There seems to be no function that simply calculates the moving average on numpy/scipy, leading to convoluted solutions.
My question is two-fold:
This question is related to
python
numpy
time-series
moving-average
rolling-computation
talib contains a simple moving average tool, as well as other similar averaging tools (i.e. exponential moving average). Below compares the method to some of the other solutions.
%timeit pd.Series(np.arange(100000)).rolling(3).mean()
2.53 ms ± 40.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit talib.SMA(real = np.arange(100000.), timeperiod = 3)
348 µs ± 3.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit moving_average(np.arange(100000))
638 µs ± 45.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
One caveat is that the real must have elements of dtype = float
. Otherwise the following error is raised
Exception: real is not double
moving average
iterator method
reverse the array at i, and simply take the mean from i to n.
use list comprehension to generate mini arrays on the fly.
x = np.random.randint(10, size=20)
def moving_average(arr, n):
return [ (arr[:i+1][::-1][:n]).mean() for i, ele in enumerate(arr) ]
d = 5
moving_average(x, d)
tensor convolution
moving_average = np.convolve(x, np.ones(d)/d, mode='valid')
This answer using Pandas is adapted from above, as rolling_mean
is not part of Pandas anymore
# 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)
Now, just call the function rolling
on the dataframe with a window size, which in my example below is 10 days.
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
Starting in Numpy 1.20
, the sliding_window_view
provides a way to slide/roll through windows of elements. Windows that you can then individually average.
For instance, for a 4
-element window:
from numpy.lib.stride_tricks import sliding_window_view
# values = np.array([5, 3, 8, 10, 2, 1, 5, 1, 0, 2])
np.average(sliding_window_view(values, window_shape = 4), axis=1)
# array([6.5, 5.75, 5.25, 4.5, 2.25, 1.75, 2])
Note the intermediary result of sliding_window_view
:
# values = np.array([5, 3, 8, 10, 2, 1, 5, 1, 0, 2])
sliding_window_view(values, window_shape = 4)
# array([[ 5, 3, 8, 10],
# [ 3, 8, 10, 2],
# [ 8, 10, 2, 1],
# [10, 2, 1, 5],
# [ 2, 1, 5, 1],
# [ 1, 5, 1, 0],
# [ 5, 1, 0, 2]])
By comparing the solution below with the one that uses cumsum of numpy, This one takes almost half the time. This is because it does not need to go through the entire array to do the cumsum and then do all the subtraction. Moreover, the cumsum can be "dangerous" if the array is huge and the number are huge (possible overflow). Of course, also here the danger exists but at least are summed together only the essential numbers.
def moving_average(array_numbers, n):
if n > len(array_numbers):
return []
temp_sum = sum(array_numbers[:n])
averages = [temp_sum / float(n)]
for first_index, item in enumerate(array_numbers[n:]):
temp_sum += item - array_numbers[first_index]
averages.append(temp_sum / float(n))
return averages
I use either the accepted answer's solution, slightly modified to have same length for output as input, or pandas
' version as mentioned in a comment of another answer. I summarize both here with a reproducible example for future reference:
import numpy as np
import pandas as pd
def moving_average(a, n):
ret = np.cumsum(a, dtype=float)
ret[n:] = ret[n:] - ret[:-n]
return ret / n
def moving_average_centered(a, n):
return pd.Series(a).rolling(window=n, center=True).mean().to_numpy()
A = [0, 0, 1, 2, 4, 5, 4]
print(moving_average(A, 3))
# [0. 0. 0.33333333 1. 2.33333333 3.66666667 4.33333333]
print(moving_average_centered(A, 3))
# [nan 0.33333333 1. 2.33333333 3.66666667 4.33333333 nan ]
I actually wanted a slightly different behavior than the accepted answer. I was building a moving average feature extractor for an sklearn
pipeline, so I required that the output of the moving average have the same dimension as the input. What I want is for the moving average to assume the series stays constant, ie a moving average of [1,2,3,4,5]
with window 2 would give [1.5,2.5,3.5,4.5,5.0]
.
