I have a sparse matrix and I need to fill certain entries with a specific value, I am using a for loop right now but I know its not the correct way to do it so I was wondering if its possible to vectorise this for loop?
K = sparse(N);
for i=vectorofrandomintegers
K(i,i) = 1;
end
If I vectorise it normally as so:
K(A,A) = 1;
then it fills all the entries in each row denoted by A whereas I want individual entries (i.e. K(1,1) = 1 or K(6,6)=1).
Also, the entries are not diagonally adjacent so I can't plop the identity matrix into it.
If you are going to use a vectorized method, you would need to get the linear indices to be set. The issue is that if you define your sparse matrix as K = sparse(N) and then linearly index into K, it would extend the size of it in one direction only and not along both row and column. Thus, you need to specify to MATLAB that you are
looking to use this sparse to store a 2D array. Thus, it would be -
K = sparse(N,N);
Get the linear indices to index into K using sub2ind and set them -
ind1 = sub2ind([N N],vectorofrandomintegers,vectorofrandomintegers);
K(ind1) = 1;
It's fairly simple
i'd use
K((A-1)*N+A))=1;
i believe that should fix your problem by treating the matrix as a vector
Instead of declaring and then filling a sparse matrix, you can fill it at the same time you define it:
i = vectorofrandomintegers; j = i;
K = sparse(i,j,1,N,N)
Related
For a current project, I have to discretize quasi-continuous values into bins defined by some pre-defined binning resolution. For this purpose, I have written a function, which I expected to be highly efficient as it is able to both process scalar inputs as well as vector inputs using bsxfun. However, after some profiling, I found out that almost all processing time of my much larger project is produced in this function, and within the function, it's mainly the bsxfun part that takes time, with the min-query following on second place. Long story short, I am looking for advice on how to solve this task MUCH faster in MATLAB. Side note: I am usually passing vectors with some 50k elements.
Here's the code:
function sampleNo = value2sample(value,bins)
%Make sure both vectors have orientations fitting bsxfun
value = value(:);
bins = bins(:)';
%Recover bin resolution (avoids passing another parameter)
delta = median(diff(bins));
%Calculate distance matrix between all combinations
dist = abs(bsxfun(#minus,value,bins));
%What we really want to know is the minimum distance per row
[minval,ind] = min(dist,[],2);
%Make sure we don't accidentally further process NaNs as 1st bin
ind(isnan(minval))=NaN;
sampleNo = ind;
sampleNo(minval>delta) = NaN;
end
The reason that your function is slow is because you are computing the distance between every element of values and bins and storing them all in an array - if there are N values and M bins then you will require NM elements to store all the distances, and this is probably a really big number (e.g. if each input has 50,000 elements then you need 2.5 billion elements in the output array).
Moreover, since your bins are sorted (you didn't state this, but it looks like you are assuming it in your code) you do not need to compute the distance from every value to every bin. You can be much smarter,
function ind = value2sample(value, bins)
% Find median bin distance
delta = median(diff(bins));
% Bucket into 'nearest' bin by using midpoints
bins = bins(:);
mids = [-Inf; 0.5 * (bins(1:end-1) + bins(2:end))];
[~, ind] = histc(value, mids);
% Ensure that NaN values and points that aren't near any bin are returned as NaN
ind(isnan(value)) = NaN;
ind(abs(value - bins(ind)) > delta) = NaN;
end
In my tests, with values = randn(10000, 1) and bins = -50:50 it takes around 4.5 milliseconds to run the original function, and 485 microseconds to run the code above, so you are getting around a 10x speedup (and the speedup will be even greater as you increase the size of the inputs).
