内部范围IF/其他范围的中位数

Question

我正在使用两个堆在Python中实现一种正在运行的中间算法。但是，即使我推向它们，堆也不会变大...

我怀疑这与if/else语句中的范围有关。我真的不明白如何解决此问题。

import os
import numpy
import functools
import heapq
from heapq import heapify, heappush, heappop 

@functools.total_ordering
class ReverseCompare(object):
    def __init__(self, obj):
        self.obj = obj
    def __eq__(self, other):
        return isinstance(other, ReverseCompare) and self.obj == other.obj
    def __le__(self, other):
        return isinstance(other, ReverseCompare) and self.obj >= other.obj
    def __str__(self):
        return str(self.obj)
    def __repr__(self):
        return '%s(%r)' % (self.__class__.__name__, self.obj)

curMedian = 0
leftHeap = map(ReverseCompare, [])
rightHeap = []
heapq.heapify(rightHeap)
heapq.heapify(leftHeap)
def runningMed(n):
    #importing the global variables
    global curMedian 
    global leftHeap
    global rightHeap
    #The first time
    if (curMedian == 0):
        curMedian = n
        return curMedian
    #the second +... time
    # print "debug"
    # print rightHeap
    # print leftHeap
    # heapq.heappush(leftHeap, 3)
    # heapq.heappush(rightHeap, 3)
    # print rightHeap
    print "length of heaps"
    print len(rightHeap)
    print len(leftHeap)
    if (len(rightHeap) > len(leftHeap) + 2):
        print "Debug long right heap"
        if(n >= curMedian):
            heapq.heappush(leftHeap, curMedian)
            curMedian = heapq.heappop(rightHeap)
            heappop.heappush(rightHeap, n)
        else:
            heapq.heappush(leftHeap, n)
    elif (len(leftHeap) > len(rightHeap) + 2):
        print "Debug long"
        if(n <= curMedian):
            heapq.heappush(rightHeap, curMedian)
            curMedian = heapq.heappop(leftHeap)
            heappop.heappush(leftHeap, n)
        else:
            heapq.heappush(rightHeap,n)
    else:
        print "Debug curMedian"
        print n
        print curMedian
        if (n > curMedian):
            heapq.heappush(rightHeap, n)
        else:
            heapq.heappush(leftHeap,n)
    #TReturn the median:
    if (len(leftHeap) == len(rightHeap)):
        return curMedian
    elif (len(leftHeap) > len(rightHeap)):
        return (heapq.heappop(leftHeap) + curMedian)/2
    else:
        return (heapq.heappop(rightHeap) + curMedian)/2

if __name__ == "__main__":
    #TODO: Input/output names must be changed
    inputFile = open('numbers.txt', 'r')
    outputFile = open('output.txt', 'w')
    for line in inputFile:
        num = int(line.rstrip('n'))
        med = runningMed(num)
        outputFile.write(str(med) + 'n')
    inputFile.close()
    outputFile.close()

Answer 1

与范围无关。堆不会生长，因为您在最后立即弹出了刚刚添加的元素：

    return (heapq.heappop(leftHeap) + curMedian)/2
else:
    return (heapq.heappop(rightHeap) + curMedian)/2

只需查看最大/分钟元素而不会弹出：

    return (leftHeap[0] + curMedian)/2
else:
    return (rightHeap[0] + curMedian)/2

---

我在评论中提到的我自己的版本：

from heapq import heappush, heappop
left, right = [], []
def runmed(n):
    global left, right
    if len(left) <= len(right):
        heappush(left, -n)
    else:
        heappush(right, n)
    if right and -left[0] > right[0]:
        heappush(left, -heappop(right))
        heappush(right, -heappop(left))
    if len(left) > len(right):
        return -left[0]
    return (right[0] - left[0]) / 2.0

• left是数字较小的最大一半，其中包含这些数字以获得最大 - 高位功能。
• right是数字的较大一半的一半。
• left总是与right或更大的大小。

测试代码：

import random
numbers = []
for _ in range(15):
    n = random.randrange(100)
    numbers.append(n)
    print '{:<4} is median of {}'.format(runmed(n), sorted(numbers))

输出：

38   is median of [38]
27.5 is median of [17, 38]
38   is median of [17, 38, 79]
27.5 is median of [4, 17, 38, 79]
17   is median of [4, 12, 17, 38, 79]
27.5 is median of [4, 12, 17, 38, 63, 79]
38   is median of [4, 12, 17, 38, 63, 69, 79]
35.0 is median of [4, 12, 17, 32, 38, 63, 69, 79]
38   is median of [4, 12, 17, 32, 38, 39, 63, 69, 79]
38.5 is median of [4, 12, 17, 32, 38, 39, 63, 69, 79, 82]
39   is median of [4, 12, 17, 32, 38, 39, 47, 63, 69, 79, 82]
38.5 is median of [4, 12, 17, 21, 32, 38, 39, 47, 63, 69, 79, 82]
38   is median of [4, 12, 17, 21, 25, 32, 38, 39, 47, 63, 69, 79, 82]
35.0 is median of [4, 12, 14, 17, 21, 25, 32, 38, 39, 47, 63, 69, 79, 82]
38   is median of [4, 12, 14, 17, 21, 25, 32, 38, 39, 47, 62, 63, 69, 79, 82]

Answer 2

python全局范围可能有点挑剔。如果您将其包裹在每个实例都有自己的Leftheap，Rightheap和Curmedian的班级中，生活可能会更轻松。