似乎没有函数可以简单地计算numpy/scipy的移动平均值,这导致了复杂的解决方案。

我的问题有两个方面:

用numpy(正确地)实现移动平均的最简单方法是什么? 既然这似乎不是小事,而且容易出错,有没有一个很好的理由不包括电池在这种情况下?


当前回答

所有的答案似乎都集中在预先计算的列表的情况下。对于实际运行的用例,数字一个接一个地进来,这里有一个简单的类,它提供了对最后N个值求平均的服务:

import numpy as np
class RunningAverage():
    def __init__(self, stack_size):
        self.stack = [0 for _ in range(stack_size)]
        self.ptr = 0
        self.full_cycle = False
    def add(self,value):
        self.stack[self.ptr] = value
        self.ptr += 1
        if self.ptr == len(self.stack):
            self.full_cycle = True
            self.ptr = 0
    def get_avg(self):
        if self.full_cycle:
            return np.mean(self.stack)
        else:
            return np.mean(self.stack[:self.ptr])

用法:

N = 50  # size of the averaging window
run_avg = RunningAverage(N)
for i in range(1000):
    value = <my computation>
    run_avg.add(value)
    if i % 20 ==0: # print once in 20 iters:
        print(f'the average value is {run_avg.get_avg()}')

其他回答

如果你只想要一个简单的非加权移动平均,你可以很容易地用np实现它。cumsum,可能比基于FFT的方法更快:

修正了Bean在代码中发现的偏离一的错误索引。编辑

def moving_average(a, n=3) :
    ret = np.cumsum(a, dtype=float)
    ret[n:] = ret[n:] - ret[:-n]
    return ret[n - 1:] / n

>>> a = np.arange(20)
>>> moving_average(a)
array([  1.,   2.,   3.,   4.,   5.,   6.,   7.,   8.,   9.,  10.,  11.,
        12.,  13.,  14.,  15.,  16.,  17.,  18.])
>>> moving_average(a, n=4)
array([  1.5,   2.5,   3.5,   4.5,   5.5,   6.5,   7.5,   8.5,   9.5,
        10.5,  11.5,  12.5,  13.5,  14.5,  15.5,  16.5,  17.5])

所以我猜答案是:它真的很容易实现,也许numpy已经有了一些专门的功能。

所有的答案似乎都集中在预先计算的列表的情况下。对于实际运行的用例,数字一个接一个地进来,这里有一个简单的类,它提供了对最后N个值求平均的服务:

import numpy as np
class RunningAverage():
    def __init__(self, stack_size):
        self.stack = [0 for _ in range(stack_size)]
        self.ptr = 0
        self.full_cycle = False
    def add(self,value):
        self.stack[self.ptr] = value
        self.ptr += 1
        if self.ptr == len(self.stack):
            self.full_cycle = True
            self.ptr = 0
    def get_avg(self):
        if self.full_cycle:
            return np.mean(self.stack)
        else:
            return np.mean(self.stack[:self.ptr])

用法:

N = 50  # size of the averaging window
run_avg = RunningAverage(N)
for i in range(1000):
    value = <my computation>
    run_avg.add(value)
    if i % 20 ==0: # print once in 20 iters:
        print(f'the average value is {run_avg.get_avg()}')

如果你已经有一个已知大小的数组

import numpy as np                                         
M=np.arange(12)
                                                               
avg=[]                                                         
i=0
while i<len(M)-2: #for n point average len(M) - (n-1)
        avg.append((M[i]+M[i+1]+M[i+2])/3) #n is denominator                       
        i+=1     
                                                                                                    
print(avg)

NumPy缺乏特定领域的函数可能是由于核心团队的纪律和对NumPy主要指令的忠实:提供n维数组类型,以及用于创建和索引这些数组的函数。像许多基本目标一样,这个目标并不小,NumPy出色地完成了它。

更大的SciPy包含更大的特定于领域的库集合(被SciPy开发人员称为子包)——例如,数值优化(optimize)、信号处理(signal)和积分(integrate)。

我的猜测是,您要查找的函数至少在SciPy子包中的一个(SciPy。也许信号);然而,我将首先在SciPy scikit集合中查找,确定相关的scikit并在其中寻找感兴趣的函数。

Scikits是基于NumPy/SciPy独立开发的包,并针对特定的技术规程(例如,Scikits -image, Scikits -learn等),其中几个(特别是用于数值优化的令人钦佩的OpenOpt)在选择位于相对较新的Scikits主题之下很久以前就得到了高度重视,成熟的项目。Scikits主页上列出了大约30个这样的Scikits,尽管其中至少有几个已经不再处于积极的开发中。

按照这个建议,你会发现scikits-timeseries;但是,该软件包已不再处于积极开发阶段;实际上,Pandas已经成为AFAIK,事实上的基于numpy的时间序列库。

Pandas有几个函数可以用来计算移动平均线;其中最简单的可能是rolling_mean,你可以这样使用:

>>> # the recommended syntax to import pandas
>>> import pandas as PD
>>> import numpy as NP

>>> # prepare some fake data:
>>> # the date-time indices:
>>> t = PD.date_range('1/1/2010', '12/31/2012', freq='D')

>>> # the data:
>>> x = NP.arange(0, t.shape[0])

>>> # combine the data & index into a Pandas 'Series' object
>>> D = PD.Series(x, t)

现在,只需调用函数rolling_mean,传入Series对象和窗口大小,在下面的例子中是10天。

>>> d_mva = PD.rolling_mean(D, 10)

>>> # d_mva is the same size as the original Series
>>> d_mva.shape
    (1096,)

>>> # though obviously the first w values are NaN where w is the window size
>>> d_mva[:3]
    2010-01-01         NaN
    2010-01-02         NaN
    2010-01-03         NaN

验证它是否有效。,将原系列中的值10 - 15与用滚动平均值平滑的新系列进行比较

>>> D[10:15]
     2010-01-11    2.041076
     2010-01-12    2.041076
     2010-01-13    2.720585
     2010-01-14    2.720585
     2010-01-15    3.656987
     Freq: D

>>> d_mva[10:20]
      2010-01-11    3.131125
      2010-01-12    3.035232
      2010-01-13    2.923144
      2010-01-14    2.811055
      2010-01-15    2.785824
      Freq: D

The function rolling_mean, along with about a dozen or so other function are informally grouped in the Pandas documentation under the rubric moving window functions; a second, related group of functions in Pandas is referred to as exponentially-weighted functions (e.g., ewma, which calculates exponentially moving weighted average). The fact that this second group is not included in the first (moving window functions) is perhaps because the exponentially-weighted transforms don't rely on a fixed-length window

这个使用Pandas的答案是从上面改编的,因为rolling_mean不再是Pandas的一部分了

# the recommended syntax to import pandas
import pandas as pd
import numpy as np

# prepare some fake data:
# the date-time indices:
t = pd.date_range('1/1/2010', '12/31/2012', freq='D')

# the data:
x = np.arange(0, t.shape[0])

# combine the data & index into a Pandas 'Series' object
D = pd.Series(x, t)

现在,只需要在窗口大小的数据框架上调用滚动函数,在下面的例子中,窗口大小是10天。

d_mva10 = D.rolling(10).mean()

# d_mva is the same size as the original Series
# though obviously the first w values are NaN where w is the window size
d_mva10[:11]

2010-01-01    NaN
2010-01-02    NaN
2010-01-03    NaN
2010-01-04    NaN
2010-01-05    NaN
2010-01-06    NaN
2010-01-07    NaN
2010-01-08    NaN
2010-01-09    NaN
2010-01-10    4.5
2010-01-11    5.5
Freq: D, dtype: float64