如何获取 orderedDict 中具有最小值的对象？

Question

我有一个有序的对象字典，

 [<f.Packet object at 0x07AD7090>, <f.Packet object at 0x07ACA8F0>, <f.Packet object at 0x07ACAC90>, <f.Packet object at 0x07A5F5D0>, <f.Packet object at 0x07ACA410>, <f.Packet object at 0x07ABBF50>, <f.Packet object at 0x07ACA830>]

这些对象中的每一个都具有名称、年龄和来源等属性。我怎样才能得到年龄最小的对象？

Answer 1

试试这个，假设你有 d as orderedDict

import operator
#...
person=sorted(d.values(), key=operator.attrgetter('age'))[0]
print person.age

Answer 2

根据你的需要，你可以使用简单的遍历：

min(orderedDict.values(), key=lambda x: x.age)

但是如果您需要 O(1) 的方式来做到这一点，您需要创建自己的 class，因为 OrderedDict 仅根据插入顺序对项目进行排序。例如，您可以使用 Sorted Containers 中的 SortedDict（或自己编写）并执行如下操作（假设您仍希望能够根据插入顺序获取项目）：

from collections import OrderedDict
from sortedcontainers import SortedDict

class MyOrderedDict(OrderedDict):
     def __init__(self):
             super(MyOrderedDict, self).__init__(self)
             self.sorted = SortedDict()
     def __setitem__(self, key, value):
             super(MyOrderedDict, self).__setitem__(key, value)
             self.sorted[value.age] = value
     def __delitem__(self, key):
             age = super(MyOrderedDict, self).__getitem__(key).age
             super(MyOrderedDict, self).__delitem__(key)
             self.sorted.__delitem__(age)
     def ageIterator(self):
         for age in self.sorted:
             yield (age, self.sorted[age])

orderedDict = MyOrderedDict()
#...

for item in orderedDict:
    # Items in the order of insertions

for age, item in orderedDict.ageIterator():
    # Items by age

如何获取 orderedDict 中具有最小值的对象？

how get the object which has the minimum value in an orderedDict?

python

attributes

object

ordereddict