Kdtyjb

I am trying to figure out what combination of clothing customers are buying together. I can figure out the exact combination, but the problem I can't figure out is the count that includes the combination + others.

For example, I have:

Cust_num  Item    Rev

Cust1     Shirt1  $40

Cust1     Shirt2  $40

Cust1     Shorts1 $40

Cust2     Shirt1  $40

Cust2     Shorts1 $40

This should result in:

Combo                  Count

Shirt1,Shirt2,Shorts1    1

Shirt1,Shorts1           2

The best I can do is unique combinations:

Combo                 Count

Shirt1,Shirt2,Shorts1   1

Shirt1,Shorts1          1

I tried:

df = df.pivot(index='Cust_num',columns='Item').sum()

df[df.notnull()] = "x"

df = df.loc[:,"Shirt1":].replace("x", pd.Series(df.columns, df.columns))

col = df.stack().groupby(level=0).apply(','.join)

df2 = pd.DataFrame(col)

df2.groupby([0]).size().reset_index(name='counts')

But that is just the unique counts.

Using pandas.DataFrame.groupby:

grouped_item = df.groupby('Cust_num')['Item']

subsets = grouped_item.apply(lambda x: set(x)).tolist()

Count = [sum(s2.issubset(s1) for s1 in subsets) for s2 in subsets]

combo = grouped_item.apply(lambda x:','.join(x))

combo = combo.reset_index()

combo['Count']=Count

Output:

  Cust_num                   Item  Count

0    Cust1  Shirt1,Shirt2,Shorts1      1

1    Cust2         Shirt1,Shorts1      2

Late answer, but you can use:

df = df.groupby(['Cust_num'], as_index=False).agg(','.join).drop(columns=['Rev']).set_index(['Item']).rename_axis("combo").rename(columns={"Cust_num": "Count"})

df['Count'] = df['Count'].str.replace(r'Cust','')

combo                   Count                 

Shirt1,Shirt2,Shorts1     1

Shirt1,Shorts1            2

I think you need to create a combination of items first.

How to get all possible combinations of a list’s elements?

I used the function from Dan H's answer.

from itertools import chain, combinations

def all_subsets(ss):

    return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))



uq_items = df.Item.unique()



list(all_subsets(uq_items))



[(),

 ('Shirt1',),

 ('Shirt2',),

 ('Shorts1',),

 ('Shirt1', 'Shirt2'),

 ('Shirt1', 'Shorts1'),

 ('Shirt2', 'Shorts1'),

 ('Shirt1', 'Shirt2', 'Shorts1')]

And use groupby each customer to get their items combination.

ls = []



for _, d in df.groupby('Cust_num', group_keys=False):

    # Get all possible subset of items

    pi = np.array(list(all_subsets(d.Item)))



    # Fliter only > 1

    ls.append(pi[[len(l) > 1 for l in pi]])

Then convert to Series and use value_counts().

pd.Series(np.concatenate(ls)).value_counts()



(Shirt1, Shorts1)            2

(Shirt2, Shorts1)            1

(Shirt1, Shirt2, Shorts1)    1

(Shirt1, Shirt2)             1

My version which I believe is easier to understand

new_df = df.groupby("Cust_num").agg({lambda x: ''.join(x.unique())})



new_df ['count'] = range(1, len(new_df ) + 1)

Output:

                            Item      Rev count

                        <lambda> <lambda>      

Cust_num                                       

Cust1      Shirt1 Shirt2 Shorts1      $40     1

Cust2             Shirt1 Shorts1      $40     2

Since you do not need the Rev column, you can drop it:

new_df = new_df = new_df.drop(columns=["Rev"]).reset_index()



new_df

Output:

  Cust_num                    Item count

                          <lambda>      

0    Cust1   Shirt1 Shirt2 Shorts1     1

1    Cust2          Shirt1 Shorts1     2

This edit is to respond to @Chris by looking at his approach written using list comprehension. He created an list of sets:

[{' Shirt1', ' Shirt2', ' Shorts1'}, {' Shirt1', ' Shorts1'}]

Then next step finds the subsets:

for s1 in subsets:

    for s2 in subsets:

        if s2.issubset(s1):

            print("{}: {}".format(s2,s2.issubset(s1)))

Output:

{' Shirt2', ' Shorts1', ' Shirt1'}: True

{' Shorts1', ' Shirt1'}: True

{' Shorts1', ' Shirt1'}: True

You asked me to explain myself and I did. However after thinking about it, I realized your approach was wrong too. As such I was not mocking you, but thanking you for making me think about my solution. Also thanks to @ResidentSleeper for his solution.