最近我注意到，当我转换一个列表来设置元素的顺序是改变的，并按字符排序。
想想这个例子:
x=[1,2,20,6,210]
print(x)
# [1, 2, 20, 6, 210] # the order is same as initial order
set(x)
# set([1, 2, 20, 210, 6]) # in the set(x) output order is sorted
我的问题是
为什么会这样? 如何才能在不丢失初始顺序的情况下进行设置操作(特别是设置差异)?
当前回答
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)])
其他回答
回答你的第一个问题，集合是为集合操作优化的数据结构。像数学集合一样，它不强制或维护元素的任何特定顺序。集合的抽象概念并不强制执行顺序，因此不需要实现。当你从列表中创建一个set时，Python可以根据它用于set的内部实现的需要自由地改变元素的顺序，这能够有效地执行set操作。
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']
基于Sven的回答，我发现使用集合。OrderedDict像这样帮助我完成你想要的，并允许我添加更多的项目到dict:
import collections
x=[1,2,20,6,210]
z=collections.OrderedDict.fromkeys(x)
z
OrderedDict([(1, None), (2, None), (20, None), (6, None), (210, None)])
如果你想添加项目，但仍然把它当作一个集合，你可以这样做:
z['nextitem']=None
你可以在dict上执行类似z.keys()的操作并获得集合:
list(z.keys())
[1, 2, 20, 6, 210]
正如在其他答案中指出的那样，集合是不保留元素顺序的数据结构(和数学概念)
然而，通过使用集合和字典的组合，你可以实现任何你想要的-尝试使用这些片段:
# save the element order in a dict:
x_dict = dict(x,y for y, x in enumerate(my_list) )
x_set = set(my_list)
#perform desired set operations
...
#retrieve ordered list from the set:
new_list = [None] * len(new_set)
for element in new_set:
new_list[x_dict[element]] = element
另一种更简单的方法是创建一个空列表，例如“unique_list”，用于从原始列表中添加唯一的元素，例如:
unique_list=[]
for i in original_list:
if i not in unique_list:
unique_list.append(i)
else:
pass
这将为您提供所有独特的元素，并保持顺序。
