Python itertools对象的组合

Question

Can the python itertools combinations library by used on objects rather than lists?

For instance, how may I use it on the following data?

Rahul - 20,000 - Mumbai

Shivani - 30,000 - Mumbai

Akash - 40,000 - Bangalore

I want all the possible combinations on the names and the combined salary value.

How can I do this with combinations ?
Assuming the data is read using pd.read_csv and is stored.

Code so far -

import pandas as pd
import itertools
df = pd.read_csv('stack.csv')

print (df)

for L in range(0, len(df)+1):
    for subset in itertools.combinations(df['Name'], L):
        print (subset)

Output

      Name  Salary       City
0    Rahul   20000     Mumbai
1  Shivani   30000     Mumbai
2    Akash   40000  Bangalore
()
('Rahul',)
('Shivani',)
('Akash',)
('Rahul', 'Shivani')
('Rahul', 'Akash')
('Shivani', 'Akash')
('Rahul', 'Shivani', 'Akash')

Process finished with exit code 0

How do I add salary to these combinations?

Answer 1

First, get your indices:

idx = [j for i in range(1, len(df) + 1) for j in list(itertools.combinations(df.index, i))]
# [(0,), (1,), (2,), (0, 1), (0, 2), (1, 2), (0, 1, 2)]

Get your dataframes for each group:

dfs = [df.iloc[list(i)] for i in idx]

Finally, join and sum:

out = [(', '.join(i.name.values), sum(i.salary.values)) for i in dfs]

Output:

[('Rahul', 20000),
 ('Shivani', 30000),
 ('Akash', 40000),
 ('Rahul, Shivani', 50000),
 ('Rahul, Akash', 60000),
 ('Shivani, Akash', 70000),
 ('Rahul, Shivani, Akash', 90000)]

If you want this as a dataframe, it's quite simple:

df1 = pd.DataFrame(out, columns=['names', 'salaries'])

                   names  salaries
0                  Rahul     20000
1                Shivani     30000
2                  Akash     40000
3         Rahul, Shivani     50000
4           Rahul, Akash     60000
5         Shivani, Akash     70000
6  Rahul, Shivani, Akash     90000

To query this dataframe to find the closest value to a given salary, we can write a helper function:

def return_closest(val):
    return df1.iloc[(df1.salaries - val).abs().idxmin()]


>>> return_closest(55000)
names       Rahul, Shivani
salaries             50000
Name: 3, dtype: object

I intentionally broke this down so you could understand what was going on at each step. Once you do understand, you could combine this into a one-liner to create your dataframe:

pd.DataFrame(
    [(', '.join(d.name.values), sum(d.salary.values))
    for i in [j for i in range(1, len(df) + 1)
    for j in list(itertools.combinations(df.index, i))]
    for d in [df.iloc[list(i)]]], columns=['names', 'salaries']
)

Answer 2

you can use zip to iterate through both columns at the same time and use a list comprehension to generate the output dataframe such as:

df_ouput = pd.DataFrame( [[', '.join(subset), sum(salaries)] for L in range(1, len(df)+1)
                           for subset, salaries in zip(itertools.combinations(df['Name'], L),
                                                       itertools.combinations(df['Salary'], L))], 
                         columns = ['Names','Sum Salaries'])

and you get:

                   Names  Sum Salaries
0                  Rahul         20000
1                Shivani         30000
2                  Akash         40000
3         Rahul, Shivani         50000
4           Rahul, Akash         60000
5         Shivani, Akash         70000
6  Rahul, Shivani, Akash         90000

Answer 3

How about like this?

nameList = list()
sumList = list()
for L in range(0, len(df)+1):
    for x in itertools.combinations(df['Name'], L):
        nameList.append(x)
    for y in itertools.combinations(df['Salary'], L):
        sumList.append(sum(y))

newDf = pd.DataFrame()
newDf['Names'] = nameList
newDf['Salary Sum'] = sumList

Output:

                     Names  Salary Sum
0                       ()           0
1                 (Rahul,)       20000
2               (Shivani,)       30000
3                 (Akash,)       40000
4         (Rahul, Shivani)       50000
5           (Rahul, Akash)       60000
6         (Shivani, Akash)       70000
7  (Rahul, Shivani, Akash)       90000