,n) wherein the referred third dimension to A is taken from the corresponding choice. For example:
idn(1) = 1;
idn(2) = k;
idn(j) = k-1;
Then:
B(1,:) = A(1,:,idn(1)) = A(1,:,1);
B(2,:) = A(2,:,idn(2)) = A(2,:,k);
B(j,:) = A(j,:,idn(j)) = A(j,:,k-1);
Since idn is not constant, a simple squeeze could not help.
I have also tried the below code, but it does not work either.
B = A(:,:,idn(:));
It is very much appreciated if anyone could give me a solution.
This could be done with sub2ind and permute, but the simplest way I can think of is using linear indexing manually:
A = rand(3, 4, 5); % example data
idn = [5; 1; 2]; % example data
ind = (1:size(A,1)).' + size(A,1)*size(A,2)*(idn(:)-1); % 1st and 3rd dimensions
ind = ind + size(A,1)*(0:size(A,2)-1); % include 2nd dimension using implicit expansion
B = A(ind); % index into A to get result
Consider n row vectors in Matlab, each of size 1xU. For example,
U=20;
n=3;
sU=[U U U];
vectors = arrayfun(#(x) {1:x}, sU);
where vector{1} is the first row vector, vector{2} is the second row vector,..., vector{n} is the last row vector.
We create the matrix Tcoord of size U^n x n reporting all the possible n-tuples from the n row vectors. For each row i of Tcoord, Tcoord(i,1) is an element of the first row vector, Tcoord(i,2) is an element of the second row vector, ... , Tcoord(i,n) is an element of the last row vector.
Tcoord_temp = cell(1,n);
[Tcoord_temp{:}] = ndgrid(vectors{:});
Tcoord_temp = cat(n+1, Tcoord_temp{:});
Tcoord = reshape(Tcoord_temp,[],n);
Suppose now that I augment each of the n row vectors of 3 elements. For example,
vectors_augmented{1}=[vectors{1} 8 9 10];
vectors_augmented{2}=[vectors{2} 11 12 13];
vectors_augmented{3}=[vectors{3} 14 15 16];
I then create a matrix similar to Tcoord but now using vectors_augmented.
Tcoord_temp = cell(1,n);
[Tcoord_temp{:}] = ndgrid(vectors_augmented{:});
Tcoord_temp = cat(n+1, Tcoord_temp{:});
Tcoord_augmented = reshape(Tcoord_temp,[],n); %(U+3)^nxn
I would like your help to re-order the rows of the matrix Tcoord_augmented in a matrix Tcoord_augmented_reshape such that
Tcoord_augmented_reshape(1:U^n,:) is equal to Tcoord.
The remaining rows of Tcoord_augmented_reshape contains the other left rows of Tcoord_augmented.
The simplest approach is to build an auxiliary zero-one matrix the same size as Tcoord_augmented and sort rows based on that:
aug_size = [3 3 3]; % augment size of each vector. Not necessarily equal
vectors_aux = arrayfun(#(a) {[false(1,U) true(1, a)]}, aug_size);
T_aux = cell(1,n);
[T_aux{:}] = ndgrid(vectors_aux{:});
T_aux = cat(n+1, T_aux{:});
T_aux = reshape(T_aux,[],n);
[~, ind] = sortrows(T_aux, n:-1:1); % indices of stably sorting the rows.
% Most significant column is rightmost, as per your code
Tcoord_augmented_reorder = Tcoord_augmented(ind, :);
For instance, I have the following cell array:
a = [1,2,3; 1,5,8; 6,5,0; 0,0,2];
A = cell(3,4);
for i = 1:3
for j = 1:4
A{i,j} = (j-i)*a;
end
end
How could I sum up all the elements i.e. A{1,1} + A{1,2} + ... + A{3,4}?
Concatenate the matrices in each cell along the third dimension and then sum along the third dimension.
TD = cat(3, A{:}); %Converting the cell array to a 3D array
result = sum(TD, 3); %Summation of 3D slices
Hi guys I want to solve sodoku puzzles in matlab. My problem is that I should find same value in every row and every column and every 3*3 sub array.
Our 2d array is 9*9 and populated with value 1-9 randomly.
