[Python-checkins] Add a minor `Fraction.__hash__()` optimization (GH-15313)
Tim Peters
webhook-mailer at python.org
Fri Aug 16 22:09:32 EDT 2019
https://github.com/python/cpython/commit/29bb227a0ce6d355a2b3e5d6a25872e3702ba9bb
commit: 29bb227a0ce6d355a2b3e5d6a25872e3702ba9bb
branch: master
author: Tim Peters <tim.peters at gmail.com>
committer: GitHub <noreply at github.com>
date: 2019-08-16T21:09:16-05:00
summary:
Add a minor `Fraction.__hash__()` optimization (GH-15313)
* Add a minor `Fraction.__hash__` optimization that got lost in the shuffle.
Document the optimizations.
files:
M Lib/fractions.py
diff --git a/Lib/fractions.py b/Lib/fractions.py
index c922c38e2441..2e7047a81844 100644
--- a/Lib/fractions.py
+++ b/Lib/fractions.py
@@ -564,10 +564,25 @@ def __hash__(self):
try:
dinv = pow(self._denominator, -1, _PyHASH_MODULUS)
except ValueError:
- # ValueError means there is no modular inverse
+ # ValueError means there is no modular inverse.
hash_ = _PyHASH_INF
else:
- hash_ = hash(abs(self._numerator)) * dinv % _PyHASH_MODULUS
+ # The general algorithm now specifies that the absolute value of
+ # the hash is
+ # (|N| * dinv) % P
+ # where N is self._numerator and P is _PyHASH_MODULUS. That's
+ # optimized here in two ways: first, for a non-negative int i,
+ # hash(i) == i % P, but the int hash implementation doesn't need
+ # to divide, and is faster than doing % P explicitly. So we do
+ # hash(|N| * dinv)
+ # instead. Second, N is unbounded, so its product with dinv may
+ # be arbitrarily expensive to compute. The final answer is the
+ # same if we use the bounded |N| % P instead, which can again
+ # be done with an int hash() call. If 0 <= i < P, hash(i) == i,
+ # so this nested hash() call wastes a bit of time making a
+ # redundant copy when |N| < P, but can save an arbitrarily large
+ # amount of computation for large |N|.
+ hash_ = hash(hash(abs(self._numerator)) * dinv)
result = hash_ if self._numerator >= 0 else -hash_
return -2 if result == -1 else result
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