问题描述
我尝试重新实现 heapify 方法,以便使用 _siftup
和 _siftdown
更新或删除堆中的任何节点并保持 O(log(n)) 的时间复杂度。>
我做了一些努力来优化我的代码,但与 heapq.heapify
(就所用的总时间而言)相比,它们被证明更糟。所以我决定研究source code。并将复制的代码与模块代码进行比较。
# heap invariant.
def _siftdown(heap,startpos,pos):
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 newitem < parent:
heap[pos] = parent
pos = parentpos
continue
break
heap[pos] = newitem
def _siftup(heap,pos):
endpos = len(heap)
startpos = pos
newitem = heap[pos]
# Bubble up the smaller child until hitting a leaf.
childpos = 2*pos + 1 # leftmost child position
while childpos < endpos:
# Set childpos to index of smaller child.
rightpos = childpos + 1
if rightpos < endpos and not heap[childpos] < heap[rightpos]:
childpos = rightpos
# Move the smaller 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(heap,pos)
def heapify(x):
"""Transform list into a heap,in-place,in O(len(x)) time."""
n = len(x)
# Transform bottom-up. The largest index there's any point to looking at
# is the largest with a child index in-range,so must have 2*i + 1 < n,# or i < (n-1)/2. If n is even = 2*j,this is (2*j-1)/2 = j-1/2 so
# j-1 is the largest,which is n//2 - 1. If n is odd = 2*j+1,this is
# (2*j+1-1)/2 = j so j-1 is the largest,and that's again n//2-1.
for i in reversed(range(n//2)):
_siftup(x,i)
a = list(reversed(range(1000000)))
b = a.copy()
import heapq
import time
cp1 = time.time()
heapq.heapify(a)
cp2 = time.time()
heapify(b)
cp3 = time.time()
print(a == b)
print(cp3 - cp2,cp2 - cp1)
而且我发现总是 cp3 - cp2
>= cp2 - cp1
并且不一样,heapify
比 heapq.heaify
花费的时间更长,即使两者相同。
在某些情况下,heapify
用了 3 秒而 heapq.heapify
用了 0.1 秒
heapq.heapfy 模块的执行速度比相同的 heapify 更快,它们仅通过导入有所不同。
请告诉我原因,如果我犯了一些愚蠢的错误,我很抱歉。
解决方法
public class RotateObjects : MonoBehaviour
{
[SerializeField] Button rotateButton;
// Start is called before the first frame update
void Start()
{
rotateButton.onClick.RemoveAllListeners();
rotateButton.onClick.AddListener(RotateClockwise);
}
void RotateClockwise()
{
float newRotation = (360 / transform.childCount);
transform.Rotate(new Vector3(0,newRotation,0));
}
}
模块中的 heapify
实际上是一个内置函数:
heapq
>>> import heapq
>>> heapq
<module 'heapq' from 'python3.9/heapq.py'>
>>> heapq.heapify
<built-in function heapify>
说:
关于模块 help(heapq.heapify)
... 中内置函数 heapify
的帮助
所以它实际上是导入内置模块 _heapq
并因此运行 C 代码,而不是 Python。
如果进一步滚动 _heapq
代码,您将看到 this:
heapq.py
这将使用其 C 实现覆盖 # If available,use C implementation
try:
from _heapq import *
except ImportError:
pass
等函数。例如,heapify
是 here。