Elementwise array-operations and performance

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I’ve annotated your method below with where the array allocations occur:

function myImg(pts::Integer)::Array{Float64,2}
  # variables
  # rotation of ellipse
  aEll = 35.0/180.0*pi
  axes = [1/6,1/25]     # allocates a 2-element array
  values = collect(linspace(-0.5,0.5,pts))  # allocates a `pts`-element array
  # meshes
  gridX = [i for i in values, j in values] # allocates a `pts`-by-`pts` matrix
  gridY = [j for i in values, j in values] # allocates a `pts`-by-`pts` matrix
  # generate ellipse
  # rotate by alpha
  Xr = cos(aEll).*gridX - sin(aEll).*gridY  # allocates 3 `pts`-by-`pts` matrices
  Yr = cos(aEll).*gridY + sin(aEll).*gridX # allocates 3 `pts`-by-`pts` matrices
  img = ((1/axes[1]*Xr.^2 + 1/axes[2]*Yr.^2).<=1).*( 10*pi*Yr); # allocates 5 `pts`-by-`pts` matrices
  return mod(img-pi,2*pi)+pi # allocates 3 `pts`-by-`pts` matrices
end

Now #17623 will help a little with the Xr, Yr and img lines, reducing them each to one array allocation, but @kristoffer.carlsson’s can eliminate these directly.

A few other minor points (not really performance related, but just good practice):

  • Use sind and cosd instead of sin(x/180.0*pi): these handle cases like sind(180) correctly (though it won’t make a huge difference in this case).
  • Similarly there is a mod2pi function, though in this case you could move the multiplication by pi outside, so the last few lines of the above solution would look like:
k = v <= 1 ? 10*Yr : 0.0
img[i,j] = (mod(k-1,2)+1)*pi
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