你可以使用scipy. nmage .uniform_filter1d:

import numpy as np
from scipy.ndimage import uniform_filter1d
N = 1000
x = np.random.random(100000)
y = uniform_filter1d(x, size=N)

uniform_filter1d:

给出具有相同numpy形状的输出(即点数) 允许多种方式处理边界，其中'reflect'是默认的，但在我的情况下，我更想要'nearest'

它也相当快(比np快近50倍)。卷积，比上述cumsum方法快2-5倍):

%timeit y1 = np.convolve(x, np.ones((N,))/N, mode='same')
100 loops, best of 3: 9.28 ms per loop

%timeit y2 = uniform_filter1d(x, size=N)
10000 loops, best of 3: 191 µs per loop

这里有3个函数可以让你比较不同实现的错误/速度:

from __future__ import division
import numpy as np
import scipy.ndimage as ndi
def running_mean_convolve(x, N):
    return np.convolve(x, np.ones(N) / float(N), 'valid')
def running_mean_cumsum(x, N):
    cumsum = np.cumsum(np.insert(x, 0, 0))
    return (cumsum[N:] - cumsum[:-N]) / float(N)
def running_mean_uniform_filter1d(x, N):
    return ndi.uniform_filter1d(x, N, mode='constant', origin=-(N//2))[:-(N-1)]

虽然这里有这个问题的解决方案，但请看看我的解决方案。这是非常简单和工作良好。

import numpy as np
dataset = np.asarray([1, 2, 3, 4, 5, 6, 7])
ma = list()
window = 3
for t in range(0, len(dataset)):
    if t+window <= len(dataset):
        indices = range(t, t+window)
        ma.append(np.average(np.take(dataset, indices)))
else:
    ma = np.asarray(ma)

我还没有检查这有多快，但你可以试试:

from collections import deque

cache = deque() # keep track of seen values
n = 10          # window size
A = xrange(100) # some dummy iterable
cum_sum = 0     # initialize cumulative sum

for t, val in enumerate(A, 1):
    cache.append(val)
    cum_sum += val
    if t < n:
        avg = cum_sum / float(t)
    else:                           # if window is saturated,
        cum_sum -= cache.popleft()  # subtract oldest value
        avg = cum_sum / float(n)

你可以使用scipy. nmage .uniform_filter1d:

import numpy as np
from scipy.ndimage import uniform_filter1d
N = 1000
x = np.random.random(100000)
y = uniform_filter1d(x, size=N)

uniform_filter1d:

给出具有相同numpy形状的输出(即点数) 允许多种方式处理边界，其中'reflect'是默认的，但在我的情况下，我更想要'nearest'

它也相当快(比np快近50倍)。卷积，比上述cumsum方法快2-5倍):

%timeit y1 = np.convolve(x, np.ones((N,))/N, mode='same')
100 loops, best of 3: 9.28 ms per loop

%timeit y2 = uniform_filter1d(x, size=N)
10000 loops, best of 3: 191 µs per loop

这里有3个函数可以让你比较不同实现的错误/速度:

from __future__ import division
import numpy as np
import scipy.ndimage as ndi
def running_mean_convolve(x, N):
    return np.convolve(x, np.ones(N) / float(N), 'valid')
def running_mean_cumsum(x, N):
    cumsum = np.cumsum(np.insert(x, 0, 0))
    return (cumsum[N:] - cumsum[:-N]) / float(N)
def running_mean_uniform_filter1d(x, N):
    return ndi.uniform_filter1d(x, N, mode='constant', origin=-(N//2))[:-(N-1)]

如果你必须为非常小的数组(少于200个元素)重复这样做，我发现只用线性代数就能得到最快的结果。 最慢的部分是建立你的乘法矩阵y，你只需要做一次，但之后可能会更快。

import numpy as np
import random 

N = 100      # window size
size =200     # array length

x = np.random.random(size)
y = np.eye(size, dtype=float)

# prepare matrix
for i in range(size):
  y[i,i:i+N] = 1./N
  
# calculate running mean
z = np.inner(x,y.T)[N-1:]

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