I've a matrix of order 100*10 . Now the objective is to fill each columns of the matrix with random integer within a specific range. Now the problem is for every column the range of the random number changes. For instance, for the first column, the range is [1,100] , for the second its -10 to 1 and so on till 10th column.
This is what I've tried:
b = [0,100;-10,1;0,1;-1,1;10,20]
a = []
for i=1 to 10
a[] = [(i:100)' randi(1,100)]
end
How do I generate a matrix of this form?
I don't have matlab installed right now, but i would do something like this.
m = 100;
n = size(b, 1);
range = b(:, 2) - b(:, 1);
offset = b(:, 1);
A = round(bsxfun(#minus, bsxfun(#times, rand(m, n), range), offset);
Without loop it would become:
M = 100;
N = size(b, 1);
A = zeros(m, n); % preallocate to avoid matrix expansion
for ii = 1:n
A(:, ii) = randi(b(ii,:), m, 1);
end
Related
I would like to generate all the possible adjacency matrices (zero diagonale) of an undirected graph of n nodes.
For example, with no relabeling for n=3 we get 23(3-1)/2 = 8 possible network configurations (or adjacency matrices).
One solution that works for n = 3 (and which I think is quite stupid) would be the following:
n = 3;
A = [];
for k = 0:1
for j = 0:1
for i = 0:1
m = [0 , i , j ; i , 0 , k ; j , k , 0 ];
A = [A, m];
end
end
end
Also I though of the following which seems to be faster but something is wrong with my indexing since 2 matrices are missing:
n = 3
C = [];
E = [];
A = zeros(n);
for i = 1:n
for j = i+1:n
A(i,j) = 1;
A(j,i) = 1;
C = [C,A];
end
end
B = ones(n);
B = B- diag(diag(ones(n)));
for i = 1:n
for j = i+1:n
B(i,j) = 0;
B(j,i) = 0;
E = [E,B];
end
end
D = [C,E]
Is there a faster way of doing this?
I would definitely generate the off-diagonal elements of the adjacency matrices with binary encoding:
n = 4; %// number of nodes
m = n*(n-1)/2;
offdiags = dec2bin(0:2^m-1,m)-48; %//every 2^m-1 possible configurations
If you have the Statistics and Machine Learning Toolbox, then squareform will easily create the matrices for you, one by one:
%// this is basically a for loop
tmpcell = arrayfun(#(k) squareform(offdiags(k,:)),1:size(offdiags,1),...
'uniformoutput',false);
A = cat(2,tmpcell{:}); %// concatenate the matrices in tmpcell
Although I'd consider concatenating along dimension 3, then you can see each matrix individually and conveniently.
Alternatively, you can do the array synthesis yourself in a vectorized way, it's probably even quicker (at the cost of more memory):
A = zeros(n,n,2^m);
%// lazy person's indexing scheme:
[ind_i,ind_j,ind_k] = meshgrid(1:n,1:n,1:2^m);
A(ind_i>ind_j) = offdiags.'; %'// watch out for the transpose
%// copy to upper diagonal:
A = A + permute(A,[2 1 3]); %// n x n x 2^m matrix
%// reshape to n*[] matrix if you wish
A = reshape(A,n,[]); %// n x (n*2^m) matrix
Let's say I've a large N x M -sized matrix A (e.g. 1000 x 1000). Selecting k random elements without replacement from A is relatively straightforward in MATLAB:
A = rand(1000,1000); % Generate random data
k = 5; % Number of elements to be sampled
sizeA = numel(A); % Number of elements in A
idx = randperm(sizeA); % Random permutation
B = A(idx(1:k)); % Random selection of k elements from A
However, I'm looking for a way to expand the above concept so that I could randomly select k non-overlapping n x m -sized sub-matrices (e.g. 5 x 5) from A. What would be the most convenient way to achieve this? I'd very much appreciate any help!
This probably isn't the most efficient way to do this. I'm sure if I (or somebody else) gave it more thought there would be a better way but it should help you get started.