Thanks to #Chris Taylor, I was able to solve the problem very efficiently. The code now runs almost 400 times faster than before. The only changes I had to make from his version are reflected in the code below. Main issue was to replace histc (whose use is not encouraged anymore) by discretize.
function ind = value2sample(value, bins)
% Make sure the vectors are standing
value = value(:);
bins = bins(:);
% Bucket into 'nearest' bin by using midpoints
mids = [eps; 0.5 * (bins(1:end-1) + bins(2:end))];
ind = discretize(value, mids);
The only thing is, that in this implementation your bins must be non-negative. Other than that, this code does exactly what I want, including the fact that ind has the same size as value and contains NaNs whenever a value is NaN or out of the range of bins.
I am a new student learning to use Matlab.
Could anyone please tell me is there a faster way possibly without loops:
to assign for each row only two values 1, -1 into different positions of a big sparse matrix.
My code to build a bimatrix or bibimatrix for the MILP problem of condition :
f^k_{ij} <= y_{ij} for every arc (i,j) and all k ~=r; in a multi-commodity flow model.
Naive approach:
bimatrix=[];
% create each row and then add to bimatrix
newrow4= zeros(1,n*(n+1)^2);
for k=1:n
for i=0:n
for j=1: n
if j~=i
%change value of some positions to -1 and 1
newrow4(i*n^2+(j-1)*n+k)=1;
newrow4((n+1)*n^2+i*n+j)=-1;
% add to bimatrix
bimatrix=[bimatrix; newrow4];
% change newrow4 back to zeros row.
newrow4(i*n^2+(j-1)*n+k)=0;
newrow4((n+1)*n^2+i*n+j)=0;
end
end
end
end
OR:
% Generate the big sparse matrix first.
bibimatrix=zeros(n^3 ,n*(n+1)^2);
t=1;
for k=1:n
for i=0:n
for j=1: n
if j~=i
%Change 2 positions in each row to -1 and 1 in each row.
bibimatrix(t,i*n^2+(j-1)*n+k)=1;
bibimatrix(t,(n+1)*n^2+i*n+j)=-1;
t=t+1
end
end
end
end
With these above code in Matlab, the time to generate this matrix, with n~12, is more than 3s. I need to generate a larger matrix in less time.
Thank you.
Suggestion: Use sparse matrices.
You should be able to create two vectors containing the column number where you want your +1 and -1 in each row. Let's call these two vectors vec_1 and vec_2. You should be able to do this without loops (if not, I still think the procedure below will be faster).
Let the size of your matrix be (max_row X max_col). Then you can create your matrix like this:
bibimatrix = sparse(1:max_row,vec_1,1,max_row,max_col);
bibimatrix = bibimatrix + sparse(1:max_row, vec_2,-1,max_row,max_col)
If you want to see the entire matrix (which you don't, since it's huge) you can write: full(bibimatrix).
EDIT:
You may also do it this way:
col_vec = [vec_1, vec_2];
row_vec = [1:max_row, 1:max_row];
s = [ones(1,max_row), -1*ones(1,max_row)];
bibimatrix = sparse(row_vec, col_vec, s, max_row, max_col)
Disclaimer: I don't have MATLAB available, so it might not be error-free.
Background
I have a large set of vectors (orientation data in an axis-angle representation... the axis is the vector). I want to apply a clustering algorithm to. I tried kmeans but the computational time was too long (never finished). So instead I am trying to implement KFCG algorithm which is faster (Kirke 2010):
Initially we have one cluster with the entire training vectors and the codevector C1 which is centroid. In the first iteration of the algorithm, the clusters are formed by comparing first element of training vector Xi with first element of code vector C1. The vector Xi is grouped into the cluster 1 if xi1< c11 otherwise vector Xi is grouped into cluster2 as shown in Figure 2(a) where codevector dimension space is 2. In second iteration, the cluster 1 is split into two by comparing second element Xi2 of vector Xi belonging to cluster 1 with that of the second element of the codevector. Cluster 2 is split into two by comparing the second element Xi2 of vector Xi belonging to cluster 2 with that of the second element of the codevector as shown in Figure 2(b). This procedure is repeated till the codebook size is reached to the size specified by user.
