I have a matrix, matrix_logical(50000,100000), that is a sparse logical matrix (a lot of falses, some true). I have to produce a matrix, intersect(50000,50000), that, for each pair, i,j, of rows of matrix_logical(50000,100000), stores the number of columns for which rows i and j have both "true" as the value.
Here is the code I wrote:
% store in advance the nonzeros cols
for i=1:50000
nonzeros{i} = num2cell(find(matrix_logical(i,:)));
end
intersect = zeros(50000,50000);
for i=1:49999
a = cell2mat(nonzeros{i});
for j=(i+1):50000
b = cell2mat(nonzeros{j});
intersect(i,j) = numel(intersect(a,b));
end
end
Is it possible to further increase the performance? It takes too long to compute the matrix. I would like to avoid the double loop in the second part of the code.
matrix_logical is sparse, but it is not saved as sparse in MATLAB because otherwise the performance become the worst possible.
Since the [i,j] entry counts the number of non zero elements in the element-wise multiplication of rows i and j, you can do it by multiplying matrix_logical with its transpose (you should convert to numeric data type first, e.g matrix_logical = single(matrix_logical)):
inter = matrix_logical * matrix_logical';
And it works both for sparse or full representation.
EDIT
In order to calculate numel(intersect(a,b))/numel(union(a,b)); (as asked in your comment), you can use the fact that for two sets a and b, you have
length(union(a,b)) = length(a) + length(b) - length(intersect(a,b))
so, you can do the following:
unLen = sum(matrix_logical,2);
tmp = repmat(unLen, 1, length(unLen)) + repmat(unLen', length(unLen), 1);
inter = matrix_logical * matrix_logical';
inter = inter ./ (tmp-inter);
If I understood you correctly, you want a logical AND of the rows:
intersct = zeros(50000, 50000)
for ii = 1:49999
for jj = ii:50000
intersct(ii, jj) = sum(matrix_logical(ii, :) & matrix_logical(jj, :));
intersct(jj, ii) = intersct(ii, jj);
end
end
Doesn't avoid the double loop, but at least works without the first loop and the slow find command.
Elaborating on my comment, here is a distance function suitable for pdist()
function out = distfun(xi,xj)
out = zeros(size(xj,1),1);
for i=1:size(xj,1)
out(i) = sum(sum( xi & xj(i,:) )) / sum(sum( xi | xj(i,:) ));
end
In my experience, sum(sum()) is faster for logicals than nnz(), thus its appearance above.
You would also need to use squareform() to reshape the output of pdist() appropriately:
squareform(pdist(martrix_logical,#distfun));
Note that pdist() includes a 'jaccard' distance measure, but it is actually the Jaccard distance and not the Jaccard index or coefficient, which is the value you are apparently after.
Related
It is a straightforward question: Is there a faster alternative to all(a(:,i)==a,1) in MATLAB?
I'm thinking of a implementation that benefits from short-circuit evaluations in the whole process. I mean, all() definitely benefits from short-circuit evaluations but a(:,i)==a doesn't.
I tried the following code,
% example for the input matrix
m = 3; % m and n aren't necessarily equal to those values.
n = 5000; % It's only possible to know in advance that 'm' << 'n'.
a = randi([0,5],m,n); % the maximum value of 'a' isn't necessarily equal to
% 5 but it's possible to state that every element in
% 'a' is a positive integer.
% all, equal solution
tic
for i = 1:n % stepping up the elapsed time in orders of magnitude
%%%%%%%%%% all and equal solution %%%%%%%%%
ax_boo = all(a(:,i)==a,1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
toc
% alternative solution
tic
for i = 1:n % stepping up the elapsed time in orders of magnitude
%%%%%%%%%%% alternative solution %%%%%%%%%%%
ax_boo = a(1,i) == a(1,:);
for k = 2:m
ax_boo(ax_boo) = a(k,i) == a(k,ax_boo);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
toc
but it's intuitive that any "for-loop-solution" within the MATLAB environment will be naturally slower. I'm wondering if there is a MATLAB built-in function written in a faster language.
EDIT:
After running more tests I found out that the implicit expansion does have a performance impact in evaluating a(:,i)==a. If the matrix a has more than one row, all(repmat(a(:,i),[1,n])==a,1) may be faster than all(a(:,i)==a,1) depending on the number of columns (n). For n=5000 repmat explicit expansion has proved to be faster.
