This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
A similar function to R's rep in Matlab [duplicate]
(4 answers)
Closed 8 years ago.
Let's say, I have:
A=[1 2; 3 4];
I want to use repmat that return:
B = [1 1 2 2; 1 1 2 2; 3 3 4 4; 3 3 4 4]
Kindly need your help. Thank you
I do not know a method using repmat but here is a method using kron
kron([1 2 ; 3 4],[1 1;1 1])
ans =
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
An alternative which uses repmat is
A=[1 2; 3 4];
cell2mat(arrayfun(#(x)repmat(x,2,2),A,'UniformOutput',false))
ans =
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
arrayfun is used to evaluate each element in A using the anonymous function #(x)repmat(x,2,2) which replicates that single element into a 2x2 matrix.
The result of arrayfun is a 2x2 cell array where each element is a 2x2 matrix. We then convert this cell array into a matrix via cell2mat.
Let the data be defined as
A = [1 2; 3 4];
R = 2; %// number of repetitions of each row
C = 2; %// number of repetitions of each column. May be different from R
Two possible approaches are as follows:
The simplest method is to use indexing:
B = A(ceil(1/R:1/R:size(A,1)), ceil(1/C:1/C:size(A,2)));
If you really want to do it with repmat, you need to play with dimensions using permute and reshape: move original dimensions 1, 2 to dimensions 2, 4 (permute); do the repetition along new dimensions 1, 3 (repmat); collapse dimensions 1, 2 into one dimension and 3, 4 into another dimension (reshape):
[r c] = size(A);
B = reshape(repmat(permute(A, [3 1 4 2]), [R 1 C 1]), [r*R c*C]);
Example result for R=2, C=3 (obtained with any of the two approaches):
B =
1 1 1 2 2 2
1 1 1 2 2 2
3 3 3 4 4 4
3 3 3 4 4 4
Related
For example, I have a matrix A (Figure 1). When the variable n = 2, I want it to be transformed to the matrix B. The red rectangle shows the transformation rule of every column. According to this rule, when the n = 3, it can become the matrix C.
I have written a script using a for loop method, but it is a waste of time when the matrix A is very large (e.g. 11688* 140000). Is there an efficient way to solve this problem?
Figure 1:
Here is a way using reshape and implicit expansion:
result = reshape(A((1:size(A,1)-n+1) + (0:n-1).', :), n, []);
For example assume that n = 3. Implicit expansion is used to extract indices of rows:
row_ind = (1:size(A,1)-n+1) + (0:n-1).';
The following matrix is created:
1 2
2 3
3 4
Extract the desired rows of A:
A_expanded = A(row_ind, :)
When the matrix row_ind is used as an index it behaves like a vector:
1
2
1 2 3
2 3 -> 2
3 4 3
4
A_expanded =
3 5 7
6 8 9
2 6 3
6 8 9
2 6 3
1 2 1
Now A_expanded can be reshaped to the desired size:
result = reshape(A_expanded, n, []);
>>result =
3 6 5 8 7 9
6 2 8 6 9 3
2 1 6 2 3 1
If you have the Image Processing Toolbox you can use im2col as follows:
result = im2col(A, [n 1], 'sliding');
I am trying to generate random numbers between 1 and 5 using Matlab's randperm and calling randperm = 5.
Each time this gives me a different array let's say for example:
x = randperm(5)
x = [3 2 4 1 5]
I need the vector to be arranged such that 4 and 5 are always next to each other and 2 is always between 1 and 3... so for e.g. [3 2 1 4 5] or [4 5 1 2 3].
So essentially I have two "blocks" of unequal length - 1 2 3 and 4 5. The order of the blocks is not so important, just that 4 & 5 end up together and 2 in between 1 and 3.
I can basically only have 4 possible combinations:
[1 2 3 4 5]
[3 2 1 4 5]
[4 5 1 2 3]
[4 5 3 2 1]
Does anyone know how I can do this?
Thanks
I'm not sure if you want a solution that would somehow generalize to a larger problem, but based on how you've described your problem above it looks like you are only going to have 8 possible combinations that satisfy your constraints:
possible = [1 2 3 4 5; ...
1 2 3 5 4; ...
3 2 1 4 5; ...
3 2 1 5 4; ...
4 5 1 2 3; ...
5 4 1 2 3; ...
4 5 3 2 1; ...
5 4 3 2 1];
You can now randomly select one or more of these rows using randi, and can even create an anonymous function to do it for you:
randPattern = #(n) possible(randi(size(possible, 1), [1 n]), :)
This allows you to select, for example, 5 patterns at random (one per row):
>> patternMat = randPattern(5)
patternMat =
4 5 3 2 1
3 2 1 4 5
4 5 3 2 1
1 2 3 5 4
5 4 3 2 1
You can generate each block and shuffle each one then and set them as members of a cell array and shuffle the cell array and finally convert the cell array to a vector.
b45=[4 5]; % block 1
b13=[1 3]; % block 2
r45 = randperm(2); % indices for shuffling block 1
r13 = randperm(2); % indices for shuffling block 2
r15 = randperm(2); % indices for shuffling the cell
blocks = {b45(r45) [b13(r13(1)) 2 b13(r13(2))]}; % shuffle each block and set them a members of a cell array
result = [blocks{r15}] % shuffle the cell and convert to a vector
This question already has answers here:
Octave / Matlab: Extend a vector making it repeat itself?
