I have the above loop running on the above variables:
A is a 2d array of size mxn.
mask is a 1d logical array of size 1xn
results is a 1d array of size 1xn
B is a vector of the form mx1
C is a mxm matrix, m is the same as the above.
Edit: expanded foo(x) into the function.
here is the code:
temp = (B.'*C*B);
for k = 1:n
x = A(:,k);
if(mask(k) == 1)
result(k) = (B.'*C*x)^2 / (temp*(x.'*C*x)); %returns scalar
end
end
take note, I am already successfully using the above code as a parfor loop instead of for. I was hoping you would be able to suggest some way to use meshgrid or the sort to yield better performance improvement. I don't think I have RAM problems so a solution can also be expensive memory wise.
Many thanks.
try this:
result=(B.'*C*A).^2./diag(temp*(A.'*C*A))'.*mask;
This vectorization via matrix multiplication will also make sure that result is a 1xn vector. In the code you provided there can be a case where the last elements in mask are zeros, in this case your code will truncate result to a smaller length, whereas, in the answer it'll keep these elements zero.
If your foo admits matrix input, you could do:
result = zeros(1,n); % preallocate result with zeros
mask = logical(mask); % make mask logical type
result(mask) = foo(A(mask),:); % compute foo for all selected columns
Related
I ran into a big of a problem with a tetris program I'm writing currently in C.
I am trying to use a 4D multi-dimensional array e.g.
uint8_t shape[7][4][4][4]
but I keep getting seg faults when I try that, I've read around and it seems to be that I'm using up all the stack memory with this kind of array (all I'm doing is filling the array with 0s and 1s to depict a shape so I'm not inputting a ridiculously high number or something).
Here is a version of it (on pastebin because as you can imagine its very ugly and long).
If I make the array smaller it seems to work but I'm trying to avoid a way around it as theoretically each "shape" represents a rotation as well.
https://pastebin.com/57JVMN20
I've read that you should use dynamic arrays so they end up on the heap but then I run into the issue how someone would initialize a dynamic array in such a way as linked above. It seems like it would be a headache as I would have to go through loops and specifically handle each shape?
I would also be grateful for anybody to let me pick their brain on dynamic arrays how best to go about them and if it's even worth doing normal arrays at all.
Even though I have not understood why do you use 4D arrays to store shapes for a tetris game, and I agree with bolov's comment that such an array should not overflow the stack (7*4*4*4*1 = 448 bytes), so you should probably check other code you wrote.
Now, to your question on how to manage 4D (N-Dimensional)dynamically sized arrays. You can do this in two ways:
The first way consists in creating an array of (N-1)-Dimensional arrays. If N = 2 (a table) you end up with a "linearized" version of the table (a normal array) which dimension is equal to R * C where R is the number of rows and C the number of columns. Inductively speaking, you can do the very same thing for N-Dimensional arrays without too much effort. This method has some drawbacks though:
You need to know beforehand all the dimensions except one (the "latest") and all the dimensions are fixed. Back to the N = 2 example: if you use this method on a table of C columns and R rows, you can change the number of rows by allocating C * sizeof(<your_array_type>) more bytes at the end of the preallocated space, but not the number of columns (not without rebuilding the entire linearized array). Moreover, different rows must have the same number of columns C (you cannot have a 2D array that looks like a triangle when drawn on paper, just to get things clear).
You need to carefully manage the indicies: you cannot simply write my_array[row][column], instead you must access that array with my_array[row*C + column]. If N is not 2, then this formula gets... interesting
You can use N-1 arrays of pointers. That's my favourite solution because it does not have any of the drawbacks from the previous solution, although you need to manage pointers to pointers to pointers to .... to pointers to a type (but that's what you do when you access my_array[7][4][4][4].
Solution 1
Let's say you want to build an N-Dimensional array in C using the first solution.
You know the length of each dimension of the array up to the (N-1)-th (let's call them d_1, d_2, ..., d_(N-1)). We can build this inductively:
We know how to build a dynamic 1-dimensional array
Supposing we know how to build a (N-1)-dimensional array, we show that we can build a N-Dimensional array by putting each (N-1)-dimensional array we have available in a 1-Dimensional array, thus increasing the available dimensions by 1.
Let's also assume that the data type that the arrays must hold is called T.
Let's suppose we want to create an array with R (N-1)-dimensional arrays inside it. For that we need to know the size of each (N-1)-dimensional array, so we need to calculate it.
