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
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']
其他回答
我们可以使用集合。计数器:
# tested on python 3.7
>>> from collections import Counter
>>> lst = ["1", "2", "20", "6", "210"]
>>> for i in Counter(lst):
>>> print(i, end=" ")
1 2 20 6 210
>>> for i in set(lst):
>>> print(i, end=" ")
20 6 2 1 210
迟了，但你可以用熊猫，pd。转换列表，同时保持顺序:
import pandas as pd
x = pd.Series([1, 2, 20, 6, 210, 2, 1])
print(pd.unique(x))
输出: 数组([1,2,20,6,210])
适用于字符串列表
x = pd.Series(['c', 'k', 'q', 'n', 'p','c', 'n'])
print(pd.unique(x))
输出 ['c' 'k' 'q' 'n' 'p']
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)])
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。
如果愿意，可以删除重复的值并保持插入的列表顺序
lst = [1,2,1,3]
new_lst = []
for num in lst :
if num not in new_lst :
new_lst.append(num)
# new_lst = [1,2,3]
如果你想要的是“order”，不要使用“sets”来删除重复，
使用集合进行搜索。 X在列表中 花费O(n)时间 在哪里 集合中的X 在大多数情况下需要O(1)时间*