(3 answers)
Closed 6 years ago.
I'm trying to take:
a = [1 2 3]
and repeat it 5 times to get:
b = [1 2 3 1 2 3 1 2 3 1 2 3 1 2 3]
but when I try:
b = repmat(a, 5, 1)
instead I get:
b =
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
I could probably do it with a for loop but I'd like to do it correctly if possible. Any suggestions? Thanks in advance
Use the following code:
b = repmat(a,1,5)
The numbers '1' and '5' refer to the amount of rows and columns that you want to repeat the matrix a. The order is important.
Suppose A is a 3-D matrix as below (2 rows-2 columns-2 pages).
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
I want to have a vector, say "a", whose inputs are the average of diagonal elements of matrices on each page. So in this simple case, a=[(1+4)/2;(5+8)/2].
But I have difficulties in matlab to do so. I tried the codes below but failed.
mean(A(1,1,:),A(2,2,:))
You can use "partially linear indexing" in the two dimensions that define the diagonal, as follows:
Since partially linear indexing can only be applied on trailing dimensions, you first need to apply permute to rearrange dimensions, so that the first and second dimensions become second and third.
Now you leave the first dimension untouched, linearly-index the diagonals in the second and third dimensions (which effectly reduces those two dimensions to one), and apply mean along the (combined) second dimension.
Code:
B = permute(A, [3 1 2]); %// step 1: permute
result = mean(B(:,1:size(A,1)+1:size(A,1)*size(A,2)), 2); %// step 2: index and mean
In your example,
A(:,:,1)=[1,2;3,4];
A(:,:,2)=[5,6;7,8];
this gives
result =
2.5000
6.5000
You can use bsxfun for a generic solution -
[m,n,r] = size(A)
mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
Sample run -
>> A
A(:,:,1) =
8 4 1
7 6 3
1 5 8
A(:,:,2) =
1 7 6
8 5 2
1 2 7
A(:,:,3) =
6 2 8
1 1 6
1 4 5
A(:,:,4) =
8 1 6
1 5 1
9 2 7
>> [m,n,r] = size(A);
>> sum(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
22 13 12 20
>> mean(A(bsxfun(#plus,[1:n+1:n^2]',[0:r-1]*m*n)),1)
ans =
7.3333 4.3333 4 6.6667
How can I remove elements in a matrix, that aren't all in a straight line, without going through a row at a time in a for loop?
Example:
[1 7 3 4;
1 4 4 6;
2 7 8 9]
Given a vector (e.g. [2,4,3]) How could I remove the elements in each row (where each number in the vector corresponds to the column number) without going through each row at a time and removing each element?
The example output would be:
[1 3 4;
1 4 4;
2 7 9]
It can be done using linear indexing at follows. Note that it's better to work down columns (because of Matlab's column-major order), which implies transposing at the beginning and at the end:
A = [ 1 7 3 4
1 4 4 6
2 7 8 9 ];
v = [2 4 3]; %// the number of elements of v must equal the number of rows of A
B = A.'; %'// transpose to work down columns
[m, n] = size(B);
ind = v + (0:n-1)*m; %// linear index of elements to be removed
B(ind) = []; %// remove those elements. Returns a vector
B = reshape(B, m-1, []).'; %'// reshape that vector into a matrix, and transpose back
Here's one approach using bsxfun and permute to solve for a 3D array case, assuming you want to remove indexed elements per row across all 3D slices -
%// Inputs
A = randi(9,3,4,3)
idx = [2 4 3]
%// Get size of input array, A
[M,N,P] = size(A)
%// Permute A to bring the columns as the first dimension
Ap = permute(A,[2 1 3])
%// Per 3D slice offset linear indices
offset = bsxfun(#plus,[0:M-1]'*N,[0:P-1]*M*N) %//'
%// Get 3D array linear indices and remove those from permuted array
Ap(bsxfun(#plus,idx(:),offset)) = []
%// Permute back to get the desired output
out = permute(reshape(Ap,3,3,3),[2 1 3])
Sample run -
>> A
A(:,:,1) =
4 4 1 4
2 9 7 5
5 9 3 9
A(:,:,2) =
4 7 7 2
9 6 6 9
3 5 2 2
A(:,:,3) =
1 7 5 8
6 2 9 6
8 4 2 4
>> out
out(:,:,1) =
4 1 4
2 9 7
5 9 9
out(:,:,2) =
4 7 2
9 6 6
3 5 2
out(:,:,3) =
1 5 8
6 2 9
8 4 4