Please correct me if there are somethings unclear in this question. I have two matrices pop, and ben of 3 dimensions. Call these dimensions as c,t,w . I want to repeat the exact same process I describe below for all of the c dimensions, without using a for loop as that is slow. For the discussion below, fix a value of the dimension c, to explain my thinking, later I will give a MWE. So when c is fixed I have a 2D matrix with dimension t,w.
Now I repeat the entire process (coming below!) for all of the w dimension.
If the value of u is zero, then I find the next non zero entry in this same t dimension. I save both this entry as well as the corresponding t index. If the value of u is non zero, I simply store this value and the corresponding t index. Call the index as i - note i would be of dimension (c,t,w). The last entry of every u(c,:,w) is guaranteed to be non zero.
Example if the u(c,:,w) vector is [ 3 0 4 2 0 1], then the corresponding i values are [1,3,3,4,6,6].
Now I take these entries and define a new 3d array of dimension (c,t,w) as follows. I take my B array and do the following what is not a correct syntax but to explain you: B(c,t,w)/u(c,i(c,t,w),w). Meaning I take the B values and divide it by the u values corresponding to the non zero indices of u from i that I computed.
For the above example, the denominator would be [3,4,4,2,1,1]. I hope that makes sense!!
QUESTION:
To do this, as this process simply repeats for all c, I can do a very fast vectorizable calculation for a single c. But for multiple c I do not know how to avoid the for loop. I don't knw how to do vectorizable calculations across dimensions.
Here is what I did, where c_size is the dimension of c.
for c=c_size:-1:1
uu=squeeze(pop(c,:,:)) ; % EXTRACT A 2D MATRIX FROM pop.
BB=squeeze(B(c,:,:)) ; % EXTRACT A 2D MATRIX FROM B
ii = nan(size(uu)); % Start with all nan values
[dum_row, ~] = find(uu); % Get row indices of non-zero values
ii(uu ~= 0) = dum_row; % Place row indices in locations of non-zero values
ii = cummin(ii, 1, 'reverse'); % Column-wise cumulative minimum, starting from bottomi
dum_i = ii+(time_size+1).*repmat(0:(scenario_size-1), time_size+1, 1); % Create linear index
ben(c,:,:) = BB(dum_i)./uu(dum_i);
i(c,:,:) = ii ;
clear dum_i dum_row uu BB ii
end
The central question is to avoid this for loop.
Related questions:
Vectorizable FIND function with if statement MATLAB
Efficiently finding non zero numbers from a large matrix
Vectorizable FIND function with if statement MATLAB
Related
I am new to Octave and I wanted to know if there is a way to parse each row of a matrix and use it individually. Ultimately I want to use the rows to check if they are all vertical to each other (the dot product have to be equal to 0 for two vectors to be vertical to each other) so if you have some ideas about that I would love to hear them. Also I wanted to know if there is a function to determine the length (or the amplitude) of a vector.
Thank you in advance.
If by "parse each row" you mean a loop that takes each row one by one, you only need a for loop over the transposed matrix. This works because the for loop takes successive columns of its argument.
Example:
A = [10 20; 30 40; 50 60];
for row = A.'; % loop over columns of transposed matrix
row = row.'; % transpose back to obtain rows of the original matrix
disp(row); % do whatever you need with each row
end
However, loops can often be avoided in Matlab/Octave, in favour of vectorized code. For the specific case you mention, computing the dot product between each pair of rows of A is the same as computing the matrix product of A times itself transposed:
A*A.'
However, for the general case of a complex matrix, the dot product is defined with a complex conjugate, so you should use the complex-conjugate transpose:
P = A*A';
Now P(m,n) contains the dot product of the n-th and m-th rows of A. The condition you want to test is equivalent to P being a diagonal matrix:
result = isdiag(P); % gives true of false
Given two random variables X and Y, where X=(x1,..,xn) and Y=(y1,...,yn) in a nx2 matrix A, so A=[X Y], i need to perform the next operation:
median((x-median(x))(y-median(y)))
I'm trying to obtain an estimator of the covariance matrix using the median instead the mean, for a nxt matrix where t represents the number of random variables and n the length of the data set.