For column vectors (my use case) we get
def moving_average_col(X, n):
z2 = np.cumsum(np.pad(X, ((n,0),(0,0)), 'constant', constant_values=0), axis=0)
z1 = np.cumsum(np.pad(X, ((0,n),(0,0)), 'constant', constant_values=X[-1]), axis=0)
return (z1-z2)[(n-1):-1]/n
And for arrays
def moving_average_array(X, n):
z2 = np.cumsum(np.pad(X, (n,0), 'constant', constant_values=0))
z1 = np.cumsum(np.pad(X, (0,n), 'constant', constant_values=X[-1]))
return (z1-z2)[(n-1):-1]/n
Of course, one doesn't have to assume constant values for the padding, but doing so should be adequate in most cases.
for i in range(len(Data)):
Data[i, 1] = Data[i-lookback:i, 0].sum() / lookback
Try this piece of code. I think it's simpler and does the job. lookback is the window of the moving average.
In the Data[i-lookback:i, 0].sum()
I have put 0
to refer to the first column of the dataset but you can put any column you like in case you have more than one column.
If you just want a straightforward non-weighted moving average, you can easily implement it with np.cumsum
, which may be is faster than FFT based methods:
EDIT Corrected an off-by-one wrong indexing spotted by Bean in the code. EDIT
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])
So I guess the answer is: it is really easy to implement, and maybe numpy is already a little bloated with specialized functionality.
Here is a fast implementation using numba (mind the types). Note it does contain nans where shifted.
import numpy as np
import numba as nb
@nb.jit(nb.float64[:](nb.float64[:],nb.int64),
fastmath=True,nopython=True)
def moving_average( array, window ):
ret = np.cumsum(array)
ret[window:] = ret[window:] - ret[:-window]
ma = ret[window - 1:] / window
n = np.empty(window-1); n.fill(np.nan)
return np.concatenate((n.ravel(), ma.ravel()))
NumPy's lack of a particular domain-specific function is perhaps due to the Core Team's discipline and fidelity to NumPy's prime directive: provide an N-dimensional array type, as well as functions for creating, and indexing those arrays. Like many foundational objectives, this one is not small, and NumPy does it brilliantly.
The (much) larger SciPy contains a much larger collection of domain-specific libraries (called subpackages by SciPy devs)--for instance, numerical optimization (optimize), signal processsing (signal), and integral calculus (integrate).
My guess is that the function you are after is in at least one of the SciPy subpackages (scipy.signal perhaps); however, i would look first in the collection of SciPy scikits, identify the relevant scikit(s) and look for the function of interest there.
Scikits are independently developed packages based on NumPy/SciPy and directed to a particular technical discipline (e.g., scikits-image, scikits-learn, etc.) Several of these were (in particular, the awesome OpenOpt for numerical optimization) were highly regarded, mature projects long before choosing to reside under the relatively new scikits rubric. The Scikits homepage liked to above lists about 30 such scikits, though at least several of those are no longer under active development.
Following this advice would lead you to scikits-timeseries; however, that package is no longer under active development; In effect, Pandas has become, AFAIK, the de facto NumPy-based time series library.
Pandas has several functions that can be used to calculate a moving average; the simplest of these is probably rolling_mean, which you use like so:
>>> # 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)
Now, just call the function rolling_mean passing in the Series object and a window size, which in my example below is 10 days.
>>> 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
verify that it worked--e.g., compared values 10 - 15 in the original series versus the new Series smoothed with rolling mean
>>> 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
Here are a variety of ways to do this, along with some benchmarks. The best methods are versions using optimized code from other libraries. The bottleneck.move_mean
method is probably best all around. The scipy.convolve
approach is also very fast, extensible, and syntactically and conceptually simple, but doesn't scale well for very large window values. The numpy.cumsum
method is good if you need a pure numpy
approach.