First I take the original idx(1:k) and reshape it into a 3D matrix reshape(idx(1:k), 1, 1, k). Then I extend it to the length required, padding with zeros, idx(k, k, 1) = 0; % Extend padding with zeros and lastly I use 2 for loops to create the correct indices
for n = 1:k
for m = 1:k
idx(m, 1:k, n) = size(A)*(m - 1) + idx(1, 1, n):size(A)*(m - 1) + idx(1, 1, n) + k - 1;
end
end
The complete script built onto the end of yours
A = rand(1000, 1000);
k = 5;
idx = randperm(numel(A));
B = A(idx(1:k));
idx = reshape(idx(1:k), 1, 1, k);
idx(k, k, 1) = 0; % Extend padding with zeros
for n = 1:k
for m = 1:k
idx(m, 1:k, n) = size(A)*(m - 1) + idx(1, 1, n):size(A)*(m - 1) + idx(1, 1, n) + k - 1;
end
end
C = A(idx);
I was wondering if there was a way to vectorize the nested for loop in this function which is filling up the entries of the 2D dynamic programming table DP. I believe that at the very least the inner loop could be vectorized as each row only depends on the previous row. I'm not sure how to do it though. Note this function is called on large 2D arrays (images) so the nested for loop really doesn't cut it.
function [cols] = compute_seam(energy)
[r, c, ~] = size(energy);
cols = zeros(r);
DP = padarray(energy, [0, 1], Inf);
BP = zeros(r, c);
for i = 2 : r
for j = 1 : c
[x, l] = min([DP(i - 1, j), DP(i - 1, j + 1), DP(i - 1, j + 2)]);
DP(i, j + 1) = DP(i, j + 1) + x;
BP(i, j) = j + (l - 2);
end
end
[~, j] = min(DP(r, :));
j = j - 1;
for i = r : -1 : 1
cols(i) = j;
j = BP(i, j);
end
end
Vectorization of the innermost nested loop
You were right in postulating that at least the inner loop is vectorizable. Here's the modified code for the nested loops part -
rows_DP = size(DP,1); %// rows in DP
%// Get first row linear indices for a group of neighboring three columns,
%// which would be incremented as we move between rows with the row iterator
start_ind1 = bsxfun(#plus,[1:rows_DP:2*rows_DP+1]',[0:c-1]*rows_DP); %//'
for i = 2 : r
ind1 = start_ind1 + i-2; %// setup linear indices for the row of this iteration
[x,l] = min(DP(ind1),[],1); %// get x and l values in one go
DP(i,2:c+1) = DP(i,2:c+1) + x; %// set DP values of a row in one go
BP(i,1:c) = [1:c] + l-2; %// set BP values of a row in one go
end
Benchmarking
Benchmarking Code -
N = 3000; %// Datasize
energy = rand(N);
[r, c, ~] = size(energy);
disp('------------------------------------- With Original Code')
DP = padarray(energy, [0, 1], Inf);
BP = zeros(r, c);
tic
for i = 2 : r
for j = 1 : c
[x, l] = min([DP(i - 1, j), DP(i - 1, j + 1), DP(i - 1, j + 2)]);
DP(i, j + 1) = DP(i, j + 1) + x;
BP(i, j) = j + (l - 2);
end
end
toc,clear DP BP x l
disp('------------------------------------- With Vectorized Code')
DP = padarray(energy, [0, 1], Inf);
BP = zeros(r, c);
tic
rows_DP = size(DP,1); %// rows in DP
start_ind1 = bsxfun(#plus,[1:rows_DP:2*rows_DP+1]',[0:c-1]*rows_DP); %//'
for i = 2 : r
ind1 = start_ind1 + i-2; %// setup linear indices for the row of this iteration
[x,l] = min(DP(ind1),[],1); %// get x and l values in one go
DP(i,2:c+1) = DP(i,2:c+1) + x; %// set DP values of a row in one go
BP(i,1:c) = [1:c] + l-2; %// set BP values of a row in one go
end
toc
Results -
------------------------------------- With Original Code
Elapsed time is 44.200746 seconds.
------------------------------------- With Vectorized Code
Elapsed time is 1.694288 seconds.
Thus, you might enjoy a good 26x speedup improvement in performance with that little vectorization tweak.
More tweaks
Few more optimization tweaks could be tried into your code for performance -
cols = zeros(r) could be replaced with col(r,r) = 0.
DP = padarray(energy, [0, 1], Inf) could be replaced with
DP(1:size(energy,1),1:size(energy,2)+2)=Inf;
DP(:,2:end-1) = energy;
BP = zeros(r, c) could be replaced with BP(r, c) = 0.
The pre-allocation tweaks used here are inspired by this blog post.
