How to repeat
A = [ 1 2 ;
3 4 ]
repeated by
B = [ 1 2 ;
2 1 ]
So I want my answer like matrix C:
C = [ 1 2 2;
3 3 4 ]
Thanks for your help.
Just for the fun of it, another solution making use of arrayfun:
res = cell2mat(arrayfun(#(a,b) ones(b,1).*a, A', B', 'uniformoutput', false))'
This results in:
res =
1 2 2
3 3 4
To make this simple, I assume that you're only going to add more columns, and that you've checked that you have the same number of columns for each row.
Then it becomes a simple combination of repeating elements and reshaping.
EDIT I've modified the code so that it also works if A and B are 3D arrays.
%# get the number of rows from A, transpose both
%# A and B so that linear indexing works
[nRowsA,~,nValsA] = size(A);
A = permute(A,[2 1 3]);
B = permute(B,[2 1 3]);
%# create an index vector from B
%# so that we know what to repeat
nRep = sum(B(:));
repIdx = zeros(1,nRep);
repIdxIdx = cumsum([1 B(1:end-1)]);
repIdx(repIdxIdx) = 1;
repIdx = cumsum(repIdx);
%# assemble the array C
C = A(repIdx);
C = permute(reshape(C,[],nRowsA,nValsA),[2 1 3]);
C =
1 2 2
3 3 4
Related
I am trying to perform a row-wise "ismember" in MATLAB in order to find out where, in Set, each element of Input is.
Here is my work so far.
function lia = ismemberrow(Input, Set)
lia = false(size(Input)); % Pre-assign the output lia
for J = 1 : size(Input,1)
% The J-th row of "lia" is the list of locations of the
% elements of Input(J,:) in Set
lia(J,:) = find(ismember(Set, Input(J,:)));
end
end
For example, if the variables Input and Set are defined as follows
Input = [1 4;
4 2;
4 3;
2 4;
1 2;
3 2];
Set = [3 2 4 1];
The the output lia of lia = ismemberrow(Input,Set) will be:
lia = [4 3;
3 2;
3 1;
2 3;
4 2;
1 2];
My function works accurately so far, but this function is called many times in my project so I am thinking of that if I can reduce the for-loop so that it spends less time. May I have some opinions on it?
A single call to ismember (no loop necessary) will give you what you want in the second output argument:
>> [~, lia] = ismember(Input, Set)
lia =
4 3
3 2
3 1
2 3
4 2
1 2
I'd go with ismember as in #gnovice's answer. But here are some alternatives, just for the fun of it.
If the values in Input are guaranteed to be in Set:
[ind, ~] = find(bsxfun(#eq, Set(:), Input(:).'));
result = reshape(ind, size(Input));
If they are not guaranteed to be:
[ind, val] = max(bsxfun(#eq, Set(:), permute(Input, [3 1 2])));
result = permute(ind.*val, [2 3 1]);
If your inputs are positive integers you simply can use indexing
m(Set)=1:numel(Set);
result = m(Input)
If the range of input is large you can use sparse matrix:
s = sparse(Set,1,1:numel(Set));
result = s(Input)
Result:
4 3
3 2
3 1
2 3
4 2
1 2
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,[])
if you have an array from a for loop
for a =1:3;
for b=1:3;
for c=1:3;
disp([a(:) b(:) c(:)])
end
end
end
and you want to separate each column of the output
to separate each column you need this code: assuming the data stored in 'A' array:
yy = A(:,1)
yx = A(:,2)
yz = A(:,3)
and so on...
but what if you don't know the size of your array or if you have m x n array? what is the general code to separate column? thanks
You can do it like this:
y = [1 1 1 1 1
1 2 3 1 1
1 0 2 2 1]
[m,n] = size(y)
for i=1:n
C{i} = y(:,i);
end
Now you can access each column via C(1), C(2), C(3), etc. C looks like this:
C =
{
[1,1] =
1
1
1
[1,2] =
1
2
0
[1,3] =
1
3
2
}
The output of just C(2) is:
ans =
{
[1,1] =
1
2
0
}
That said, it is probably better to stick with the code that you have to directly access the column from the matrix (y(:,1)), instead of creating a bunch of dynamic variables.
How would you manage these columns separated ? It's not a good idea to name them as y1, y2, ... .
If you want to pick up the i-th column, just use y(:, i). That's a lot easier to use.
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 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)