For N = 1 the size is just sizeof(T)
For N = 2 the size is d_1 * sizeof(T)
For N = 3 the size is d_2 * d_1 * sizeof(T)
You can easily inductively prove that the number of bytes required to store R (N-1)-dimensional arrays is R*(d_1 * d_2 * ... * d_(n-1) * sizeof(T)). And that's done.
Now, we need to access a random element inside this massive N-dimensional array. Let's say we want to access the item with indicies (i_1, i_2, ..., i_N). For this we are going to repeat the inductive reasoning:
For N = 1, the index of the i_1 element is just my_array[i_1]
For N = 2, the index of the (i_1, i_2) element can be calculated by thinking that each d_1 elements, a new array begins, so the element is my_array[i_1 * d_1 + i_2].
For N = 3, we can repeat the same process and end up having the element my_array[d_2 * ((i_1 * d_1) + i_2) + i_3]
And so on.
Solution 2
The second solution wastes a bit more memory, but it's more straightforward, both to understand and to implement.
Let's just stick to the N = 2 case so that we can think better. Imagine to have a table and to split it row by row and to place each row in its own memory slot. Now, a row is a 1-dimensional array, and to make a 2-dimensional array we only need to be able to have an ordered array with references to each row. Something like the following drawing shows (the last row is the R-th row):
+------+
| R1 -------> [1,2,3,4]
|------|
| R2 -------> [2,4,6,8]
|------|
| R3 -------> [3,6,9,12]
|------|
| .... |
|------|
| RR -------> [R, 2*R, 3*R, 4*R]
+------+
In order to do that, you need to first allocate the references array (R elements long) and then, iterate through this array and assign to each entry the pointer to a newly allocated memory area of size d_1.
We can easily extend this for N dimensions. Simply build a R dimensional array and, for each entry in this array, allocate a new 1-Dimensional array of size d_(N-1) and do the same for the newly created array until you get to the array with size d_1.
Notice how you can easily access each element by simply using the expression my_array[i_1][i_2][i_3]...[i_N].
For example, let's suppose N = 3 and T is uint8_t and that d_1, d_2 and d_3 are known (and not uninitialized) in the following code:
size_t d1 = 5, d2 = 7, d3 = 3;
int ***my_array;
my_array = malloc(d1 * sizeof(int**));
for(size_t x = 0; x<d1; x++){
my_array[x] = malloc(d2 * sizeof(int*));
for (size_t y = 0; y < d2; y++){
my_array[x][y] = malloc(d3 * sizeof(int));
}
}
//Accessing a random element
size_t x1 = 2, y1 = 6, z1 = 1;
my_array[x1][y1][z1] = 32;
I hope this helps. Please feel free to comment if you have questions.
I need to multiply parts of a column vector with a fixed row vector. I solved this problem using a for-loop. However, I am wondering if the performance can be improved as I have to perform this kind of computation around 50 million times. Here's my code so far:
multMat = 1:5;
mat = randi(5,10,1);
windowSize = 5;
vout = nan(10,1);
for r = windowSize : 10
vout(r) = multMat * mat( (r - windowSize + 1) : r);
end
I was thinking about uisng arrayfun. However, first I don't know how to adress the cell range (i.e. the previous five cells including the current cell), and second, I am not sure if arrayfun will be any faster than using the loop?
This sliding vector multiplication you're describing is an example of what is known as convolution. The following produces the same result as the loop in your example:
vout = [nan(windowSize-1,1);
conv(mat,flip(multMat),'valid')];
If your output doesn't really need the leading NaN values which aren't overwritten in your loop then the conv expression is sufficient without concatenating the NaN elements to it.
For sufficiently large vectors this is of course not guaranteed to be as fast as you'd like it to be, but MATLAB's built-in convolution implementation is likely to be pretty close to an optimal tool for the job.
I have an issue with a code performing some array operations. It is too slow, because I use loops and input data are quite big. It was the easiest way for me, but now I am looking for something faster than for loops. I was trying to optimize or rewrite code, but unsuccessful. I really aprecciate Your help.
In my code I have three arrays x1, y1 (coordinates of points in grid), g1 (values in the points) and for example their size is 300 x 300. I treat each matrix as composition of 9 and I make calculation for points in the middle one. For example I start with g1(101,101), but I am using data from g1(1:201,1:201)=g2. I need to calculate distance from each point of g1(1:201,1:201) to g1(101,101) (ll matrix), then I calculate nn as it is in the code, next I find value for g1(101,101) from nn and put it in N array. Then I go to g1(101,102) and so on until g1(200,200), where in this last case g2=g1(99:300,99:300).