I wrote this for finding same value in rows, but I don't know how I should do it for columns and 3*3 sub arrays.
conflict_row = 0;
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
Sorry I'm a newbie.
Find values that are present in all columns:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,1), [3 1 2])),1),2));
Find values that are present in all rows:
v = find(all(any(bsxfun(#eq, A, permute(1:size(A,2), [3 1 2])),2),1));
Find values that are present in all 3x3 blocks: reshape the matrix as in this answer by A. Donda to transform each block into a 3D-slice; then reshape each block into a column; and apply 1:
m = 3; %// columns per block
n = 3; %// rows per block
B = permute(reshape(permute(reshape(A, size(A, 1), n, []), [2 1 3]), n, m, []), [2 1 3]);
B = reshape(B,m*n,[]);
v = find(all(any(bsxfun(#eq, B, permute(1:size(B,1), [3 1 2])),1),2));
Probably not the fastest solution but why don't you make a function of it and use it once for rows and once for columns
[conflict_row ] = get_conflict(A)
for i=1:9
temp = 0;
for j=1:9
if (temp==A(i,j))
conflict_row = conflict_row+1;
end
temp = A(i,j);
end
end
And then you call it twice
conflict_row = get_conflict(A); % Rows
Transpose A to get the columns
Convert the columns to rows and use the same code as before
conflict_col = get_conflict(A.');
If you want to work within the same column then you should do something like this (also sorry this is in C# I don't know what language you are working in):
int currentCol = 0;
foreach (var item in myMultiArray)
{
int currentColValue = item[currentCol];
}
This works because myArray is a array of arrays thus to select a specific column can easily be picked out by just allowing the foreach to perform your row iteration, and you just have to select the column you need with the currentCol value.
I have 64 characters in a 4*4 matrix.I need to convert it into a cell array such that cell has 4 characters.For eg
Consider A=[TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT]
i need to get a cell array as below
B={[TCTG][CTCT][CGGT][TATA]
[TACA][CTGC][CCAG][AACA]
[CGTC][AACA][AGGC][CAGT]
[GTAT][CCTT][CTTT][GTGT]}
i tried using the mat2cell function but im not able to understand it.please help.
Using a for-loop:
clc
clear
A = 'TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT';
B = cell(4,4);
currentIdx = 0; % Use index to increment by steps of 4 when going through A
for k = 1:16
B{k} = A(currentIdx+1:currentIdx+4);
currentIdx = currentIdx+4;
end
B = B'
B =
'TCTG' 'CTCT' 'CGGT' 'TATA'
'TACA' 'CTGC' 'CCAG' 'AACA'
'CGTC' 'AACA' 'AGGC' 'CAGT'
'GTAT' 'CCTT' 'CTTT' 'GTGT'
You are starting with a 1xN matrix and want to convert it to a 1xN/4 cell array of 1x4 matrices. Your command should then be:
N = length(A);
M = 4;
B = mat2cell(A,1,ones(1,N/M)*M);
The first dimension is the 1, the second dimension is a string of 4's the size of the output cell array. The result:
B =
Columns 1 through 12
'TCTG' 'CTCT' 'CGGT' 'TATA' 'TACA' 'CTGC' 'CCAG' 'AACA' 'CGTC' 'AACA' 'AGGC' 'CAGT'
Columns 13 through 16
'GTAT' 'CCTT' 'CTTT' 'GTGT'
You can use method vec2mat that breaks your input vector to matrix
M = vec2mat(A, numberOfColumns)
(In your case numberOfColumns would be 16) and then use mat2cell. In your case, it would be:
C = mat2cell(M, [1,1,1,1], [4,4,4,4])
It means that all cels will have one row and 4 columns).
Effect of function c = mat2cell(x, [10, 20, 30], [25, 25]) would be:
The image shows why you have to convert vector to matrix. (example from matlab documentation)
You can also (ab)use the very versatile accumarray for this task:
A = 'TCTGCTCTCGGTTATATACACTGCCCAGAACACGTCAACAAGGCCAGTGTATCCTTCTTTGTGT';
n = 4;
B = accumarray(ceil(1/n:1/n:numel(A)/n).', A(:), [], #(x) {x.'}).'