So far, I made the next code:
for i=1:n
for j=1:n
a1=median(A(:,i));
a2=median(A(:,j));
SMM(i,j)=median(((A(:,i)-a1(ones(t,1),:)).*(A(:,j)-a2(ones(t,1),:))));
end
end
However, theoretically I must obtain a semidefinite (positive or negative) symmetric matrix, however that's not the case with this code.
Am I making any mistake in the code formulation?
Various points:
For each of your columns of A (x, y), the median (a1, a2) doesn't change. You should compute these outside the loops.
The loops go over n, rather than t, which are the variables and the indices to the output matrix.
I would first subtract the median from each column, to avoid repeatedly doing the same computations:
A = A - median(A,1); % be explicit about which dimension to take the median over!
Next, we'd loop over the txt output elements of the covariance matrix, and compute each of the elements:
t = size(A,2);
SMM = zeros(t,t); % always preallocate output arrays before a loop
for j=1:t
for i=1:t
SMM(i,j) = median(A(:,i).*A(:,j));
end
end
The loop can likely be vectorized, but that leads to a large intermediate matrix, which slow down code also. So it might not be worth the effort to vectorize. Only try it if this code is too slow!
It should also be possible to run the inner loop from i=j:t, to skip computing the redundant half of the symmetric matrix, instead copying over the previously computed values.
I have a big matrix (1,000 rows and 50,000 columns). I know some columns are correlated (the rank is only 100) and I suspect some columns are even proportional. How can I find such proportional columns? (one way would be looping corr(M(:,j),M(:,k))), but is there anything more efficient?
If I am understanding your problem correctly, you wish to determine those columns in your matrix that are linearly dependent, which means that one column is proportional or a scalar multiple of another. There's a very basic algorithm based on QR Decomposition. For QR decomposition, you can take any matrix and decompose it into a product of two matrices: Q and R. In other words:
A = Q*R
Q is an orthogonal matrix with each column as being a unit vector, such that multiplying Q by its transpose gives you the identity matrix (Q^{T}*Q = I). R is a right-triangular or upper-triangular matrix. One very useful theory by Golub and Van Loan in their 1996 book: Matrix Computations is that a matrix is considered full rank if all of the values of diagonal elements of R are non-zero. Because of the floating point precision on computers, we will have to threshold and check for any values in the diagonal of R that are greater than this tolerance. If it is, then this corresponding column is an independent column. We can simply find the absolute value of all of the diagonals, then check to see if they're greater than some tolerance.
We can slightly modify this so that we would search for values that are less than the tolerance which would mean that the column is not independent. The way you would call up the QR factorization is in this way:
[Q,R] = qr(A, 0);
Q and R are what I just talked about, and you specify the matrix A as input. The second parameter 0 stands for producing an economy-size version of Q and R, where if this matrix was rectangular (like in your case), this would return a square matrix where the dimensions are the largest of the two sizes. In other words, if I had a matrix like 5 x 8, producing an economy-size matrix will give you an output of 5 x 8, where as not specifying the 0 will give you an 8 x 8 matrix.
Now, what we actually need is this style of invocation:
[Q,R,E] = qr(A, 0);
In this case, E would be a permutation vector, such that:
A(:,E) = Q*R;
The reason why this is useful is because it orders the columns of Q and R in such a way that the first column of the re-ordered version is the most probable column that is independent, followed by those columns in decreasing order of "strength". As such, E would tell you how likely each column is linearly independent and those "strengths" are in decreasing order. This "strength" is exactly captured in the diagonals of R corresponding to this re-ordering. In fact, the strength is proportional to this first element. What you should do is check to see what diagonals of R in the re-arranged version are greater than this first coefficient scaled by the tolerance and you use these to determine which of the corresponding columns are linearly independent.
However, I'm going to flip this around and determine the point in the R diagonals where the last possible independent columns are located. Anything after this point would be considered linearly dependent. This is essentially the same as checking to see if any diagonals are less than the threshold, but we are using the re-ordering of the matrix to our advantage.