As i said, this code is not very efficient, even I have to use larger arrays than I gave in the example, it takes too much time. I hope I explain enough clearly what I expect from the code. I was thinking of using arrayfun, but I have never worked with this function, so I don't know how should use it, however it seems to me it won't handle. Maybe there are other solutions, however I couldn't find anything apropriate.
tic
x1=randn(300,300);
y1=randn(300,300);
g1=randn(300,300);
m=size(g1,1);
n=size(g1,2);
w=1/3*m;
k=1/3*n;
N=zeros(w,k);
for i=w+1:2*w
for j=k+1:2*k
x=x1(i,j);
y=y1(i,j);
x2=y1(i-k:i+k,j-w:j+w);
y2=y1(i-k:i+k,j-w:j+w);
g2=g1(i-k:i+k,j-w:j+w);
ll=1./sqrt((x2-x).^2+(y2-y).^2);
ll(isinf(ll))=0;
nn=ifft2(fft2(g2).*fft2(ll));
N(i-w,j-k)=nn(w+1,k+1);
end
end
czas=toc;
For what it's worth, arrayfun() is just a wrapper for a for loop, so it wouldn't lead to any performance improvements. Also, you probably have a typo in the definition of x2, I'll assume that it depends on x1. Otherwise it would be a superfluous variable. Also, your i<->w/k, j<->k/w pairing seems inconsistent, you should check that as well. Also also, just timing with tic/toc is rarely accurate. When profiling your code, put it in a function and run the timing multiple times, and exclude the variable generation from the timing. Even better: use the built-in profiler.
Disclaimer: this solution will likely not help for your actual problem due to its huge memory need. For your input of 300x300 matrices this works with arrays of size 300x300x100x100, which is usually a no-go. Still, it's here for reference with a smaller input size. I wanted to add a solution based on nlfilter(), but your problem seems to be too convoluted to be able to use that.
As always with vectorization, you can do it faster if you can spare the memory for it. You are trying to work with matrices of size [2*k+1,2*w+1] for each [i,j] index. This calls for 4d arrays, of shape [2*k+1,2*w+1,w,k]. For each element [i,j] you have a matrix with indices [:,:,i,j] to treat together with the corresponding elements of x1 and y1. It also helps that fft2 accepts multidimensional arrays.
Here's what I mean:
tic
x1 = randn(30,30); %// smaller input for tractability
y1 = randn(30,30);
g1 = randn(30,30);
m = size(g1,1);
n = size(g1,2);
w = 1/3*m;
k = 1/3*n;
%// these will be indexed on the fly:
%//x = x1(w+1:2*w,k+1:2*k); %// size [w,k]
%//y = x1(w+1:2*w,k+1:2*k); %// size [w,k]
x2 = zeros(2*k+1,2*w+1,w,k); %// size [2*k+1,2*w+1,w,k]
y2 = zeros(2*k+1,2*w+1,w,k); %// size [2*k+1,2*w+1,w,k]
g2 = zeros(2*k+1,2*w+1,w,k); %// size [2*k+1,2*w+1,w,k]
%// manual definition for now, maybe could be done smarter:
for ii=w+1:2*w %// don't use i and j as variables
for jj=k+1:2*k %// don't use i and j as variables
x2(:,:,ii-w,jj-k) = x1(ii-k:ii+k,jj-w:jj+w); %// check w vs k here
y2(:,:,ii-w,jj-k) = y1(ii-k:ii+k,jj-w:jj+w); %// check w vs k here
g2(:,:,ii-w,jj-k) = g1(ii-k:ii+k,jj-w:jj+w); %// check w vs k here
end
end
%// use bsxfun to operate on [2*k+1,2*w+1,w,k] vs [w,k]-sized arrays
%// need to introduce leading singletons with permute() in the latter
%// in order to have shape [1,1,w,k] compatible with the first array
ll = 1./sqrt(bsxfun(#minus,x2,permute(x1(w+1:2*w,k+1:2*k),[3,4,1,2])).^2 ...
