I'm trying to multiply various matrices by constants, and usually vectorizing makes things like this faster, but I can't quite figure this one out. Let's say I'm doing a polynomial expansion and I have a list of constants for different functions.
for i in range(10):
expansions[i,:,:] += c[i,k]*(matrix)
Now, this is the simple way to do it, but I was wondering if there was some way to do it like this:
expansions[:,:,:] += c[:,k]*(matrix)
Where I apply each constant individually to "matrix". I tried to stack lots of "matrix" on top of itself, to create something of size (10xNxN), but this didn't work. Any ideas?
Related
The problem here is: I have to add two numbers and then do 3 operations with that sum. So either i add them both and put their value inside one variable and compute operations or i re-add like(a+b<c) while doing any operation. So which way is more memory efficient and fast?
val sum = k+d
if(sum<=b && sum>spend){
spend = sum
}
or,
if(k+d<=b && k+d>spend){
spend = k+d
}
It depends on the context. If k and d happen to be compile-time constants then the compiler can just replace k+d with the sum so it won't matter how many times you write k+d. Also if the second form would turn out to be faster and the sum variable would not escape the function (not returned or used as parameter to other functions) then the compiler could replace sum with k+d and again it would make no difference. Compiler optimization is usually pretty good so I don't think you should worry about it.
Since the first is clearer you should go for that, but you can always benchmark (make sure not to use compile-time constants then, as the sums will get optimized away) and tune if that piece of could would turn out to be a bottleneck.
I am struggling a bit with the API of the Eigen Library, namely the SimplicialLLT class for Cholesky factorization of sparse matrices.
I have three matrices that I need to factor and later use to solve many equation systems (changing only the right side) - therefore I would like to factor these matrices only once and then just re-use them. Moreover, they all have the same sparcity pattern, so I would like to do the symbolic decomposition only once and then use it for the numerical decomposition for all three matrices. According to the documentation, this is exactly what the SimplicialLLT::analyzePattern and SimplicialLLT::factor methods are for. However, I can't seem to find a way to keep all three factors in the memory.
This is my code:
I have these member variables in my class I would like to fill with the factors:
Eigen::SimplicialLLT<Eigen::SparseMatrix<double>> choleskyA;
Eigen::SimplicialLLT<Eigen::SparseMatrix<double>> choleskyB;
Eigen::SimplicialLLT<Eigen::SparseMatrix<double>> choleskyC;
Then I create the three sparse matrices A, B and C and want to factor them:
choleskyA.analyzePattern(A);
choleskyA.factorize(A);
choleskyB.analyzePattern(B); // this has already been done!
choleskyB.factorize(B);
choleskyC.analyzePattern(C); // this has already been done!
choleskyC.factorize(C);
And later I can use them for solutions over and over again, changing just the b vectors of right sides:
xA = choleskyA.solve(bA);
xB = choleskyB.solve(bB);
xC = choleskyC.solve(bC);
This works (I think), but the second and third call to analyzePattern are redundant. What I would like to do is something like:
choleskyA.analyzePattern(A);
choleskyA.factorize(A);
choleskyB = choleskyA.factorize(B);
choleskyC = choleskyA.factorize(C);
But that is not an option with the current API (we use Eigen 3.2.3, but if I see correctly there is no change in this regard in 3.3.2). The problem here is that the subsequent calls to factorize on the same instance of SimplicialLLT will overwrite the previously computed factor and at the same time, I can't find a way to make a copy of it to keep. I took a look at the sources but I have to admit that didn't help much as I can't see any simple way to copy the underlying data structures. It seems to me like a rather common usage, so I feel like I am missing something obvious, please help.
What I have tried:
I tried using simply choleskyB = choleskyA hoping that the default copy constructor will get things done, but I have found out that the base classes are designed to be non-copyable.
