common elements of a list

[Skaperen]

given 2 or more lists, i'd like to get a list of the common elements.  if the order gets re-arranged that is ok.

a=['apple','banana','cherry','lemon','orange']
b=['car','lemon',80,443,'python','banana','yellow']
c=commonlist(a,b)
print(c)

Output:
['banana','lemon']

is there an easy way to do that?

[buran]

Convert lists to sets and check for intersection

a=['apple','banana','cherry','lemon','orange']
b=['car','lemon',80,443,'python','banana','yellow']
c=list(set(a) & set(b)) # of course converting back to list is optional
print c

Output:
['lemon', 'banana']

[Skaperen]

set might be a better way to store certain "lists" where order does not matter.

[snippsat]

One with list comprehension.

>>> a = ['apple','banana','cherry','lemon','orange']
>>> b = ['car','lemon',80,443,'python','banana','yellow']
>>> [i for i in a if i in b]
['banana', 'lemon']

not change to difference.

>>> [i for i in a if i not in b]
['apple', 'cherry', 'orange']
# sets
>>> list(set(a) - set(b))
['cherry', 'apple', 'orange']

[zivoni]

List comprehension is more universal, as sets work only with hashable items. Sometimes one could be bitten with a fact that immutability doesnt guarantee hashability (usually with tuples - while (1, [2]) is tuple and therefore immutable, it is not hashable).

[buran]

List comprehension is more universal, as sets work only with hashable items. Sometimes one could be bitten with a fact that immutability doesnt guarantee hashability (usually with tuples - while (1, [2]) is tuple and therefore immutable, it is not hashable).

Good point regarding hashability of elements of the set