获取groupby中的对应值

Question

我有一个类似的数据集

Serial    A        B       
1         12               
1         31
1
1         12
1         31       203
1         10    
1         2
2         32       100     
2         32       242
2         3
3         2                
3         23       100     
3
3         23

我根据 Serial 对数据帧进行分组，并通过df['A_MAX'] = df.groupby('Serial')['A'].transform('max').values 找到每个A 列的最大值，并通过df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated(), '') 保留第一个值

Serial    A        B       A_MAX    B_corresponding
1         12               31       203
1         31
1
1         12
1         31       203
1         10    
1         2
2         32       100     32       100
2         32       242
2         3
3         2                23       100
3         23       100     
3
3         23

现在对于B_corresponding 列，我想获取A_MAX 对应的B 值。我想在A 中找到A_MAX 值，但每组有类似的最大值A 值。附加条件，例如在Serial 2 中，我还希望获得32 之间的最小B 值

Answer 1

想法是使用DataFrame.sort_values 获取每个组的最大值，然后通过DataFrame.dropna 删除缺失值并通过Serial 通过DataFrame.drop_duplicates 获取第一行。通过DataFrame.set_index 创建Series 并最后使用Series.map：

df['A_MAX'] = df.groupby('Serial')['A'].transform('max')
df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated())

s = (df.sort_values(['Serial','A'], ascending=[True, False])
       .dropna(subset=['B'])
       .drop_duplicates('Serial')
       .set_index('Serial')['B'])
df['B_corresponding'] = df['Serial'].map(s).mask(df['Serial'].duplicated())
print (df)
    Serial     A      B  A_MAX  B_corresponding
0        1  12.0    NaN   31.0            203.0
1        1  31.0    NaN    NaN              NaN
2        1   NaN    NaN    NaN              NaN
3        1  12.0    NaN    NaN              NaN
4        1  31.0  203.0    NaN              NaN
5        1  10.0    NaN    NaN              NaN
6        1   2.0    NaN    NaN              NaN
7        2  32.0  100.0   32.0            100.0
8        2  32.0  242.0    NaN              NaN
9        2   3.0    NaN    NaN              NaN
10       3   2.0    NaN   23.0            100.0
11       3  23.0  100.0    NaN              NaN
12       3   NaN    NaN    NaN              NaN
13       3  23.0    NaN    NaN              NaN

将缺失值转换为空字符串是可能的，但是得到混合值 - 数字和字符串，所以接下来的处理应该是有问题的：

df['A_MAX'] = df.groupby('Serial')['A'].transform('max')
df['A_MAX'] = df['A_MAX'].mask(df['Serial'].duplicated(), '')

s = (df.sort_values(['Serial','A'], ascending=[True, False])
       .dropna(subset=['B'])
       .drop_duplicates('Serial')
       .set_index('Serial')['B'])

df['B_corresponding'] = df['Serial'].map(s).mask(df['Serial'].duplicated(), '')
print (df)
    Serial     A      B A_MAX B_corresponding
0        1  12.0    NaN    31             203
1        1  31.0    NaN                      
2        1   NaN    NaN                      
3        1  12.0    NaN                      
4        1  31.0  203.0                      
5        1  10.0    NaN                      
6        1   2.0    NaN                      
7        2  32.0  100.0    32             100
8        2  32.0  242.0                      
9        2   3.0    NaN                      
10       3   2.0    NaN    23             100
11       3  23.0  100.0                      
12       3   NaN    NaN                      
13       3  23.0    NaN                      

Answer 2

如果您不那么倾向于只使用 pandas，您也可以使用字典来达到同样的效果。

a_to_b_mapping = df.groupby('A')['B'].min().to_dict()
series_to_a_mapping = df.groupby('Series')['A'].max().to_dict()
agg_df = {}
for series, a in series_to_a_mapping.items():
    agg_df.append((series, a, a_to_b_mapping.get(a, None)))
  
agg_df = pd.DataFrame(agg_df, columns=['Series', 'A_max', 'B_corresponding'])
agg_df.head()

    Series  A_max   B_corresponding
0   1   31.0    203.0
1   2   32.0    100.0
2   3   23.0    100.0

如果需要，您可以将其加入原始数据框并屏蔽重复项。

dft = df.join(final_df.set_index('Serial'), on='Serial', how='left')
dft['A_max'] = dft['A_max'].mask(dft['A_max'].duplicated(), '')
dft['B_corresponding'] = dft['B_corresponding'].mask(dft['B_corresponding'].duplicated(), '')
dft