But I think that a generalization of Kenneth Boyd's answer is the "ultimate solution" if all elements of a are positive integers. Instead of dealing with a (m x n matrix) in its original form, I will store and deal with adec (1 x n matrix):
exps = ((0):(m-1)).';
base = max(a,[],[1,2]) + 1;
adec = sum( a .* base.^exps , 1 );
In other words, each column will be encoded to one integer. And of course adec(i)==adec is faster than all(a(:,i)==a,1).
EDIT 2:
I forgot to mention that adec approach has a functional limitation. At best, storing adec as uint64, the following inequality must hold base^m < 2^64 + 1.
Since your goal is to count the number of columns that match, my example converts the binary encoding to integer decimals, then you just loop over the possible values (with 3 rows that are 8 possible values) and count the number of matches.
a_dec = 2.^(0:(m-1)) * a;
num_poss_values = 2 ^ m;
num_matches = zeros(num_poss_values, 1);
for i = 1:num_poss_values
num_matches(i) = sum(a_dec == (i - 1));
end
On my computer, using 2020a, Here are the execution times for your first 2 options and the code above:
Elapsed time is 0.246623 seconds.
Elapsed time is 0.553173 seconds.
Elapsed time is 0.000289 seconds.
So my code is 853 times faster!
I wrote my code so it will work with m being an arbitrary integer.
The num_matches variable contains the number of columns that add up to 0, 1, 2, ...7 when converted to a decimal.
As an alternative you can use the third output of unique:
[~, ~, iu] = unique(a.', 'rows');
for i = 1:n
ax_boo = iu(i) == iu;
end
As indicated in a comment:
ax_boo isolates the indices of the columns I have to sum in a row vector b. So, basically the next line would be something like c = sum(b(ax_boo),2);
It is a typical usage of accumarray:
[~, ~, iu] = unique(a.', 'rows');
C = accumarray(iu,b);
for i = 1:n
c = C(i);
end
I am looking for an optimal way to program this summation ratio. As input I have two vectors v_mn and x_mn with (M*N)x1 elements each.
The ratio is of the form:
The vector x_mn is 0-1 vector so when x_mn=1, the ration is r given above and when x_mn=0 the ratio is 0.
The vector v_mn is a vector which contain real numbers.
I did the denominator like this but it takes a lot of times.
function r_ij = denominator(v_mn, M, N, i, j)
%here x_ij=1, to get r_ij.
S = [];
for m = 1:M
for n = 1:N
if (m ~= i)
if (n ~= j)
S = [S v_mn(i, n)];
else
S = [S 0];
end
else
S = [S 0];
end
end
end
r_ij = 1+S;
end
Can you give a good way to do it in matlab. You can ignore the ratio and give me the denominator which is more complicated.
EDIT: I am sorry I did not write it very good. The i and j are some numbers between 1..M and 1..N respectively. As you can see, the ratio r is many values (M*N values). So I calculated only the value i and j. More precisely, I supposed x_ij=1. Also, I convert the vectors v_mn into a matrix that's why I use double index.
If you reshape your data, your summation is just a repeated matrix/vector multiplication.
Here's an implementation for a single m and n, along with a simple speed/equality test:
clc
%# some arbitrary test parameters
M = 250;
N = 1000;
v = rand(M,N); %# (you call it v_mn)
x = rand(M,N); %# (you call it x_mn)
m0 = randi(M,1); %# m of interest
n0 = randi(N,1); %# n of interest
%# "Naive" version
tic
S1 = 0;
for mm = 1:M %# (you call this m')
if mm == m0, continue; end
for nn = 1:N %# (you call this n')
if nn == n0, continue; end
S1 = S1 + v(m0,nn) * x(mm,nn);
end
end
r1 = v(m0,n0)*x(m0,n0) / (1+S1);
toc
%# MATLAB version: use matrix multiplication!
