使用 pandas 识别两列之间的关系

Question

I have two columns in a dataframe as follows, namely Letter and Number

I want to do following

1. In the above table letter A is repeated two times in column "Letter" which I want to classify as "One to Many" in a new column.
2. 15 is repeated two times in number column which i want to classify as "many to one".
3. Letter B, C and Number 5, 6 occurred only one time in each column therefore should be classified as one to one.
4. For other should be classified as many to many.

Expected output is shown below.

5. I tried using groupby function by shifting the column name, it helped to identify item 1 and item 2 separately.

I want to do it in single function, Please help.....

Answer 1

You could write a function like this:

import pandas as pd

letter = ['A', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'F', 'G']
number = [10,11,5,6,15,15,20,20,25,28]
data = {'letter': letter, 'number': number}    
df = pd.DataFrame(data)

def relationship(letter, number):
    number_of_letters = {}
    number_of_numbers = {}
    relationship = [] 

    for i in letter:
        if i in number_of_letters:
            number_of_letters[i] += 1
        else:
            number_of_letters[i] = 1    
    for i in number:
        if i in number_of_numbers:
            number_of_numbers[i] += 1
        else:
            number_of_numbers[i] = 1    
    for i in range(len(letter)):
        if number_of_letters[letter[i]] == 1 and number_of_numbers[number[i]] == 1:
            relationship.append('One to One')
        elif number_of_letters[letter[i]] > 1 and number_of_numbers[number[i]] == 1:
            relationship.append('One to Many')
        elif number_of_letters[letter[i]] == 1 and number_of_numbers[number[i]] > 1:
            relationship.append('Many to One') 
        elif number_of_letters[letter[i]] > 1 and number_of_numbers[number[i]] > 1:
            relationship.append('Many to Many') 

    return relationship 

df['relationship'] = relationship(letter, number)

Answer 2

This could be your solution

import pandas as pd

d1 = ['A','A','B','C','D','E','F','G','F','G']
d2 = [10,11,5,6,15,15,20,20,25,28]

df = pd.DataFrame(list(zip(d1,d2)), columns = ['col1', 'col2'])


df['one to one'] = (df.groupby('col2')['col1'].transform(lambda x:x.nunique()==1) & df.groupby('col1')['col2'].transform(lambda x:x.nunique()==1))


df['many to one'] = (df.groupby('col2')['col1'].transform(lambda x:x.nunique()>1) & df.groupby('col1')['col2'].transform(lambda x:x.nunique()==1))


df['one to many'] = (df.groupby('col1')['col2'].transform(lambda x:x.nunique()>1) & df.groupby('col2')['col1'].transform(lambda x:x.nunique()==1))



df['many to many'] = (df.groupby('col1')['col2'].transform(lambda x:x.nunique()>1) & df.groupby('col2')['col1'].transform(lambda x:x.nunique()>1))


import numpy as np

conditions = [
    (df['one to one'] == True), (df['one to many'] == True),(df['many to one'] == True),(df['many to many'] == True)]
choices = ['one to one', 'one to many', 'many to one','many to many']
df['relation'] = np.select(conditions, choices)


df.drop(['one to one', 'one to many', 'many to one','many to many'], axis = 1)