I'm unsure what ratio is appropriate for the codebook, but it shouldn't matter for the code optimization. Also note mine is 3-D so the same process is done for the 3rd dimension.
My code attempts
I've tried implementing the above algorithm into Matlab 2013 (Student Version). Here's some different structures I've tried - BUT take way too long (have never seen it completed):
%training vectors:
Atgood = Nx4 vector (see test data below if want to test);
vecA = Atgood(:,1:3);
roA = size(vecA,1);
%Codebook size, Nsel, is ratio of data
remainFrac2=0.5;
Nseltemp = remainFrac2*roA; %codebook size
%Ensure selected size after nearest power of 2 is NOT greater than roA
if 2^round(log2(Nseltemp)) < roA
NselIter = round(log2(Nseltemp));
else
NselIter = ceil(log2(Nseltemp)-1);
end
Nsel = 2^NselIter; %power of 2 - for LGB and other algorithms
MAIN BLOCK TO OPTIMIZE:
%KFCG:
%%cluster = cell(1,Nsel); %Unsure #rows - Don't know how to initialize if need mean...
codevec(1,1:3) = mean(vecA,1);
count1=1;
count2=1;
ind=1;
for kk = 1:NselIter
hh2 = 1:2:size(codevec,1)*2;
for hh1 = 1:length(hh2)
hh=hh2(hh1);
% for ii = 1:roA
% if vecA(ii,ind) < codevec(hh1,ind)
% cluster{1,hh}(count1,1:4) = Atgood(ii,:); %want all 4 elements
% count1=count1+1;
% else
% cluster{1,hh+1}(count2,1:4) = Atgood(ii,:); %want all 4
% count2=count2+1;
% end
% end
%EDIT: My ATTEMPT at optimizing above for loop:
repcv=repmat(codevec(hh1,ind),[size(vecA,1),1]);
splitind = vecA(:,ind)>=repcv;
splitind2 = vecA(:,ind)<repcv;
cluster{1,hh}=vecA(splitind,:);
cluster{1,hh+1}=vecA(splitind2,:);
end
clear codevec
%Only mean the 1x3 vector portion of the cluster - for centroid
codevec = cell2mat((cellfun(#(x) mean(x(:,1:3),1),cluster,'UniformOutput',false))');
if ind < 3
ind = ind+1;
else
ind=1;
end
end
if length(codevec) ~= Nsel
warning('codevec ~= Nsel');
end
Alternatively, instead of cells I thought 3D Matrices would be faster? I tried but it was slower using my method of appending the next row each iteration (temp=[]; for...temp=[temp;new];)
Also, I wasn't sure what was best to loop with, for or while:
%If initialize cell to full length
while length(find(~cellfun('isempty',cluster))) < Nsel
Well, anyways, the first method was fastest for me.
Questions
Is the logic standard? Not in the sense that it matches with the algorithm described, but from a coding perspective, any weird methods I employed (especially with those multiple inner loops) that slows it down? Where can I speed up (you can just point me to resources or previous questions)?
My array size, Atgood, is 1,000,000x4 making NselIter=19; - do I just need to find a way to decrease this size or can the code be optimized?
Should this be asked on CodeReview? If so, I'll move it.
Testing Data
Here's some random vectors you can use to test:
for ii=1:1000 %My size is ~ 1,000,000
omega = 2*rand(3,1)-1;
omega = (omega/norm(omega))';
Atgood(ii,1:4) = [omega,57];
end
Your biggest issue is re-iterating through all of vecA FOR EACH CODEVECTOR, rather than just the ones that are part of the corresponding cluster. You're supposed to split each cluster on it's codevector. As it is, your cluster structure grows and grows, and each iteration is processing more and more samples.
Your second issue is the loop around the comparisons, and the appending of samples to build up the clusters. Both of those can be solved by vectorizing the comparison operation. Oh, I just saw your edit, where this was optimized. Much better. But codevec(hh1,ind) is just a scalar, so you don't even need the repmat.