tic
ninds = [1:m0-1 m0+1:M];
minds = [1:n0-1 n0+1:N];
S2 = sum( x(minds, ninds) * v(m0, ninds).' );
r2 = v(m0,n0)*x(m0,n0) / (1+S2);
toc
%# Test if values are equal
abs(r1-r2) < 1e-12
Outputs on my machine:
Elapsed time is 0.327004 seconds. %# loop-version
Elapsed time is 0.002455 seconds. %# version with matrix multiplication
ans =
1 %# and yes, both are equal
So the speedup is ~133×
Now that's for a single value of m and n. To do this for all values of m and n, you can use an (optimized) double loop around it:
r = zeros(M,N);
for m0 = 1:M
xx = x([1:m0-1 m0+1:M], :);
vv = v(m0,:).';
for n0 = 1:N
ninds = [1:n0-1 n0+1:N];
denom = 1 + sum( xx(:,ninds) * vv(ninds) );
r(m0,n0) = v(m0,n0)*x(m0,n0)/denom;
end
end
which completes in ~15 seconds on my PC for M = 250, N= 1000 (R2010a).
EDIT: actually, with a little more thought, I was able to reduce it all down to this:
denom = zeros(M,N);
for mm = 1:M
xx = x([1:mm-1 mm+1:M],:);
denom(mm,:) = sum( xx*v(mm,:).' ) - sum( bsxfun(#times, xx, v(mm,:)) );
end
denom = denom + 1;
r_mn = x.*v./denom;
which completes in less than 1 second for N = 250 and M = 1000 :)
For a start you need to pre-alocate your S matrix. It changes size every loop so put
S = zeros(m*n, 1)
at the start of your function. This will also allow you to do away with your else conditional statements, ie they will reduce to this:
if (m ~= i)
if (n ~= j)
S(m*M + n) = v_mn(i, n);
Otherwise since you have to visit every element im afraid it may not be able to get much faster.
If you desperately need more speed you can look into doing some mex coding which is code in c/c++ but run in matlab.
http://www.mathworks.com.au/help/matlab/matlab_external/introducing-mex-files.html
Rather than first jumping into vectorization of the double loop, you may want modify the above to make sure that it does what you want. In this code, there is no summing of the data, instead a vector S is being resized at each iteration. As well, the signature could include the matrices V and X so that the multiplication occurs as in the formula (rather than just relying on the value of X to be zero or one, let us pass that matrix in).
The function could look more like the following (I've replaced the i,j inputs with m,n to be more like the equation):
function result = denominator(V,X,m,n)
% use the size of V to determine M and N
[M,N] = size(V);
% initialize the summed value to one (to account for one at the end)
result = 1;
% outer loop
for i=1:M
% ignore the case where m==i
if i~=m
for j=1:N
% ignore the case where n==j
if j~=n
result = result + V(m,j)*X(i,j);
end
end
end
end
Note how the first if is outside of the inner for loop since it does not depend on j. Try the above and see what happens!
You can vectorize from within Matlab to speed up your calculations. Every time you use an operation like ".^" or ".*" or any matrix operation for that matter, Matlab will do them in parallel, which is much, much faster than iterating over each item.
In this case, look at what you are doing in terms of matrices. First, in your loop you are only dealing with the mth row of $V_{nm}$, which we can use as a vector for itself.
If you look at your formula carefully, you can figure out that you almost get there if you just write this row vector as a column vector and multiply the matrix $X_{nm}$ to it from the left, using standard matrix multiplication. The resulting vector contains the sums over all n. To get the final result, just sum up this vector.
function result = denominator_vectorized(V,X,m,n)
% get the part of V with the first index m
Vm = V(m,:)';
% remove the parts of X you don't want to iterate over. Note that, since I
% am inside the function, I am only editing the value of X within the scope
% of this function.
X(m,:) = 0;
X(:,n) = 0;
%do the matrix multiplication and the summation at once
result = 1-sum(X*Vm);
To show you how this optimizes your operation, I will compare it to the code proposed by another commenter:
function result = denominator(V,X,m,n)
% use the size of V to determine M and N
[M,N] = size(V);
% initialize the summed value to one (to account for one at the end)
result = 1;
% outer loop
for i=1:M
% ignore the case where m==i
if i~=m
for j=1:N
% ignore the case where n==j
if j~=n
result = result + V(m,j)*X(i,j);
end
end
end
end
The test:
V=rand(10000,10000);
X=rand(10000,10000);
disp('looped version')
tic
denominator(V,X,1,1)
toc
disp('matrix operation')
tic
denominator_vectorized(V,X,1,1)
toc
The result:
looped version
ans =
2.5197e+07
Elapsed time is 4.648021 seconds.
matrix operation
ans =
2.5197e+07
Elapsed time is 0.563072 seconds.
