Common Elements Comparison Between 2 Lists

Common elements between two lists with no duplicates

1. You are appending a list containing i to c, so i not in c will always return True. You should append i on its own: c.append(i)

Or

2. Simply use sets (if order is not important):

   a = [1, 1, 2, 2, 3, 5, 8, 13, 21, 34, 55, 89]
   b = [1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
   c = set(a) & set(b)  #  & calculates the intersection.
   print(c)
   #  {1, 2, 3, 5, 8, 13}
   

EDIT As @Ev. Kounis suggested in the comment, you will gain some speed by using

c = set(a).intersection(b).

How to find the common elements of two lists?

Use sets, its like a list, but can only hold unique items:

set(lista).intersection(listb)

Compare lists within a python to list to find common elements

One way is to run a counter over the flattened list and choose those that appeared more than once:

from collections import Counter
from itertools import chain

flattened = chain.from_iterable(many_lists)

result = [elem
          for elem, count in Counter(flattened).items()
          if count > 1]

to get

>>> result
[1, 11, 800, 7620]

Fastest way to find the common item between two lists in Python

This is very inefficient, and it only acquires the first common item in the two lists, not all of them, a better compact solution is by using sets.
Like this,

def common_item(l1, l2):
    return list(set(l1).intersection(set(l2)))

OR

def common_item(l1, l2):
    return list(set(l1) & set(l2))

which should return a list of all the elements that are common in the two lists, given that all the elements are unique.

If you have repeated elements in the lists, then you can try this approach which removes the element from the list if you encounter it, at the cost of runtime, which is insignificant when it is small.

def common_item(l1, l2):
    res = []
    for x in l1:
        if x in l2:
            res.append(x)
            l2.remove(x)
    return res

Compare two list of sets

You can enumerate over the zipped lists and filter the indices based on whether the pair of sets is disjoint:

list1 = [{'a','b'}, {'c','d'}, {'a','b','c'}, {'c','f'}]
list2 = [{'c','d','e'}, {'e','f'}, {'a','b','d'}, {'c','f'}]

indices = [i for i, (a, b) in enumerate(zip(list1, list2)) if a.isdisjoint(b)]

# [0, 1]

find common elements in two lists in linear time complexity

To find them in linear time you should use some kind of hashing. The easiest way in Python is to use a dict:

list1 = [1, 8, 5, 3, 4]
list2 = [5, 4, 1, 3, 8]

common = set(list1).intersection(list2)
dict2 = {e: i for i, e in enumerate(list2) if e in common}
result = [(i, dict2[e]) for i, e in enumerate(list1) if e in common]

The result will be

[(0, 2), (1, 4), (2, 0), (3, 3), (4, 1)]

You can use something like this to format and print it:

for i1, i2 in result:
    print(f"list1[{i1}] with list2[{i2}]")

you get:

list1[0] with list2[2]
list1[1] with list2[4]
list1[2] with list2[0]
list1[3] with list2[3]
list1[4] with list2[1]

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