Can I check if a list contains any item from another list? [duplicate]

Question

using Python, I want to check if a list contains an item/value that is also present in another list. For example, here is what I am trying to do:

list1 = ['item1','item2','item3']
list2 = ['item4','item5','item3']

if list1 contains any items also in list2:
    print("Duplicates found.")
else:
    print("No duplicates found.")

As you can see, list2 contains an item that is also found in list1, which is 'item3'. Is there any way I can detect if this occurs?

Thanks.

Overbid

Does this answer your question? How to find list intersection? – Addie

Answer 1

Using any() along with a generator expression:

list1 = ['item1','item2','item3']
list2 = ['item4','item5','item3']

if any(x in list1 for x in list2):
    print("Duplicates found.")
else:
    print("No duplicates found.")

Coition

Answer 2

You could use a set. isdisjoint is a method which returns True if two sets have nothing in common and False if the sets overlap. After converting list1 and list2 into sets, they both contain 'item3' so isdisjoint returns False.

set(list1).isdisjoint(set(list2))

>>> list1 = ['item1','item2','item3']
>>> list2 = ['item4','item5','item3']
>>> set(list1).isdisjoint(set(list2))
False

You can combine this with the not operator to do what you want.

list1 = ['item1','item2','item3']
list2 = ['item4','item5','item3']

if not set(list1).isdisjoint(set(list2)):
    print("Duplicates found.")
else:
    print("No duplicates found.")

Chinookan

I prefer this as it is congruent as to the purpose of sets and to the clarity of reading the code. – Buoyant

Answer 3

Try using sets:

Following https://mcmap.net/q/86508/-best-way-to-find-the-intersection-of-multiple-sets,

i=set.intersection(set(list1),set(list2))
if len(i) > 0:
    print('duplicates found')
else:
    print('no duplicates found')

Harleyharli

The example provided contains a syntax error. – Chinookan

Answer 4

list1.extend(list2)
if len(set(list1)) != len(list1) + len(list2):
    print("Duplicates Found")
else:
    print("Duplicates Not Found")

Time Analysis

>>>def fun():
       list1 = [1,2,3,4]
       list2 = [3,4,2,1]
       if len(set(list1.extend(list2)) != len(list1) + len(list2):
           return "Duplicates Found"
       else:
           return "Duplicates Not Found"
>>>%timeit fun()
>>>3.59 µs ± 213 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>>def fun1():
       list1 = [1,2,3,4]
       list2 = [3,4,2,1]

       if any(x in list1 for x in list2):
           return "Duplicates Found"
       else:
           return "Duplicates Not Found"
>>>%%timeit fun1():
>>>3.93 µs ± 696 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Conclusion

Both methods take nearly equal time but 'set + extend' method take slight less time to execute.

Roughrider

This is super expensive imo. – Heinrich

your fun won't work simply because list.extend returns None – Translator

Yes..Actually it is a new list with both of list merged together. – Roughrider

this is unfortunately less readable than any(x in list1 for x in list2) – Maclean