create a bigram from a column in pandas df

Question

i have this test table in pandas dataframe

   Leaf_category_id  session_id  product_id
0               111           1         987
3               111                


        

Answer 1

try this code

from itertools import combinations
import pandas as pd

df = pd.DataFrame.from_csv("data.csv")
#consecutive
grouped_consecutive_product_ids = df.groupby(['Leaf_category_id','session_id'])['product_id'].apply(lambda x: [tuple(sorted(pair)) for pair in zip(x,x[1:])]).reset_index()

df1=pd.DataFrame(grouped_consecutive_product_ids)
s=df1.product_id.apply(lambda x: pd.Series(x)).unstack()
df2=pd.DataFrame(s.reset_index(level=0,drop=True)).dropna()
df2.rename(columns = {0:'Bigram'}, inplace = True)
df2["freq"] = df2.groupby('Bigram')['Bigram'].transform('count')
bigram_frequency_consecutive = df2.drop_duplicates(keep="first").sort_values("Bigram").reset_index()
del bigram_frequency_consecutive["index"]

for combinations (all possible bi-grams)

from itertools import combinations
import pandas as pd

df = pd.DataFrame.from_csv("data.csv")
#combinations
grouped_combination_product_ids = df.groupby(['Leaf_category_id','session_id'])['product_id'].apply(lambda x: [tuple(sorted(pair)) for pair in combinations(x,2)]).reset_index()

df1=pd.DataFrame(grouped_combination_product_ids)
s=df1.product_id.apply(lambda x: pd.Series(x)).unstack()
df2=pd.DataFrame(s.reset_index(level=0,drop=True)).dropna()
df2.rename(columns = {0:'Bigram'}, inplace = True)
df2["freq"] = df2.groupby('Bigram')['Bigram'].transform('count')
bigram_frequency_combinations = df2.drop_duplicates(keep="first").sort_values("Bigram").reset_index()
del bigram_frequency_combinations["index"]

where data.csv contains

Leaf_category_id,session_id,product_id
0,111,1,111
3,111,4,987
4,111,1,741
1,222,2,654
2,333,3,321
5,111,1,87
6,111,1,34
7,111,1,12
8,111,1,987
9,111,4,1232
10,222,2,12
11,222,2,324
12,222,2,465
13,222,2,342
14,222,2,32
15,333,3,321
16,333,3,741
17,333,3,987
18,333,3,324
19,333,3,654
20,333,3,862
21,222,1,123
22,222,1,987
23,222,1,741
24,222,1,34
25,222,1,12

The resultant bigram_frequency_consecutive will be

         Bigram  freq
0      (12, 34)     2
1     (12, 324)     1
2     (12, 654)     1
3     (12, 987)     1
4     (32, 342)     1
5      (34, 87)     1
6     (34, 741)     1
7     (87, 741)     1
8    (111, 741)     1
9    (123, 987)     1
10   (321, 321)     1
11   (321, 741)     1
12   (324, 465)     1
13   (324, 654)     1
14   (324, 987)     1
15   (342, 465)     1
16   (654, 862)     1
17   (741, 987)     2
18  (987, 1232)     1

The resultant bigram_frequency_combinations will be

           Bigram  freq
0      (12, 32)     1
1      (12, 34)     2
2      (12, 87)     1
3     (12, 111)     1
4     (12, 123)     1
5     (12, 324)     1
6     (12, 342)     1
7     (12, 465)     1
8     (12, 654)     1
9     (12, 741)     2
10    (12, 987)     2
11    (32, 324)     1
12    (32, 342)     1
13    (32, 465)     1
14    (32, 654)     1
15     (34, 87)     1
16    (34, 111)     1
17    (34, 123)     1
18    (34, 741)     2
19    (34, 987)     2
20    (87, 111)     1
21    (87, 741)     1
22    (87, 987)     1
23   (111, 741)     1
24   (111, 987)     1
25   (123, 741)     1
26   (123, 987)     1
27   (321, 321)     1
28   (321, 324)     2
29   (321, 654)     2
30   (321, 741)     2
31   (321, 862)     2
32   (321, 987)     2
33   (324, 342)     1
34   (324, 465)     1
35   (324, 654)     2
36   (324, 741)     1
37   (324, 862)     1
38   (324, 987)     1
39   (342, 465)     1
40   (342, 654)     1
41   (465, 654)     1
42   (654, 741)     1
43   (654, 862)     1
44   (654, 987)     1
45   (741, 862)     1
46   (741, 987)     3
47   (862, 987)     1
48  (987, 1232)     1

in the above case it groups by both