如何检查两个列表是否具有相同顺序的相同对象？

Question

我有两个列表，需要检查它们是否包含相同顺序的相同对象。

list1 = [object1 , object2 , object3]
list2 = [object1 , object2 , object3]
list3 = [object2 , object3 , object1]

比较列表list1和list2 ，结果应为true 。

比较列表list1和list3 ，结果应为false 。

编辑：示例列表示例： list = [[<object1 at 0x04130AB0>], [<object2 at 0x04130210>, <object3 at 0x04130A10>]]

Answer 1

一个基本==将只检查如果一个每个元素list等于在另一相应的元素list 。 但是，假设你的例子是lst = [[<object1 at 0x04130AB0>], [<object2 at 0x04130210>, <object3 at 0x04130A10>]] ，你试图检查身份。 具有不同身份的两个对象可以比较为“相等”，这取决于该类的定义。 以下是一些示例，其中==会告诉您两个list在不同时包含相同顺序的相同对象：

>>> import collections
>>> a = collections.Counter()
>>> b = collections.Counter()
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> l1 == l2
True
>>> class Person:
...     def __eq__(self, other):
...         return True
...
>>> a = Person()
>>> b = Person()
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> l1 == l2
True
>>> a = []
>>> b = []
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> l1 == l2
True
>>> a.append(0)
>>> l1 == l2
False

如果要检查相应的元素实际上是同一个对象（标识，而不是相等），则需要以某种方式手动比较标识，例如：

>>> import collections
>>> a = collections.Counter()
>>> b = collections.Counter()
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> all(x is y for x,y in zip(l1, l2))
False
>>> a = Person()
>>> b = Person()
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> all(x is y for x,y in zip(l1, l2))
False
>>> a = []
>>> b = []
>>> a is b
False
>>> l1 = [a, b]
>>> l2 = [b, a]
>>> all(x is y for x,y in zip(l1, l2))
False

Answer 2

很简单，只需使用==来共同绘制两个列表。 如果两个列表具有相同的元素并且应该以相同的顺序放置，则它应返回True ，否则返回False

list1 == list2

实现上述逻辑的函数应该是这样的，

def check_lists(list1, list2):
    return list1 == list2

例如：

>>> ['object1' , 'object2' , 'object3'] == ['object1' , 'object2' , 'object3']
True
>>> ['object1' , 'object2' , 'object3'] == ['object3', 'object1' , 'object2' ]
False
>>> [['object1' , 'object2'] , ['object3']] == [['object1' , 'object2'] ,['object3']]
True
>>> [['object1' , 'object2'] , ['object3']] == [['object1' , 'object2'] , 'object3']
False

Answer 3

假设T在平等上的身份区别很重要。

def check_id(li1, li2):
    return [id(o) for o in li1] == [id(o) for o in li2]