8000 WIP: implement __rop__ logic for scalar operators by ewmoore · Pull Request #7459 · numpy/numpy · GitHub
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WIP: implement __rop__ logic for scalar operators #7459

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1163594
BUG: longdouble op None could cause infinite recursion
ewmoore Mar 31, 2016
f2fc804
ENH: implement __rop__ logic for mixed type scalar-scalar operators
ewmoore Apr 1, 2016
223ec73
ENH/BUG: try hard not to fall back to a ufunc call for scalar ops
ewmoore Apr 1, 2016
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ENH: implement __rop__ logic for mixed type scalar-scalar operators 
Previously it took 10x longer to do np.type1(1) op np.type2(2) when
type2 could not be safely cast to type1 than the equivalent operation
with type1 == type1 or when type2 could be safely upcast to type1. This
was due to falling back to calling the equivalent ufunc without trying
to defer the call to the scalar operator of type2 (and safely upcasting
type1)

so previously:

In [2]: a = np.float32(4)

In [3]: b = np.float64(4)

In [4]: timeit a * a
10000000 loops, best of 3: 69.1 ns per loop

In [5]: timeit b * b
10000000 loops, best of 3: 69.5 ns per loop

In [6]: timeit a * b
1000000 loops, best of 3: 1.29 µs per loop

In [7]: timeit b * a
10000000 loops, best of 3: 116 ns per loop

and with these changes:

In [2]: a = np.float32(4)

In [3]: b = np.float64(4)

In [4]: timeit a * a
10000000 loops, best of 3: 74 ns per loop

In [5]: timeit b * b
10000000 loops, best of 3: 73.7 ns per loop

In [6]: timeit a * b
10000000 loops, best of 3: 181 ns per loop

In [7]: timeit b * a
10000000 loops, best of 3: 125 ns per loop

Operations on mixed type scalars that result in a scalar of a new type
still use the ufunc fallback and hit the speed penalty e.g. F op d -> D.

In [2]: a = np.complex64(1+2j)

In [3]: b = np.double(4)

In [4]: timeit a * b
1000000 loops, best of 3: 1.22 µs per loop
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@ewmoore
ewmoore committed Apr 18, 2016
commit f2fc804fd9e271a762458af9518d095c2572d072
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