I have 2 matrices = X in R^(n*m) and W in R^(k*m) where k<<n.
Let x_i be the i-th row of X and w_j be the j-th row of W.
I need to find, for each x_i what is the j that maximizes <w_j,x_i>
I can't see a way around iterating over all the rows in X, but it there a way to find the maximum dot product without iterating every time over all of W?
A naive implementation would be:
n = 100;
m = 50;
k = 10;
X = rand(n,m);
W = rand(k,m);
Y = zeros(n, 1);
for i = 1 : n
max_ind = 1;
max_val = dot(W(1,:), X(i,:));
for j = 2 : k
cur_val = dot(W(j,:),X(i,:));
if cur_val > max_val
max_val = cur_val;
max_ind = j;
end
end
Y(i,:) = max_ind;
end
Dot product is essentially matrix multiplication:
[~, Y] = max(W*X');
bsxfun based approach to speed-up things for you -
[~,Y] = max(sum(bsxfun(#times,X,permute(W,[3 2 1])),2),[],3)
On my system, using your dataset I am getting a 100x+ speedup with this.
One can think of two more "closeby" approaches, but they don't seem to give any huge improvement over the earlier one -
[~,Y] = max(squeeze(sum(bsxfun(#times,X,permute(W,[3 2 1])),2)),[],2)
and
[~,Y] = max(squeeze(sum(bsxfun(#times,X',permute(W,[2 3 1]))))')
I have two matrices of big sizes, which are something similar to the following matrices.
m; with size 1000 by 10
n; with size 1 by 10.
I would like to subtract each element of n from all elements of m to get ten different matrices, each has size of 1000 by 10.
I started as follows
clc;clear;
nrow = 10000;
ncol = 10;
t = length(n)
for i = 1:nrow;
for j = 1:ncol;
for t = 1:length(n);
m1(i,j) = m(i,j)-n(1);
m2(i,j) = m(i,j)-n(2);
m3(i,j) = m(i,j)-n(3);
m4(i,j) = m(i,j)-n(4);
m5(i,j) = m(i,j)-n(5);
m6(i,j) = m(i,j)-n(6);
m7(i,j) = m(i,j)-n(7);
m8(i,j) = m(i,j)-n(8);
m9(i,j) = m(i,j)-n(9);
m10(i,j) = m(i,j)-n(10);
end
end
end
can any one help me how can I do it without writing the ten equations inside the loop? Or can suggest me any convenient way especially when the two matrices has many columns.
Why can't you just do this:
m01 = m - n(1);
...
m10 = m - n(10);
What do you need the loop for?
Even better:
N = length(n);
m2 = cell(N, 1);
for k = 1:N
m2{k} = m - n(k);
end
Here we go loopless:
nrow = 10000;
ncol = 10;
%example data
m = ones(nrow,ncol);
n = 1:ncol;
M = repmat(m,1,1,ncol);
N = permute( repmat(n,nrow,1,ncol) , [1 3 2] );
result = bsxfun(#minus, M, N );
%or just
result = M-N;
Elapsed time is 0.018499 seconds.
or as recommended by Luis Mendo:
M = repmat(m,1,1,ncol);
result = bsxfun(#minus, m, permute(n, [1 3 2]) );
Elapsed time is 0.000094 seconds.
please make sure that your input vectors have the same orientation like in my example, otherwise you could get in trouble. You should be able to obtain that by transposements or you have to modify this line:
permute( repmat(n,nrow,1,ncol) , [1 3 2] )
according to your needs.
You mentioned in a comment that you want to count the negative elements in each of the obtained columns:
A = result; %backup results
A(A > 0) = 0; %set non-negative elements to zero
D = sum( logical(A),3 );
which will return the desired 10000x10 matrix with quantities of negative elements. (Please verify it, I may got a little confused with the dimensions ;))
Create the three dimensional result matrix. Store your results, for example, in third dimension.
clc;clear;
nrow = 10000;
ncol = 10;
N = length(n);
resultMatrix = zeros(nrow, ncol, N);
neg = zeros(ncol, N); % amount of negative values
for j = 1:ncol
for i = 1:nrow
for t = 1:N
resultMatrix(i,j,t) = m(i,j) - n(t);
end
end
for t = 1:N
neg(j,t) = length( find(resultMatrix(:,j,t) < 0) );
end
end