Question

I need to compare list_a against many others. my problem starts when there’s a duplicated item in the other lists (two k’s in other_b).

my goal is to filter out all the lists with the same items (up to three matching items).

list_a =  ['j','k','a','7']
other_b =  ['k', 'j', 'k', 'q']
other_c = ['k','k','9','k']

>>>filter(lambda x: not x in list_a,other_b)
['q']

I need a way that would return [‘k’, ‘q’], because ‘k’ appears only once in list_a.

comparing list_a and other_c with set() isn’t good for my purpose since it will return only one element: k. while I need [‘k’,’9′,’k’]

I hope I was clear enough.

Thank you

Answer 1

The following function returns the new list that you want, when giving list_a as the first argument and other_b as the second:

def my_comp(my_list, other_list):
    other_cop = other_list[:]
    for x in my_list:
        try:
            other_cop.remove(x)
        except ValueError:
            pass
    return other_cop

It copys the other_list and removes every elemnt from my_list from the copy. Because if the element to remove is not in other_list, remove raises a ValueError that we have to catch and do nothing on this exception.