Pandas - find rows sharing two out the three common values, order-independent, and collect values pairs - pandas

Question

Given a dataframe, I am looking for rows where two out of three values are in common, regardless of the columns, hence order, in which they appear. I would like to then collect those common pairs.
Please note
a couple of values can appear at most in two rows
a value can appear only once in a row
I would like to know what the most efficient/elegant way is in numpy or pandas to solve this problem.
For example, taking as input the dataframe
d = {'col1': [1, 2,5,1], 'col2': [1, 7,1,2],'col3': [3, 3,1,7]}
df = pd.DataFrame(data=d)
col1 col2 col3
0 1 2 3
1 2 7 3
2 5 1 2
3 9 2 7
I expect as result an array, list, something as
1 2
2 3
2 7
as the values (1,2) , (2,3) and (2,7) are present in two rows (first and third, first and second, and second and forth respectively).
I cannot find a concise solution.
At the moment I skecthed a numpy solution such as
def func(x):
rows, columns = x.shape[0], x.shape[1]
res = []
for i in range(0,rows):
for j in range(i+1, rows):
aux = np.intersect1d(x[i,:], x[j,:])
if aux.size>1:
res.append(aux)
return res
which outputs
func(df.values)
Out: [array([2, 3]), array([1, 2]), array([2, 7])]
It looks well cumbersome, how could get it done with one of those cool numpy/pandas one-liners?

Answer 1

I would suggest using python built in set operations to do most of the heavy lifting, just apply them with pandas:
import itertools
import pandas as pd
d = {'col1': [1, 2,5,9], 'col2': [2, 7,1,2],'col3': [3, 3,2,7]}
df = pd.DataFrame(data=d)
pairs = df.apply(set, axis=1).apply(lambda x: set(itertools.combinations(x, 2))).explode()
out = set(pairs[pairs.duplicated()])
Output:
{(2, 3), (1, 2), (2, 7)}
Optionally to get it in list[np.ndarray] format:
out = list(map(np.array, out))

Answer 2

Similar approach to that of #Chrysophylaxs but in pure python:
from itertools import combinations
from collections import Counter
c = Counter(s for x in df.to_numpy().tolist() for s in set(combinations(set(x), r=2)))
out = [k for k,v in c.items() if v>1]
# [(2, 3), (1, 2), (2, 7)]

Answer 3

df=df.assign(col4=df.index)
def function1(ss:pd.Series):
ss1=ss.value_counts().loc[lambda ss:ss>=2]
return ss1.index.tolist() if ss1.size>=2 else None
df.merge(df,how='cross',suffixes=('','_2')).query("col4!=col4_2").filter(regex=r'col[^4]', axis=1)\
.apply(function1,axis=1).dropna().drop_duplicates()
out
1 [2, 3]
2 [1, 2]
7 [2, 7]