Note: Some of these (e.g. bottleneck.move_mean
) are not centered, and will shift your data.
import numpy as np
import scipy as sci
import scipy.signal as sig
import pandas as pd
import bottleneck as bn
import time as time
def rollavg_direct(a,n):
'Direct "for" loop'
assert n%2==1
b = a*0.0
for i in range(len(a)) :
b[i]=a[max(i-n//2,0):min(i+n//2+1,len(a))].mean()
return b
def rollavg_comprehension(a,n):
'List comprehension'
assert n%2==1
r,N = int(n/2),len(a)
return np.array([a[max(i-r,0):min(i+r+1,N)].mean() for i in range(N)])
def rollavg_convolve(a,n):
'scipy.convolve'
assert n%2==1
return sci.convolve(a,np.ones(n,dtype='float')/n, 'same')[n//2:-n//2+1]
def rollavg_convolve_edges(a,n):
'scipy.convolve, edge handling'
assert n%2==1
return sci.convolve(a,np.ones(n,dtype='float'), 'same')/sci.convolve(np.ones(len(a)),np.ones(n), 'same')
def rollavg_cumsum(a,n):
'numpy.cumsum'
assert n%2==1
cumsum_vec = np.cumsum(np.insert(a, 0, 0))
return (cumsum_vec[n:] - cumsum_vec[:-n]) / n
def rollavg_cumsum_edges(a,n):
'numpy.cumsum, edge handling'
assert n%2==1
N = len(a)
cumsum_vec = np.cumsum(np.insert(np.pad(a,(n-1,n-1),'constant'), 0, 0))
d = np.hstack((np.arange(n//2+1,n),np.ones(N-n)*n,np.arange(n,n//2,-1)))
return (cumsum_vec[n+n//2:-n//2+1] - cumsum_vec[n//2:-n-n//2]) / d
def rollavg_roll(a,n):
'Numpy array rolling'
assert n%2==1
N = len(a)
rolling_idx = np.mod((N-1)*np.arange(n)[:,None] + np.arange(N), N)
return a[rolling_idx].mean(axis=0)[n-1:]
def rollavg_roll_edges(a,n):
# see https://stackoverflow.com/questions/42101082/fast-numpy-roll
'Numpy array rolling, edge handling'
assert n%2==1
a = np.pad(a,(0,n-1-n//2), 'constant')*np.ones(n)[:,None]
m = a.shape[1]
idx = np.mod((m-1)*np.arange(n)[:,None] + np.arange(m), m) # Rolling index
out = a[np.arange(-n//2,n//2)[:,None], idx]
d = np.hstack((np.arange(1,n),np.ones(m-2*n+1+n//2)*n,np.arange(n,n//2,-1)))
return (out.sum(axis=0)/d)[n//2:]
def rollavg_pandas(a,n):
'Pandas rolling average'
return pd.DataFrame(a).rolling(n, center=True, min_periods=1).mean().to_numpy()
def rollavg_bottlneck(a,n):
'bottleneck.move_mean'
return bn.move_mean(a, window=n, min_count=1)
N = 10**6
a = np.random.rand(N)
functions = [rollavg_direct, rollavg_comprehension, rollavg_convolve,
rollavg_convolve_edges, rollavg_cumsum, rollavg_cumsum_edges,
rollavg_pandas, rollavg_bottlneck, rollavg_roll, rollavg_roll_edges]
print('Small window (n=3)')
%load_ext memory_profiler
for f in functions :
print('\n'+f.__doc__+ ' : ')
%timeit b=f(a,3)
print('\nLarge window (n=1001)')
for f in functions[0:-2] :
print('\n'+f.__doc__+ ' : ')
%timeit b=f(a,1001)
print('\nMemory\n')
print('Small window (n=3)')
N = 10**7
a = np.random.rand(N)
%load_ext memory_profiler
for f in functions[2:] :
print('\n'+f.__doc__+ ' : ')
%memit b=f(a,3)
print('\nLarge window (n=1001)')
for f in functions[2:-2] :
print('\n'+f.__doc__+ ' : ')
%memit b=f(a,1001)
Timing, Small window (n=3)
Direct "for" loop :
4.14 s ± 23.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
List comprehension :
3.96 s ± 27.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
scipy.convolve :
1.07 ms ± 26.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
scipy.convolve, edge handling :
4.68 ms ± 9.69 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
numpy.cumsum :
5.31 ms ± 5.11 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
numpy.cumsum, edge handling :
8.52 ms ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Pandas rolling average :
