I'm running a double for loop for the calculation of solar radiation, something like:
for b=1:365 %amount of days
for n=1:24 %amount of hours
solar(b,n)=sin(ht(b,n))+...
end
end
However, instead of the creation of a 365x24 martrix, I'd like a 8760x1 array where all the values are plotted under each other in 1 colomn. It is important that this happens within the for loop, as some other calculation need to be done on the array in this loop.
Thanks!
MATLAB matrices are inherently accessible as if they are 1-dimensional vectors via linear indexing. The linear order of matrix elements goes in ascending order of dimension, so to make your matrix order by hour and then by day, simply swap the dimensions you index into solar:
solar(n,b)=sin(ht(b,n))+...
If you explicitly need a 8760x1 array, you can then obtain this as solar(:). However if you simply needed to iterate over all elements in a single loop you can rely on linear indexing without reshaping the matrix:
for n = 1:numel(solar)
% doSomething(solar(n));
end
Related
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
In Matlab, I have two m-by-n matrices X and Y, with n>m.
I need to define a 3D m-by-m-by-n matrix Z whose components can be computed as
for i=2:m
for j=i+1:m
for k=1:n
Z(i,j,k) = (Y(j-1,k)-Y(i-1,k))*X(j-1,k);
end
end
end
As these nested loops require a long computational time, I have been looking for a way to define the matrix Z using matrices multiplication, but I have not managed so far. Any suggestion?
You can simply remove the inner loop (over k) by writing
Z(i,j,:) = (Y(j-1,:)-Y(i-1,:)).*X(j-1,:);
Note the .* element-wise multiplication. You can then proceed to remove additional loops in a similar manner.
But note that, most likely, your loop is slow because you don't preallocate the output array. Do this before the loop:
Z = zeros(m,m,n);
You can also gain a bit of speed by reversing the loop order, such that the first index is iterated over in the innermost loop, and the last index is iterated over in the outermost loop. This accesses the matrix data in memory order, making your cache more efficient.
I'm currently attempting to select a subset of 0's in a very large matrix (about 400x300 elements) and change their value to 1. I am able to do this, but it requires using a loop where each instance it selects the next value in a randperm vector. In other words, 50% of the 0's in the matrix are randomly selected, one-at-a-time, and changed to 1:
z=1;
for z=1:(.5*numberofzeroes)
A(zeroposition(rpnumberofzeroes(z),1),zeroposition(rpnumberofzeroes(z),2))=1;
z=z+1;
end
Where 'A' is the matrix, 'zeroposition' is a 2-column-wide matrix with the positions of the 0's in the matrix (the "coordinates" if you like), and 'rpnumberofzeros' is a randperm vector from 1 to the number of zeroes in the matrix.
So for example, for z=20, the code might be something like this:
A(3557,2684)=1;
...so that the 0 which appears in this location within A will now be a 1.
It performs this loop thousands of times, because .5*numberofzeroes is a very big number. This inevitably takes a long time, so my question is can this be done without using a loop? Or at least, in some way that takes less processing resources/time?
As I said, the only thing that needs to be done is an entirely random selection of 50% (or whatever proportion) of the 0's changed to 1.
Thanks in advance for the help, and let me know if I can clear anything up! I'm new here, so apologies in advance if I've made any faux pa's.
That's very easy. I'd like to introduce you to my friend sub2ind. sub2ind allows you to take row and column coordinates of a matrix and convert them into linear column-major indices so that you can access multiple values in a matrix simultaneously in a single call. As such, the equivalent code you want is:
%// First access the values in rpnumberofzeroes
vals = rpnumberofzeroes(1:0.5*numberofzeroes, :);
%// Now, use the columns of these to determine which rows and columns we want
%// to access A
rows = zeroposition(vals(:,1), 1);
cols = zeroposition(vals(:,2), 2);
%// Get linear indices via sub2ind
ind1 = sub2ind(size(A), rows, cols);
%// Now set these locations to 1
A(ind1) = 1;
The first statement gets the first half of your matrix of coordinates stored in rpnumberofzeroes. The first column is the row coordinates, the second column is the column coordinates. Notice that in your code, you wish to use the values in zeroposition to access the locations in A. As such, extract out the corresponding rows and columns from rpnumberofzeroes to figure out the right rows and columns from zeroposition. Once that's done, we wish to use these new rows and columns from zeroposition and index into A. sub2ind requires three inputs - the size of the matrix you are trying to access... so in our case, that's A, the row coordinates and the column coordinates. The output is a set of column major indices that are computed for each row and column pair.