I can get the L and U matrices (there's a getter for them) from choleskyA, make a copy of them and store only those and then basically copy-paste the content of SimplicialCholeskyBase::_solve_impl() (copy-pasted below) to write the method for solving myself using the previously stored L and U directly.
template<typename Rhs,typename Dest>
void _solve_impl(const MatrixBase<Rhs> &b, MatrixBase<Dest> &dest) const
{
eigen_assert(m_factorizationIsOk && "The decomposition is not in a valid state for solving, you must first call either compute() or symbolic()/numeric()");
eigen_assert(m_matrix.rows()==b.rows());
if(m_info!=Success)
return;
if(m_P.size()>0)
dest = m_P * b;
else
dest = b;
if(m_matrix.nonZeros()>0) // otherwise L==I
derived().matrixL().solveInPlace(dest);
if(m_diag.size()>0)
dest = m_diag.asDiagonal().inverse() * dest;
if (m_matrix.nonZeros()>0) // otherwise U==I
derived().matrixU().solveInPlace(dest);
if(m_P.size()>0)
dest = m_Pinv * dest;
}
...but that's quite an ugly solution plus I would probably screw it up since I don't have that good understanding of the process (I don't need the m_diag from the above code since I am doing LLT, right? that would be relevant only if I was using LDLT?). I hope this is not what I need to do...
A final note - adding the necessary getters/setters to the Eigen classes and compiling "my own" Eigen is not an option (well, not a good one) as this code will (hopefully) be further redistributed as open source, so it would be troublesome.
This is a quite unusual pattern. In practice the symbolic factorization is very cheap compared to the numerical factorization, so I'm not sure it's worth bothering much. The cleanest solution to address this pattern would be to let SimplicialL?LT to be copiable.
Having left Fortran for several years, now I have to pick it up and start to work with it again.
I'd like to construct a matrix with entry(i,j) in the form f(x_i,y_j), where f is a function of two variables, e.g., f(x,y)=cos(x-y). In Matlab or Python(Numpy), there are efficient ways to handle this kind of specific issue. I wonder whether there is such optimization in Fortran.
BTW, is it also true in Fortran that a vectorized operation is faster than a do/for loop (as is the case in Matlab and Numpy) ?
If you mean by vectorized the same as you mean in Matlab and Python, the short form you call on whole array then no, these forms are often slower, because they mey be harder to optimize than simple loops. What is faster is when the compiler actually uses the vector instructions of the CPU, but that is something else. And it is easier for the compiler to use them for simple loops.
Fortran has elemental functions, do concurrent, forall and where constructs, implied loops and array constructors. There is no point repeating them here, they have been described many times on this site or in tutorials.
Your example is most simply done using a loop
do j = 1, ny
do i = 1, nx
entry(i,j) = f(x(i), y(j))
end do
end do
One of the short ways, you probably meant by Python-like vectorization, would be the whole-array operations, e.g.,
A = cos(B)
C = A * B
D = f(A*B)
and similar. The function (which is called on each element of the array), must be elemental. These operations are not necessarily efficient. For example, the last call may require a temporary array to be created, which would be avoided when using a loop.
I would like to verify whether an element is present in a MATLAB matrix.
At the beginning, I implemented as follows:
if ~isempty(find(matrix(:) == element))
which is obviously slow. Thus, I changed to:
if sum(matrix(:) == element) ~= 0
but this is again slow: I am calling a lot of times the function that contains this instruction, and I lose 14 seconds each time!
Is there a way of further optimize this instruction?
Thanks.
If you just need to know if a value exists in a matrix, using the second argument of find to specify that you just want one value will be slightly faster (25-50%) and even a bit faster than using sum, at least on my machine. An example:
matrix = randi(100,1e4,1e4);
element = 50;
~isempty(find(matrix(:)==element,1))
However, in recent versions of Matlab (I'm using R2014b), nnz is finally faster for this operation, so:
matrix = randi(100,1e4,1e4);
element = 50;
nnz(matrix==element)~=0
On my machine this is about 2.8 times faster than any other approach (including using any, strangely) for the example provided. To my mind, this solution also has the benefit of being the most readable.
