A set is an unordered data structure, so it does not preserve the insertion order. This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension: >>> a = [1, 2, 20, 6, 210] >>> b = set([6, 20, 1]) >>> [x for x in a if x not in b] [2, 210] If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order: >>> a = dict.fromkeys([1, 2, 20, 6, 210]) >>> b = dict.fromkeys([6, 20, 1]) >>> dict.fromkeys(x for x in a if x not in b) {2: None, 210: None} b doesn't really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won't preserve the insertion order. In older versions of Python, you can use collections.OrderedDict instead: >>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210]) >>> b = collections.OrderedDict.fromkeys([6, 20, 1]) >>> collections.OrderedDict.fromkeys(x for x in a if x not in b) OrderedDict([(2, None), (210, None)])

A set is an unordered data structure, so it does not preserve the insertion order. This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension: >>> a = [1, 2, 20, 6, 210] >>> b = set([6, 20, 1]) >>> [x for x in a if x not in b] [2, 210] If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order: >>> a = dict.fromkeys([1, 2, 20, 6, 210]) >>> b = dict.fromkeys([6, 20, 1]) >>> dict.fromkeys(x for x in a if x not in b) {2: None, 210: None} b doesn't really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won't preserve the insertion order. In older versions of Python, you can use collections.OrderedDict instead: >>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210]) >>> b = collections.OrderedDict.fromkeys([6, 20, 1]) >>> collections.OrderedDict.fromkeys(x for x in a if x not in b) OrderedDict([(2, None), (210, None)])

在Python 3.6中，set()现在应该保持顺序，但Python 2和3有另一个解决方案:

>>> x = [1, 2, 20, 6, 210]
>>> sorted(set(x), key=x.index)
[1, 2, 20, 6, 210]

如果你有少量的元素在你的两个初始列表上，你想做集差操作，而不是使用集合。OrderedDict使实现复杂化，使其可读性较差，您可以使用:

# initial lists on which you want to do set difference
>>> nums = [1,2,2,3,3,4,4,5]
>>> evens = [2,4,4,6]
>>> evens_set = set(evens)
>>> result = []
>>> for n in nums:
...   if not n in evens_set and not n in result:
...     result.append(n)
... 
>>> result
[1, 3, 5]

它的时间复杂度不是很好，但它很简洁，易于阅读。

你可以用一行代码删除重复的值并保持插入的列表顺序，Python 3.8.2

mylist = ['b', 'b', 'a', 'd', 'd', 'c']


results = list({value:"" for value in mylist})

print(results)

>>> ['b', 'a', 'd', 'c']

results = list(dict.fromkeys(mylist))

print(results)

>>> ['b', 'a', 'd', 'c']

上面的最高分概念的实现，将它带回一个列表:

def SetOfListInOrder(incominglist):
    from collections import OrderedDict
    outtemp = OrderedDict()
    for item in incominglist:
        outtemp[item] = None
    return(list(outtemp))

在Python 3.6和Python 2.7上测试(简要)。