更新:已经提出了更有效的解决方案，scipy的uniform_filter1d可能是“标准”第三方库中最好的，还有一些更新的或专门的库可用。

你可以用np。卷积得到:

np.convolve(x, np.ones(N)/N, mode='valid')

解释

The running mean is a case of the mathematical operation of convolution. For the running mean, you slide a window along the input and compute the mean of the window's contents. For discrete 1D signals, convolution is the same thing, except instead of the mean you compute an arbitrary linear combination, i.e., multiply each element by a corresponding coefficient and add up the results. Those coefficients, one for each position in the window, are sometimes called the convolution kernel. The arithmetic mean of N values is (x_1 + x_2 + ... + x_N) / N, so the corresponding kernel is (1/N, 1/N, ..., 1/N), and that's exactly what we get by using np.ones(N)/N.

边缘

np的模态参数。Convolve指定如何处理边缘。我在这里选择有效模式，因为我认为这是大多数人期望的运行方式，但您可能有其他优先级。下面是一个图表，说明了模式之间的差异:

import numpy as np
import matplotlib.pyplot as plt
modes = ['full', 'same', 'valid']
for m in modes:
    plt.plot(np.convolve(np.ones(200), np.ones(50)/50, mode=m));
plt.axis([-10, 251, -.1, 1.1]);
plt.legend(modes, loc='lower center');
plt.show()

从其他答案来看，我不认为这是问题所要求的，但我需要保持一个不断增长的值列表的运行平均值。

因此，如果你想保持从某个地方(站点，测量设备等)获取的值的列表和最近n个值更新的平均值，你可以使用下面的代码，这将最大限度地减少添加新元素的工作:

class Running_Average(object):
    def __init__(self, buffer_size=10):
        """
        Create a new Running_Average object.

        This object allows the efficient calculation of the average of the last
        `buffer_size` numbers added to it.

        Examples
        --------
        >>> a = Running_Average(2)
        >>> a.add(1)
        >>> a.get()
        1.0
        >>> a.add(1)  # there are two 1 in buffer
        >>> a.get()
        1.0
        >>> a.add(2)  # there's a 1 and a 2 in the buffer
        >>> a.get()
        1.5
        >>> a.add(2)
        >>> a.get()  # now there's only two 2 in the buffer
        2.0
        """
        self._buffer_size = int(buffer_size)  # make sure it's an int
        self.reset()

    def add(self, new):
        """
        Add a new number to the buffer, or replaces the oldest one there.
        """
        new = float(new)  # make sure it's a float
        n = len(self._buffer)
        if n < self.buffer_size:  # still have to had numbers to the buffer.
            self._buffer.append(new)
            if self._average != self._average:  # ~ if isNaN().
                self._average = new  # no previous numbers, so it's new.
            else:
                self._average *= n  # so it's only the sum of numbers.
                self._average += new  # add new number.
                self._average /= (n+1)  # divide by new number of numbers.
        else:  # buffer full, replace oldest value.
            old = self._buffer[self._index]  # the previous oldest number.
            self._buffer[self._index] = new  # replace with new one.
            self._index += 1  # update the index and make sure it's...
            self._index %= self.buffer_size  # ... smaller than buffer_size.
            self._average -= old/self.buffer_size  # remove old one...
            self._average += new/self.buffer_size  # ...and add new one...
            # ... weighted by the number of elements.

    def __call__(self):
        """
        Return the moving average value, for the lazy ones who don't want
        to write .get .
        """
        return self._average

    def get(self):
        """
        Return the moving average value.
        """
        return self()

    def reset(self):
        """
        Reset the moving average.

        If for some reason you don't want to just create a new one.
        """
        self._buffer = []  # could use np.empty(self.buffer_size)...
        self._index = 0  # and use this to keep track of how many numbers.
        self._average = float('nan')  # could use np.NaN .

    def get_buffer_size(self):
        """
        Return current buffer_size.
        """
        return self._buffer_size

    def set_buffer_size(self, buffer_size):
        """
        >>> a = Running_Average(10)
        >>> for i in range(15):
        ...     a.add(i)
        ...
        >>> a()
        9.5
        >>> a._buffer  # should not access this!!
        [10.0, 11.0, 12.0, 13.0, 14.0, 5.0, 6.0, 7.0, 8.0, 9.0]

        Decreasing buffer size:
        >>> a.buffer_size = 6
        >>> a._buffer  # should not access this!!
        [9.0, 10.0, 11.0, 12.0, 13.0, 14.0]
        >>> a.buffer_size = 2
        >>> a._buffer
        [13.0, 14.0]

        Increasing buffer size:
        >>> a.buffer_size = 5
        Warning: no older data available!
        >>> a._buffer
        [13.0, 14.0]