In my opinion, there are several things you could try to improve performance:
following your initial idea, i would go for the function any to test is any of the equality tests had a success:
if any(matrix(:) == element)
I tested this on a 1000 by 1000 matrix and it is faster than the solutions you have tested.
I do not think that the unfolding matrix(:) is penalizing since it is equivalent to a reshape and Matlab does this in a smart way where it does not actually allocate and move memory since you are not modifying the temporary object matrix(:)
If your does not change between the calls to the function or changes rarely you could simply use another vector containing all the elements of your matrix, but sorted. This way you could use a more efficient search algorithm O(log(N)) test for the presence of your element.
I personally like the ismember function for this kind of problems. It might not be the fastest but for non critical parts of the code it greatly improves readability and code maintenance (and I prefer to spend one hour coding something that will take day to run than spending one day to code something that will run in one hour (this of course depends on how often you use this program, but it is something one should never forget)
If you can have a sorted copy of the elements of your matrix, you could consider using the undocumented Matlab function ismembc but remember that inputs must be sorted non-sparse non-NaN values.
If performance really is critical you might want to write your own mex file and for this task you could even include some simple parallelization using openmp.
Hope this helps,
Adrien.
Suppose that I have these Three variables in matlab Variables
I want to extract diverse values in NewGrayLevels and sum rows of OldHistogram that are in the same rows as one diverse value is.
For example you see in NewGrayLevels that the six first rows are equal to zero. It means that 0 in the NewGrayLevels has taken its value from (0 1 2 3 4 5) of OldGrayLevels. So the corresponding rows in OldHistogram should be summed.
So 0+2+12+38+113+163=328 would be the frequency of the gray level 0 in the equalized histogram and so on.
Those who are familiar with image processing know that it's part of the histogram equalization algorithm.
Note that I don't want to use built-in function "histeq" available in image processing toolbox and I want to implement it myself.
I know how to write the algorithm with for loops. I'm seeking if there is a faster way without using for loops.
The code using for loops:
for k=0:255
Condition = NewGrayLevels==k;
ConditionMultiplied = Condition.*OldHistogram;
NewHistogram(k+1,1) = sum(ConditionMultiplied);
end
I'm afraid if this code gets slow for high resolution big images.Because the variables that I have uploaded are for a small image downloaded from the internet but my code may be used for sattellite images.
I know you say you don't want to use histeq, but it might be worth your time to look at the MATLAB source file to see how the developers wrote it and copy the parts of their code that you would like to implement. Just do edit('histeq') or edit('histeq.m'), I forget which.
Usually the MATLAB code is vectorized where possible and runs pretty quick. This could save you from having to reinvent the entire wheel, just the parts you want to change.
I can't think a way to implement this without a for loop somewhere, but one optimisation you could make would be using indexing instead of multiplication:
for k=0:255
Condition = NewGrayLevels==k; % These act as logical indices to OldHistogram
NewHistogram(k+1,1) = sum(OldHistogram(Condition)); % Removes a vector multiplication, some additions, and an index-to-double conversion
end
Edit:
On rereading your initial post, I think that the way to do this without a for loop is to use accumarray (I find this a difficult function to understand, so read the documentation and search online and on here for examples to do so):
NewHistogram = accumarray(1+NewGrayLevels,OldHistogram);
This should work so long as your maximum value in NewGrayLevels (+1 because you are starting at zero) is equal to the length of OldHistogram.
Well I understood that there's no need to write the code that #Hugh Nolan suggested. See the explanation here:
%The green lines are because after writing the code, I understood that
%there's no need to calculate the equalized histogram in
%"HistogramEqualization" function and after gaining the equalized image
%matrix you can pass it to the "ExtractHistogram" function
% (which there's no loops in it) to acquire the
%equalized histogram.
%But I didn't delete those lines of code because I had tried a lot to
%understand the algorithm and write them.
For more information and studying the code, please see my next question.