Related
I have a 3D matrix X of size a x b x c.
I want to create a 3D matrix Y in MATLAB as follows:
X = rand(10, 10, 5);
[a, b, c] = size(X);
for i = 1 : c
for j = 1 : a
for k = 1 : b
if j<a && k<b
Y(j, k, i) = X(j+1, k, i) + X(j, k+1, i).^4;
else
Y(j, k, i) = X(a, b, i) + X(a, b, i).^4;
end
end
end
end
How can I do that by avoiding using a lot of for loops? In other words, how can I rewrite the above code in a faster way without using a lot of loops?
Indexing Arrays/Matrices
Below I added a portion to your script that creates an array Z that is identical to array Y by using indexing that covers the equivalent indices and element-wise operations that are indicated by the dot . preceding the operation. Operations such as multiplication * and division / can be specified element-wise as .* and ./ respectively. Addition and subtraction act in the element-wise fashion and do not need the dot .. I also added an if-statement to check that the arrays are the same and that the for-loops and indexing methods give equivalent results. Indexing using end refers to the last index in the corresponding/respective dimension.
Snippet:
Y = zeros(a,b,c);
Y(1:end-1,1:end-1,:) = X(2:end,1:end-1,:) + X(1: end-1, 2:end,:).^4;
Y(end,1:end,:) = repmat(X(a,b,:) + X(a,b,:).^4,1,b,1);
Y(1:end,end,:) = repmat(X(a,b,:) + X(a,b,:).^4,a,1,1);
Full Script: Including Both Methods and Checking
X = rand(10, 10, 5);
[a, b, c] = size(X);
%Initialed for alternative result%
Z = zeros(a,b,c);
%Looping method%
for i = 1 : c
for j = 1 : a
for k = 1 : b
if j < a && k < b
Y(j, k, i) = X(j+1, k, i) + X(j, k+1, i).^4;
else
Y(j, k, i) = X(a, b, i) + X(a, b, i).^4;
end
end
end
end
%Indexing and element-wise method%
Z(1:end-1,1:end-1,:) = X(2:end,1:end-1,:) + X(1: end-1, 2:end,:).^4;
Z(end,1:end,:) = repmat(X(a,b,:) + X(a,b,:).^4,1,b,1);
Z(1:end,end,:) = repmat(X(a,b,:) + X(a,b,:).^4,a,1,1);
%Checking if results match%
if(Z == Y)
fprintf("Matched result\n");
end
Ran using MATLAB R2019b
I work primarily in MATLAB but I think the answer should not be too hard to carry over from one language to another.
I have a multi-dimensional array X with dimensions [n, p, 3].
I would like to calculate the following multi-dimensional array.
T = zeros(p, p, p)
for i = 1:p
for j = 1:p
for k = 1:p
T(i, j, k) = sum(X(:, i, 1) .* X(:, j, 2) .* X(:, k, 3));
end
end
end
The sum is of the elements of a length-n vector. Any help is appreciated!
You only need some permuting of dimensions and multiplication with singleton expansion:
T = sum(bsxfun(#times, bsxfun(#times, permute(X(:,:,1), [2 4 5 3 1]), permute(X(:,:,2), [4 2 5 3 1])), permute(X(:,:,3), [4 5 2 3 1])), 5);
From R2016b onwards, this can be written more easily as
T = sum(permute(X(:,:,1), [2 4 5 3 1]) .* permute(X(:,:,2), [4 2 5 3 1]) .* permute(X(:,:,3), [4 5 2 3 1]), 5);
As I mentioned in a comment, vectorization is not always a huge advantage any more. Therefore there are vectorization methods that slow down the code rather than speed it up. You must always time your solutions. Vectorization often involves the creation of large temporary arrays, or the copy of large amounts of data, which are avoided in loop code. It depends on the architecture, the size of the input, and many other factors if such a solution is going to be faster.
Nonetheless, in this case it seems vectorization approaches can yield a large speedup.
