I have a list of tuples:

[('fruit', 'O'), ('is', 'O'), ('the', 'O'), ('subject', 'O'), ('of', 'O'), ('a', 'O'), ('Roald', 'PERSON'), ('Dahl', 'PERSON'), ('children', 'O'), ("'s", 'O'), ('book', 'O'), ('?', 'O')]

I want to reduce this list to:

[('fruit', 'O'), ('is', 'O'), ('the', 'O'), ('subject', 'O'), ('of', 'O'), ('a', 'O'), ('Roald Dahl', PERSON'), ('children', 'O'), ("'s", 'O'), ('book', 'O'), ('?', 'O')]

That is, any consecutive tuples whose second value is not 'O' should have their first value concatenated. This should work for list of any length, and for any number of consecutive tuples to be joined.

You can use itertools.groupby to group utilizing the last element in each tuple:

import itertools
s = [(1, 2), (3, 4), (5, 4), (10, 4), (7, 8)]
s = [(a, list(b)) for a, b in itertools.groupby(s, key=lambda x:x[-1])]
final_s = [(sum(i[0] for i in b), a) for a, b in s] 

Output:

[(1, 2), (18, 4), (7, 8)]

Edit: regarding your new, non numeric list of tuples, you can try this:

from functools import reduce
def remove(data, to_avoid='O'):s = [(a, list(b)) for a, b in itertools.groupby(data, key=lambda x:x[-1])]final_s = [x for i in [b if a == to_avoid else [(reduce(lambda c, d: "{} {}".format(c, d), [h[0] for h in b]), a)] for a, b in s] for x in i]return final_s>>remove([('fruit', 'O'), ('is', 'O'), ('the', 'O'), ('subject', 'O'), ('of', 'O'), ('a', 'O'), ('Roald', 'PERSON'), ('Dahl', 'PERSON'), ('children', 'O'), ("'s", 'O'), ('book', 'O'), ('?', 'O')])

Output:

[('fruit', 'O'), ('is', 'O'), ('the', 'O'), ('subject', 'O'), ('of', 'O'), ('a', 'O'), ('Roald Dahl', 'PERSON'), ('children', 'O'), ("'s", 'O'), ('book', 'O'), ('?', 'O')]

For those of us less comprehension aware, and using operator.itemgetter instead of the lamda's

import itertools, operator
item0 = operator.itemgetter(0)
item1 = operator.itemgetter(1)
result = []
for k, g in itertools.groupby(s, key=item1):if k != 'O':result.append((' '.join(map(item0, g)),k))else:result.extend(g)