我想在 python 中创建一个有效的循环缓冲区(目标是取缓冲区中整数值的平均值)。
这是使用列表收集值的有效方法吗?
def add_to_buffer( self, num ):
self.mylist.pop( 0 )
self.mylist.append( num )
什么会更有效(为什么)?
我想在 python 中创建一个有效的循环缓冲区(目标是取缓冲区中整数值的平均值)。
这是使用列表收集值的有效方法吗?
def add_to_buffer( self, num ):
self.mylist.pop( 0 )
self.mylist.append( num )
什么会更有效(为什么)?
我会collections.deque
使用maxlen
arg
>>> import collections
>>> d = collections.deque(maxlen=10)
>>> d
deque([], maxlen=10)
>>> for i in xrange(20):
... d.append(i)
...
>>> d
deque([10, 11, 12, 13, 14, 15, 16, 17, 18, 19], maxlen=10)
文档中有一个与您想要的相似的配方。deque
我断言它是最有效的完全取决于这样一个事实,即它是由一个非常熟练的团队用 C 实现的,他们习惯于编写一流的代码。
尽管这里已经有很多很好的答案,但我找不到所提到的选项的任何时间直接比较。因此,请在下面的比较中找到我的卑微尝试。
list
仅出于测试目的,该类可以在基于 - 的缓冲区、collections.deque
基于 - 的缓冲区和基于 - 的缓冲区之间切换Numpy.roll
。
请注意,该update
方法一次只添加一个值,以保持简单。
import numpy
import timeit
import collections
class CircularBuffer(object):
buffer_methods = ('list', 'deque', 'roll')
def __init__(self, buffer_size, buffer_method):
self.content = None
self.size = buffer_size
self.method = buffer_method
def update(self, scalar):
if self.method == self.buffer_methods[0]:
# Use list
try:
self.content.append(scalar)
self.content.pop(0)
except AttributeError:
self.content = [0.] * self.size
elif self.method == self.buffer_methods[1]:
# Use collections.deque
try:
self.content.append(scalar)
except AttributeError:
self.content = collections.deque([0.] * self.size,
maxlen=self.size)
elif self.method == self.buffer_methods[2]:
# Use Numpy.roll
try:
self.content = numpy.roll(self.content, -1)
self.content[-1] = scalar
except IndexError:
self.content = numpy.zeros(self.size, dtype=float)
# Testing and Timing
circular_buffer_size = 100
circular_buffers = [CircularBuffer(buffer_size=circular_buffer_size,
buffer_method=method)
for method in CircularBuffer.buffer_methods]
timeit_iterations = 1e4
timeit_setup = 'from __main__ import circular_buffers'
timeit_results = []
for i, cb in enumerate(circular_buffers):
# We add a convenient number of convenient values (see equality test below)
code = '[circular_buffers[{}].update(float(j)) for j in range({})]'.format(
i, circular_buffer_size)
# Testing
eval(code)
buffer_content = [item for item in cb.content]
assert buffer_content == range(circular_buffer_size)
# Timing
timeit_results.append(
timeit.timeit(code, setup=timeit_setup, number=int(timeit_iterations)))
print '{}: total {:.2f}s ({:.2f}ms per iteration)'.format(
cb.method, timeit_results[-1],
timeit_results[-1] / timeit_iterations * 1e3)
在我的系统上,这会产生:
list: total 1.06s (0.11ms per iteration)
deque: total 0.87s (0.09ms per iteration)
roll: total 6.27s (0.63ms per iteration)
从列表的头部弹出会导致整个列表被复制,因此效率低下
您应该使用固定大小的列表/数组和在添加/删除项目时在缓冲区中移动的索引
根据MoonCactus 的回答,这里有一个circularlist
类。与他的版本不同的是,这里 c[0]
总是会给出最旧的附加元素,c[-1]
最新的附加元素,c[-2]
倒数第二个......这对于应用程序来说更自然。
c = circularlist(4)
c.append(1); print(c, c[0], c[-1]) #[1] (1/4 items) 1 1
c.append(2); print(c, c[0], c[-1]) #[1, 2] (2/4 items) 1 2