In any case, putting what I have mentioned in code, this is what you should do, assuming your matrix is stored in A:
%// Step #1 - Define tolerance
tol = 1e-10;
%// Step #2 - Do QR Factorization
[Q, R, E] = qr(A,0);
diag_R = abs(diag(R)); %// Extract diagonals of R
%// Step #3 -
%// Find the LAST column in the re-arranged result that
%// satisfies the linearly independent property
r = find(diag_R >= tol*diag_R(1), 1, 'last');
%// Step #4
%// Anything after r means that the columns are
%// linearly dependent, so let's output those columns to the
%// user
idx = sort(E(r+1:end));
Note that E will be a permutation vector, and I'm assuming you want these to be sorted so that's why we sort them after the point where the vectors fail to become linearly independent anymore. Let's test out this theory. Suppose I have this matrix:
A =
1 1 2 0
2 2 4 9
3 3 6 7
4 4 8 3
You can see that the first two columns are the same, and the third column is a multiple of the first or second. You would just have to multiply either one by 2 to get the result. If we run through the above code, this is what I get:
idx =
1 2
If you also take a look at E, this is what I get:
E =
4 3 2 1
This means that column 4 was the "best" linearly independent column, followed by column 3. Because we returned [1,2] as the linearly dependent columns, this means that columns 1 and 2 that both have [1,2,3,4] as their columns are a scalar multiple of some other column. In this case, this would be column 3 as columns 1 and 2 are half of column 3.
Hope this helps!
Alternative Method
If you don't want to do any QR factorization, then I can suggest reducing your matrix into its row-reduced Echelon form, and you can determine the basis vectors that make up the column space of your matrix A. Essentially, the column space gives you the minimum set of columns that can generate all possible linear combinations of output vectors if you were to apply this matrix using matrix-vector multiplication. You can determine which columns form the column space by using the rref command. You would provide a second output to rref that gives you a vector of elements that tell you which columns are linearly independent or form a basis of the column space for that matrix. As such:
[B,RB] = rref(A);
RB would give you the locations of which columns for the column space and B would be the row-reduced echelon form of the matrix A. Because you want to find those columns that are linearly dependent, you would want to return a set of elements that don't contain these locations. As such, define a linearly increasing vector from 1 to as many columns as you have, then use RB to remove these entries in this vector and the result would be those linearly dependent columns you are seeking. In other words:
[B,RB] = rref(A);
idx = 1 : size(A,2);
idx(RB) = [];
By using the above code, this is what we get:
idx =
2 3
Bear in mind that we identified columns 2 and 3 to be linearly dependent, which makes sense as both are multiples of column 1. The identification of which columns are linearly dependent are different in comparison to the QR factorization method, as QR orders the columns based on how likely that particular column is linearly independent. Because columns 1 to 3 are related to each other, it shouldn't matter which column you return. One of these forms the basis of the other two.
I haven't tested the efficiency of using rref in comparison to the QR method. I suspect that rref does Gaussian row eliminations, where the complexity is worse compared to doing QR factorization as that algorithm is highly optimized and efficient. Because your matrix is rather large, I would stick to the QR method, but use rref anyway and see what you get!
If you normalize each column by dividing by its maximum, proportionality becomes equality. This makes the problem easier.
Now, to test for equality you can use a single (outer) loop over columns; the inner loop is easily vectorized with bsxfun. For greater speed, compare each column only with the columns to its right.
Also to save some time, the result matrix is preallocated to an approximate size (you should set that). If the approximate size is wrong, the only penalty will be a little slower speed, but the code works.
As usual, tests for equality between floating-point values should include a tolerance.
The result is given as a 2-column matrix (S), where each row contains the indices of two rows that are proportional.
