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?
The pop() method removes an element at a given index, and will also return the removed item. You can also use the del keyword in Python to remove an element or slice from a list.
In Python, use list methods clear() , pop() , and remove() to remove items (elements) from a list. It is also possible to delete items using del statement by specifying a position or range with an index or slice.
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.
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 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')]
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