        Keeping buffer size:
        >>> a = Running_Average(10)
        >>> for i in range(15):
        ...     a.add(i)
        ...
        >>> a()
        9.5
        >>> a._buffer  # should not access this!!
        [10.0, 11.0, 12.0, 13.0, 14.0, 5.0, 6.0, 7.0, 8.0, 9.0]
        >>> a.buffer_size = 10  # reorders buffer!
        >>> a._buffer
        [5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0]
        """
        buffer_size = int(buffer_size)
        # order the buffer so index is zero again:
        new_buffer = self._buffer[self._index:]
        new_buffer.extend(self._buffer[:self._index])
        self._index = 0
        if self._buffer_size < buffer_size:
            print('Warning: no older data available!')  # should use Warnings!
        else:
            diff = self._buffer_size - buffer_size
            print(diff)
            new_buffer = new_buffer[diff:]
        self._buffer_size = buffer_size
        self._buffer = new_buffer

    buffer_size = property(get_buffer_size, set_buffer_size)

你可以测试它，例如:

def graph_test(N=200):
    import matplotlib.pyplot as plt
    values = list(range(N))
    values_average_calculator = Running_Average(N/2)
    values_averages = []
    for value in values:
        values_average_calculator.add(value)
        values_averages.append(values_average_calculator())
    fig, ax = plt.subplots(1, 1)
    ax.plot(values, label='values')
    ax.plot(values_averages, label='averages')
    ax.grid()
    ax.set_xlim(0, N)
    ax.set_ylim(0, N)
    fig.show()

这使:

比起numpy或scipy，我建议熊猫们更快地做到这一点:

df['data'].rolling(3).mean()

这取列“数据”的3个周期的移动平均值(MA)。你也可以计算移位的版本，例如排除当前单元格的版本(向后移位一个)可以很容易地计算为:

df['data'].shift(periods=1).rolling(3).mean()

Python标准库解决方案

这个生成器函数接受一个可迭代对象和一个窗口大小为N的值，并生成窗口内当前值的平均值。它使用了deque，这是一种类似于列表的数据结构，但针对在两端进行快速修改(弹出、追加)进行了优化。

from collections import deque
from itertools import islice

def sliding_avg(iterable, N):        
    it = iter(iterable)
    window = deque(islice(it, N))        
    num_vals = len(window)

    if num_vals < N:
        msg = 'window size {} exceeds total number of values {}'
        raise ValueError(msg.format(N, num_vals))

    N = float(N) # force floating point division if using Python 2
    s = sum(window)
    
    while True:
        yield s/N
        try:
            nxt = next(it)
        except StopIteration:
            break
        s = s - window.popleft() + nxt
        window.append(nxt)
        

下面是函数的运行情况:

>>> values = range(100)
>>> N = 5
>>> window_avg = sliding_avg(values, N)
>>> 
>>> next(window_avg) # (0 + 1 + 2 + 3 + 4)/5
>>> 2.0
>>> next(window_avg) # (1 + 2 + 3 + 4 + 5)/5
>>> 3.0
>>> next(window_avg) # (2 + 3 + 4 + 5 + 6)/5
>>> 4.0

如果你选择自己生成，而不是使用现有的库，请注意浮点错误并尽量减少其影响:

class SumAccumulator:
    def __init__(self):
        self.values = [0]
        self.count = 0

    def add( self, val ):
        self.values.append( val )
        self.count = self.count + 1
        i = self.count
        while i & 0x01:
            i = i >> 1
            v0 = self.values.pop()
            v1 = self.values.pop()
            self.values.append( v0 + v1 )

    def get_total(self):
        return sum( reversed(self.values) )

    def get_size( self ):
        return self.count

如果所有的值都是大致相同的数量级，那么这将通过始终添加大致相似的数量级值来帮助保持精度。

我的解决方案是基于维基百科上的“简单移动平均”。

from numba import jit
@jit
def sma(x, N):
    s = np.zeros_like(x)
    k = 1 / N
    s[0] = x[0] * k
    for i in range(1, N + 1):
        s[i] = s[i - 1] + x[i] * k
    for i in range(N, x.shape[0]):
        s[i] = s[i - 1] + (x[i] - x[i - N]) * k
    s = s[N - 1:]
    return s

与之前建议的解决方案相比，它比scipy最快的解决方案“uniform_filter1d”快两倍，并且具有相同的错误顺序。 速度测试:

import numpy as np    
x = np.random.random(10000000)
N = 1000

from scipy.ndimage.filters import uniform_filter1d
%timeit uniform_filter1d(x, size=N)
95.7 ms ± 9.34 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit sma(x, N)
47.3 ms ± 3.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

错误的比较:

np.max(np.abs(np.convolve(x, np.ones((N,))/N, mode='valid') - uniform_filter1d(x, size=N, mode='constant', origin=-(N//2))[:-(N-1)]))
8.604228440844963e-14
np.max(np.abs(np.convolve(x, np.ones((N,))/N, mode='valid') - sma(x, N)))
1.41886502547095e-13