有效地平均numpy数组重复行

Question

我有这样的 numpy 数组：

old=([[5.00000000e+00, 3.39622642e-03],
       [5.00000000e+00, 5.84905660e-04],
       [1.00000000e+01, 4.15094340e-04],
       [1.50000000e+01, 2.26415094e-03],
       [2.00000000e+01, 4.90566038e-02],
       [2.50000000e+01, 4.90566038e-01],
       [3.00000000e+01, 4.90566038e-01],
       [4.00000000e+01, 8.11320755e-01],
       [4.00000000e+01, 1.00000000e+00]])

第一列是深度，第二列是一些测量值。我需要有效地（因为它们很多）平均测量重复行（具有相同深度的行）并用新计算的平均值替换旧行，如下所示：

new=([[5.00000000e+00, 1.99056600e-03],
       [1.00000000e+01, 4.15094340e-04],
       [1.50000000e+01, 2.26415094e-03],
       [2.00000000e+01, 4.90566038e-02],
       [2.50000000e+01, 4.90566038e-01],
       [3.00000000e+01, 4.90566038e-01],
       [4.00000000e+01, 9.05660377e-01]])

Answer 1

IIUC，您可以使用pandas 来处理这些数据。

df = pd.DataFrame(old)

给予

    0       1
0   5.0     0.003396
1   5.0     0.000585
2   10.0    0.000415
3   15.0    0.002264
4   20.0    0.049057
5   25.0    0.490566
6   30.0    0.490566
7   40.0    0.811321
8   40.0    1.000000

然后只需通过将第一列和aggregating 分组到mean 来利用groupby（使用.mean() 或.agg('mean')。最后，访问您的.values 以获取np.array 与结果值

df.groupby(0).mean().reset_index().values

array([[  5.00000000e+00,   1.99056604e-03],
       [  1.00000000e+01,   4.15094340e-04],
       [  1.50000000e+01,   2.26415094e-03],
       [  2.00000000e+01,   4.90566038e-02],
       [  2.50000000e+01,   4.90566038e-01],
       [  3.00000000e+01,   4.90566038e-01],
       [  4.00000000e+01,   9.05660378e-01]])

Answer 2

这当然不是向量化的，但是可以像这样获得带有列表理解的纯 numpy 方式：

new = np.stack([np.mean(old[np.where(old[:,0] == i)], axis=0) for i in np.unique(old[:,0])])


>>> new
array([[5.00000000e+00, 1.99056604e-03],
       [1.00000000e+01, 4.15094340e-04],
       [1.50000000e+01, 2.26415094e-03],
       [2.00000000e+01, 4.90566038e-02],
       [2.50000000e+01, 4.90566038e-01],
       [3.00000000e+01, 4.90566038e-01],
       [4.00000000e+01, 9.05660378e-01]])