c.append(3); print(c, c[0], c[-1]) #[1, 2, 3] (3/4 items) 1 3
c.append(8); print(c, c[0], c[-1]) #[1, 2, 3, 8] (4/4 items) 1 8
c.append(10); print(c, c[0], c[-1]) #[2, 3, 8, 10] (4/4 items) 2 10
c.append(11); print(c, c[0], c[-1]) #[3, 8, 10, 11] (4/4 items) 3 11
d = circularlist(4, [1, 2, 3, 4, 5]) #[2, 3, 4, 5]
班级:
class circularlist(object):
def __init__(self, size, data = []):
"""Initialization"""
self.index = 0
self.size = size
self._data = list(data)[-size:]
def append(self, value):
"""Append an element"""
if len(self._data) == self.size:
self._data[self.index] = value
else:
self._data.append(value)
self.index = (self.index + 1) % self.size
def __getitem__(self, key):
"""Get element by index, relative to the current index"""
if len(self._data) == self.size:
return(self._data[(key + self.index) % self.size])
else:
return(self._data[key])
def __repr__(self):
"""Return string representation"""
return (self._data[self.index:] + self._data[:self.index]).__repr__() + ' (' + str(len(self._data))+'/{} items)'.format(self.size)
可以使用 deque 类,但是对于问题的要求(平均),这是我的解决方案:
>>> from collections import deque
>>> class CircularBuffer(deque):
... def __init__(self, size=0):
... super(CircularBuffer, self).__init__(maxlen=size)
... @property
... def average(self): # TODO: Make type check for integer or floats
... return sum(self)/len(self)
...
>>>
>>> cb = CircularBuffer(size=10)
>>> for i in range(20):
... cb.append(i)
... print "@%s, Average: %s" % (cb, cb.average)
...
@deque([0], maxlen=10), Average: 0
@deque([0, 1], maxlen=10), Average: 0
@deque([0, 1, 2], maxlen=10), Average: 1
@deque([0, 1, 2, 3], maxlen=10), Average: 1
@deque([0, 1, 2, 3, 4], maxlen=10), Average: 2
@deque([0, 1, 2, 3, 4, 5], maxlen=10), Average: 2
@deque([0, 1, 2, 3, 4, 5, 6], maxlen=10), Average: 3
@deque([0, 1, 2, 3, 4, 5, 6, 7], maxlen=10), Average: 3
@deque([0, 1, 2, 3, 4, 5, 6, 7, 8], maxlen=10), Average: 4
@deque([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], maxlen=10), Average: 4
@deque([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], maxlen=10), Average: 5
@deque([2, 3, 4, 5, 6, 7, 8, 9, 10, 11], maxlen=10), Average: 6
@deque([3, 4, 5, 6, 7, 8, 9, 10, 11, 12], maxlen=10), Average: 7
@deque([4, 5, 6, 7, 8, 9, 10, 11, 12, 13], maxlen=10), Average: 8
@deque([5, 6, 7, 8, 9, 10, 11, 12, 13, 14], maxlen=10), Average: 9
@deque([6, 7, 8, 9, 10, 11, 12, 13, 14, 15], maxlen=10), Average: 10
@deque([7, 8, 9, 10, 11, 12, 13, 14, 15, 16], maxlen=10), Average: 11
@deque([8, 9, 10, 11, 12, 13, 14, 15, 16, 17], maxlen=10), Average: 12
@deque([9, 10, 11, 12, 13, 14, 15, 16, 17, 18], maxlen=10), Average: 13
@deque([10, 11, 12, 13, 14, 15, 16, 17, 18, 19], maxlen=10), Average: 14
Python 的双端队列很慢。您也可以改用 numpy.roll 如何旋转形状为 (n,) 或 (n,1) 的 numpy 数组中的数字?
在这个基准测试中,双端队列是 448 毫秒。Numpy.roll 为 29 毫秒 http://scimusing.wordpress.com/2013/10/25/ring-buffers-in-pythonnumpy/
Python Cookbook 中的解决方案怎么样,包括当环形缓冲区实例变满时对其进行重新分类?
