Question

我有一个名为lt。

的元组列表

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

我想从列表中删除所有连续的重复元组。结果应如下所示。

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

Answer 1

使用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]

这为您提供了一个删除了所有连续元组的新列表。您还可以使用itertools.groupby() function和默认身份密钥：

from itertools import groupby

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

演示：

>>> 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)]

Answer 2

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)

<强> RESULT

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

Answer 3

错综复杂的列表理解怎么样？

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

Answer 4

您可以尝试创建新列表，并将元素添加到该列表中。如果元素已经在新列表中，请不要添加它。

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)]

编辑：新解决方案，这应该工作。它显然非常基础。

如何从元组列表中删除连续的重复元组？

4 个答案: