Good day,
I have got something similar as:
A = [1 2; 3 4];
B = [2 3; 4 5; 6 7];
C{1} = A;
C{2} = B;
clear A;
clear B;
Now I would like to select element (2,1) from matrix B, that is, element (2,1) from C{2}. However, matrix B itself does not exist any more.
One possibility is:
B = C{2};
B(2,1)
However, is there a more direct way to access elements from matrices which are stored in a cell?
Cells allow you to chain subscripts in the following manner
>> C{2}(2, 1)
ans =
4
Related
I would like to know if there is an easy way to find the indices of a vector in another vector in matlab:
a = [1 2 3 5 7 10 2 3 6 8 7 5 2 4 7 2 3]
b = [2 3]
So how to get the indices of a when comparing it with b (index of first element is needed)
In this case:
ans = [2 7 16]
Thanks in advance
find(a(1:end-1) == b(1) & a(2:end) == b(2) == 1)
You can re-purpose strfind by converting the elements of both vectors to byte arrays (uint8) with typecast:
bytesPerEl = numel(typecast(a(1),'uint8'));
byteLocs = strfind(char(typecast(a,'uint8')),char(typecast(b,'uint8')));
locsb = (byteLocs-1)/bytesPerEl + 1
locsb =
2 7 16
Just make sure a and b are of the same type. Also note that this works for 1D vectors, not matrixes or higher dimensional arrays.
General approach with length of b arbitrary (not necessarily 2 as in the example), and avoiding the use of strings:
match1 = bsxfun(#eq, a(:), b(:).'); %'// now we just need to make the diagonals
%// horizontal (in order to apply "all" row-wise). For that we'll use indices
%// ind, ind1, ind2
ind = reshape(1:numel(match1), numel(a), numel(b));
ind1 = nonzeros(tril(ind)); %// source indices
ind2 = sort(nonzeros(tril(flipud(ind)))); %// destination indices
match2 = zeros(size(match1));
match2(ind2) = match1(ind1); %// diagonals have become horizontal
result = find(all(match2.'));
I have two 2-dimensional matrices A,B, where B is produced by a (row-wise) permutation of A. There are a few repetitive records in A (and so in B). I want to find the mapping that produced B. I am using Matlab. Only one solution is sufficient for me.
Example:
A = [ 2 3 4; 4 5 6; 2 3 4];
B = [ 4 5 6; 2 3 4; 2 3 4];
The mapping would be:
p = [3 1 2] // I want this mapping, however the solution p= [2 1 3] is also correct and acceptable
where A = B(p,:) in Matlab. // EDITED
Regards
low hanging fruits first.
Suppose there are no duplicate rows:
% compute the permutation matrix
P = all( bsxfun( #eq, permute( A, [1 3 2]),permute(B,[3 1 2]) ), 3 );
[~, p] = max(P, [], 2 ); % gives you what you want
If there are duplicates, we need to "break ties" in the rows/columns of P:
n = size(A,1);
bt = abs( bsxfun(#minus, 1:n, (1:n)' ) )/n; %//'
[~, p] = max( P+bt, [], 2 );
Since we know that A and B always have the same rows, let's look for a transformation that will convert each one to a common identical representation. How about sort?
[As, Ai] = sortrows(A);
[Bs, Bi] = sortrows(B);
Now A(Ai,:) == B(Bi,:), so all we have to do is find the indices for Bi that match Ai. Bi is a forward mapping, Ai is a reverse mapping. So:
p = zeros(size(A,1),1);
p(Ai) = Bi;
(Answer edited to match edit of problem statement)
Here is a solution using sort() to get around the problem of needing to generate all permutations.
The idea is to sort both A and B which will produce the same sorted matrix. The permutation can now be found by using the indices IA and IB that produce the two sorted matrices.
A = [ 2 3 4; 4 5 6; 2 3 4];
B = [ 4 5 6; 2 3 4; 2 3 4];
[CA,IA]=sort(A,1)
[CB,IB]=sort(B,1)
idxA = IA(:,1)
idxB = IB(:,1)
[~, idxB_inverse] = sort(idxB)
idxA(idxB_inverse)
Consider having the following multidimensional array:
A = [1 2;3 4];
B = [5 6;7 8];
C = cat(3, A, B);
Well it is like a cube, I want to slice the first row, slice the first column and that's it.
When I do:
C(1,1,:)
I get two separate answers:
C(1,1,1) = 1
C(1,1,2) = 5
And if I do
D = C(1,1,:)
I get D to be a multidimensional.
