信号的 Numpy 均方根 (RMS) 平滑

2022-01-24 00:00:00 numpy scipy iteration smoothing moving-average

问题描述

我有一个肌电数据信号，我应该(科学论文的明确建议)使用 RMS 进行平滑处理.

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，产生所需的输出，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

我已经看到了一些关于移动窗口循环优化的 deque 和 itertools 建议，以及来自 numpy 的 convolve，但我不能弄清楚如何使用它们来完成我想要的.

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.

此外，我不再关心避免边界问题，因为我最终得到了大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.

感谢阅读

解决方案

可以使用卷积来执行您所指的操作.为了处理脑电信号，我也做了几次.

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'))

分解后，np.power(a, 2) 部分创建了一个与 a 维度相同的新数组，但每个值都是平方的.np.ones(window_size)/float(window_size) 产生一个数组或长度window_size，其中每个元素是1/window_size.所以卷积有效地产生了一个新数组，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

这是移动窗口内数组元素的 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) 会生成一个相同维度的 new 数组.如果 a 是 really 大，我的意思是足够大以至于它不能在内存中容纳两次，您可能需要一个策略来修改每个元素.此外，'valid' 参数指定丢弃边框效果，从而生成由 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 文档).

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).

相关文章