问检查python/大熊猫中列之间的关系类型？(一对一，一对多，或多对多)

EN

df = pd.DataFrame({
'Column1': [1, 2, 3, 4, 5, 6, 7, 8, 9],
'Column2': [4, 3, 6, 8, 3, 4, 1, 4, 3],
'Column3': [7, 3, 3, 1, 2, 2, 3, 2, 7],
'Column4': [9, 8, 7, 6, 5, 4, 3, 2, 1],
'Column5': [1, 1, 1, 1, 1, 1, 1, 1, 1]})

def get_relation(df, col1, col2):        
    first_max = df[[col1, col2]].groupby(col1).count().max()[0]
    second_max = df[[col1, col2]].groupby(col2).count().max()[0]
    if first_max==1:
        if second_max==1:
            return 'one-to-one'
        else:
            return 'one-to-many'
    else:
        if second_max==1:
            return 'many-to-one'
        else:
            return 'many-to-many'

from itertools import product
for col_i, col_j in product(df.columns, df.columns):
    if col_i == col_j:
        continue
    print(col_i, col_j, get_relation(df, col_i, col_j))

Column1 Column2 one-to-many
Column1 Column3 one-to-many
Column1 Column4 one-to-one
Column1 Column5 one-to-many
Column2 Column1 many-to-one
Column2 Column3 many-to-many
Column2 Column4 many-to-one
Column2 Column5 many-to-many
Column3 Column1 many-to-one
Column3 Column2 many-to-many
Column3 Column4 many-to-one
Column3 Column5 many-to-many
Column4 Column1 one-to-one
Column4 Column2 one-to-many
Column4 Column3 one-to-many
Column4 Column5 one-to-many
Column5 Column1 many-to-one
Column5 Column2 many-to-many
Column5 Column3 many-to-many
Column5 Column4 many-to-one

a = df.nunique()
is9, is1 = a==9, a==1
one_one = is9[:, None] & is9
one_many = is1[:, None]
many_one = is1[None, :]
many_many = (~is9[:,None]) & (~is9)

pd.DataFrame(np.select([one_one, one_many, many_one],
                       ['one-to-one', 'one-to-many', 'many-to-one'],
                       'many-to-many'),
             df.columns, df.columns)

              Column1       Column2       Column3       Column4      Column5
Column1    one-to-one  many-to-many  many-to-many    one-to-one  many-to-one
Column2  many-to-many  many-to-many  many-to-many  many-to-many  many-to-one
Column3  many-to-many  many-to-many  many-to-many  many-to-many  many-to-one
Column4    one-to-one  many-to-many  many-to-many    one-to-one  many-to-one
Column5   one-to-many   one-to-many   one-to-many   one-to-many  one-to-many

from itertools import product

def check_cardinality(df):

    combinations_lst = list(product(df.columns, df.columns))
    relations = ['one_to_one', 'one_to_many', 'many_to_one']

    output = []
    for col1, col2 in combinations_lst:
        for relation in relations:
            try:
                pd.merge(df[[col1]], df[[col2]], left_on=col1, right_on=col2, validate=relation)
                output.append([col1, col2, relation])
            except:
                continue

    return output

cardinality = (pd.DataFrame(check_cardinality(df), columns=['first_column', 'second_column', 'cardinality'])
               .drop_duplicates(['first_column', 'second_column'])
               .reset_index(drop=True))

   first_column second_column  cardinality
0       Column1       Column1   one_to_one
1       Column1       Column2  one_to_many
2       Column1       Column3  one_to_many
3       Column1       Column4   one_to_one
4       Column1       Column5  one_to_many
5       Column2       Column1  many_to_one
6       Column2       Column4  many_to_one
7       Column3       Column1  many_to_one
8       Column3       Column4  many_to_one
9       Column4       Column1   one_to_one
10      Column4       Column2  one_to_many
11      Column4       Column3  one_to_many
12      Column4       Column4   one_to_one
13      Column4       Column5  one_to_many
14      Column5       Column1  many_to_one
15      Column5       Column4  many_to_one