That is almost ten times the speed of the loop iteration. So, always look out for possible matrix operations in your code. If you have the Parallel Computing Toolbox installed and a CUDA-enabled graphics card installed, Matlab will even perform these operations on your graphics card without any further effort on your part!
EDIT: That last bit is not entirely true. You still need to take a few steps to do operations on CUDA hardware, but they aren't a lot. See Matlab documentation.
I am running the following code to obtain the values of the inverse EDF of a data Matrix at the data points:
function [mOUT] = InvEDF (data)
% compute inverse of EDF at data values
% function takes T*K matrix of data and returns T*K matrix of transformed
% data, keepin the order of the original series
T = rows(data);
K = cols(data);
mOUT=zeros(T,K);
for j = 1:K
for i = 1:T
temp = data(:,j)<=data(i,j);
mOUT(i,j) = 1/(T+1)*sum(temp);
end
end
The data Matrix is usually of size 1000*10 or even 1000*30 and I am calling this function a few thousand times. Is there a faster way of doinf this? Any answers are appreciated. Thanks!
You can sort the values and use the index in the sorted matrix as the count of values less or equal. We treat each column by itself, so I will illustrate on a Mx1 matrix.
A = rand(M,1);
[B,I] = sort(A);
C(I) = 1:M;
C(i) will now contain the count of values less or equal to A(i). If you can have duplicate values you need to take that into account.
The advantage of this approach is that we can do it in O(M log M) time, whereas your original inner loop is O(M^2)
Try this -
mOUT=zeros(T,K);
for j = 1:K
d1 = data(:,j);
mOUT(:,j) = sum(bsxfun(#ge,d1,d1'),2); %%//'
end
mOUT = mOUT./(T+1);
I have been given an assignment in which I am supposed to write an algorithm which performs polynomial interpolation by the barycentric formula. The formulas states that:
p(x) = (SIGMA_(j=0 to n) w(j)*f(j)/(x - x(j)))/(SIGMA_(j=0 to n) w(j)/(x - x(j)))
I have written an algorithm which works just fine, and I get the polynomial output I desire. However, this requires the use of some quite long loops, and for a large grid number, lots of nastly loop operations will have to be done. Thus, I would appreciate it greatly if anyone has any hints as to how I may improve this, so that I will avoid all these loops.
In the algorithm, x and f stand for the given points we are supposed to interpolate. w stands for the barycentric weights, which have been calculated before running the algorithm. And grid is the linspace over which the interpolation should take place:
function p = barycentric_formula(x,f,w,grid)
%Assert x-vectors and f-vectors have same length.
if length(x) ~= length(f)
sprintf('Not equal amounts of x- and y-values. Function is terminated.')
return;
end
n = length(x);
m = length(grid);
p = zeros(1,m);
% Loops for finding polynomial values at grid points. All values are
% calculated by the barycentric formula.
for i = 1:m
var = 0;
sum1 = 0;
sum2 = 0;
for j = 1:n
if grid(i) == x(j)
p(i) = f(j);
var = 1;
else
sum1 = sum1 + (w(j)*f(j))/(grid(i) - x(j));
sum2 = sum2 + (w(j)/(grid(i) - x(j)));
end
end
if var == 0
p(i) = sum1/sum2;
end
end
This is a classical case for matlab 'vectorization'. I would say - just remove the loops. It is almost that simple. First, have a look at this code:
function p = bf2(x, f, w, grid)
m = length(grid);
p = zeros(1,m);
for i = 1:m
var = grid(i)==x;
if any(var)
p(i) = f(var);
else
sum1 = sum((w.*f)./(grid(i) - x));
sum2 = sum(w./(grid(i) - x));
p(i) = sum1/sum2;
end
end
end
I have removed the inner loop over j. All I did here was in fact removing the (j) indexing and changing the arithmetic operators from / to ./ and from * to .* - the same, but with a dot in front to signify that the operation is performed on element by element basis. This is called array operators in contrast to ordinary matrix operators. Also note that treating the special case where the grid points fall onto x is very similar to what you had in the original implementation, only using a vector var such that x(var)==grid(i).
Now, you can also remove the outermost loop. This is a bit more tricky and there are two major approaches how you can do that in MATLAB. I will do it the simpler way, which can be less efficient, but more clear to read - using repmat:
function p = bf3(x, f, w, grid)
% Find grid points that coincide with x.
