I have a NxNx4 matrix(A) and a 4x4 matrix (B). I need to multiply the vector a composed by the four elements of the first matrix A, let's say
a = A(1,1,1)
A(1,1,2)
A(1,1,3)
A(1,1,4)
by the 4x4 matrix B but I'm not sure if there is a faster and clever solution than using a for loop to build the vector a. Does exist a way to do this computation with few lines of code?
I built A like
A(:,:,1) = rand(20);
A(:,:,2) = rand(20);
A(:,:,3) = rand(20);
A(:,:,4) = rand(20);
and the matrix B
B = rand(4);
now I want to multiply B with
B*[A(1,1,1);A(1,1,2);A(1,1,3);A(1,1,4)]
This, for each element of A
B*[A(1,2,1);A(1,2,2);A(1,2,3);A(1,2,4)]
B*[A(1,3,1);A(1,3,2);A(1,3,3);A(1,3,4)]
...
You can do this with a simple loop, note loops aren't necessarily slow in newer MATLAB versions. Mileage may vary.
Loops have the advantage of improving code readability, it's extremely clear what's happening here:
% For matrix A of size N*N*4
C = zeros( size( A ) );
for ii = 1:N
for jj = 1:N
C( ii, jj, : ) = B * reshape( A( ii, jj, : ), [], 1 );
end
end
A loop solution that has good performance specially when N is large:
s = size(A, 3);
C = A(:,:,1) .* reshape(B(:,1),1,1,[]);
for k = 2:s
C = C + A(:,:,k) .* reshape(B(:,k),1,1,[]);
end
I think this does what you want:
C = permute(sum(bsxfun(#times, permute(B, [3 4 2 1]), A), 3), [1 2 4 3]);
Check:
>> C(1,2,:)
ans(:,:,1) =
1.501739582138850
ans(:,:,2) =
1.399465238902816
ans(:,:,3) =
0.715531734553844
ans(:,:,4) =
1.617019921519029
>> B*[A(1,2,1);A(1,2,2);A(1,2,3);A(1,2,4)]
ans =
1.501739582138850
1.399465238902816
0.715531734553844
1.617019921519029
Related
I want to write this matrix in matlab,
s=[0 ..... 0
B 0 .... 0
AB B .... 0
. . .
. . .
. . . 0 ....
A^(n-1)*B ... AB B ]
I have tried this below code but giving error,
N = 50;
A=[2 3;4 1];
B=[3 ;2];
[nx,ny] = size(A);
s(nx,ny,N) = 0;
for n=1:1:N
s(:,:,n)=A.^n;
end
s_x=cat(3, eye(size(A)) ,s);
for ii=1:1:N-1
su(:,:,ii)=(A.^ii).*B ;
end
z= zeros(1,60,1);
su1 = [z;su] ;
s_u=repmat(su1,N);
seems like the concatenation of matrix is not being done.
I am a beginner so having serious troubles,please help.
Use cell arrays and the answer to your previous question
A = [2 3; 4 1];
B = [3 ;2 ];
N = 60;
[cs{1:(N+1),1:N}] = deal( zeros(size(B)) ); %// allocate space, setting top triangle to zero
%// work on diagonals
x = B;
for ii=2:(N+1)
[cs{ii:(N+2):((N+1)*(N+2-ii))}] = deal(x); %//deal to diagonal
x = A*x;
end
s = cell2mat(cs); %// convert cells to a single matrix
For more information you can read about deal and cell2mat.
Important note about the difference between matrix operations and element-wise operations
In your question (and in your previous one) you confuse between matrix power: A^2 and element-wise operation A.^2:
matrix power A^2 = [16 9;12 13] is the matrix product of A*A
element-wise power A.^2 takes each element separately and computes its square: A.^2 = [4 9; 16 1]
In yor question you ask about matrix product A*b, but the code you write is A.*b which is an element-by-element product. This gives you an error since the size of A and the size of b are not the same.
I often find that Matlab gives itself to a coding approach of "write what it says in the equation". That also leads to code that is easy to read...
