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  1. PO
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    <p>Following the other suggestions, using <code>p=np.asfortranarray(p)</code> before the timer indeed puts the performance on par with numpy when I tested it. I extended the range for the bivariate bench to <code>n_bi = np.array([2**i for i in xrange(1, 15)])</code>, so that the p matrix would be larger than my L3 cache size.</p> <p>To further optimize this, I don't think automatic compiler options will be much help, since the inner loop has a dependency. Only if you manually unroll it, does <code>ifort</code> vectorize the innermost loop. With <code>gfortran</code>, <code>-O3</code> and <code>-ffast-math</code> were needed. For matrix sizes limited by main memory bandwidth, this increase the performance benefit over numpy from a factor of 1 to 3.</p> <p><strong>Update</strong>: after applying this also to the univariate code and compiling with <code>f2py --opt='-O3 -ffast-math' -c -m polynomial polynomial.f90</code>, I get the following for the source and results for benchmark.py:</p> <pre><code>subroutine polyval(p, x, pval, nx) implicit none real*8, dimension(nx), intent(in) :: p real*8, intent(in) :: x real*8, intent(out) :: pval integer, intent(in) :: nx integer, parameter :: simd = 8 real*8 :: tmp(simd), vecx(simd), xfactor integer :: i, j, k ! precompute factors do i = 1, simd vecx(i)=x**(i-1) end do xfactor = x**simd tmp = 0.0d0 do i = 1, nx, simd do k = 1, simd tmp(k) = tmp(k)*xfactor + p(nx-(i+k-1)+1)*vecx(simd-k+1) end do end do pval = sum(tmp) end subroutine polyval subroutine polyval2(p, x, y, pval, nx, ny) implicit none real*8, dimension(nx, ny), intent(in) :: p real*8, intent(in) :: x, y real*8, intent(out) :: pval integer, intent(in) :: nx, ny integer, parameter :: simd = 8 real*8 :: tmp(simd), vecx(simd), xfactor integer :: i, j, k ! precompute factors do i = 1, simd vecx(i)=x**(i-1) end do xfactor = x**simd ! horner pval=0.0d0 do i = 1, ny tmp = 0.0d0 do j = 1, nx, simd ! inner vectorizable loop do k = 1, simd tmp(k) = tmp(k)*xfactor + p(nx-(j+k-1)+1,ny-i+1)*vecx(simd-k+1) end do end do pval = pval*y + sum(tmp) end do end subroutine polyval2 </code></pre> <p><strong>Update 2</strong>: As pointed out, this code is not correct, at least when sizes are not divisible by <code>simd</code>. It's just showing the concept of manually helping the compiler, so don't just use it like this. If the sizes are not powers of two, a small remainder loop has to take care of the dangling indices. It's not so difficult to do this, here is the correct procedure for the univariate case, should be straightforward to extend it to bivariate:</p> <pre><code>subroutine polyval(p, x, pval, nx) implicit none real*8, dimension(nx), intent(in) :: p real*8, intent(in) :: x real*8, intent(out) :: pval integer, intent(in) :: nx integer, parameter :: simd = 4 real*8 :: tmp(simd), vecx(simd), xfactor integer :: i, j, k, nr ! precompute factors do i = 1, simd vecx(i)=x**(i-1) end do xfactor = x**simd ! check remainder nr = mod(nx, simd) ! horner tmp = 0.0d0 do i = 1, nx-nr, simd do k = 1, simd tmp(k) = tmp(k)*xfactor + p(nx-(i+k-1)+1)*vecx(simd-k+1) end do end do pval = sum(tmp) ! do remainder pval = pval * x**nr do i = 1, nr pval = pval + p(i) * vecx(i) end do end subroutine polyval </code></pre> <p><img src="https://i.stack.imgur.com/QBN6R.png" alt="univariate"></p> <p><img src="https://i.stack.imgur.com/rVYkn.png" alt="bivariate"></p> <p>Also, one should be careful with very small sizes, as the time will be too small to have an accurate performance profile. Also, relative times with respect to <code>numpy</code> could be deceiving, as the absolute time with numpy could be very bad. So below are timings for the largest case:</p> <p>For univariate with nx=2<strong>20, time is 1.21 s for numpy, and 1.69e-3 s for the custom fortran version. For bivariate with nx<em>ny=2</strong>20, time is 8e-3 s for numpy, and 1.68e-3 s for the custom version. The fact that the time for both univariate and bivariate is the same when the total nx</em>ny size is the same is very important, as it supports the fact that the code is performing near the memory bandwidth limit.</p> <p><strong>Update 3</strong>: with the new python script for smaller sizes, and <code>simd=4</code> I get the following performance:</p> <p><img src="https://i.stack.imgur.com/0Zz2t.png" alt="enter image description here"></p> <p><img src="https://i.stack.imgur.com/Xdxkf.png" alt="enter image description here"></p> <p><strong>Update 4</strong>: As for correctness, the results are the same within double precision accuracy, which you can see if you run this python code for the univariate example:</p> <pre><code>import polynomial as P import numpy.polynomial.polynomial as PP import numpy as np for n in xrange(2,100): poly1n = np.random.rand(n) poly1f = np.asfortranarray(poly1n) x = 2 print "%18.14e" % P.polyval(poly1f, x) print "%18.14e" % PP.polyval(x, poly1n) print (P.polyval(poly1f, x) - PP.polyval(x, poly1n))/PP.polyval(x,poly1n), '\n' </code></pre>
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    1. POWhy is univariate Horner in Fortran faster than NumPy counterpart while bivariate Horner is not
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    1. USsteabert
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    1. COI ran your code, but for me it does not improve performance. I have applied a lot of the comments posted here. I will soon update my code and benchmark script. ps: I am limiting the size of p to max 100 (univariate), 100x100 (bivariate) and 100x100x100 (trivariate). This will be the range of my actual uses.
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      1. USAeronaelius
    2. COit only improves performance with `gfortran` if you use `f2py --opt='-O3 -ffast-math' -c -m polynomial polynomial.f95` (at least with version 4.8 which I use...). Which processor do you use and what is your memory bandwidth?
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    3. COIntel i7-2670QM, 4 cores, 8 threads, 32 KB dedicated L1 cache, 256 KB dedicated L2 cache, 6 MB shared L3 cache. 8 GB RAM. See this page for more information: http://www.cpu-world.com/CPUs/Core_i7/Intel-Core%20i7%20Mobile%20i7-2670QM.html. I have updated my benchmark script to execute each function several times in order to deal with inaccurate performance profile for a polynomial with a small number of coefficients
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