I want to perform matrix related operations such as multiplication, transpose and inversion of a matrix. I could find out matrix support in Lua here
I have a table which I want to convert to matrix. The table has following structure-
for i=1,myTableSize[1],1 do
str=''
for j=1,myTableSize[2],1 do
if #str~=0 then
str=str..', '
end
str=str..string.format("%.1e",myTable[(j-1)*myTableSize[1]+i])
end
print(str)
end
I am looking for something like myMatrix=matrix(myTable) or myMatrix=matrix.init(myTable), which is compatible with Lua Matrix.
-
Thanks
Try (not tested)
local function tableToMatrix(table, rows cols)
local myMatrix = matrix:new(rows, cols) -- function returns matrix of size rows x cols
for i=1, rows do
for j=1, cols do
matrix.setelement(myMatrix, i, j, table[(i - 1) * cols + j] )
end
end
return matrix
end
Related
I have 200 vectors; each one has a length of 10000.
I want to fill a matrix such that each line represents a vector.
If your vectors are already stored in an array then you can use vcat( ) here:
A = [rand(10000)' for idx in 1:200]
B = vcat(A...)
Julia stores matrices in column-major order so you are going to have to adapt a bit to that
If you have 200 vectors of length 100000 you should make first an empty vector, a = [], this will be your matrix
Then you have to vcat the first vector to your empty vector, like so
v = your vectors, however they are defined
a = []
a = vcat(a, v[1])
Then you can iterate through vectors 2:200 by
for i in 2:200
a = hcat(a,v[i])
end
And finally transpose a
a = a'
Alternatively, you could do
a = zeros(200,10000)
for i in 1:length(v)
a[i,:] = v[i]
end
but I suppose that wont be as fast, if performance is at all an issue, because as I said, julia stores in column major order so access will be slower
EDIT from reschu's comment
a = zeros(10000,200)
for i in 1:length(v)
a[:,i] = v[i]
end
a = a'
I have 5000 vectors; each one has a length of 10000 (values)
I want to fill a matrix column by column such that each columns represents a vector.
5000 columns and 10000 rows.
it did't work in this way . I have this structure:
vector = Vector()
for i in 1:5000
println(vector[i])
end
for example
julia> vector[502]
10000-element Array{Float64,1}: -3.0 1.0 . . . -2.0
when I do
a = zeros(10000,0)
for v in vector
a = hcat(a,v)
end
it doesn't work.
This is almost identical to the question you asked previously on populating a matrix row by row. The solution likewise is:
A = [rand(10000) for idx in 1:5000];
B = hcat(A...);
a = zeros(10000,0)
vector = #whatever is here
for v in vector
a = hcat(a,v)
end
Note
If you get a
**ERROR: ArgumentError: number of rows of each array must match**
the loop must be
a = hcat(a,v')
I am find the rank of binary matrix in GF(2)( Galois Field). The rank function in matlab cannot find it. For example, Given a matrix 400 by 400 as here. If you use the rank function as
rank(A)
ans=357
However, the correct ans. in GF(2) must be 356 by this code
B=gf(A);
rank(B);
ans=356;
But this way spends a lot a time (about 16s). Hence, I used Gaussian elimination to find the rank in GF(2) with small time. But, it does not works well. Sometime, it returns the true value, but sometime it returns wrong. Please see my code and let me know the problem in my code. Note that, it spend very small time compare with above code
function rankA =GaussEliRank(A)
tic
mat = A;
[m n] = size(A); % read the size of the original matrix A
for i = 1 : n
j = find(mat(i:m, i), 1); % finds the FIRST 1 in i-th column starting at i
if isempty(j)
mat = mat( sum(mat,2)>0 ,:);
rankA=rank(mat);
return;
else
j = j + i - 1; % we need to add i-1 since j starts at i
temp = mat(j, :); % swap rows
mat(j, :) = mat(i, :);
mat(i, :) = temp;
% add i-th row to all rows that contain 1 in i-th column
% starting at j+1 - remember up to j are zeros
for k = find(mat( (j+1):m, i ))'
mat(j + k, :) = bitxor(mat(j + k, :), mat(i, :));
end
end
end
%remove all-zero rows if there are some
mat = mat( sum(mat,2)>0 ,:);
if any(sum( mat(:,1:n) ,2)==0) % no solution because matrix A contains
error('No solution.'); % all-zero row, but with nonzero RHS
end
rankA=sum(sum(mat,2)>0);
end
Let use the gfrank function. It is suitable for your matrix.
Use:
gfrank(A)
ans=
356
More detail: How to find the row rank of matrix in Galois fields?
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 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.