The last piece of the puzzle is to use these to index into A and set the locations to 1.
This can be accomplished with linear indexing as well:
% find linear position of all zeros in matrix
ix=find(abs(A)<eps);
% set one half of those, selected at random, to one.
A(ix(randperm(round(numel(ix)*.5)))=1;
Im am currently writing a code to implement a numerical approximation to the 3D steady state heat equation using finite difference matrix methods. This involves discritising the 2nd order PDE into the matrix A and solving Ax=b. where x is temperature at each of the specified grid points. Further information on this type of question can be found here:
http://people.nas.nasa.gov/~pulliam/Classes/New_notes/Matrix_ODE.pdf
To complete this problem, I have represented the 3D matrix A by a 2D array calling the values in the 1D array b using an indexing function of the form:
i+(j-1)*Nx+Nx*Ny*(k-1)
for the (i,j,k)th element of the 3D matrix where Nx, Ny, Nz are the number of points in the x,y,z coordinates. There ends up being a lot of loop computation in order to create the matrix A and b and I was wondering what is the most computationally efficient and less memory exhaustive way to run these loops, i.e. is it better to use something like
for j=1:Ny
for i=2:Nx-1
b(i+(j-1)*Nx)=D4;
end
end
for j=1:Ny
for i=2:Nx-1
b(i+(j-1)*Nx+Nx*Ny*(Nz-1))=D3;
end
end
or should I condense these into a single loop like:
for j=1:Ny
for i=2:Nx-1
b(i+(j-1)*Nx)=D4;
b(i+(j-1)*Nx+Nx*Ny*(Nz-1))=D3;
end
end
I have preallocated both the arrays A and b. Is there a vectorised way to do this also?
Assuming Nx, Ny, Nz, D3 and D4 to be scalars and that you are using pre-allocation for b with zeros, you may try this vectorized approach -
I = 2:Nx-1; %// Vectors to represent i
J = 1:Ny; %// Vectors to represent j
ind1 = bsxfun(#plus,I,[(J-1)*Nx]'); %//' Indices, 1st set of nested loops
ind2 = bsxfun(#plus,I,[(J-1)*Nx+Nx*Ny*(Nz-1)]'); %//' Indices, 2nd set of loops
b(ind1) = D4; %// Assign values for 1st set
b(ind2) = D3; %// Assign values for 2nd set
The second method should be slightly faster since it performs the same number of calculations with fewer increments of the loop variables. You can look into MATLAB's built-in stopwatch commands tic and toc to time your code. http://www.mathworks.com/help/matlab/ref/tic.html
Something more vectorized might be possible but I would need to know more about the format of the arrays that contain D3 and D4. The reshape() function might be able to help.
I have to two evenly sized very large vectors (columns) A and B. I would like to divide vector A by vector B. This will give me a large matrix AxB filled with zeros, except the last column. This column contains the values I'm interested in. When I simple divide the vectors in a Matlab script, I run out of memory. Probably because the matrix AxB becomes very large. Probably I can prevent this from happening by repeating the following:
calculating the first row of matrix AxB
filter the last value and put it into another vector C.
delete the used row of matrix AxB
redo step 1-4 for all rows in vector A
How can I make a loop which does this?
You're question doesn't make it clear what you are trying to do, although it sounds like you want to do an element wise division.
Try:
C = A./B
"Matrix product AxB" and "dividing vectors" are distinct operations.
If we understood this correctly, what you do want to calculate is "C = last column from AxB", such that:
lastcolsel=zeros(size(B,2),1)
C=(A*B)*lastcolsel
If that code breaks your memory limit, recall that matrix product is associative (MxN)xP = Mx(NxP). Simplifying your example, we get:
lastcolsel=zeros(size(B,2),1)
simplifier=B*lastcolsel
C=A*simplifier