Common elements between two lists and preserving the order of elements in the two lists

Question

I have two lists list1 and list2. I've found on stackoverflow a very simple method to get the common elements in this two lists as follows result = list(set(list1) & set(list2)). Unfortunately, with this, the order of elements in the resulting list, is not preserved.

For instance:

list1 = ['a', 'e', 't', 'b', 'c']
list2 = ['e', 'b', 'a', 'c', 'n', 's']

I want the result (common elements) to be ['e', 'a', 'b', 'c'] in this order. Because, for instance, 'e' is in list1 and in list2 and is in position 2 in list1 and position 1 in list2, while 'a' is in list1 and in list2 and is in position 1 in list1 and position 3 in list2, so 'e' is before 'a' since 2+1 < 1+3.

So, is there any simple way to have the common elements between two lists and preserving the order of elements ?

Answer 1

You could use a list comprehension to filter all elements from list1 that don't also belong to list2:

list1 = ['a', 'e', 't', 'b', 'c']
list2 = ['e', 'b', 'a', 'c', 'n', 's']

result = [item for item in list1 if item in list2]
print result

Result:

['a', 'e', 'b', 'c']

Although this does not conform to the desired result in your main post, from your follow-up comment it seems like this is an acceptable result.

---

You could also continue using the set approach, and then sort the results after the fact using the positioning algorithm you described:

list1 = ['a', 'e', 't', 'b', 'c']
list2 = ['e', 'b', 'a', 'c', 'n', 's']

items = set(list1) & set(list2)
result = sorted(items, key=lambda element: list1.index(element) + list2.index(element))

print result

Result:

['e', 'a', 'b', 'c']

Answer 2

list1 = ['a', 'e', 't', 'b', 'c']
list2 = ['e', 'b', 'a', 'c', 'n', 's']

weights = defaultdict(int)

for i, e in enumerate(list1):
   weights[e] += i

for i, e in enumerate(list2):
   weights[e] += i

>>> result = sorted(set(list1) & set(list2), key=lambda i: weights[i])
>>> result
['e', 'a', 'b', 'c']