计算 dataframe 列中最常见的值组合

Question

I have DataFrame in the following form:

ID Product
1   A
1   B
2   A 
3   A
3   C 
3   D 
4   A
4   B

I would like to count the most common combination of two values from Product column grouped by ID . So for this example expected result would be:

Combination Count
A-B          2
A-C          1
A-D          1
C-D          1

Is this output possible with pandas?

Answer 1

We can merge within ID and filter out duplicate merges (I assume you have a default RangeIndex ). Then we sort so that the grouping is regardless of order:

import pandas as pd
import numpy as np

df1 = df.reset_index()
df1 = df1.merge(df1, on='ID').query('index_x > index_y')

df1 = pd.DataFrame(np.sort(df1[['Product_x', 'Product_y']].to_numpy(), axis=1))
df1.groupby([*df1]).size()

---

0  1
A  B    2
   C    1
   D    1
C  D    1
dtype: int64

Answer 2

You can use combinations from itertools along with groupby and apply

from itertools import combinations

def get_combs(x):
    return pd.DataFrame({'Combination': list(combinations(x.Product.values, 2))})

(df.groupby('ID').apply(get_combs)
 .reset_index(level=0)
 .groupby('Combination')
 .count()
)

             ID
Combination    
(A, B)        2
(A, C)        1
(A, D)        1
(C, D)        1

Answer 3

Use itertools.combinations , explode and value_counts

import itertools

(df.groupby('ID').Product.agg(lambda x: list(itertools.combinations(x,2)))
                 .explode().str.join('-').value_counts())

Out[611]:
A-B    2
C-D    1
A-D    1
A-C    1
Name: Product, dtype: int64

---

Or:

import itertools

(df.groupby('ID').Product.agg(lambda x: list(map('-'.join, itertools.combinations(x,2))))
                 .explode().value_counts())

Out[597]:
A-B    2
C-D    1
A-D    1
A-C    1
Name: Product, dtype: int64

Answer 4

Using itertools and Counter .

import itertools
from collections import Counter

agg_ = lambda x: tuple(itertools.combinations(x, 2))
product = list(itertools.chain(*df.groupby('ID').agg({'Product': lambda x: agg_(sorted(x))}).Product))
# You actually do not need to wrap product with list. The generator is ok
counts = Counter(product)

Output

Counter({('A', 'B'): 2, ('A', 'C'): 1, ('A', 'D'): 1, ('C', 'D'): 1})

You could also do the following to get a dataframe

pd.DataFrame(list(counts.items()), columns=['combination', 'count'])

  combination  count
0      (A, B)      2
1      (A, C)      1
2      (A, D)      1
3      (C, D)      1

Answer 5

Another trick with itertools.combinations function:

from itertools import combinations
import pandas as pd

test_df = ... # your df
counts_df = test_df.groupby('ID')['Product'].agg(lambda x: list(combinations(x, 2)))\
    .apply(pd.Series).stack().value_counts().to_frame()\
    .reset_index().rename(columns={'index': 'Combination', 0:'Count'})
print(counts_df)

The output:

  Combination  Count
0      (A, B)      2
1      (A, C)      1
2      (A, D)      1
3      (C, D)      1