The first thing to notice about the original code is that X(:, i, 1) .* X(:, j, 2) gets re-computed in the inner loop, though it is a constant value there. Rewriting the inner loop as this will save time:
Y = X(:, i, 1) .* X(:, j, 2);
for k = 1:p
T(i, j, k) = sum(Y .* X(:, k, 3));
end
Now we notice that the inner loop is a dot product, and can be written as follows:
Y = X(:, i, 1) .* X(:, j, 2);
T(i, j, :) = Y.' * X(:, :, 3);
The .' transpose on Y does not copy the data, as Y is a vector. Next, we notice that X(:, :, 3) is indexed repeatedly. Let's move this out of the outer loop. Now I'm left with the following code:
T = zeros(p, p, p);
X1 = X(:, :, 1);
X2 = X(:, :, 2);
X3 = X(:, :, 3);
for i = 1:p
for j = 1:p
Y = X1(:, i) .* X2(:, j);
T(i, j, :) = Y.' * X3;
end
end
It is likely that removing the loop over j is equally easy, which would leave a single loop over i. But this is where I stop.
This is the timings I see (R2017a, 3-year old iMac with 4 cores). For n=10, p=20:
original: 0.0206
moving Y out the inner loop: 0.0100
removing inner loop: 0.0016
moving indexing out of loops: 7.6294e-04
Luis' answer: 1.9196e-04
For a larger array with n=50, p=100:
original: 2.9107
moving Y out the inner loop: 1.3488
removing inner loop: 0.0910
moving indexing out of loops: 0.0361
Luis' answer: 0.1417
"Luis' answer" is this one. It is by far fastest for small arrays, but for larger arrays it shows the cost of the permutation. Moving the computation of the first product out of the inner loop saves a bit over half the computation cost. But removing the inner loop reduces the cost quite dramatically (which I hadn't expected, I presume the single matrix product can use parallelism better than the many small element-wise products). We then get a further time reduction by reducing the amount of indexing operations within the loop.
This is the timing code:
function so()
n = 10; p = 20;
%n = 50; p = 100;
X = randn(n,p,3);
T1 = method1(X);
T2 = method2(X);
T3 = method3(X);
T4 = method4(X);
T5 = method5(X);
assert(max(abs(T1(:)-T2(:)))<1e-13)
assert(max(abs(T1(:)-T3(:)))<1e-13)
assert(max(abs(T1(:)-T4(:)))<1e-13)
assert(max(abs(T1(:)-T5(:)))<1e-13)
timeit(#()method1(X))
timeit(#()method2(X))
timeit(#()method3(X))
timeit(#()method4(X))
timeit(#()method5(X))
function T = method1(X)
p = size(X,2);
T = zeros(p, p, p);
for i = 1:p
for j = 1:p
for k = 1:p
T(i, j, k) = sum(X(:, i, 1) .* X(:, j, 2) .* X(:, k, 3));
end
end
end
function T = method2(X)
p = size(X,2);
T = zeros(p, p, p);
for i = 1:p
for j = 1:p
Y = X(:, i, 1) .* X(:, j, 2);
for k = 1:p
T(i, j, k) = sum(Y .* X(:, k, 3));
end
end
end
function T = method3(X)
p = size(X,2);
T = zeros(p, p, p);
for i = 1:p
for j = 1:p
Y = X(:, i, 1) .* X(:, j, 2);
T(i, j, :) = Y.' * X(:, :, 3);
end
end
function T = method4(X)
p = size(X,2);
T = zeros(p, p, p);
X1 = X(:, :, 1);
X2 = X(:, :, 2);
X3 = X(:, :, 3);
for i = 1:p
for j = 1:p
Y = X1(:, i) .* X2(:, j);
T(i, j, :) = Y.' * X3;
end
end
function T = method5(X)
T = sum(permute(X(:,:,1), [2 4 5 3 1]) .* permute(X(:,:,2), [4 2 5 3 1]) .* permute(X(:,:,3), [4 5 2 3 1]), 5);
You have mentioned you are open to other languages and NumPy by its syntax is very close to MATLAB, so we will try to have a NumPy based solution on this.