A = [2 3; 4 1];
B = [3; 2];
Q = 4;
%// don't need to...
s = [];
%// ...but better to pre-allocate s for performance
s = zeros((Q+1)*2, Q);
X = B;
for m = 2:Q+1
for n = m:Q+1
s(n*2+(-1:0), n-m+1) = X;
end
X = A * X;
end
I have three matrices in Matlab, A which is n x m, B which is p x m and C which is r x n.
Say we initialize some matrices using:
A = rand(3,4);
B = rand(2,3);
C = rand(5,4);
The following two are equivalent:
>> s=0;
>> for i=1:3
for j=1:4
s = s + A(i,j)*B(:,i)*C(:,j)';
end;
end
>> s
s =
2.6823 2.2440 3.5056 2.0856 2.1551
2.0656 1.7310 2.6550 1.5767 1.6457
>> B*A*C'
ans =
2.6823 2.2440 3.5056 2.0856 2.1551
2.0656 1.7310 2.6550 1.5767 1.6457
The latter being much more efficient.
I can't find any efficient version for the following variant of the loop:
s=0;
for i=1:3
for j=1:4
x = A(i,j)*B(:,i)*C(:,j)';
s = s + x/sum(sum(x));
end;
end
Here, the matrices being added are normalized by the sum of their values after each step.
Any ideas how to make this efficient like the matrix multiplication above? I thought maybe accumarray could help, but not sure how.
You can do it efficiently with bsxfun:
aux1 = bsxfun(#times, permute(B,[1 3 2]), permute(C,[3 1 4 2]));
aux2 = sum(sum(aux1,1),2);
s = sum(sum(bsxfun(#rdivide, aux1, aux2),3),4);
Note that, because of the normalization, the result is independent of A, assuming it doesn't contain any zero entries (if it does the result is undefined).
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Closed 10 years ago.
Possible Duplicate:
How to Add a row vector to a column vector like matrix multiplication
I have a nx1 vector and a 1xn vector. I want to add them in a special manner like matrix multiplication in an efficient manner (vectorized):
Example:
A=[1 2 3]'
B=[4 5 6]
A \odd_add B =
[1+4 1+5 1+6
2+4 2+5 2+6
3+4 3+5 3+6
]
I have used bsxfun in MATLAB, but I think it is slow. Please help me...
As mentioned by #b3. this would be an appropriate place to use repmat. However in general, and especially if you are dealing with very large matrices, bsxfun normally makes a better substitute. In this case:
>> bsxfun(#plus, [1,2,3]', [4,5,6])
returns the same result, using about a third the memory in the large-matrix limit.
bsxfun basically applies the function in the first argument to every combination of items in the second and third arguments, placing the results in a matrix according to the shape of the input vectors.
I present a comparison of the different methods mentioned here. I am using the TIMEIT function to get robust estimates (takes care of warming up the code, average timing on multiple runs, ..):
function testBSXFUN(N)
%# data
if nargin < 1
N = 500; %# N = 10, 100, 1000, 10000
end
A = (1:N)';
B = (1:N);
%# functions
f1 = #() funcRepmat(A,B);
f2 = #() funcTonyTrick(A,B);
f3 = #() funcBsxfun(A,B);
%# timeit
t(1) = timeit( f1 );
t(2) = timeit( f2 );
t(3) = timeit( f3 );
%# time results
fprintf('N = %d\n', N);
fprintf('REPMAT: %f, TONY_TRICK: %f, BSXFUN: %f\n', t);
%# validation
v{1} = f1();
v{2} = f2();
v{3} = f3();
assert( isequal(v{:}) )
end
where
function C = funcRepmat(A,B)
N = numel(A);
C = repmat(A,1,N) + repmat(B,N,1);
end
function C = funcTonyTrick(A,B)
N = numel(A);
C = A(:,ones(N,1)) + B(ones(N,1),:);
end
function C = funcBsxfun(A,B)
C = bsxfun(#plus, A, B);
end
The timings:
>> for N=[10 100 1000 5000], testBSXFUN(N); end
N = 10
REPMAT: 0.000065, TONY_TRICK: 0.000013, BSXFUN: 0.000031
N = 100
REPMAT: 0.000120, TONY_TRICK: 0.000065, BSXFUN: 0.000085
N = 1000
REPMAT: 0.032988, TONY_TRICK: 0.032947, BSXFUN: 0.010185
N = 5000
REPMAT: 0.810218, TONY_TRICK: 0.824297, BSXFUN: 0.258774
BSXFUN is a clear winner.