class RingBuffer:
""" class that implements a not-yet-full buffer """
def __init__(self,size_max):
self.max = size_max
self.data = []
class __Full:
""" class that implements a full buffer """
def append(self, x):
""" Append an element overwriting the oldest one. """
self.data[self.cur] = x
self.cur = (self.cur+1) % self.max
def get(self):
""" return list of elements in correct order """
return self.data[self.cur:]+self.data[:self.cur]
def append(self,x):
"""append an element at the end of the buffer"""
self.data.append(x)
if len(self.data) == self.max:
self.cur = 0
# Permanently change self's class from non-full to full
self.__class__ = self.__Full
def get(self):
""" Return a list of elements from the oldest to the newest. """
return self.data
# sample usage
if __name__=='__main__':
x=RingBuffer(5)
x.append(1); x.append(2); x.append(3); x.append(4)
print(x.__class__, x.get())
x.append(5)
print(x.__class__, x.get())
x.append(6)
print(x.data, x.get())
x.append(7); x.append(8); x.append(9); x.append(10)
print(x.data, x.get())
实现中值得注意的设计选择是,由于这些对象在其生命周期中的某个时刻经历了不可逆的状态转换——从非完整缓冲区到完整缓冲区(以及在那一点的行为变化)——我通过更改
self.__class__
. 这甚至在 Python 2.2 中也有效,只要两个类具有相同的插槽(例如,它适用于两个经典类,例如 RingBuffer 和__Full
本秘籍)。在许多语言中更改实例的类可能很奇怪,但它是一种 Python 风格的替代方案,可以替代其他表示偶尔、大规模、不可逆和离散的状态变化的方式,这些变化会极大地影响行为,如本节所示。好在 Python 支持所有类型的类。
学分:塞巴斯蒂安·凯姆
您还可以看到这个相当古老的Python 配方。
这是我自己的带有 NumPy 数组的版本:
#!/usr/bin/env python
import numpy as np
class RingBuffer(object):
def __init__(self, size_max, default_value=0.0, dtype=float):
"""initialization"""
self.size_max = size_max
self._data = np.empty(size_max, dtype=dtype)
self._data.fill(default_value)
self.size = 0
def append(self, value):
"""append an element"""
self._data = np.roll(self._data, 1)
self._data[0] = value
self.size += 1
if self.size == self.size_max:
self.__class__ = RingBufferFull
def get_all(self):
"""return a list of elements from the oldest to the newest"""
return(self._data)
def get_partial(self):
return(self.get_all()[0:self.size])
def __getitem__(self, key):
"""get element"""
return(self._data[key])
def __repr__(self):
"""return string representation"""
s = self._data.__repr__()
s = s + '\t' + str(self.size)
s = s + '\t' + self.get_all()[::-1].__repr__()
s = s + '\t' + self.get_partial()[::-1].__repr__()
return(s)
class RingBufferFull(RingBuffer):
def append(self, value):
"""append an element when buffer is full"""
self._data = np.roll(self._data, 1)
self._data[0] = value
来自 Github:
class CircularBuffer:
def __init__(self, size):
"""Store buffer in given storage."""
self.buffer = [None]*size
self.low = 0
self.high = 0
self.size = size
self.count = 0
def isEmpty(self):
"""Determines if buffer is empty."""
return self.count == 0
def isFull(self):
"""Determines if buffer is full."""
return self.count == self.size
def __len__(self):
"""Returns number of elements in buffer."""
return self.count
def add(self, value):
"""Adds value to buffer, overwrite as needed."""
if self.isFull():
self.low = (self.low+1) % self.size
else:
self.count += 1
self.buffer[self.high] = value
self.high = (self.high + 1) % self.size
def remove(self):
"""Removes oldest value from non-empty buffer."""
if self.count == 0:
raise Exception ("Circular Buffer is empty");
value = self.buffer[self.low]
self.low = (self.low + 1) % self.size
self.count -= 1
return value
def __iter__(self):
"""Return elements in the circular buffer in order using iterator."""
idx = self.low
num = self.count
while num > 0:
yield self.buffer[idx]
idx = (idx + 1) % self.size
num -= 1
def __repr__(self):
"""String representation of circular buffer."""
