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)?
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*
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)])