9.85 ms ± 9.63 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
bottleneck.move_mean :
1.3 ms ± 12.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Numpy array rolling :
31.3 ms ± 91.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Numpy array rolling, edge handling :
61.1 ms ± 55.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Timing, Large window (n=1001)
Direct "for" loop :
4.67 s ± 34 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
List comprehension :
4.46 s ± 14.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
scipy.convolve :
103 ms ± 165 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
scipy.convolve, edge handling :
272 ms ± 1.23 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
numpy.cumsum :
5.19 ms ± 12.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
numpy.cumsum, edge handling :
8.7 ms ± 11.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Pandas rolling average :
9.67 ms ± 199 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
bottleneck.move_mean :
1.31 ms ± 15.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Memory, Small window (n=3)
The memory_profiler extension is already loaded. To reload it, use:
%reload_ext memory_profiler
scipy.convolve :
peak memory: 362.66 MiB, increment: 73.61 MiB
scipy.convolve, edge handling :
peak memory: 510.24 MiB, increment: 221.19 MiB
numpy.cumsum :
peak memory: 441.81 MiB, increment: 152.76 MiB
numpy.cumsum, edge handling :
peak memory: 518.14 MiB, increment: 228.84 MiB
Pandas rolling average :
peak memory: 449.34 MiB, increment: 160.02 MiB
bottleneck.move_mean :
peak memory: 374.17 MiB, increment: 75.54 MiB
Numpy array rolling :
peak memory: 661.29 MiB, increment: 362.65 MiB
Numpy array rolling, edge handling :
peak memory: 1111.25 MiB, increment: 812.61 MiB
Memory, Large window (n=1001)
scipy.convolve :
peak memory: 370.62 MiB, increment: 71.83 MiB
scipy.convolve, edge handling :
peak memory: 521.98 MiB, increment: 223.18 MiB
numpy.cumsum :
peak memory: 451.32 MiB, increment: 152.52 MiB
numpy.cumsum, edge handling :
peak memory: 527.51 MiB, increment: 228.71 MiB
Pandas rolling average :
peak memory: 451.25 MiB, increment: 152.50 MiB
bottleneck.move_mean :
peak memory: 374.64 MiB, increment: 75.85 MiB
In case you want to take care the edge conditions carefully (compute mean only from available elements at edges), the following function will do the trick.
import numpy as np
def running_mean(x, N):
out = np.zeros_like(x, dtype=np.float64)
dim_len = x.shape[0]
for i in range(dim_len):
if N%2 == 0:
a, b = i - (N-1)//2, i + (N-1)//2 + 2
else:
a, b = i - (N-1)//2, i + (N-1)//2 + 1
#cap indices to min and max indices
a = max(0, a)
b = min(dim_len, b)
out[i] = np.mean(x[a:b])
return out
>>> running_mean(np.array([1,2,3,4]), 2)
array([1.5, 2.5, 3.5, 4. ])
>>> running_mean(np.array([1,2,3,4]), 3)
array([1.5, 2. , 3. , 3.5])
I feel this can be easily solved using bottleneck
See basic sample below:
import numpy as np
import bottleneck as bn
a = np.random.randint(4, 1000, size=(5, 7))
mm = bn.move_mean(a, window=2, min_count=1)
This gives move mean along each axis.
"mm" is the moving mean for "a".
"window" is the max number of entries to consider for moving mean.
"min_count" is min number of entries to consider for moving mean (e.g. for first element or if the array has nan values).
The good part is Bottleneck helps to deal with nan values and it's also very efficient.
Source: Stackoverflow.com