Python: How to chain tuple in a list by the next item value

You could use itertools.groupby to group by the second element, then extract the first element of each group:

from itertools import groupby
from operator import itemgetter

data = [('I', 'N'), ('love', 'V'), ('Miley', 'N'), ('Cyrus', 'N'), ('but', 'ADV'), ('hate', 'V'), ('Liam', 'N'),
        ('Hemsworth', 'N')]

result = [' '.join(t[0] for t in group) for k, group in groupby(data, key=itemgetter(1))]

print(result)

Output

['I', 'love', 'Miley Cyrus', 'but', 'hate', 'Liam Hemsworth']

The above list comprehension is equivalent to the following for loop:

result = []
for k, group in groupby(data, key=itemgetter(1)):
    result.append(' '.join(t[0] for t in group))

Using itertools.groupby

Ex:

from itertools import groupby
data = [('I', 'N'),('love', 'V'),('Miley', 'N'),('Cyrus', 'N'),('but', 'ADV'),('hate', 'V'),('Liam', 'N'),('Hemsworth', 'N')]

result = [" ".join(m for m, _ in v) for k, v in groupby(data, lambda x: x[1])]
print(result)

Output:

['I', 'love', 'Miley Cyrus', 'but', 'hate', 'Liam Hemsworth']