Finding moving average from data points in Python

扶醉桌前 提交于 2019-11-26 21:39:43

Before reading this answer, bear in mind that there is another answer below, from Roman Kh, which uses numpy.cumsum and is MUCH MUCH FASTER than this one.


Best One common way to apply a moving/sliding average (or any other sliding window function) to a signal is by using numpy.convolve().

def movingaverage(interval, window_size):
    window = numpy.ones(int(window_size))/float(window_size)
    return numpy.convolve(interval, window, 'same')

Here, interval is your x array, and window_size is the number of samples to consider. The window will be centered on each sample, so it takes samples before and after the current sample in order to calculate the average. Your code would become:

plot(x,y)
xlim(0,1000)

x_av = movingaverage(interval, r)
plot(x_av, y)

xlabel("Months since Jan 1749.")
ylabel("No. of Sun spots")
show()

Hope this helps!

As numpy.convolve is pretty slow, those who need a fast performing solution might prefer an easier to understand cumsum approach. Here is the code:

cumsum_vec = numpy.cumsum(numpy.insert(data, 0, 0)) 
ma_vec = (cumsum_vec[window_width:] - cumsum_vec[:-window_width]) / window_width

where data contains your data, and ma_vec will contain moving averages of window_width length.

On average, cumsum is about 30-40 times faster than convolve.

reptilicus

A moving average is a convolution, and numpy will be faster than most pure python operations. This will give you the 10 point moving average.

import numpy as np
smoothed = np.convolve(data, np.ones(10)/10)

I would also strongly suggest using the great pandas package if you are working with timeseries data. There are some nice moving average operations built in.

ravgs = [sum(data[i:i+5])/5. for i in range(len(data)-4)]

This isn't the most efficient approach but it will give your answer and I'm unclear if your window is 5 points or 10. If its 10, replace each 5 with 10 and the 4 with 9.

There is a problem with the accepted answer. I think we need to use "valid" instead of "same" here - return numpy.convolve(interval, window, 'same') .

As an Example try out the MA of this data-set = [1,5,7,2,6,7,8,2,2,7,8,3,7,3,7,3,15,6] - the result should be [4.2,5.4,6.0,5.0,5.0,5.2,5.4,4.4,5.4,5.6,5.6,4.6,7.0,6.8], but having "same" gives us an incorrect output of [2.6,3.0,4.2,5.4,6.0,5.0,5.0,5.2,5.4,4.4,5.4,5.6,5.6, 4.6,7.0,6.8,6.2,4.8]

Rusty code to try this out -:

result=[]
dataset=[1,5,7,2,6,7,8,2,2,7,8,3,7,3,7,3,15,6]
window_size=5
for index in xrange(len(dataset)):
    if index <=len(dataset)-window_size :
        tmp=(dataset[index]+ dataset[index+1]+ dataset[index+2]+ dataset[index+3]+ dataset[index+4])/5.0
        result.append(tmp)
    else:
      pass

result==movingaverage(y, window_size) 

Try this with valid & same and see whether the math makes sense.

See also -: http://sentdex.com/sentiment-analysisbig-data-and-python-tutorials-algorithmic-trading/how-to-chart-stocks-and-forex-doing-your-own-financial-charting/calculate-simple-moving-average-sma-python/

Armanda_An

My Moving Average function, without numpy function:

from __future__ import division  # must be on first line of script

class Solution:
    def Moving_Avg(self,A):
        m = A[0]
        B = []
        B.append(m)
        for i in range(1,len(A)):
            m = (m * i + A[i])/(i+1)
            B.append(m)
        return B

I think something like:

aves = [sum(data[i:i+6]) for i in range(0, len(data), 5)]

But I always have to double check the indices are doing what I expect. The range you want is (0, 5, 10, ...) and data[0:6] will give you data[0]...data[5]

ETA: oops, and you want ave rather than sum, of course. So actually using your code and the formula:

r = 5
x = data[:,0]
y1 = data[:,1]
y2 = [ave(y1[i-r:i+r]) for i in range(r, len(y1), 2*r)]
y = [y1, y2]
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