I would like to do an operation on a 2-D matrix which somehow looks like the outer product of a vector. I already have written some codes for this task, but it is pretty slow, so I would like to know if there is anything I can do to accelerate it.
I would like to show the code I wrote first, followed by an example to illustrate the task I wanted to do.
My code, version row-by-row
function B = outer2D(A)
B = zeros(size(A,1),size(A,2),size(A,2)); %Pre-allocate the output array
for J = 1 : size(A,1)
B(J,:,:) = transpose(A(J,:))*A(J,:); %Perform outer product on each row of A and assign to the J-th layer of B
end
end
Using the matrix A = randn(30000,20) as the input for testing, it spends 0.317 sec.
My code, version page-by-page
function B = outer2D(A)
B = zeros(size(A,1),size(A,2),size(A,2)); %Pre-allocate the output array
for J = 1 : size(A,2)
B(:,:,J) = repmat(A(:,J),1,size(A,2)).*A; %Evaluate B page-by-page
end
end
Using the matrix A = randn(30000,20) as the input for testing, it spends 0.146 sec.
Example 1
A = [3 0; 1 1; 1 0; -1 1; 0 -2]; %A is the input matrix.
B = outer2D(A);
disp(B)
Then I would expect
(:,:,1) =
9 0
1 1
1 0
1 -1
0 0
(:,:,2) =
0 0
1 1
0 0
-1 1
0 4
The first row of B, [9 0; 0 0], is the outer product of [3 0],
i.e. [3; 0]*[3 0] = [9 0; 0 0].
The second row of B, [1 1; 1 1], is the outer product of [1 1],
i.e. [1; 1]*[1 1] = [1 1; 1 1].
The third row of B, [1 0; 0 0], is the outer product of [1 0],
i.e. [1; 0]*[1 0] = [1 0; 0 0].
And the same for the remaining rows.
Example 2
A =
0 -1 -2
0 1 0
-3 0 2
0 0 0
1 0 0
B = outer2D(A)
disp(B)
Then, similar to the example 1, the expected output is
(:,:,1) =
0 0 0
0 0 0
9 0 -6
0 0 0
1 0 0
(:,:,2) =
0 1 2
0 1 0
0 0 0
0 0 0
0 0 0
(:,:,3) =
0 2 4
0 0 0
-6 0 4
0 0 0
0 0 0
Because the real input in my project is like in the size of 30000 × 2000 and this task is to be performed for many times. So the acceleration of this task is quite essential for me.
I am thinking of eliminating the for-loop in the function. May I have some opinions on this problem?
With auto expansion:
function B = outer2D(A)
B=permute(permute(A,[3 1 2]).*A',[2 3 1]);
end
Without auto expansion:
function B = outer2Dold(A)
B=permute(bsxfun(#times,permute(A,[3 1 2]),A'),[2 3 1]);
end
Outer products are not possible in the matlab language.
Related
I have a code for a 1D heat equation. Im trying to format a for loop so that the A matrix will follow a certain pattern of 1 -2 1 down the entire diagonal of a matrix that could be infinite. The pattern starts to take shape when I mess around with the initialized count at the beginning of the for loop but this changes the size of the matrix which fails the rest of the code.
My current code is below. The commented A matrix edits are what it should be.
N = 5;
%A(2,1:3) = [1 -2 1];
%A(3,2:4) = [1 -2 1];
%A(4,3:5) = [1 -2 1];
%A(5,4:6) = [1 -2 1];
A = zeros(N+1,N+1);
A(1,1) = 1;
for count=N:N+1
A(count+1,count:count+2) = [1 -2 1];
end
A(N+1,N+1) = 1;
In Matlab you can often avoid loops. In this case you can get the desired result with 2D convolution:
>> N = 6;
>> A = [1 zeros(1,N-1); conv2(eye(N-2), [1 -2 1]); zeros(1,N-1) 1]
A =
1 0 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 0 1
Or, depending on what you want,
>> A = conv2(eye(N), [1 -2 1], 'same')
A =
-2 1 0 0 0 0
1 -2 1 0 0 0
0 1 -2 1 0 0
0 0 1 -2 1 0
0 0 0 1 -2 1
0 0 0 0 1 -2
There are many simple ways of creating this matrix.
Your loop can be amended as follows:
N = 5;
A = zeros(N+1,N+1);
A(1,1) = 1;
for row = 2:N
A(row, row-1:row+1) = [1 -2 1];
end
A(N+1,N+1) = 1;
I've renamed count to row, we're indexing each row (from 2 to N, skipping the first and last rows), then finding with row-1:row+1 the three indices for that row that you want to address.
