Q:

How to remove list items depending on predecessor in python

Given a Python list, I want to remove consecutive 'duplicates'. The duplicate value however is a attribute of the list item (In this example, the tuple's first element).
Input:
[(1, 'a'), (2, 'b'), (2, 'b'), (2, 'c'), (3, 'd'), (2, 'e')]

Desired Output:
[(1, 'a'), (2, 'b'), (3, 'd'), (2, 'e')]

Cannot use set or dict, because order is important.
Cannot use list comprehension [x for x in somelist if not determine(x)], because the check depends on predecessor.
What I want is something like:
mylist = [...]

for i in range(len(mylist)):
    if mylist[i-1].attr == mylist[i].attr:
        mylist.remove(i)

What is the preferred way to solve this in Python?

A:

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.

A:

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')]

A:

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')]