A = [1 5 2 6 3 1
2 5 4 7 6 1
3 5 6 8 9 1]; %// example data matrix
tol = 1e-6; %// relative tolerance
A = bsxfun(#rdivide, A, max(A,[],1)); %// normalize A
C = size(A,2);
S = NaN(round(C^1.5),2); %// preallocate result to *approximate* size
used = 0; %// number of rows of S already used
for c = 1:C
ind = c+find(all(abs(bsxfun(#rdivide, A(:,c), A(:,c+1:end))-1)<tol));
u = numel(ind); %// number of columns proportional to column c
S(used+1:used+u,1) = c; %// fill in result
S(used+1:used+u,2) = ind; %// fill in result
used = used + u; %// update number of results
end
S = S(1:used,:); %// remove unused rows of S
In this example, the result is
S =
1 3
1 5
2 6
3 5
meaning column 1 is proportional to column 3; column 1 is proportional to column 5 etc.
If the determinant of a matrix is zero, then the columns are proportional.
There are 50,000 Columns, or 2 times 25,000 Columns.
It is easiest to solve the determinant of a 2 by 2 matrix.
Hence: to Find the proportional matrix, the longest time solution is to
define the big-matrix on a spreadsheet.
Apply the determinant formula to a square beginning from 1st square on the left.
Copy it for every row & column to arrive at the answer in the next spreadsheet.
Find the Columns with Determinants Zero.
This is Quite basic,not very time consuming and should be result oriented.
Manual or Excel SpreadSheet(Efficient)
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.
I have a sparse logical matrix, which is quite large. I would like to draw random non-zero elements from it without storing all of its non-zero elements in a separate vector (eg. by using find command). Is there an easy way to do this?
Currently I am implementing rejection sampling, which is drawing a random element and checking whether that is non-zero or not. But it is not efficient when the ratio of non-zero elements is small.
A sparse logical matrix is not a very practical representation of your data if you want to pick random locations. Rejection sampling and find are the only two ways that make sense to me. Here's how you can do them efficiently (assuming you want to get 4 random locations):
%# using find
idx = find(S);
%# draw 4 without replacement
fourRandomIdx = idx(randperm(length(idx),4));
%# draw 4 with replacement
fourRandomIdx = idx(randi(1,length(idx),4));
%# get row, column values
[row,col] = ind2sub(size(S),fourRandomIdx);
%# using rejection sampling
density = nnz(S)/prod(size(S));
%# estimate how many samples you need to get at least 4 hits
%# and multiply by 2 (or 3)
n = ceil( 1 / (1-(1-density)^4) ) * 2;
%# random indices w/ replacement
randIdx = randi(1,n,prod(size(S)));
%# identify the first four non-zero elements
[row,col] = find(S(randIdx),4,'first');
An n x m matrix with nnz non-zero elements requires nnz + n + 1 integers to store the locations of its non-zero entries. For a logical matrix there is no need to store the value of the non-zero entries: these are all true. Correspondingly, you would do best to convert your logical sparse matrix into a list of the linear indices of its non-zero entries, together with n and m, which requires only nnz + 2 integers of storage. From these (and ind2sub) you can readily reconstruct the subscripts corresponding to any non-zero entry that you choose randomly using randi over the range 1..nnz
find is the standard interface to get the non-zero elements in a sparse matrix. Have a look here http://www.mathworks.se/help/techdoc/math/f6-9182.html#f6-13040
[i,j,s] = find(S)
find returns the row indices of nonzero values in vector i, the column indices in vector j, and the nonzero values themselves in the vector s.
No need to get s. Just pick a random index in i,j.
By representing the entries in a 3 column format, aka a coordinate list (i, j, value), you can simply select the items from the list. To get this, you can either use your original method for creating the sparse matrix (i.e. the precursor to sparse()), or use the find command, a la [i,j,s] = find(S);
If you don't need the entries, and it seems you don't, you can just extract i and j.
If, for some reason, your matrix is massive and your RAM limitations are severe, you can simply divide the matrix into regions, and let the probability of selecting a given sub-matrix be proportional to the number of non-zero elements (using nnz) in that sub-matrix. You could go so far as to divide the matrix into individual columns, and the rest of the calculation is trivial. NB: by applying sum to the matrix, you can get the per-column counts (assuming your entries are just 1s).
In this way, you need not even bother with rejection sampling (which seems pointless to me in this case, since Matlab knows where all of the non-zero entries are).