我的基本想法是创建一个链表,当每个新值进入时,加上新值的1 / N倍并减去第一个值的1 / N倍,然后将指针移到第一个并且自由与第一个相关的记忆。
这最终不会在Python中实现,只是为了让我的脑海中清楚地看到这个过程,我试着用Python写它,但是我的实现是有缺陷的。我需要一个双重链表吗?是否有更好的替代方法(不基于链表)?
到目前为止,这是我的尝试:
class Link:
def __init__(self,val):
self.next = None
self.value = val
class LinkedList:
def __init__(self,maxlength):
self.current_link = None
self.maxlength = maxlength
self.sum = 0.
self.average = None
self.length = 0
self._first_link = None
def add_link(self,val):
new_link = Link(val)
new_link.next = self.current_link
self.current_link = new_link
if self._first_link is None:
self._first_link = self.current_link
self.sum += val
if self.length < self.maxlength:
self.length += 1
else:
self.sum -= self._first_link.value
self._first_link = self._first_link.next # this line is flawed
self.average = self.sum/self.length
def get_first(self):
return self._first_link.value
# Main
ll = LinkedList(5)
for ii in xrange(10):
ll.add_link(ii)
print ii,ll.get_first(),ll.average
问题是_first_link被设置为没有下一个的值。也就是说,_first_link被设置为添加的第一个项目,但是它的下一个是None,所以我不知道如何将它按照我想要的方式移动1。这让我想知道是否需要一个双重链表。
我很感激任何建议。
答案 0 :(得分:1)
我认为最简单的实现是使用circular linked list(a.k.a. a ring ):
class Link(object):
def __init__(self, value=0.0):
self.next = None
self.value = value
class LinkedRing(object):
def __init__(self, length):
self.sum = 0.0
self.length = length
self.current = Link()
# Initialize all the nodes:
last = self.current
for i in xrange(length-1): # one link is already created
last.next = Link()
last = last.next
last.next = self.current # close the ring
def add_val(self, val):
self.sum -= current.value
self.sum += val
self.current.value = val
self.current = self.current.next
def average(self):
return self.sum / self.length
# Test example:
rolling_sum = LinkedRing(5)
while True:
x = float(raw_input())
rolling_sum.add_val(x)
print(">> Average: %f" % rolling_sum.average())
答案 1 :(得分:1)
您可以使用collections.deque
和数值稳定的数学来实现此目的,以维持运行平均值:
import collections
class AveragingBuffer(object):
def __init__(self, maxlen):
assert( maxlen>1)
self.q=collections.deque(maxlen=maxlen)
self.xbar=0.0
def append(self, x):
if len(self.q)==self.q.maxlen:
# remove first item, update running average
d=self.q.popleft()
self.xbar=self.xbar+(self.xbar-d)/float(len(self.q))
# append new item, update running average
self.q.append(x)
self.xbar=self.xbar+(x-self.xbar)/float(len(self.q))
if __name__=="__main__":
import scipy
ab=AveragingBuffer(10)
for i in xrange(32):
ab.append(scipy.rand())
print ab.xbar, scipy.average(ab.q), len(ab.q)
答案 2 :(得分:0)
好的,我想到了一个在O [1]时间内有效的解决方案。我仍然很好奇是否有人有基于链表的解决方案,但这个解决方案完全避免了LL:
class Recent:
def __init__(self,maxlength):
self.maxlength = maxlength
self.length = 0
self.values = [0 for ii in xrange(maxlength)]
self.index = 0
self.total = 0.
self.average = 0.
def add_val(self,val):
last = self.values[self.index%self.maxlength]
self.values[self.index%self.maxlength] = val
self.total += val
self.total -= last
if self.length < self.maxlength:
self.length += 1
self.average = self.total / self.length
self.index += 1
def print_vals(self):
print ""
for ii in xrange(self.length):
print ii,self.values[ii%self.maxlength]
print "average:",self.average
# Example to show it works
rr = Recent(5)
for ii in xrange(3):
rr.add_val(ii)
rr.print_vals()
for ii in xrange(13):
rr.add_val(ii)
rr.print_vals()