Common elements comparison between 2 lists.

Question

def common_elements(list1, list2):
    """
    Return a list containing the elements which are in both list1 and list2

    >>> common_elements([1,2,3,4,5,6], [3,5,7,9])
    [3, 5]
    >>> common_elements(['this','this','n','that'],['this','not','that','that'])
    ['this', 'that']
    """
    for element in list1:
        if element in list2:
            return list(element)

Got that so far, but can't seem to get it to work! Thanks

Answer 1

>>> list1 = [1,2,3,4,5,6]
>>> list2 = [3, 5, 7, 9]
>>> list(set(list1).intersection(list2))
[3, 5]

Answer 2

use set intersections, set(list1) & set(list2)

>>> def common_elements(list1, list2):
...     return list(set(list1) & set(list2))
...
>>>
>>> common_elements([1,2,3,4,5,6], [3,5,7,9])
[3, 5]
>>>
>>> common_elements(['this','this','n','that'],['this','not','that','that'])
['this', 'that']
>>>
>>>

Note that result list could be different order with original list.

Answer 3

The solutions suggested by S.Mark and SilentGhost generally tell you how it should be done in a Pythonic way, but I thought you might also benefit from knowing why your solution doesn't work. The problem is that as soon as you find the first common element in the two lists, you return that single element only. Your solution could be fixed by creating a result list and collecting the common elements in that list:

def common_elements(list1, list2):
    result = []
    for element in list1:
        if element in list2:
            result.append(element)
    return result

An even shorter version using list comprehensions:

def common_elements(list1, list2):
    return [element for element in list1 if element in list2]

However, as I said, this is a very inefficient way of doing this -- Python's built-in set types are way more efficient as they are implemented in C internally.