Question

感谢您的帮助。我对熊猫还比较陌生，在搜索结果中没有观察到这种特定的查询。

我有一个熊猫数据框：

+-----+---------+----------+
| id  |  value  | match_id |
+-----+---------+----------+
| A10 | grass   |        1 |
| B45 | cow     |        3 |
| B98 | bird    |        6 |
| B17 | grass   |        1 |
| A20 | tree    |        2 |
| A87 | farmer  |        5 |
| B11 | grass   |        1 |
| A33 | chicken |        4 |
| B56 | tree    |        2 |
| A23 | farmer  |        5 |
| B65 | cow     |        3 |
+-----+---------+----------+

我需要过滤此数据框以查找包含匹配的match_id值的行，条件是id列还必须包含两个字符串A < strong> 和 B。

这是预期的输出：

+-----+-------+----------+
| id  | value | match_id |
+-----+-------+----------+
| A10 | grass |        1 |
| B17 | grass |        1 |
| A20 | tree  |        2 |
| B11 | grass |        1 |
| B56 | tree  |        2 |
+-----+-------+----------+

我如何用一行熊猫代码做到这一点？下面的可复制程序：

import pandas as pd

data_example = {'id': ['A10', 'B45', 'B98', 'B17', 'A20', 'A87', 'B11', 'A33', 'B56', 'A23', 'B65'], 
                'value': ['grass', 'cow', 'bird', 'grass', 'tree', 'farmer', 'grass', 'chicken', 'tree', 'farmer', 'cow'], 
                'match_id': [1, 3, 6, 1, 2, 5, 1, 4, 2, 5, 3]}
df_example = pd.DataFrame(data=data_example)

data_expected = {'id': ['A10', 'B17', 'A20', 'B11', 'B56'], 
                'value': ['grass', 'grass', 'tree', 'grass', 'tree'], 
                'match_id': [1, 1, 2, 1, 2]}
df_expected = pd.DataFrame(data=data_expected)

谢谢！

Answer 1

单行似乎很困难，但是您可以str.extract从id中获得两个字符串，然后groupby match_id并使用any来查看每个match_id是否至少有一行所需的字符串，然后将all与轴1结合使用，将True赋予match_id这两个字符串。然后，您可以使用刚刚创建的系列在map match_id列之后仅选择True match_id。

s = df_example['id'].str.extract('(A)|(B)').notna()\
                    .groupby(df_example['match_id']).any().all(1)
df_expected = df_example.loc[df_example['match_id'].map(s), :]

print (df_expected)
    id  value  match_id
0  A10  grass         1
3  B17  grass         1
4  A20   tree         2
6  B11  grass         1
8  B56   tree         2

Answer 2

@ Ben.T解决方案的另一种说法：

#create a helper column that combines the letters per gropu
res = (df_example
        #the id column starts with a letter
       .assign(letter = lambda x: x.id.str[0])
       .groupby('match_id')
       .letter.transform(','.join)
      )

df['grp'] = res
df

    id  value   match_id    grp
0   A10 grass   1          A,B,B
1   B45 cow     3          B,B
2   B98 bird    6           B
3   B17 grass   1           A,B,B
4   A20 tree    2           A,B
5   A87 farmer  5          A,A
6   B11 grass   1         A,B,B
7   A33 chicken 4          A
8   B56 tree    2          A,B
9   A23 farmer  5          A,A
10  B65 cow     3          B,B

#filter for grps that contain A and B, and keep only relevant columns
df.loc[df.grp.str.contains('A,B'), "id":"match_id"]

     id value   match_id
0   A10 grass   1
3   B17 grass   1
4   A20 tree    2
6   B11 grass   1
8   B56 tree    2

#or u could use a list comprehension that assures u of both A and B (not just A following B)

filtered = [True if ("A" in ent) and ("B" in ent) else False for ent in df.grp.array]
df.loc[filtered,"id":"match_id"]

     id value   match_id
0   A10 grass   1
3   B17 grass   1
4   A20 tree    2
6   B11 grass   1
8   B56 tree    2

如何通过字符串值和行中匹配的整数来过滤熊猫数据框？

2 个答案: