Check if two unordered lists are equal

Question

I m looking for an easy  and quick  way to determine if two unordered lists contain the same elements   For example     one    two    three        one    two    three      true   one    two    three        one    three    two      true   one    two    three        one    two    three    three      false   one    two    three        one    two    three    four      false   one    two    three        one    two    four      false   one    two    three        one      false   I m hoping to do this without using a map

User · Answer

You want to see if they contain the same elements, but don't care about the order.

You can use a set:

>>> set(['one', 'two', 'three']) == set(['two', 'one', 'three'])
True

But the set object itself will only contain one instance of each unique value, and will not preserve order.

>>> set(['one', 'one', 'one']) == set(['one'])
True

So, if tracking duplicates/length is important, you probably want to also check the length:

def are_eq(a, b):
    return set(a) == set(b) and len(a) == len(b)

User · Answer

One liner answer to the above question is  -  let the two lists be list1 and list2  and your requirement is to ensure whether two lists have the same elements  then as per me  following will be the best approach  -  if   len list1     len list2   and     all i in list2 for i in list1         print  True  else      print  False    The above piece of code will work per your need i e  whether all the elements of list1 are in list2 and vice-verse   But if you want to just check whether all elements of list1 are present in list2 or not  then you need to use the below code piece only  -  if all i in list2 for i in list1       print  True  else      print  False    The difference is  the later will print True  if list2 contains some extra elements along with all the elements of list1  In simple words  it will ensure that all the elements of list1 should be present in list2  regardless of whether list2 has some extra elements or not

User · Answer

sorted x     sorted y    Copying from here  Check if two unordered lists are equal  I think this is the best answer for this question because   It is better than using counter as pointed in this answer x sort   sorts x  which is a side effect  sorted x  returns a new list

User · Answer

If elements are always nearly sorted as in your example then builtin  sort    timsort  should be fast    gt  gt  gt  a    1 1 2   gt  gt  gt  b    1 2 2   gt  gt  gt  a sort    gt  gt  gt  b sort    gt  gt  gt  a    b False   If you don t want to sort inplace you could use sorted     In practice it might always be faster then collections Counter    despite asymptotically O n  time being better then O n log n   for  sort     Measure it  If it is important

User · Answer

if you do not want to use the collections library  you can always do something like this  given that a and b are your lists  the following returns the number of matching elements  it considers the order    sum  1 for i j in zip a b  if i  j     Therefore    len a   len b  and len a   sum  1 for i j in zip a b  if i  j     will be True if both lists are the same  contain the same elements and in the same order  False otherwise   So  you can define the compare function like the first response above but without the collections library    compare   lambda a b  len a   len b  and len a   sum  1 for i j in zip a b  if i  j     and   gt  gt  gt  compare  1 2 3    1 2 3 3   False  gt  gt  gt  compare  1 2 3    1 2 3   True  gt  gt  gt  compare  1 2 3    1 2 4   False

User · Answer

What about getting the string representation of the lists and comparing them     gt  gt  gt  l1     one    two    three    gt  gt  gt  l2     one    two    three    gt  gt  gt  l3     one    three    two    gt  gt  gt  print str l1     str l2  True  gt  gt  gt  print str l1     str l3  False

User · Answer

Assuming you already know lists are of equal size  the following will guarantee True if and only if two vectors are exactly the same  including order   functools reduce lambda b1 b2  b1 and b2  map lambda e1 e2  e1  e2  listA  ListB   True    Example    gt  gt  gt  from functools import reduce  gt  gt  gt  def compvecs a b           return reduce lambda b1 b2  b1 and b2  map lambda e1 e2  e1  e2  a  b   True        gt  gt  gt  compvecs a  1 2 3 4   b  1 2 4 3   False  gt  gt  gt  compvecs a  1 2 3 4   b  1 2 3 4   True  gt  gt  gt  compvecs a  1 2 3 4   b  1 2 4 3   False  gt  gt  gt  compare vectors a  1 2 3 4   b  1 2 2 4   False  gt  gt  gt

User · Answer

Python has a built-in datatype for an unordered collection of  hashable  things  called a set  If you convert both lists to sets  the comparison will be unordered   set x     set y    Documentation on set    EDIT   mdwhatcott points out that you want to check for duplicates  set ignores these  so you need a similar data structure that also keeps track of the number of items in each list  This is called a multiset  the best approximation in the standard library is a collections Counter    gt  gt  gt  import collections  gt  gt  gt  compare   lambda x  y  collections Counter x     collections Counter y   gt  gt  gt    gt  gt  gt  compare  1 2 3    1 2 3 3   False  gt  gt  gt  compare  1 2 3    1 2 3   True  gt  gt  gt  compare  1 2 3 3    1 2 2 3   False  gt  gt  gt

[python] Check if two unordered lists are equal

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