检查python / pandas中列之间的关系类型？（一对一，一对多或多对多）

Question

假设我有5列。

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]})

是否有一个函数可以知道每个par列具有的关系类型？（一对一，一对多，多对一，多对多）

输出如下：

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

Answer 1

这应该适合你：

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))

输出：

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

Answer 2

这可能不是一个完美的答案，但它应该可以进行一些进一步的修改：

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)

输出：

              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