+ bsxfun(#minus,y2,permute(y1(w+1:2*w,k+1:2*k),[3,4,1,2])).^2);
ll(isinf(ll)) = 0;
%// compute fft2, operating on [2*k+1,2*w+1,w,k]
%// will return fft2 for each index in the [w,k] subspace
nn = ifft2(fft2(g2).*fft2(ll));
%// we need nn(w+1,k+1,:,:) which is exactly of size [w,k] as needed
N = reshape(nn(w+1,k+1,:,:),[w,k]); %// quicker than squeeze()
N = real(N); %// this solution leaves an imaginary part of around 1e-12
czas=toc;
I have a three-dimensional matrix of these sizes, approximately
A = rand(20, 1000, 20);
where the first and third dimensions are always the same length. I want to zero the elements in a main diagonal slice. This does what I mean
for ii = 1:size(A, 1)
A(ii, :, ii) = 0;
end
Is there a vectorized or otherwise faster way to do this? This code runs about 100,000 times, with these approximate sizes, but not the exact same sizes each time.
You can use logical indexing for multible tailing dimensions while using subscript indexing for all previous dimensions individually. This way you can easily do the operation on an 1000 20 20 matrix. To apply this to your matrix, permute is required which might be slow:
n=size(A,3)
A=permute(A,[2,1,3]);
A(:,diag(true(n,1)))=0;
A=permute(A,[2,1,3]);
If it would be possible to permanently swap the dimensions of A in your code and avoid the permute, this would lead to the fastest solution.
Alternatively you can use repmat to expand the index to the dimensions of A
ix=repmat(reshape(diag(true(n,1)),n,1,n),[1,size(A,2),1])
A(ix)=0
For matrices of the same size you could keep ix. Not having access to MATLAB right now, I don't know which solution is faster.
You can use bsxfun to build a linear index of the elements to be zeroed:
ind = bsxfun(#plus, (0:size(A,2)-1).'*size(A,1), 1:size(A,1)*size(A,2)+1:numel(A) );
A(ind) = 0;
I want to create 3d arrays that are functions of 2d arrays and apply matrix operations on each of the 2D arrays. Right now I am using for loop to create a series of 2d arrays, as in the code below:
for i=1:50
F = [1 0 0; 0 i/10 0; 0 0 1];
B=F*F';
end
Is there a way to do this without the for loop? I tried things such as:
F(2,2) = 0:0.1:5;
and:
f=1:0.1:5;
F=[1 0 0; 0 f 0; 0 0 1];
to create them without the loop, but both give errors of dimension inconsistency.
I also want to perform matrix operations on F in my code, such as
B=F*F';
and want to plot certain components of F as a function of something else. Is it possible to completely eliminate the for loop in such a case?
If I understand what you want correctly, you want 50 2D matrices stacked into a 3D matrix where the middle entry varies from 1/10 to 50/10 = 5 in steps of 1/10. You almost have it correct. What you would need to do is first create a 3D matrix stack, then assign a 3D vector to the middle entry.
Something like this would do:
N = 50;
F = repmat(eye(3,3), [1 1 N]);
F(2,2,:) = (1:N)/10; %// This is 1/10 to 5 in steps of 1/10... or 0.1:0.1:5
First pre-allocate a matrix F that is the identity matrix for all slices, then replace the middle row and middle column of each slice with i/10 for i = 1, 2, ..., 50.
Therefore, to get the ith slice, simply do:
out = F(:,:,i);
Minor Note
I noticed that what you want to do in the end is a matrix multiplication of the 3D matrices. That operation is not defined in MATLAB nor anywhere in a linear algebra context. If you want to multiply each 2D slice independently, you'd be better off using a for loop. Doing this vectorized with native operations isn't supported in this context.
To do it in a loop, you'd do something like this for each slice:
B = zeros(size(F));
for ii = 1 : size(B,3)
B(:,:,ii) = F(:,:,ii)*F(:,:,ii).';
end
... however, examining the properties of your matrix, the only thing that varies is the middle entry. If you perform a matrix multiplication, all of the entries per slice are going to be the same... except for the middle, where the entry is simply itself squared. It doesn't matter if you multiple one slice by the transpose of the other. The transpose of the identity is still the identity.
If your matrices are going to be like this, you can just perform an element-wise multiplication with itself:
B = F.*F;
This will not work if F is anything else but what you have above.
Creating the matrix would be easy:
N = 50;
S = cell(1,N);
S(:) = {eye(3,3)};
F = cat(3, S{:});
F(2,2,:) = (1:N)/10;
Another (faster) way would be:
N = 50;
F = zeros(3,3,N);
F(1,1,:) = 1;
F(2,2,:) = (1:N)/10;
F(3,3,:) = 1;
You then can get the 3rd matrix (for example) by:
F(:,:,3)