python - 递归删除列表中相邻的重复项

Question

我查找并找到了一个接近的示例，但在此链接中找到了答案:Remove adjacent duplicate elements from a list不会针对此问题运行测试用例。这就是我到目前为止所拥有的一切:

def remove_dups(thelist):
    """Returns: a COPY of thelist with adjacent duplicates removed.

    Example: for thelist = [1,2,2,3,3,3,4,5,1,1,1],
    the answer is [1,2,3,4,5,1]

    Precondition: thelist is a list of ints"""
    i = 1
    if len(thelist) == 0:
        return []
    elif len(thelist) == 1:
        return thelist
    elif thelist[i] == thelist[i-1]:
        del thelist[i]
    return remove_dups(thelist[i:])


def test_remove_dups():
    assert_equals([], remove_dups([]))
    assert_equals([3], remove_dups([3,3]))
    assert_equals([4], remove_dups([4]))
    assert_equals([5], remove_dups([5, 5]))
    assert_equals([1,2,3,4,5,1], remove_dups([1,2,2,3,3,3,4,5,1,1,1]))

# test for whether the code is really returning a copy of the original list
    mylist = [3]
    assert_equals(False, mylist is remove_dups(mylist))

编辑虽然我确实明白上面使用 itertools.groupby 链接的接受答案是可行的，但我认为它不会告诉我我的代码有什么问题，并且如果我从 itertools 导入 grouby ，就会违背练习的目的。

Answer 1

from itertools import groupby

def remove_dups(lst):
    return [k for k,items in groupby(lst)]

如果你真的想要一个递归解决方案，我会建议类似的东西

def remove_dups(lst):
    if lst:
        firstval = lst[0]

        # find lowest index of val != firstval
        for index, value in enumerate(lst):
            if value != firstval:
                return [firstval] + remove_dups(lst[index:])

        # no such value found
        return [firstval]
    else:
        # empty list
        return []