Try this version:
% (preallocs added in edit)
cluster = cell(1,Nsel);
codevec = zeros(Nsel, 3);
codevec(1,:) = mean(Atgood(:,1:3),1);
cluster{1} = Atgood;
nClusters = 1;
ind = 1;
while nClusters < Nsel
for c = 1:nClusters
lower_cluster_logical = cluster{c}(:,ind) < codevec(c,ind);
cluster{nClusters+c} = cluster{c}(~lower_cluster_logical,:);
cluster{c} = cluster{c}(lower_cluster_logical,:);
codevec(c,:) = mean(cluster{c}(:,1:3), 1);
codevec(nClusters+c,:) = mean(cluster{nClusters+c}(:,1:3), 1);
end
ind = rem(ind,3) + 1;
nClusters = nClusters*2;
end
I have an N by 2 matrix A of indices of elements I want to get from a 2D matrix B, each row of A being the row and column index of an element of B that I want to get. I would like to get all of those elements stacked up as an N by 1 vector.
B is a square matrix, so I am currently using
N = size(B,1);
indices = arrayfun(#(i) A(i,1) + N*(A(i,2)-1), 1:size(A,1));
result = B(indices);
but, while it works, this is probing to be a huge bottleneck and I need to speed up the code in order for it to be useful.
What is the fastest way I can achieve the same result?
How about
indices = [1 N] * (A'-1) + 1;
I can never remember if B(A(:,1), A(:,2)) works the way you want it to, but I'd try that to avoid the intermediate variable. If that does not work, try subs2ind.
Also, you can look at how you generated A in the first place. if A came about from the output of find, for example, it is faster to use logical indexing. i.e if
B( B == 2 )
Is faster than finding the row,col indexes that satisfy that condition, then indexing into B.
I don't have enough memory to simply create a diagonal D-by-D matrix, since D is large. I keep getting an 'out of memory' error.
Instead of performing M x D x D operations in the first multiplication, I do M x D operations, but still my code takes ages to run.
Can anybody find a more effective way to perform the multiplication A'*B*A? Here's what I've attempted so far:
D=20000
M=25
A = floor(rand(D,M)*10);
B = floor(rand(1,D)*10);
for i=1:D
for j=1:M
result(i,j) = A(i,j) * B(1,j);
end
end
manual = result * A';
auto = A*diag(B)*A';
isequal(manual,auto)
One option that should solve your problem is using sparse matrices. Here's an example:
D = 20000;
M = 25;
A = floor(rand(D,M).*10); %# A D-by-M matrix
diagB = rand(1,D).*10; %# Main diagonal of B
B = sparse(1:D,1:D,diagB); %# A sparse D-by-D diagonal matrix
result = (A.'*B)*A; %'# An M-by-M result
Another option would be to replicate the D elements along the main diagonal of B to create an M-by-D matrix using the function REPMAT, then use element-wise multiplication with A.':
B = repmat(diagB,M,1); %# Replicate diagB to create an M-by-D matrix
result = (A.'.*B)*A; %'# An M-by-M result
And yet another option would be to use the function BSXFUN:
result = bsxfun(#times,A.',diagB)*A; %'# An M-by-M result
Maybe I'm having a bit of a brainfart here, but can't you turn your DxD matrix into a DxM matrix (with M copies of the vector you're given) and then .* the last two matrices rather than multiply them (and then, of course, normally multiply the first with the found product quantity)?
You are getting "out of memory" because MATLAB can not find a chunk of memory large enough to accommodate the entire matrix. There are different techniques to avoid this error described in MATLAB documentation.
In MATLAB you obviously do not need programming explicit loops in most cases because you can use operator *. There exists a technique how to speed up matrix multiplication if it is done with explicit loops, here is an example in C#. It has a good idea how (potentially large) matrix can be split into smaller matrices. To contain these smaller matrices in MATLAB you can use cell matrix. It is much more probably that system finds enough RAM to accommodate two smaller sub-matrices then the resulting large matrix.