将唯一 ID 分配给 Pandas 数据框中两列的组合，按其顺序独立

Question

I have a dataframe like this

col1 col2
1    2
2    1
2    3
3    2
3    4
4    3

and I would like to assign to each row a unique dataset based on col1 and col2 but independently on their order

col1 col2 id
1    2    1
2    1    1
2    3    2
3    2    2
3    4    3
4    3    3

How can I do this?

Answer 1

One approach:

df["id"] = df.groupby(df[["col1", "col2"]].apply(frozenset, axis=1)).ngroup() + 1
print(df)

Output

   col1  col2  id
0     1     2   1
1     2     1   1
2     2     3   2
3     3     2   2
4     3     4   3
5     4     3   3

Alternative using np.unique + np.sort :

_, indices = np.unique(np.sort(df.values, axis=1), return_inverse=True, axis=0)
df["id"] = indices + 1
print(df)

Output

   col1  col2  id
0     1     2   1
1     2     1   1
2     2     3   2
3     3     2   2
4     3     4   3
5     4     3   3

Answer 2

You can apply it:

import pandas as pd

df = pd.DataFrame(data={"col1":[1,2,3,1,2,3], "col2":[3,2,1,3,2,1]})
df['id'] = df.apply(lambda row: min(row.col1, row.col2), axis=1)
print(df)

output:

   col1  col2  id
0     1     3   1
1     2     2   2
2     3     1   1
3     1     3   1
4     2     2   2
5     3     1   1

Answer 3

Try np.sort :

a = np.sort(df, axis=1)
df['id'] = df.groupby([a[:,0],a[:,1]]).ngroup() + 1

Output:

   col1  col2  id
0     1     2   1
1     2     1   1
2     2     3   2
3     3     2   2
4     3     4   3
5     4     3   3

Answer 4

Can also use:

df['mask'] = df.apply(lambda x:','.join(map(str, x.sort_values())), axis=1)
df['id'] = (df['mask'] != df['mask'].shift()).cumsum()
df.drop(columns=['mask'], inplace=True)

Output:

   col1  col2  id
0     1     2   1
1     2     1   1
2     2     3   2
3     3     2   2
4     3     4   3
5     4     3   3