Numpy Root-Mean-Squared（RMS）信号平滑

问题描述

我有一个肌电数据的信号，我认为（科学论文的明确推荐）可以平滑使用RMS。

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

我有以下工作代码，产生期望的输出，但它比我想象的要慢。

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

我见过一些 deque 和 itertools 有关移动窗口循环优化的建议，以及来自numpy的卷积，但我无法弄清楚如何实现我想要的东西他们。

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.

另外，我不想再避免边界问题了，因为我最终得到了大型数组和相对较小的滑动窗口。

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

感谢您阅读

推荐答案

可以使用卷积来执行你指的是什么操作。我也做了几次处理EEG信号。

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'))

分解， np.power（a，2） part创建一个与 a 相同维度的新数组，但每个值都是平方的。 np.ones（window_size）/ float（window_size）生成一个数组或长度 window_size 其中每个元素都是 1 / WINDOW_SIZE 。因此，卷积有效地生成一个新数组，其中每个元素 i 等于

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

这是移动窗口中数组元素的RMS值。它应该以这种方式表现得非常好。

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

但是请注意， np.power（a，2）会产生一个相同维度的 new 数组。如果 真的大，我的意思是足够大以至于它不能适合内存两次，你可能需要一个策略，其中每个元素都被修改到位。此外，'valid'参数指定丢弃边框效果，从而导致由 np.convolve（）生成的较小数组。您可以通过指定'same'来保留所有内容（请参阅）。

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).

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