Removing elements that have consecutive duplicates

I was curios about the question: Eliminate consecutive duplicates of list elements, and how it should be implemented in Python.

What I came up with is this:

list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0

while i < len(list)-1:
    if list[i] == list[i+1]:
        del list[i]
    else:
        i = i+1

Output:

[1, 2, 3, 4, 5, 1, 2]

Which I guess is ok.

So I got curious, and wanted to see if I could delete the elements that had consecutive duplicates and get this output:

[2, 3, 5, 1, 2]

For that I did this:

开发者_运维知识库list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0
dupe = False

while i < len(list)-1:
    if list[i] == list[i+1]:
        del list[i]
        dupe = True
    elif dupe:
        del list[i]
        dupe = False
    else:
        i += 1

But it seems sort of clumsy and not pythonic, do you have any smarter / more elegant / more efficient way to implement this?

>>> L = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> from itertools import groupby
>>> [key for key, _group in groupby(L)]
[1, 2, 3, 4, 5, 1, 2]

For the second part

>>> [k for k, g in groupby(L) if len(list(g)) < 2]
[2, 3, 5, 1, 2]

If you don't want to create the temporary list just to take the length, you can use sum over a generator expression

>>> [k for k, g in groupby(L) if sum(1 for i in g) < 2]
[2, 3, 5, 1, 2]

Oneliner in pure Python

[v for i, v in enumerate(your_list) if i == 0 or v != your_list[i-1]]

If you use Python 3.8+, you can use assignment expression :=:

list1 = [1, 2, 3, 3, 4, 3, 5, 5]

prev = object()
list1 = [prev:=v for v in list1 if prev!=v]

print(list1)

Prints:

[1, 2, 3, 4, 3, 5]

A "lazy" approach would be to use itertools.groupby.

import itertools

list1 = [1, 2, 3, 3, 4, 3, 5, 5]
list1 = [g for g, _ in itertools.groupby(list1)]
print(list1)

outputs

[1, 2, 3, 4, 3, 5]

You can do this by using zip_longest() + list comprehension.

from itertools import zip_longest 
list1 = [1, 2, 3, 3, 4, 3, 5, 5].
     # using zip_longest()+ list comprehension       
     res = [i for i, j in zip_longest(list1, list1[1:]) 
                                                            if i != j] 
        print ("List after removing consecutive duplicates : " +  str(res)) 

Here is a solution without dependence on outside packages:

list = [1,1,1,1,1,1,2,3,4,4,5,1,2] 
L = list + [999]  # append a unique dummy element to properly handle -1 index
[l for i, l in enumerate(L) if l != L[i - 1]][:-1] # drop the dummy element

Then I noted that Ulf Aslak's similar solution is cleaner :)

To Eliminate consecutive duplicates of list elements; as an alternative, you may use itertools.zip_longest() with list comprehension as:

>>> from itertools import zip_longest

>>> my_list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> [i for i, j in zip_longest(my_list, my_list[1:]) if i!=j]
[1, 2, 3, 4, 5, 1, 2]

Plenty of better/more pythonic answers above, however one could also accomplish this task using list.pop():

my_list = [1, 2, 3, 3, 4, 3, 5, 5]
for x in my_list[:-1]:
    next_index = my_list.index(x) + 1
    if my_list[next_index] == x:
        my_list.pop(next_index)

outputs

[1, 2, 3, 4, 3, 5]

Another possible one-liner, using functools.reduce (excluding the import) - with the downside that string and list require slightly different implementations:

>>> from functools import reduce

>>> reduce(lambda a, b: a if a[-1:] == [b] else a + [b], [1,1,2,3,4,4,5,1,2], [])
[1, 2, 3, 4, 5, 1, 2]

>>> reduce(lambda a, b: a if a[-1:] == b else a+b, 'aa  bbb cc')
'a b c'