python - MATLAB在NumPy/Python中的流畅实现(n点移动平均)

Question

默认情况下，Matlab 的 smooth 函数使用 5 点移动平均线来平滑数据。在 python 中执行相同操作的最佳方法是什么？例如，如果这是我的数据

0
0.823529411764706
0.852941176470588
0.705882352941177
0.705882352941177
0.676470588235294
0.676470588235294
0.500000000000000
0.558823529411765
0.647058823529412
0.705882352941177
0.705882352941177
0.617647058823529
0.705882352941177
0.735294117647059
0.735294117647059
0.588235294117647
0.588235294117647
1
0.647058823529412
0.705882352941177
0.764705882352941
0.823529411764706
0.647058823529412
0.735294117647059
0.794117647058824
0.794117647058824
0.705882352941177
0.676470588235294
0.794117647058824
0.852941176470588
0.735294117647059
0.647058823529412
0.647058823529412
0.676470588235294
0.676470588235294
0.529411764705882
0.676470588235294
0.794117647058824
0.882352941176471
0.735294117647059
0.852941176470588
0.823529411764706
0.764705882352941
0.558823529411765
0.588235294117647
0.617647058823529
0.647058823529412
0.588235294117647
0.617647058823529
0.647058823529412
0.794117647058824
0.823529411764706
0.647058823529412
0.617647058823529
0.647058823529412
0.676470588235294
0.764705882352941
0.676470588235294
0.647058823529412
0.705882352941177
0.764705882352941
0.705882352941177
0.500000000000000
0.529411764705882
0.529411764705882
0.647058823529412
0.676470588235294
0.588235294117647
0.735294117647059
0.794117647058824
0.852941176470588
0.764705882352941

平滑后的数据应该是

0
0.558823529411765
0.617647058823530
0.752941176470588
0.723529411764706
0.652941176470588
0.623529411764706
0.611764705882353
0.617647058823530
0.623529411764706
0.647058823529412
0.676470588235294
0.694117647058824
0.700000000000000
0.676470588235294
0.670588235294118
0.729411764705882
0.711764705882353
0.705882352941177
0.741176470588235
0.788235294117647
0.717647058823529
0.735294117647059
0.752941176470588
0.758823529411765
0.735294117647059
0.741176470588235
0.752941176470588
0.764705882352941
0.752941176470588
0.741176470588235
0.735294117647059
0.711764705882353
0.676470588235294
0.635294117647059
0.641176470588236
0.670588235294118
0.711764705882353
0.723529411764706
0.788235294117647
0.817647058823530
0.811764705882353
0.747058823529412
0.717647058823530
0.670588235294118
0.635294117647059
0.600000000000000
0.611764705882353
0.623529411764706
0.658823529411765
0.694117647058824
0.705882352941176
0.705882352941176
0.705882352941176
0.682352941176471
0.670588235294118
0.676470588235294
0.682352941176471
0.694117647058824
0.711764705882353
0.700000000000000
0.664705882352941
0.641176470588236
0.605882352941177
0.582352941176471
0.576470588235294
0.594117647058824
0.635294117647059
0.688235294117647
0.729411764705882
0.747058823529412
0.803921568627451
0.764705882352941

在 Matlab 中得到这个的语法是

smooth(data)

我想在 python 中做同样的事情，但我找不到任何可以做到这一点的函数。

Answer 1

MATLAB 的 smoooth func基本上与长度 5 的滑动窗口的平均相同，除了它处理两端 2 个元素的方式。根据链接文档，这些边界情况是使用这些公式计算的 -

yy = smooth(y) smooths the data in the column vector y ..
The first few elements of yy are given by

yy(1) = y(1)
yy(2) = (y(1) + y(2) + y(3))/3
yy(3) = (y(1) + y(2) + y(3) + y(4) + y(5))/5
yy(4) = (y(2) + y(3) + y(4) + y(5) + y(6))/5
...

因此，要在 NumPy/Python 上复制相同的实现，我们可以使用 NumPy's 1D convolution用于获取滑动窗口求和并将它们除以窗口长度以获得平均结果。然后，简单地附加边界元素的特殊情况处理值。

因此，我们将有一个实现来处理通用窗口大小，就像这样 -

def smooth(a,WSZ):
    # a: NumPy 1-D array containing the data to be smoothed
    # WSZ: smoothing window size needs, which must be odd number,
    # as in the original MATLAB implementation
    out0 = np.convolve(a,np.ones(WSZ,dtype=int),'valid')/WSZ    
    r = np.arange(1,WSZ-1,2)
    start = np.cumsum(a[:WSZ-1])[::2]/r
    stop = (np.cumsum(a[:-WSZ:-1])[::2]/r)[::-1]
    return np.concatenate((  start , out0, stop  ))