英文:
Groupby if at least one element is in common
问题
以下是您要翻译的代码部分：
import pandas as pd
d1 = {'id': ["car", "car", "bus", "plane", "plane", "plane"], 'value': [["ab", "b"], ["b", "ab"], ["ab", "b"], ["cd", "df"], ["d", "cd"], ["df", "df"]]}
df = pd.DataFrame(data=d1)
df
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
3 plane [cd, df]
4 plane [d, cd]
5 plane [df, df]
我会将其翻译为英文：
I have the following data frame
```python
import pandas as pd
d1 = {'id': ["car", "car", "bus", "plane", "plane", "plane"], 'value': [["ab", "b"], ["b", "ab"], ["ab", "b"], ["cd", "df"], ["d", "cd"], ["df", "df"]]}
df = pd.DataFrame(data=d1)
df
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
3 plane [cd, df]
4 plane [d, cd]
5 plane [df, df]
<details>
<summary>英文:</summary>
I have the following data frame
```python
import pandas as pd
d1 = {'id': ["car", "car", "bus", "plane", "plane", "plane"], 'value': [["ab","b"], ["b","ab"], ["ab","b"], ["cd","df"], ["d","cd"], ["df","df"]]}
df = pd.DataFrame(data=d1)
df
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
3 plane [cd, df]
4 plane [d, cd]
5 plane [df, df]
I would like to group my ids if they have atleast one element from the value column in common. The desired output would look like this:
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
id value
0 plane [cd, df]
1 plane [d, cd]
id value
0 plane [cd, df]
1 plane [df, df]
I tried using groupby, but the problem is that some ids should be included in mutliple data frames, like
plane [cd, df]
答案1
得分: 1
你可以使用集合操作：
keep = (df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
.loc[lambda s: s.str.len()>1].unique()
)
for idx in keep:
print(df.loc[idx])
输出：
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
id value
3 plane [cd, df]
4 plane [d, cd]
id value
3 plane [cd, df]
5 plane [df, df]
工作原理
首先获取每个值对应的匹配索引：
df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
value
ab (0, 1, 2)
b (0, 1, 2)
cd (3, 4)
d (4)
df (3, 5)
Name: index, dtype: object
去除重复项，仅保留具有多个成员的组：
keep = (df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
.loc[lambda s: s.str.len()>1].unique()
)
[frozenset({0, 1, 2}), frozenset({3, 4}), frozenset({3, 5})]
最后，遍历这些组。
替代语法（相同逻辑）
s = df['value'].explode()
keep = dict.fromkeys(frozenset(x) for x in s.index.groupby(s).values() if len(x)>1)
for idx in keep:
print(df.loc[idx])
英文:
You can use set operations:
keep = (df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
.loc[lambda s: s.str.len()>1].unique()
)
for idx in keep:
print(df.loc[idx])
Output:
id value
0 car [ab, b]
1 car [b, ab]
2 bus [ab, b]
id value
3 plane [cd, df]
4 plane [d, cd]
id value
3 plane [cd, df]
5 plane [df, df]
How it works
first get the matching indices per value
df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
value
ab (0, 1, 2)
b (0, 1, 2)
cd (3, 4)
d (4)
df (3, 5)
Name: index, dtype: object
Remove duplicates, keep only groups of more than 1 member:
keep = (df.explode('value').reset_index().groupby('value')['index'].agg(frozenset)
.loc[lambda s: s.str.len()>1].unique()
)
[frozenset({0, 1, 2}), frozenset({3, 4}), frozenset({3, 5})]
Finally, loop over the groups.
alternative syntax (same logic)
s = df['value'].explode()
keep = dict.fromkeys(frozenset(x) for x in s.index.groupby(s).values() if len(x)>1)
for idx in keep:
print(df.loc[idx])
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