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PRF: Don't used MaskedArray in Aitoff transform. #9862

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Nov 28, 2017
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PRF: Don't used MaskedArray in Aitoff transform. #9862

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19 changes: 9 additions & 10 deletions lib/matplotlib/projections/geo.py
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Original file line number Diff line number Diff line change
Expand Up @@ -269,24 +269,23 @@ def __init__(self, resolution):
self._resolution = resolution

def transform_non_affine(self, ll):
longitude = ll[:, 0:1]
latitude = ll[:, 1:2]
longitude = ll[:, 0]
latitude = ll[:, 1]

# Pre-compute some values
half_long = longitude / 2.0
cos_latitude = np.cos(latitude)

alpha = np.arccos(cos_latitude * np.cos(half_long))
# Mask this array or we'll get divide-by-zero errors
alpha = ma.masked_where(alpha == 0.0, alpha)
# The numerators also need to be masked so that masked
# division will be invoked.
# Avoid divide-by-zero errors using same method as NumPy.
alpha[alpha == 0.0] = 1e-20
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@anntzer anntzer Nov 27, 2017 •
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np.maximum(alpha, 1e-20, out=alpha) should work better if you somehow have tiny alphas (between 0 and 1e-20) and also saves an extra allocation, I think.
(or alpha = np.maximum(alpha, 1e-20) if you don't want to obfuscate it :-))

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It's fast either way, but np.maximum takes more than twice as long, at least in my test with 10,000 pts. I think the 1e-20 is a completely arbitrary small number, and its only purpose is to prevent division by exactly zero. A tiny number still works:

In [21]: np.sin(1e-300) / 1e-300
1.0

# We want unnormalized sinc. numpy.sinc gives us normalized
sinc_alpha = ma.sin(alpha) / alpha
sinc_alpha = np.sin(alpha) / alpha

x = (cos_latitude * ma.sin(half_long)) / sinc_alpha
y = (ma.sin(latitude) / sinc_alpha)
return np.concatenate((x.filled(0), y.filled(0)), 1)
xy = np.empty_like(ll, float)
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I like column_stack, which I think has a very descriptive name... (but it's just personal preference)

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pre-allocating and then assigning to slices is faster, and still very readable.

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I think it's essentially the same speed (not that it really matters):

In [15]: %%timeit x = np.random.rand(10000); y = np.random.rand(10000)
    ...: z = np.column_stack([x, y])
    ...: 
37.8 µs ± 216 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [16]: %%timeit x = np.random.rand(10000); y = np.random.rand(10000)
    ...: z = np.empty((10000, 2))
    ...: z[:, 0] = x; z[:, 1] = y
    ...: 
36.7 µs ± 1.39 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

xy[:, 0] = (cos_latitude * np.sin(half_long)) / sinc_alpha
xy[:, 1] = np.sin(latitude) / sinc_alpha
return xy
transform_non_affine.__doc__ = Transform.transform_non_affine.__doc__

def transform_path_non_affine(self, path):
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