135

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)?

4

14 回答 14

251

我会collections.deque使用maxlenarg

>>> 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。

于 2010-11-11T04:29:13.860 回答
21

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)
于 2018-04-12T15:14:21.280 回答
14

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

于 2010-11-11T04:28:06.653 回答
12

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)
于 2016-11-24T11:01:53.737 回答
8

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
于 2013-09-02T08:15:34.420 回答
7

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/

于 2014-11-26T11:46:39.897 回答
4

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

于 2019-02-09T03:55:21.100 回答
3

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
于 2013-10-03T10:47:29.710 回答
3

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

于 2019-05-22T16:11:32.907 回答
2

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。

于 2014-03-14T20:59:19.133 回答
2

在进行串行编程之前,我遇到过这个问题。一年多前的那个时候,我也找不到任何有效的实现,所以我最终写了一个作为 C 扩展,它也可以在MIT 许可下的pypi 上使用。它是超级基本的,只处理 8 位有符号字符的缓冲区,但长度灵活,所以如果你需要字符以外的东西,你可以使用 Struct 或其他东西。我现在通过谷歌搜索看到这些天有几个选项,所以你可能也想看看这些。

于 2016-12-07T20:30:20.650 回答
0

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。

于 2016-05-15T22:08:21.810 回答
0

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:])  
于 2018-01-19T01:04:19.070 回答
0

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)。

于 2022-02-26T18:30:36.533 回答