Issue #16098: Update heapq.nsmallest to use the same algorithm as nlargest.

This removes the dependency on bisect and it bring the pure Python code
in-sync with the C code.
This commit is contained in:
Raymond Hettinger 2013-03-05 01:36:30 -05:00
parent 31584e30ab
commit f6b26676bc
1 changed files with 59 additions and 25 deletions

View File

@ -127,8 +127,7 @@ From all times, sorting has always been a Great Art! :-)
__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
'nlargest', 'nsmallest', 'heappushpop']
from itertools import islice, repeat, count, tee, chain
import bisect
from itertools import islice, count, tee, chain
def heappush(heap, item):
"""Push item onto heap, maintaining the heap invariant."""
@ -180,6 +179,19 @@ def heapify(x):
for i in reversed(range(n//2)):
_siftup(x, i)
def _heappushpop_max(heap, item):
"""Maxheap version of a heappush followed by a heappop."""
if heap and item < heap[0]:
item, heap[0] = heap[0], item
_siftup_max(heap, 0)
return item
def _heapify_max(x):
"""Transform list into a maxheap, in-place, in O(len(x)) time."""
n = len(x)
for i in reversed(range(n//2)):
_siftup_max(x, i)
def nlargest(n, iterable):
"""Find the n largest elements in a dataset.
@ -205,30 +217,16 @@ def nsmallest(n, iterable):
"""
if n < 0:
return []
if hasattr(iterable, '__len__') and n * 10 <= len(iterable):
# For smaller values of n, the bisect method is faster than a minheap.
# It is also memory efficient, consuming only n elements of space.
it = iter(iterable)
result = sorted(islice(it, 0, n))
if not result:
return result
insort = bisect.insort
pop = result.pop
los = result[-1] # los --> Largest of the nsmallest
for elem in it:
if elem < los:
insort(result, elem)
pop()
los = result[-1]
it = iter(iterable)
result = list(islice(it, n))
if not result:
return result
# An alternative approach manifests the whole iterable in memory but
# saves comparisons by heapifying all at once. Also, saves time
# over bisect.insort() which has O(n) data movement time for every
# insertion. Finding the n smallest of an m length iterable requires
# O(m) + O(n log m) comparisons.
h = list(iterable)
heapify(h)
return list(map(heappop, repeat(h, min(n, len(h)))))
_heapify_max(result)
_heappushpop = _heappushpop_max
for elem in it:
_heappushpop(result, elem)
result.sort()
return result
# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos
# is the index of a leaf with a possibly out-of-order value. Restore the
@ -306,6 +304,42 @@ def _siftup(heap, pos):
heap[pos] = newitem
_siftdown(heap, startpos, pos)
def _siftdown_max(heap, startpos, pos):
'Maxheap variant of _siftdown'
newitem = heap[pos]
# Follow the path to the root, moving parents down until finding a place
# newitem fits.
while pos > startpos:
parentpos = (pos - 1) >> 1
parent = heap[parentpos]
if parent < newitem:
heap[pos] = parent
pos = parentpos
continue
break
heap[pos] = newitem
def _siftup_max(heap, pos):
'Minheap variant of _siftup'
endpos = len(heap)
startpos = pos
newitem = heap[pos]
# Bubble up the larger child until hitting a leaf.
childpos = 2*pos + 1 # leftmost child position
while childpos < endpos:
# Set childpos to index of larger child.
rightpos = childpos + 1
if rightpos < endpos and not heap[rightpos] < heap[childpos]:
childpos = rightpos
# Move the larger child up.
heap[pos] = heap[childpos]
pos = childpos
childpos = 2*pos + 1
# The leaf at pos is empty now. Put newitem there, and bubble it up
# to its final resting place (by sifting its parents down).
heap[pos] = newitem
_siftdown_max(heap, startpos, pos)
# If available, use C implementation
try:
from _heapq import *