if self.isEmpty():
return 'cb:[]'
return 'cb:[' + ','.join(map(str,self)) + ']'
https://github.com/heineman/python-data-structures/blob/master/2.%20Ubiquitous%20Lists/circBuffer.py
这个不需要任何库。它增长一个列表,然后按索引循环。
占用空间非常小(没有库),它的运行速度至少是 dequeue 的两倍。这确实有助于计算移动平均值,但请注意,这些项目不会像上面那样按年龄排序。
class CircularBuffer(object):
def __init__(self, size):
"""initialization"""
self.index= 0
self.size= size
self._data = []
def record(self, value):
"""append an element"""
if len(self._data) == self.size:
self._data[self.index]= value
else:
self._data.append(value)
self.index= (self.index + 1) % self.size
def __getitem__(self, key):
"""get element by index like a regular array"""
return(self._data[key])
def __repr__(self):
"""return string representation"""
return self._data.__repr__() + ' (' + str(len(self._data))+' items)'
def get_all(self):
"""return a list of all the elements"""
return(self._data)
要获得平均值,例如:
q= CircularBuffer(1000000);
for i in range(40000):
q.record(i);
print "capacity=", q.size
print "stored=", len(q.get_all())
print "average=", sum(q.get_all()) / len(q.get_all())
结果是:
capacity= 1000000
stored= 40000
average= 19999
real 0m0.024s
user 0m0.020s
sys 0m0.000s
这大约是出列时间的 1/3。
最初的问题是:“高效的”循环缓冲区。根据所要求的效率,来自 aaronasterling 的答案似乎是绝对正确的。使用用 Python 编写的专用类并将时间处理与 collections.deque 进行比较显示,使用 deque 可实现 5.2 倍的加速!这是一个非常简单的代码来测试这个:
class cb:
def __init__(self, size):
self.b = [0]*size
self.i = 0
self.sz = size
def append(self, v):
self.b[self.i] = v
self.i = (self.i + 1) % self.sz
b = cb(1000)
for i in range(10000):
b.append(i)
# called 200 times, this lasts 1.097 second on my laptop
from collections import deque
b = deque( [], 1000 )
for i in range(10000):
b.append(i)
# called 200 times, this lasts 0.211 second on my laptop
要将双端队列转换为列表,只需使用:
my_list = [v for v in my_deque]
然后,您将获得对双端队列项目的 O(1) 随机访问。当然,这只有在设置一次后需要对双端队列进行多次随机访问时才有价值。
这将相同的原则应用于一些旨在保存最新文本消息的缓冲区。
import time
import datetime
import sys, getopt
class textbffr(object):
def __init__(self, size_max):
#initialization
self.posn_max = size_max-1
self._data = [""]*(size_max)
self.posn = self.posn_max
def append(self, value):
#append an element
if self.posn == self.posn_max:
self.posn = 0
self._data[self.posn] = value
else:
self.posn += 1
self._data[self.posn] = value
def __getitem__(self, key):
#return stored element
if (key + self.posn+1) > self.posn_max:
return(self._data[key - (self.posn_max-self.posn)])
else:
return(self._data[key + self.posn+1])
def print_bffr(bffr,bffer_max):
for ind in range(0,bffer_max):
stored = bffr[ind]
if stored != "":
print(stored)
print ( '\n' )
def make_time_text(time_value):
return(str(time_value.month).zfill(2) + str(time_value.day).zfill(2)
+ str(time_value.hour).zfill(2) + str(time_value.minute).zfill(2)
+ str(time_value.second).zfill(2))
def main(argv):
#Set things up
starttime = datetime.datetime.now()
log_max = 5
status_max = 7
log_bffr = textbffr(log_max)
status_bffr = textbffr(status_max)
scan_count = 1
#Main Loop
# every 10 secounds write a line with the time and the scan count.
while True:
time_text = make_time_text(datetime.datetime.now())
#create next messages and store in buffers
status_bffr.append(str(scan_count).zfill(6) + " : Status is just fine at : " + time_text)
log_bffr.append(str(scan_count).zfill(6) + " : " + time_text + " : Logging Text ")
#print whole buffers so far
print_bffr(log_bffr,log_max)
print_bffr(status_bffr,status_max)
time.sleep(2)
scan_count += 1
if __name__ == '__main__':
main(sys.argv[1:])
我在这里没有得到答案。显然,如果您在 NumPy 中工作,您通常希望对数组或 ndarray 进行子类化,这样(至少在循环数组已满时)您仍然可以在循环数组上使用 NumPy 数组算术运算。唯一需要注意的是,对于跨越多个组件的操作(例如移动平均线),您的窗口不会大于缓冲区中累积的值。
此外,正如所有评论者所提到的,不要使用滚动,因为这违背了效率的目的。如果您需要一个不断增长的数组,只需在每次需要调整大小时将其大小加倍(这与循环数组实现不同)。