Get intersection from list of tuples

Question

I have two list of tuples

a = [('head1','a'),('head2','b'),('head3','x'),('head4','z')]
b = [('head5','u'),('head6','w'),('head7','x'),('head8','y'),('head9','z')]

I want to take the intersection of 2nd element of each tuple for example set {a[0][0],a[0][1],a[0][2],a[0][3]} intersection with set {b[0][0],b[0][1],b[0][2],b[0][3],b[0][4]} from list a and b such that it returns the first element mapping of the tuple if intersection value exist. The resulting output should be like following:

res = [('head3','head7'),('head4','head9')]

So far I have tried this:

x = [(a[i][0],b[j][0]) for i in range(len(a)) for j in range(len(b)) if a[0][i] == b[0][j]]

but got an error IndexError: tuple index out of range

What could be the correct and fastest way of doing this ?

Answer 1

You can do the following in Python 3. Create dicts from your lists, taking the intersection of keys from both dicts, fetch the corresponding values at the key:

>>> da = {k:v for v, k in a}
>>> db = {k:v for v, k in b}
>>> [(da[k], db[k])  for k in da.keys()&db.keys()]
[('head4', 'head9'), ('head3', 'head7')]

In Python 2, you can use set(da).intersection(db) in place of da.keys()&db.keys().

Answer 2

You can use a function with a generator:

def pairs():
   a = [('head1','a'),('head2','b'),('head3','x'),('head4','z')]
   b = [('head5','u'),('head6','w'),('head7','x'),('head8','y'),('head9','z')]

   for val1, val2 in a:
       for val3, val4 in b:
           if val2 == val4:
               yield (val1, val3)

print(list(pairs()))

Output:

[('head3', 'head7'), ('head4', 'head9')]