% The below compares all grid values with all x values
% and returns a matrix of 0/1. 1 is in the (row,col)
% for which grid(row)==x(col)
var = bsxfun(#eq, grid', x);
% find the logical indexes of those x entries
varx = sum(var, 1)~=0;
% and of those grid entries
varp = sum(var, 2)~=0;
% Outer-most loop removal - use repmat to
% replicate the vectors into matrices.
% Thus, instead of having a loop over j
% you have matrices of values that would be
% referenced in the loop
ww = repmat(w, numel(grid), 1);
ff = repmat(f, numel(grid), 1);
xx = repmat(x, numel(grid), 1);
gg = repmat(grid', 1, numel(x));
% perform the calculations element-wise on the matrices
sum1 = sum((ww.*ff)./(gg - xx),2);
sum2 = sum(ww./(gg - xx),2);
p = sum1./sum2;
% fix the case where grid==x and return
p(varp) = f(varx);
end
The fully vectorized version can be implemented with bsxfun rather than repmat. This can potentially be a bit faster, since the matrices are not explicitly formed. However, the speed difference may not be large for small system sizes.
Also, the first solution with one loop is also not too bad performance-wise. I suggest you test those and see, what is better. Maybe it is not worth it to fully vectorize? The first code looks a bit more readable..
I have 2 matrices: V which is square MxM, and K which is MxN. Calling the dimension across rows x and the dimension across columns t, I need to evaluate the integral (i.e sum) over both dimensions of K times a t-shifted version of V, the answer being a function of the shift (almost like a convolution, see below). The sum is defined by the following expression, where _{} denotes the summation indices, and a zero-padding of out-of-limits elements is assumed:
S(t) = sum_{x,tau}[V(x,t+tau) * K(x,tau)]
I manage to do it with a single loop, over the t dimension (vectorizing the x dimension):
% some toy matrices
V = rand(50,50);
K = rand(50,10);
[M N] = size(K);
S = zeros(1, M);
for t = 1 : N
S(1,1:end-t+1) = S(1,1:end-t+1) + sum(bsxfun(#times, V(:,t:end), K(:,t)),1);
end
I have similar expressions which I managed to evaluate without a for loop, using a combination of conv2 and\or mirroring (flipping) of a single dimension. However I can't see how to avoid a for loop in this case (despite the appeared similarity to convolution).
Steps to vectorization
1] Perform sum(bsxfun(#times, V(:,t:end), K(:,t)),1) for all columns in V against all columns in K with matrix-multiplication -
sum_mults = V.'*K
This would give us a 2D array with each column representing sum(bsxfun(#times,.. operation at each iteration.
2] Step1 gave us all possible summations and also the values to be summed are not aligned in the same row across iterations, so we need to do a bit more work before summing along rows. The rest of the work is about getting a shifted up version. For the same, you can use boolean indexing with a upper and lower triangular boolean mask. Finally, we sum along each row for the final output. So, this part of the code would look like so -
valid_mask = tril(true(size(sum_mults)));
sum_mults_shifted = zeros(size(sum_mults));
sum_mults_shifted(flipud(valid_mask)) = sum_mults(valid_mask);
out = sum(sum_mults_shifted,2);
Runtime tests -
%// Inputs
V = rand(1000,1000);
K = rand(1000,200);
disp('--------------------- With original loopy approach')
tic
[M N] = size(K);
S = zeros(1, M);
for t = 1 : N
S(1,1:end-t+1) = S(1,1:end-t+1) + sum(bsxfun(#times, V(:,t:end), K(:,t)),1);
end
toc
disp('--------------------- With proposed vectorized approach')
tic
sum_mults = V.'*K; %//'
valid_mask = tril(true(size(sum_mults)));
sum_mults_shifted = zeros(size(sum_mults));
sum_mults_shifted(flipud(valid_mask)) = sum_mults(valid_mask);
out = sum(sum_mults_shifted,2);
toc
Output -
--------------------- With original loopy approach
Elapsed time is 2.696773 seconds.
--------------------- With proposed vectorized approach
Elapsed time is 0.044144 seconds.
This might be cheating (using arrayfun instead of a for loop) but I believe this expression gives you what you want:
S = arrayfun(#(t) sum(sum( V(:,(t+1):(t+N)) .* K )), 1:(M-N), 'UniformOutput', true)