Directly indexing the diagonal and off-diagonal elements. Diagonal elements for an NxN matrix are 1:N+1:end. This is obviously more complex, I'd prefer the loop:
N = 6;
A = zeros(N,N);
A(1:N+1:end) = -2;
A(2:N+1:end-2*N) = 1; % skip last row
A(2*N+2:N+1:end) = 1; % skip first row
A(1,1) = 1;
A(N,N) = 1;
Using diag. We need to special-case the first and last rows:
N = 6;
A = diag(-2*ones(N,1),0) + diag(ones(N-1,1),1) + diag(ones(N-1,1),-1);
A(1,1:2) = [1,0];
A(end,end-1:end) = [0,1];
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
Lets say I have a matrix A:
A =
0 1 0 0
0 0 0 0
0 0 0 1
0 0 0 0
And I want to create a new matrix B of the same dimension where all ones and accompanying neighbours are replaced by the following matrix:
X =
1 1 1
1 2 1
1 1 1
The 2 in matrix X should be placed 'on top' of the 1 values as to get:
B =
1 2 1 0
1 1 2 1
0 0 1 2
0 0 1 1
Values should be added up where elements overlap and matrix X should be 'cut off' in places where it extends the dimensions of matrix A/B The idea is to eventually replace X by a 2d gaussian distribution and matrix A will be large containing many more ones. So it's essential that the code is efficient and fast. This is the code i came up with:
A = [0 1 0 0;0 0 0 0;0 0 0 1;0 0 0 0]
X = [1 1 1;1 2 1;1 1 1]
B = zeros(4,4);
t=1;
indA = find(A==1);
indX = find(X==2);
all = find(X>0);
[iall jall] = ind2sub(size(X),all);
[ia ja] = ind2sub(size(A),indA)
[ix jx] = ind2sub(size(X),indX)
iv = ia-ix
jv = ja-jx
for t=1:numel(iv),
ib = iall+iv(t);
jb = jall+jv(t);
ibjb = [ib(:), jb(:)]
c1 = (ibjb(:,1)>4)|(ibjb(:,1)<1); c2 = (ibjb(:,2)>4)|(ibjb(:,1)<1);
ibjb((c1|c2),:)=[]
isel = ibjb(:,1)-iv(t)
jsel = ibjb(:,2)-jv(t)
B(ibjb(:,1), ibjb(:,2)) = B(ibjb(:,1), ibjb(:,2))+ X(isel, jsel)
t=t+1;
end
Is there a more efficient/faster way (minimizing the loops) to code this function?
What you want is a (2D) convolution. So use conv2:
B = conv2(A, X, 'same');
I need to vectorize the following code:
a = [1 2 3 2 3 1];
b = [1 2 3];
for i = 1:length(a)
for j = 1:length(b)
r(i, j) = (a(i) == b(j));
end
end
The output r should be a logical array:
1 0 0
0 1 0
0 0 1
0 1 0
0 0 1
1 0 0
The closest I can get is:
for j = 1:length(b)
r(:, j) = (a == b(j));
end
Iterating through the shorter vector is obviously more efficient as it generates fewer for iterations. The correct solution should have no for-loops whatsoever.
Is this possible in MATLAB/Octave?
Here's a simple solution using bsxfun.
bsxfun(#eq,b,a')
ans =
1 0 0
0 1 0
0 0 1
0 1 0
0 0 1
1 0 0
bsxfun(#eq, a', b)
I have to get the unknown matrix by changing the form of a known matrix considering the following rules:
H = [-P'|I] %'
G = [I|P]
where
H is a known matrix
G is an unknown matrix which has to be calculated
I is the identity matrix
So for example, if we had a matrix,
H = [1 1 1 1 0 0;
0 0 1 1 0 1;
1 0 0 1 1 0]
its form has to be changed to
H = [1 1 1 1 0 0;
0 1 1 0 1 0;
1 1 0 0 0 1]
So
-P' = [1 1 1;
0 1 0;
1 1 0]
and in case of binary matrices -P = P.
Therefore
G = [1 0 0 1 1 1;
0 1 0 0 1 0;
0 0 1 1 1 0]
I know how to solve it on paper by performing basic row operations but haven't figured out how to solve it using MATLAB yet.
What is the method for solving the given problem?
If the order of columns in -P' doesn't matter, here's one solution using the function ISMEMBER:
>> H = [1 1 1 1 0 0; 0 0 1 1 0 1; 1 0 0 1 1 0]; %# From above
>> pColumns = ~ismember(H',eye(3),'rows') %'# Find indices of columns that
%# are not equal to rows
pColumns = %# of the identity matrix
1
0
1
1
0
0
>> P = -H(:,pColumns)' %'# Find P
P =
-1 0 -1
-1 -1 0
-1 -1 -1
>> G = logical([eye(3) P]) %# Create the binary matrix G
G =
1 0 0 1 0 1
0 1 0 1 1 0
0 0 1 1 1 1
NOTE: This solution will work properly for integer or binary values in H. If H has floating-point values, you will likely run into an issue with floating-point comparisons when using ISMEMBER (see here and here for more discussion of this issue).