CODEWITHSUNDEEP
pythonlistunit-testing
FryMan
FryMan

Reputation: 51

How to check if two lists have the same objects in the same order?

I have two lists and need to check if they contain the same objects in the same order.

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

Comparing list list1 and list2, result should be true.

Comparing list list1 and list3, result should be false.

Edit: sample list example:list = [[<object1 at 0x04130AB0>], [<object2 at 0x04130210>, <object3 at 0x04130A10>]]

Upvotes: 1

Views: 6705

Answers (3)

TigerhawkT3
TigerhawkT3

Reputation: 49310

A basic == will only check if each element in one list is equal to the corresponding element in the other list. However, given your example of lst = [[<object1 at 0x04130AB0>], [<object2 at 0x04130210>, <object3 at 0x04130A10>]], you are trying to check for identity. Two objects with different identities may compare as "equal," depending on that class's definition of that. Here are some examples where == would tell you that two lists contain the same objects in the same order when they do not:

>>> 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

If you want to check that corresponding elements are actually the same object (identity, not equality), you will need to manually compare identities in some way, such as the following:

>>> 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

Upvotes: 2

JL Peyret
JL Peyret

Reputation: 12174

Assuming T's identity over equality distinction matters. 

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

Upvotes: 0

Avinash Raj
Avinash Raj

Reputation: 174716

Simple, Just use == to comapare two lists. This should return True if two lists having same elements and should be placed in the same order else it returns False

list1 == list2

A function which implements the above logic should look like,

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

Ex:

>>> ['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

Upvotes: 2

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