In matlab vectorization, there is no substitute for Tony's Trick in terms of speed in comparison to repmat or any other built in Matlab function for that matter. I am sure that the following code must be fastest for your purpose.
>> A = [1 2 3]';
>> B = [4 5 6];
>> AB_sum = A(:,ones(3,1)) + B(ones(3,1),:);
The speed differential will be much more apparent (at LEAST an order of magnitude) for larger size of A and B. See this test I conducted some time ago to ascertain the superiority of Tony's Trick over repmatin terms of time consumption.
REPMAT is your friend:
>> A = [1 2 3]';
>> B = [4 5 6];
>> AplusB = repmat(A, 1, 3) + repmat(B, 3, 1)
AplusB =
5 6 7
6 7 8
7 8 9
I am having trouble with the following task:
Suppose a 3x6 matrix:
A =
0.2787 0.2948 0.4635 0.8388 0.0627 0.0435
0.6917 0.1185 0.3660 0.1867 0.2383 0.7577
0.6179 0.7425 0.0448 0.4009 0.9377 0.4821
What I would like to do is to divide the matrix into blocks, like this:
A =
0.2787 0.2948 | 0.4635 0.8388 | 0.0627 0.0435
0.6917 0.1185 | 0.3660 0.1867 | 0.2383 0.7577
0.6179 0.7425 | 0.0448 0.4009 | 0.9377 0.4821
and vertically concatenate these blocks to get the final result:
0.2787 0.2948
0.6917 0.1185
0.6179 0.7425
0.4635 0.8388
0.3660 0.1867
0.0448 0.4009
0.0627 0.0435
0.2383 0.7577
0.9377 0.4821
I think if I can get help with this, then I can perhaps
do it for arbitrary matrices A. I can solve the above
problem using for-loops, but I am looking for a vectorised solution.
Thanks in advance!
N.
Consider the following:
A = rand(3,6);
blkSz = 2;
C = mat2cell(A, size(A,1), blkSz*ones(1,size(A,2)/blkSz));
C = cat(1,C{:})
This assumes that size(A,2) is evenly divisible by blkSz
This works where your matrix is A and what you want is D
C = mat2cell(A,[3],[2 2 2])
D = cat(1,C{:})
It's possible to do it without cell2mat, (only with reshapes and permute) and thus a lot faster!
You need to use the "3rd dimension". It's similar to what is described in
split long 2D matrix into the third dimension.
Here is the solution for the above matrix:
A1 = reshape(A, 3, 2, []); % 3rd dimension is numel(A)/2/3
A2 = permute(A1, [2 1 3]); % transpose 1st and 2nd dimension
Ans= reshape(A2, 2, [])' ; % note the transpose
For a matrix of this size, the difference in running time is negligible. However, for a large matrix, the difference is more than an order of magnitude:
A=rand(3, 2*10000);
%% good method
tic
A1 = reshape(A, 3, 2, []); %3rd dimension is numel(A)/2/3
A2 = permute(A1, [2 1 3]);
A3 = reshape(A2, 2, [])' ; %note the transpose'
toc
%% mat2cell method
tic
blkSz = 2;
C = mat2cell(A, size(A,1), blkSz*ones(1,size(A,2)/blkSz));
B3 = cat(1,C{:});
toc
%% make sure the answer is the same:
assert(max(A3(:)-B3(:))==0)
output:
>> Elapsed time is 0.001202 seconds.
>> Elapsed time is 0.043115 seconds.
How about this:
width = 2;
m = length(A(:))/width;
fn = #(i) reshape(A(:, i:width:end), m, 1);
B = cell2mat(arrayfun(fn, 1:width, 'UniformOutput', false));
Just specify how many columns you want at a time in the width variable.
By concatenating vertically, matrix width divisible by 3 assumed:
B = [ A(:,1:(size(A,2)/3)); A(:,size(A,2)/3+1:size(A,2)/3*2); A(:,size(A,2)/3*2+1:end) ];
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)