Now, these tensor related sum-reductions, specially matrix multiplications ones are easily expressed as einstein-notation and NumPy luckily has one function on the same as np.einsum. Under the hoods, it's implemented in C and is pretty efficient. Recently it's been optimized further to leverage BLAS based matrix-multiplication implementations.
So, a translation of the stated code onto NumPy territory keeping in mind that it follows 0-based indexing and the axes are visuallized differently than the dimensions with MATLAB, would be -
import numpy as np
# X is a NumPy array of shape : (n,p,3). So, a random one could be
# generated with : `X = np.random.rand(n,p,3)`.
T = np.zeros((p, p, p))
for i in range(p):
for j in range(p):
for k in range(p):
T[i, j, k] = np.sum(X[:, i, 0] * X[:, j, 1] * X[:, k, 2])
The einsum way to solve it would be -
np.einsum('ia,ib,ic->abc',X[...,0],X[...,1],X[...,2])
To leverage matrix-multiplication, use optimize flag -
np.einsum('ia,ib,ic->abc',X[...,0],X[...,1],X[...,2],optimize=True)
Timings (with large sizes)
In [27]: n,p = 100,100
...: X = np.random.rand(n,p,3)
In [28]: %%timeit
...: T = np.zeros((p, p, p))
...: for i in range(p):
...: for j in range(p):
...: for k in range(p):
...: T[i, j, k] = np.sum(X[:, i, 0] * X[:, j, 1] * X[:, k, 2])
1 loop, best of 3: 6.23 s per loop
In [29]: %timeit np.einsum('ia,ib,ic->abc',X[...,0],X[...,1],X[...,2])
1 loop, best of 3: 353 ms per loop
In [31]: %timeit np.einsum('ia,ib,ic->abc',X[...,0],X[...,1],X[...,2],optimize=True)
100 loops, best of 3: 10.5 ms per loop
In [32]: 6230.0/10.5
Out[32]: 593.3333333333334
Around 600x speedup there!
I have an equation for which I am trying to write a code in Matlab, but I am not sure if my code is right. The equation is as follows:
Where I think the iteration is over the superscript i.e. k, k+1 etc and the dimensions are marked by subscripts m, n, n'. The notations are not well defined in the literature so I think this is how it should be.
My code segment for this equation is as follows:
c_n = [1,2,3,4]'; % c^(0)_n (nx1) vector
K = 50;
d = [0.5,0.9]';
for k = 1:1:K
c_n = c_n.*((sum(A_mn'*d/(sum(A_mn*c_n,2)),2))./sum(A_mn',2)) ;
end
Is this code correct for the above equation?. The summations in the equation are confusing me.
If A is a matrix with m rows and n columns, then the sum is just the sum of the nth row in AT. This is the same as the corresponding sum of the nth column in A: . The matrix multiplication it represents works out nicer with the transpose because matrix multiplications are just sums of weighted rows.
Similarly, is the mth row of A, weighted element-wise by by c.
We can assume that c and d are column vectors of size n and m, respectively. (d' will be represented by just plain d in the code). In that case, most of the operations can be reduced to matrix operations:
is just the matrix product A * c, which yields a column vector of size m.
then becomes the element-wise ratio d ./ (A * c), also of size m.
The ratio is used to scale the elements of the sum of AT in the numerator, making it the matrix product A.' * (d ./ (A * c)) of size n.
Each element of that is scaled by , which can be represented by either A.' * ones(m, 1) or sum(A, 1).', so the final matrix product is just c .* (A.' * (d ./ (A * c)) ./ sum(A, 1).').
