有效的循环缓冲？

Question

我想在python中创建一个有效的循环缓冲区(目标是取缓冲区中的整数值的平均值).

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(以及为什么)？

Answer 1

我会用collections.deque一个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类似的配方deque.我断言它是最有效的,完全取决于它是由一个非常熟练的工作人员在C中实现的,这种工作人员习惯于开出顶级代码.

Answer 2

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

Answer 3

Python的deque很慢.您也可以使用numpy.roll而不是 如何在numpy数组形状(n,)或(n,1)中旋转数字？

在这个基准测试中,deque是448ms.Numpy.roll是29ms http://scimusing.wordpress.com/2013/10/25/ring-buffers-in-pythonnumpy/

Answer 4

根据MoonCactus的答案,这是一个circularlist类.与他的版本的不同之处在于,这里 c[0]将始终给出最旧的附加元素,c[-1]最新附加的元素,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, 1
c.append(2); print c, c[0], c[-1]    #[1, 2]           1, 2
c.append(3); print c, c[0], c[-1]    #[1, 2, 3]        1, 3
c.append(8); print c, c[0], c[-1]    #[1, 2, 3, 8]     1, 8
c.append(10); print c, c[0], c[-1]   #[10, 2, 3, 8]    2, 10
c.append(11); print c, c[0], c[-1]   #[10, 11, 3, 8]   3, 11

类:

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.__repr__() + ' (' + str(len(self._data))+' items)'

Answer 5

好吧使用deque类,但对于问题的重新解释(平均值),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

Answer 6

尽管这里已经有很多不错的答案，但是我找不到所提及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谦虚尝试。

仅出于测试目的，该类可以在list基于缓冲区，collections.deque基于缓冲区和Numpy.roll基于缓冲区之间切换。

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)