MATLAB 在 NumPy/Python 中的平滑实现（n 点移动平均）

Question

Matlab 的 smooth 函数默认使用 5 点移动平均线对数据进行平滑处理。在 python 中做同样的事情的最好方法是什么？ 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 基本上与长度为 5 的滑动窗口的平均相同，只是它处理两端的 2 个元素的方式不同。根据链接的文档，这些边界情况是用这些公式计算的 -

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 来获取滑动窗口总和，然后将它们除以窗口长度，得到平均结果。然后，只需为边界元素附加特殊情况处理的值。

因此，我们将有一个处理通用窗口大小的实现，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  ))