I want this
D = [1 5];
That's it, how to do this?
Use the squeeze() function to drop the extra dimensions:
>> squeeze(C(1,1,:))'
ans =
1 5
I know this is a simple question but difficult to formulate in one sentence to google the answer.So, I have a 3d matrix with size 2x2x3 like this
A(:,:,1) =[1 1; 1 1];
A(:,:,2) =[2 2; 2 2];
A(:,:,3) =[4 4; 4 4];
and matrix B with size 2x2
B = [ 1 2; 2 3];
What i need is to chose from each third dimension in A just one number using matrix B:
for i=1:2,
for j=1:2,
C(i,j) = A(i,j,B(i,j));
end
end
How to that in one line without a loop?
Not really a single line, but without a loop:
[I J] = ind2sub (size(B), 1:numel(B));
linInd = sub2ind (size (A), I, J, B(:)');
C = reshape (A(linInd), size(B));
Here is another variation:
[r,c,~] = size(A);
[J,I] = meshgrid(1:size(B,1), 1:size(B,2));
idx = reshape(I(:) + r*(J(:)-1) + r*c*(B(:)-1), size(B));
C = A(idx)
I am trying to create a function that will swap a specific number in a matrix with a specific number in the same matrix. For examlpe, if I start with A = [1 2 3;1 3 2], I want to be able to create B = [2 1 3; 2 3 1], simply by telling matlab to swap the 1's with the 2's. Any advice would be appreciated. Thanks!
If you have the following matrix:
A = [1 2 3; 1 3 2];
and you want all the ones to become twos and the twos to become ones, the following would be the simplest way to do it:
B = A;
B(find(A == 1)) = 2;
B(find(A == 2)) = 1;
EDIT:
As Kenny suggested, this can even be further simplified as:
B = A;
B(A == 1) = 2;
B(A == 2) = 1;
Another way to deal with the original problem is to create a permutation vector indicating to which numbers should the original entries be mapped to. For the example, entries [1 2 3] should be mapped respectively to [2 1 3], so that we can write
A = [1 2 3; 1 3 2];
perm = [2 1 3];
B = perm(A)
(advantage here is that everything is done in one step, and that it also works for operations more complicated than swaps ; drawback is that all elements of A must be positive integers with a known maximum)
Not sure why you would to perform that particular swap (row/column interchanges are more common). Matlab often denotes ':' to represent all of something. Here's how to swap rows and columns:
To swap rows:
A = A([New order of rows,,...], :)
To Swap columns:
A = A(:, [New order of columns,,...])
To change the entire i-th column:
A(:, i) = [New; values; for; i-th; column]
For example, to swap the 2nd and 3rd columns of A = [1 2 3;1 3 2]
A = A(:, [1, 3, 2])
A = [1 2 3; 1 3 2]
alpha = 1;
beta = 2;
indAlpha = (A == alpha);
indBeta = (A == beta);
A(indAlpha) = beta;
A(indBeta ) = alpha
I like this solution, it makes it clearer what is going on. Less magic numbers, could easily be made into a function. Recycles the same matrix if that is important.
I don't have a copy of MatLab installed, but I think you can do some thing like this;
for i=1:length(A)
if (A(i)=1), B(i) = 2, B(i)=A(i)
end
Note, that's only convert 1's to 2's and it looks like you also want to convert 2's to 1's, so you'll need to do a little more work.
There also probably a much more elegant way of doing it given you can do this sort of thing in Matlab
>> A = 1:1:3
A = [1,2,3]
>> B = A * 2
B = [2,4,6]
There might be a swapif primitive you can use, but I haven't used Matlab in a long time, so I'm not sure the best way to do it.
In reference to tarn's more elegant way of swapping values you could use a permutation matrix as follows:
>> a =[1 2 3];
>> T = [1 0 0;
0 0 1;
0 1 0];
>> b = a*T
ans =
1 3 2
but this will swap column 2 and column 3 of the vector (matrix) a; whereas the question asked about swapping the 1's and 2's.
Update
To swap elements of two different values look into the find function
ind = find(a==1);
returns the indices of all the elements with value, 1. Then you can use Mitch's suggestion to change the value of the elements using index arrays. Remeber that find returns the linear index into the matrix; the first element has index 1 and the last element of an nxm matrix has linear index n*m. The linear index is counted down the columns. For example
>> b = [1 3 5;2 4 6];
>> b(3) % same as b(1,2)
ans = 3
>> b(5) % same as b(1,3)
ans = 5
>> b(6) % same as b(2,3)
ans = 6