How to remove list items depending on predecessor in python

In order to remove consecutive duplicates, you could use itertools.groupby:

l = [(1, 'a'), (2, 'a'), (2, 'a'), (3, 'a'), (4, 'a')]
from itertools import groupby
[tuple(k) for k, _ in groupby(l)]
# [(1, 'a'), (2, 'a'), (3, 'a'), (4, 'a')]

If you just want to stick to list comprehension, you can use something like this:

>>> li = [(1, 'a'), (2, 'a'), (2, 'a'), (3, 'a'), (2, 'a')]
>>> [li[i] for i in range(len(li)) if not i or li[i] != li[i-1]]
[(1, 'a'), (2, 'a'), (3, 'a'), (2, 'a')]

Please not that not i is the pythonic way of writing i == 0.

You can use itertools.groupby (demonstration with more data):

from itertools import groupby
from operator import itemgetter

data = [(1, 'a'), (2, 'a'), (2, 'b'), (3, 'a'), (4, 'a'), (2, 'a'), (2, 'a'), (3, 'a'), (3, 'a')]

[next(group) for key, group in groupby(data, key=itemgetter(0))]

Output:

[(1, 'a'), (2, 'a'), (3, 'a'), (4, 'a'), (2, 'a'), (3, 'a')]

For completeness, an iterative approach based on other answers:

result = []

for first, second in zip(data, data[1:]):
    if first[0] != second[0]:
        result.append(first)

result

Output:

[(1, 'a'), (2, 'b'), (3, 'a'), (4, 'a'), (2, 'a')]

Note that this keeps the last duplicate, instead of the first.

If I am not mistaken, you only need to lookup the last value.

test = [(1, 'a'), (2, 'a'), (2, 'a'), (3, 'a'), (4, 'a'),(3, 'a'),(4,"a"),(4,"a")]

result = []

for i in test:
    if result and i[0] == result[-1][0]: #edited since OP considers (1,"a") and (1,"b") as duplicate
    #if result and i == result[-1]:
        continue
    else:
        result.append(i)

print (result)

Output:

[(1, 'a'), (2, 'a'), (3, 'a'), (4, 'a'), (3, 'a'), (4, 'a')]