CODEWITHSUNDEEP
pythonlistpython-3.xtuples
Fuji Komalan
Fuji Komalan

Reputation: 2047

How can i Remove consecutive duplicate tuples from a list of tuples?

I have list of tuples called lt.

lt = [ (1,2) , (1,2) , (2,3) , (3,4) , (5,6) , (7,8) , (7,8) ]

I want to remove all consecutive duplicate tuple from the list. The result should look like this.

mainList = [(1, 2), (2, 3), (3, 4), (5, 6), (7, 8)]

Upvotes: 0

Views: 239

Answers (4)

Niklas Braun
Niklas Braun

Reputation: 413

You could try to make a new list, and add the elements to that list. If an element is already in the new list, don't add it.

lt = [ (1,2) , (1,2) , (2,3) , (3,4) , (5,6) , (7,8) , (7,8) ] 
newlt = [lt[0]]
for item in range(len(lt)):
    if item == 0:
        continue
    elif lt[item] == lt[item - 1]:
        continue
    else:
        newlt.append(lt[item])
print(newlt)

[(1, 2), (2, 3), (3, 4), (5, 6), (7, 8)]

Edit: New solution, this should work. It's clearly very basic.

Upvotes: -1

szabadkai
szabadkai

Reputation: 106

How about an intricate list comprehension?

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

Upvotes: 1

Martijn Pieters
Martijn Pieters

Reputation: 1121486

Using the pairwise() function from the itertools reception section:

from itertools import tee

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

mainList = [curr for curr, next in pairwise(lt) if curr != next]

This gives you a new list with all consecutive tuples removed. You could also use the itertools.groupby() function with the default identity key:

from itertools import groupby

mainList = [k for k, g in groupby(lt)]

Demo:

>>> from itertools import tee, groupby
>>> lt = [ (1,2) , (1,2) , (2,3) , (3,4) , (5,6) , (7,8) , (7,8) ]
>>> def pairwise(iterable):
...     "s -> (s0,s1), (s1,s2), (s2, s3), ..."
...     a, b = tee(iterable)
...     next(b, None)
...     return zip(a, b)
...
>>> [curr for curr, next in pairwise(lt) if curr != next]
[(1, 2), (2, 3), (3, 4), (5, 6)]
>>> [k for k, g in groupby(lt)]
[(1, 2), (2, 3), (3, 4), (5, 6), (7, 8)]

Upvotes: 4

Fuji Komalan
Fuji Komalan

Reputation: 2047

lt = [ (1,2) , (1,2) , (2,3) , (3,4) , (5,6) , (7,8) , (7,8) ] 

mainList = []

for t in lt:
  if len(mainList) == 0:
    mainList.append(t)
  else:
    if mainList[-1] != t:
      mainList.append(t)

print(mainList)

RESULT

[(1, 2), (2, 3), (3, 4), (5, 6), (7, 8)]

Upvotes: 2

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