You can pre-calculate sum(A, 1).', call it e to get the following:
c = [1; 2; 3; 4];
d = [0.5; 0.9];
A = ... some 2x4 matrix;
e = sum(A, 1).';
k = 50;
for i = 1 : k
c = c .* (A.' * (d ./ (A * c)) ./ e);
end
If you want to retain the intermediate values of c for each k, you can allocate a matrix of size n, k + 1 and fill that in with each column representing a new iteration of c:
c = zeros(4, 51);
c(:, 1) = [1; 2; 3; 4];
for i = 1 : k
c(:, k + 1) = c(:, k) .* (A.' * (d ./ (A * c(:, k))) ./ e);
end
I have two matrices X and Y, both of order mxn. I want to create a new matrix Z of order mx1 such that each i th entry in this new matrix is computed by applying a function to ith and ith row of X and Y respectively. In my case m = 100000 and n = 2. I tried using a loop but it takes forever.
for i = 1:m
Z = function(X(1,:),Y(1,:), constant_parameters)
end
Is there an efficient way to vectorize it?
EDIT 1
This is the function
function [peso] = fxPesoTexturaCN(a,b, img, r, L)
ac = num2cell(a);
bc = num2cell(b);
imgint1 = img(sub2ind(size(img),ac{:}));
imgint2 = img(sub2ind(size(img),bc{:}));
peso = (sum((a - b) .^ 2) + (r/L) * (imgint2 - imgint1)) / (2*r^2);
Where img, r, L are constats. a is X(1,:) and b is Y(1,:)
And the call of this function is
peso = bsxfun(#(a,b) fxPesoTexturaCN(a,b,img,r,L), a, b);
I have the following code:
n = 10000;
s = 100;
Z = rand(n, 2);
x = rand(s, 1);
y = rand(s, 1);
fun = #(a) exp(a);
In principle, the anonymous function f can have a different form. I need to create two arrays.
First, I need to create an array of size n x s x s with generic elements
fun(Z(i, 1) - x(j)) * fun(Z(i, 2) - y(k))
where i=1,...n while j,k=1,...,s. What I can easily do, is to construct matrices using bsxfun, e.g.
bsxfun(#(x, y) fun(x - y), Z(:, 1), x');
bsxfun(#(x, y) fun(x - y), Z(:, 2), y');
But then I would need to combine them into 3D array by multiplying element-wise each column of those two matrices.
In the second step, I need to create an array of size n x 3 x s x s, which would look from one side as the following matrix
[ones(n, 1), Z(:, 1) - x(i), Z(:, 2) - y(j);]
where i=1,...s, j=1,...s. I could loop over the two extra dimensions with something like
A = [ones(n, 1), Z(:, 1) - x(1), Z(:, 2) - y(1)];
for i = 1:s
for j = 1:s
A(:, :, i, j) = [ones(n, 1), Z(:, 1) - x(i), Z(:, 2) - y(j);];
end
end
Is there a way to avoid loops?
In the third step, suppose that after obtaining array out1 (output from first step), I want to create a new array out3 of dimension n x n x s x s, which contains the original array out1 on the main diagonal, i.e. out3(i,i,s,s) = out1(i, s, s) and out3(i,j,s,s)=0 for all i~=j. Is there some kind of alternative of diag for creating "diagonal arrays"? Alternatively, if I create n x n x s x s array of zeros, is there a way to put out1 on the main diagonal?
Code
exp_Z_x = exp(bsxfun(#minus,Z(:,1),x.')); %//'
exp_Z_y = exp(bsxfun(#minus,Z(:,2),y.')); %//'
out1 = bsxfun(#times,exp_Z_x,permute(exp_Z_y,[1 3 2]));
Z1 = [ones(n,1) Z(:,1) Z(:,2)];
X1 = permute([ zeros(s,1) x zeros(s,1)],[3 2 1]);
Y1 = permute([ zeros(s,1) zeros(s,1) y],[4 2 3 1]);
out2 = bsxfun(#minus,bsxfun(#minus,Z1,X1),Y1);
out3 = zeros(n,n,s,s); %// out3(n,n,s,s) = 0; could be used for performance
out3(bsxfun(#plus,[1:n+1:n*n]',[0:s*s-1]*n*n)) = out1; %//'
%// out1, out2 and out3 are the desired outputs