python - 对 df 中的唯一值执行 groupby 计数的有效方法

Question

下面的代码旨在返回一个 df，用于计算引用点 (mainX, mainY) 的正负点数。这是由方向决定的。它们分为两组(I，J)。这些点位于 X,Y 中，每个点都有一个相对的 Label。

所以我将这些点分成各自的组。然后，我使用查询将 df 子集为正/负 df。然后，这些 df 按时间分组并计入单独的列。然后将这些 df 连接起来。

这一切似乎效率很低。特别是如果我在 Group 中有许多唯一值。例如，我必须向前复制查询序列以返回 Group J 的计数。

有没有更有效的方法来完成预期的输出？

import pandas as pd

df = pd.DataFrame({
        'Time' : ['09:00:00.1','09:00:00.1','09:00:00.1','09:00:00.1','09:00:00.1','09:00:00.1','09:00:00.2','09:00:00.2','09:00:00.2','09:00:00.2','09:00:00.2','09:00:00.2'],
        'Group' : ['I','J','I','J','I','J','I','J','I','J','I','J'],                  
        'Label' : ['A','B','C','D','E','F','A','B','C','D','E','F'],                 
        'X' : [8,4,3,8,7,4,2,3,3,4,6,1],
        'Y' : [3,6,4,8,5,2,8,8,2,4,5,1],
        'mainX' : [5,5,5,5,5,5,5,5,5,5,5,5],
        'mainY' : [5,5,5,5,5,5,5,5,5,5,5,5],
        'Direction' : ['Left','Right','Left','Right','Left','Right','Left','Right','Left','Right','Left','Right']
    })

# Determine amount of unique groups
Groups = df['Group'].unique().tolist()

# Subset groups into separate df's
Group_I = df.loc[df['Group'] == Groups[0]]
Group_J = df.loc[df['Group'] == Groups[1]]


# Separate into positive and negative direction for each group    
GroupI_Pos = Group_I.query("(Direction == 'Right' and X > mainX) or (Direction == 'Left' and X < mainX)").copy()
GroupI_Neg = Group_I.query("(Direction == 'Right' and X < mainX) or (Direction == 'Left' and X > mainX)").copy()

# Count of items per timestamp for Group I
GroupI_Pos['GroupI_Positive_Count'] = GroupI_Pos.groupby(['Time'])['Time'].transform('count')   
GroupI_Neg['GroupI_Negative_Count'] = GroupI_Neg.groupby(['Time'])['Time'].transform('count')   

# Combine Positive/Negative dfs
df_I = pd.concat([GroupI_Pos, GroupI_Neg], sort = False).sort_values(by = 'Time')

# Forward fill Nan grouped by time
df_I = df_I.groupby(['Time']).ffill()

预期输出:

          Time Group Label  X  Y  mainX  mainY Direction  GroupI_Positive_Count  GroupI_Negative_Count  GroupJ_Positive_Count  GroupJ_Negative_Count
0   09:00:00.1     I     A  8  3      5      5      Left                      1                      2                      1                      2
1   09:00:00.1     J     B  4  6      5      5     Right                      1                      2                      1                      2
2   09:00:00.1     I     C  3  4      5      5      Left                      1                      2                      1                      2
3   09:00:00.1     J     D  8  8      5      5     Right                      1                      2                      1                      2
4   09:00:00.1     I     E  7  5      5      5      Left                      1                      2                      1                      2
5   09:00:00.1     J     F  4  2      5      5     Right                      1                      2                      1                      2
6   09:00:00.2     I     A  2  8      5      5      Left                      2                      1                      0                      3
7   09:00:00.2     J     B  3  8      5      5     Right                      2                      1                      0                      3
8   09:00:00.2     I     C  3  2      5      5      Left                      2                      1                      0                      3
9   09:00:00.2     J     D  4  4      5      5     Right                      2                      1                      0                      3
10  09:00:00.2     I     E  6  5      5      5      Left                      2                      1                      0                      3
11  09:00:00.2     J     F  1  1      5      5     Right                      2                      1                      0                      3

Answer 1

这是我的看法

s = (((df.Direction.eq('Right') & df.X.gt(df.mainX)) | 
      (df.Direction.eq('Left')  & df.X.lt(df.mainX)))
     .replace({True: 'Pos', False: 'Neg'}))

df_count = df.groupby(['Time', 'Group', s]).size().unstack([1, 2], fill_value=0)
df_count.columns = df_count.columns.map(lambda x: f'Group{x[0]}_{x[1]}')

df_final = df.merge(df_count, left_on='Time', right_index=True)

Out[521]:
          Time Group Label  X  Y  mainX  mainY Direction  GroupI_Neg  \
0   09:00:00.1     I     A  8  3      5      5      Left           2
1   09:00:00.1     J     B  4  6      5      5     Right           2
2   09:00:00.1     I     C  3  4      5      5      Left           2
3   09:00:00.1     J     D  8  8      5      5     Right           2
4   09:00:00.1     I     E  7  5      5      5      Left           2
5   09:00:00.1     J     F  4  2      5      5     Right           2
6   09:00:00.2     I     A  2  8      5      5      Left           1
7   09:00:00.2     J     B  3  8      5      5     Right           1
8   09:00:00.2     I     C  3  2      5      5      Left           1
9   09:00:00.2     J     D  4  4      5      5     Right           1
10  09:00:00.2     I     E  6  5      5      5      Left           1
11  09:00:00.2     J     F  1  1      5      5     Right           1

    GroupI_Pos  GroupJ_Neg  GroupJ_Pos
0            1           2           1
1            1           2           1
2            1           2           1
3            1           2           1
4            1           2           1
5            1           2           1
6            2           3           0
7            2           3           0
8            2           3           0
9            2           3           0
10           2           3           0
11           2           3           0