如何使用itertools提取groupby值？

Question

data = [[12345,"AAA"],[12345,"BBB"],[12345,"CCC"],[98765,"KKK"],[98765,"MMM"],[56321,"JJJ"],[56321,"SSS"],[56321,"PPP"]]
df = pd.DataFrame(data,columns=['Sales_ID','Company_Name'])

Hi folks, I have above dataframe and I want to create a matching within each groupby Sales_ID. How can I do that in python?

I tried to groupby the df and extract all companies for each sales_ID, but don't know how to do next.

df.groupby('Sales_ID').apply(lambda x:x['Company_Name'].tolist())

Expected results:

Sales_ID Company Company
12345      AAA   BBB
12345      AAA   CCC
12345      BBB   CCC
98765      KKK   MMM
56321      JJJ   SSS
56321      JJJ   PPP
56321      SSS   PPP

Thanks for the help.

Answer 1

Edit: @brentertainer points out that a cartesian product followed by a < query is all you need to remove self-merges and duplicates irrespective of order.

df.merge(df, on='Sales_ID').query('Company_Name_x < Company_Name_y')

---

Original, more complicated solution sorted to drop duplicates irrespective of ordering

import pandas as pd
import numpy as np

res = df.merge(df, on='Sales_ID').query('Company_Name_x != Company_Name_y')

cols = ['Company_Name_x', 'Company_Name_y']
res[cols] = np.sort(res[cols].to_numpy(), axis=1)
res = res.drop_duplicates()

Output:

    Sales_ID Company_Name_x Company_Name_y
1      12345            AAA            BBB
2      12345            AAA            CCC
5      12345            BBB            CCC
10     98765            KKK            MMM
14     56321            JJJ            SSS
15     56321            JJJ            PPP
18     56321            PPP            SSS

Answer 2

I am using itertools

s=df.groupby('Sales_ID',sort=False)['Company_Name'].apply(list)
l=[list(itertools.combinations(x,2)) for x in s]
Newdf=pd.DataFrame({'Sales_ID':s.index.repeat(list(map(len,l)))})
Newdf=pd.concat([Newdf,pd.DataFrame(sum(l,[]))],axis=1)
Newdf
   Sales_ID    0    1
0     12345  AAA  BBB
1     12345  AAA  CCC
2     12345  BBB  CCC
3     98765  KKK  MMM
4     56321  JJJ  SSS
5     56321  JJJ  PPP
6     56321  SSS  PPP

Answer 3

Its is not always nescessary to use pandas *. I prefer using toolz or funcy to get the job done (that behind the screen use itertools and other python native modules and methods)

import itertools
import toolz  # pip install toolz
import toolz.curried as tc
from operator import itemgetter

grouped_data = toolz.groupby(itemgetter(0), data)

{12345: [[12345, 'AAA'], [12345, 'BBB'], [12345, 'CCC']],
 98765: [[98765, 'KKK'], [98765, 'MMM']],
 56321: [[56321, 'JJJ'], [56321, 'SSS'], [56321, 'PPP']]}

Now to get the data you'd like you need to apply a series of steps:

result = toolz.thread_first(data, # thread first pipes the data through series of functions
                            tc.groupby(itemgetter(0)), # group by first element
                            tc.valmap(tc.map(itemgetter(1))), # for each group extract the second element from a list of lists
                            tc.valmap(tc.partial(itertools.combinations, r=2)), # for each group make pairs
                            tc.valmap(list)) # this statement creates a list from the combinations generator function (it is howver not nescessary.)

The result:

{12345: [('AAA', 'BBB'), ('AAA', 'CCC'), ('BBB', 'CCC')],
 98765: [('KKK', 'MMM')],
 56321: [('JJJ', 'SSS'), ('JJJ', 'PPP'), ('SSS', 'PPP')]}

If you want to frame it into pandas you can. Otherwise you can continue with a functional programming approach if this is what you seek.

*from my own experience especially in cloud environment with serverless applications - but thats besides the point