在Pandas中，如何根据其他列的公共相互关系创建唯一ID？

Question

I have a dataframe with two IDs columns. I need to set a unique common interrelated ID with te following condition: if either ID1 or ID2 has some of them in common, they must have the same common_ID (ID_3).

The dataframe looks like:

df = pd.DataFrame({'ID_1': ['111', '111', '222', '333', '333', '444', '555', '666', '666', '777'],
               'ID_2': ['AAA', 'BBB', 'AAA', 'BBB', 'CCC', 'DDD', 'EEE', 'DDD', 'FFF', 'CCC']})

The desired output should be as follow:

ID_1	ID_2	ID_3
111	AAA	1
111	BBB	1
222	AAA	1
333	BBB	1
333	CCC	1
444	DDD	2
555	EEE	3
666	DDD	2
666	FFF	2
777	CCC	1

df_output = pd.DataFrame({'ID_1': ['111', '111', '222', '333', '333', '444', '555', '666', '666', '777'],
                      'ID_2': ['AAA', 'BBB', 'AAA', 'BBB', 'CCC', 'DDD', 'EEE', 'DDD', 'FFF', 'CCC'],
                      'ID_3': ['1', '1', '1', '1', '1', '2', '3', '2', '2', '1']})

to clarify the conditions

In 1st and 2nd row ID_1 the same, so they must have the same ID_3.

The 3rd row has the same ID_2 as 1st row, so its ID_3 must be the same as 1st row = 1.

The 4th row has the same ID_2 as 2nd row, that's why it must be set the same ID_3 as 2nd = 1.

The 5th row has the same ID_1 as 4th, so ID_3 = 1.

The 6th row has a unique combination of ID_1 and ID_2 at this moment, so it's marked as ID_3 = 2.

Than 7th row = 3.

But 8th has the same ID_2 as 6th, so ID_3 = 2.

and so on

Answer 1

I think we can use networkx to solve this:

import networkx as nx

G=nx.Graph()
G.add_edges_from(df[['ID_1','ID_2']].to_numpy().tolist())
cc = list(nx.connected_components(G))
L=[dict.fromkeys(b,a) for a, b in enumerate(cc,1)]
d={k: v for d in L for k, v in d.items()}
out = df.assign(ID_3=df['ID_2'].map(d))

---

print(out)

  ID_1 ID_2  ID_3
0  111  AAA     1
1  111  BBB     1
2  222  AAA     1
3  333  BBB     1
4  333  CCC     1
5  444  DDD     2
6  555  EEE     3
7  666  DDD     2
8  666  FFF     2
9  777  CCC     1

To see connected components:

print(cc)
[{'111', '777', '222', 'AAA', '333', 'BBB', 'CCC'}, 
 {'DDD', 'FFF', '666', '444'}, {'555', 'EEE'}]