Question

我有一个数据集，其中points的维数可变，坐标均为原始中心和相应中心：

point | c_1 | c_2 | ... | c_n | center_1 | center_2 | ... | center_n
--------------------------------------------------------------------
  p_1 | 0.1 | 0.3 | ... | 0.5 |      1.2 |      1.1 | ... |      0.7
  p_2 | 1.0 | 1.5 | ... | 1.7 |      3.1 |      2.0 | ... |      1.3
  p_3 | 0.5 | 0.8 | ... | 1.0 |      2.0 |      1.2 | ... |      3.8
  ... | ... | ... | ... | ... |      ... |      ... | ... |      ...

现在，我需要计算每个点到其中心的Euclidean距离。

例如，具有三个点的简化3D数据集将看起来像：

point | c_1 | c_2 | c_3 | center_1 | center_2 | center_3 | distance
-------------------------------------------------------------------
  p_1 | 0.0 | 0.0 | 0.0 |      1.0 |      1.0 |      1.0 |    1.732   
  p_2 | 1.0 | 1.0 | 1.0 |      3.0 |      3.0 |      3.0 |    3.464
  p_3 | 0.5 | 0.5 | 0.5 |      2.0 |      2.0 |      2.0 |    2.598

我可以在一维上执行以下操作：

import pandas as pd
import numpy as np
points = pd.DataFrame({
    "point": ("p_1", "p_2", "p_3"), 
    "c_1": (0.0, 1.0, 0.5),
    "c_2": (0.0, 1.0, 0.5),
    "c_3": (0.0, 1.0, 0.5),
    "center_1": (1.0, 3.0, 2.0),
    "center_2": (1.0, 3.0, 2.0),
    "center_3": (1.0, 3.0, 2.0)
})
points['distance'] = points.apply(lambda row:
                     np.linalg.norm(row['c_1']-row['center_1']), axis=1)

但是如何在给出范围的可变尺寸（例如10）上更好地做到这一点？

Answer 1

IIUC

from scipy.spatial import distance
a=distance.cdist(df[['c_1','c_2','c_2']].values, df[['center_1','center_2','center_3']].values)
a[np.arange(len(a)),np.arange(len(a))]
Out[249]: array([1.73205081, 3.46410162, 2.59807621])

在Python / Pandas中可变数量的列上应用函数

1 个答案: