python - 测量点之间的总数和平均距离 - python

Question

我的目标是返回特定半径内点的总数和平均距离。使用下面，半径的中心是 2，中心由 X2, Y2 确定。

总会有一个点与X2, Y2 位于同一位置。我希望在分析中忽略这一点。

注意:我希望该函数能够及时处理多个点。

import pandas as pd
import matplotlib.pyplot as plt

df = pd.DataFrame({        
    'Time' : [1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3],                               
    'Item' : ['A','B','C','D','E','F','A','B','C','D','E','F','A','B','C','D'],       
    'x' : [4,5,8,3,6,2,6,4,3.5,2,4,6,6,2,4,4],
    'y' : [-2,0,-2,0,0,4,-1,-2,-2,4,-3,2,-2,0,-2.5,4],     
    'X2' : [4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4],
    'Y2' : [-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2,-2],                          
    })

sq_dist = (df['X2'] - df['x']) ** 2 + (df['Y2'] -  df['y']) ** 2    

# count of points within radius
count = ((sq_dist <= 2 ** 2).astype(int)
                             .groupby([df['Time']])
                             .sum()
                             .reset_index()
                             .fillna(0) 
                             )

# avg_distance between points within radius
df['dist'] = np.sqrt((df['X2'] - df['x']) ** 2 + (df['Y2'] -  df['y']) ** 2)
inside = df[sq_dist <= 2 ** 2].copy()

avg_dist = (inside.groupby(['Time'])['dist']
                              .mean()
                              .reset_index()
                              .fillna(0) 
                             )

如果我合并 count 和 avg_dist 输出应该是:

   Time  count dist
0     1  0     0.0
1     2  2     0.75
2     3  1     0.5

Answer 1

你可以试试:

# distance to reference point 
dist = np.square(df[['x_ref','y_ref']] - df[['x','y']].values).sum(1) ** 0.5


(dist[dist.le(2)&dist.gt(0)]                # filter the valid points
   .groupby(df['Time'])                     # groupby Time
   .agg(['mean', 'count'])                  # count and mean
   .reindex(df.Time.unique(), fill_value=0)
)

输出:

      mean  count
Time             
1      0.0      0
2      1.0      1