Numpy Root-Mean-Squared (RMS) smoothing of a signal

I have a signal of electromyographical data that I am supposed (scientific papers' explicit recommendation) to smooth using RMS.

I have the following working code, producing the desired output, but it is way slower than I think it's possible.

#!/usr/bin/python
import numpy
def rms(interval, halfwindow):
    """ performs the moving-window smoothing of a signal using RMS """
    n = len(interval)
    rms_signal = numpy.zeros(n)
    for i in range(n):
        small_index = max(0, i - halfwindow)  # intended to avoid boundary effect
        big_index = min(n, i + halfwindow)    # intended to avoid boundary effect
        window_samples = interval[small_index:big_index]

        # here is the RMS of the window, being attributed to rms_signal 'i'th sample:
        rms_signal[i] = sqrt(sum([s**2 for s in window_samples])/len(window_samples))

    return rms_signal

I have seen some deque and itertools suggestions regarding optimization of moving window loops, and also convolve from numpy, but I couldn't figure it out how to accomplish what I want using them.

Also, I do not care to avoid boundary problems anymore, because I end up having large arrays and relatively small sliding windows.

Thanks for reading

It is possible to use convolution to perform the operation you are referring to. I did it a few times for processing EEG signals as well.

import numpy as np
def window_rms(a, window_size):
  a2 = np.power(a,2)
  window = np.ones(window_size)/float(window_size)
  return np.sqrt(np.convolve(a2, window, 'valid'))

Breaking it down, the np.power(a, 2) part makes a new array with the same dimension as a, but where each value is squared. np.ones(window_size)/float(window_size) produces an array or length window_size where each element is 1/window_size. So the convolution effectively produces a new array where each element i is equal to

(a[i]^2 + a[i+1]^2 + … + a[i+window_size]^2)/window_size

which is the RMS value of the array elements within the moving window. It should perform really well this way.

Note, though, that np.power(a, 2) produces a new array of same dimension. If a is really large, I mean sufficiently large that it cannot fit twice in memory, you might need a strategy where each element are modified in place. Also, the 'valid' argument specifies to discard border effects, resulting in a smaller array produced by np.convolve(). You could keep it all by specifying 'same' instead (see documentation).