I have a matrix X e.g = [a b; c d; e f].
I need to create another matrix listing the index positions and values of the matrix.
e.g. The output is E = [ 1 1 a ; 1 2 b ; 2 1 c ; 2 2 d ; 3 1 e ; 3 2 f ]
I have been trying to use a double for loop but even if it did work, that sounds like a bad idea.
So can anyone has a better idea to perform above task?
Here is the dumbest thing I could think of (Assuming that a,b,c,d,e,f are all scalars)
x = [1 2;3 4;5 6];
[i,j]=ind2sub(size(transpose(x)), 1:numel(x));
[j(:) i(:) reshape(transpose(x),[],1)]
However, I have a feeling that there might be an answer that is more elegant.
Nothing wrong with #Andrey's answer, but because I like to find reasons to use kron :)
A = [1 2; 3 4; 5 6];
[nrows, ncols] = size(A);
M = [kron([1 : nrows]', ones(ncols, 1))...
kron(ones(nrows, 1), [1 : ncols]')...
reshape(A', [], 1)]
Related
I have A matrix which is 16x16x155460. I have a B vector which is 12955x1. I want to multiply each 1:16x1:16x1+12*n:12+12*nwith the elements of B(n). So my goal is to find the weighted sum of the A according to B. My way to do this as follows (I don't want to use for-loop and my method gives wrong answer, I could not obtain the 1:12 vectors which is consecutive) :
B = repmat(B,[1 16 16]);
B = permute(B,[2 3 1]);
B = repmat(B,[1 1 12]);
result = B.*(A);
As a small example n=2 :
A(:,:,1)=[1 2; 3 4]
A(:,:,2)=[1 2; 3 4]
A(:,:,3)=[1 2; 3 4]
A(:,:,4)=[1 2; 3 4]
B = [2,3]
Result would be:
result(:,:,1)=A(:,:,1)*B(1);
result(:,:,2)=A(:,:,2)*B(1);
result(:,:,3)=A(:,:,1)*B(2);
result(:,:,4)=A(:,:,2)*B(2);
If I understood the problem correctly, you can use the powerful trio of bsxfun, permute and reshape to solve it, like so -
[M,N,R] = size(A);
mult_out = bsxfun(#times,reshape(A,M,N,numel(B),[]),permute(B(:),[4 3 1 2]))
out = reshape(mult_out,M,N,[])
I have a matrix:
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
I want to find all row-wise combinations of X. i.e.
A(1) = [2, 6, 1; 3, 8, 1; 4, 7, 1]
A(2) = [2, 6, 1; 3, 8, 1; 6, 2, 1]
:
:
:
Here's what I've tried:
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
p = 3
[m, n] = size(X);
comb = combnk(1:m, p);
[s, t] = size(comb);
c = [X(comb(:,1), :, :) X(comb(:,2), :, :) X(comb(:,3), :, :)];
This gives me a matrix like:
c = 2 6 1 3 8 1 4 7 1
2 6 1 3 8 1 6 2 1
2 6 1 3 8 1 6 4 1
I want to apply the concatenate matrix option to obtain c to make it dynamic depending on value of p but I'm not sure how to use it. I don't want to use For loops. Please help me out.
This is fully vectorized, so it should be fast:
n = 3; %// number of rows to pick each time
ind = reshape(nchoosek(1:size(X,1), n).', [], 1); %'// indices of combinations
A = permute(reshape(X(ind,:).', size(X,2), n, []), [2 1 3]);
The result is
A(:,:,1)
ans =
2 6 1
3 8 1
4 7 1
A(:,:,2)
ans =
2 6 1
3 8 1
6 2 1
etc.
Should you need the result in the form of a cell array, you can convert A from 3D-array to cell array this way:
A = mat2cell(A, size(A,1), size(A,2), ones(1,size(A,3)));
Your thinking is pretty close. This code does the job. I put comments in code, which should be easy to read.
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
p = 3;
%// List all combinations choosing 3 out of 1:8.
v = nchoosek(1:size(X,1), p);
%// Use each row of v to create the matrices, and put the results in an cell array.
%// This is the A matrix in your question.
A = arrayfun(#(k)X(v(k,:), :), 1:size(v,1), 'UniformOutput', false);
%// And you can concatenate A vertically to get c.
flatA = cellfun(#(x)reshape(x, 1, []), A, 'UniformOutput', false);
c = vertcat(flatA{:});
PS: From my understanding I thought the result you wanted was A, which is an easy to use cell array. But I added an extra step to get c exactly as in your question just in case.
Disclaimer: arrayfun and cellfun are pretty much equivalent to for loop in terms of performance.
You can do it using reshape and a bunch of transposes since Matlab is column-major ordered:
c = reshape(X(comb',:)',9,[])'
or if you want a 3D matrix:
A = permute(reshape(X(comb',:)',3,3,[])', [2,1,3])
I have the following two arrays:
A = [1 2;3 4] and B = [1 5 4]
I want to do the following operation:
for each element of A(call it A(i))
for each element of B~=b do
( (A(i) - 1)/(b-1) ) * ( (A(i) - 5)/(b-5) ) * ( (A(i)- 4)/(b-4) )
end
end
It means that, sometimes the numerator equals to zero, so the product should be zeros. And I want to do the operation for the elements of B which are not equal to the b in denominator to not make it Inf.
How can I do this for the whole matrix A instead of using for loop?
Code
A = [1 2;3 4];
B = [1 5 4];
m1 = bsxfun(#minus,A,permute([1 5 4],[3 1 2]));
m2 = bsxfun(#minus,B,permute([1 5 4],[3 1 2]));
for k1=1:size(A,1)
for k2=1:size(A,2)
t2 = squeeze(bsxfun(#rdivide,m1(k1,k2,:),m2));
t2(1:size(t2,1)+1:end)=1;
A1(k1,k2) = prod(t2(:)); %%// Output
end
end
Output
A1 =
0 -0.2500
-0.1111 0
You can remove the nested loops, but at least two issues there -
You would be going to 4th and 5th dimension with it, using bsxfun. So, debugging would be tough.
bsxfun with higher dimensions to my knowledge seems to get slower.
You could just do the operation, and correct later:
C = (A-1)./(B-1) .* (A-5)./(B-5) .* (A-4)./(B-4)
C(isinf(C)) = 0;
or
C(B==b) = 0;
Possibly you'd need bsxfun, I'm not clear on the size of the output you want...
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)
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