如何在此列表理解中不获取重复项？

Question

I am trying to take two lists with different lengths, and trying to make a third list which contains the same numbers using list comprehension. I want to avoid duplicates.

I attempted to use list comprehension with an if test, as I will show in the code. I also attempted an and statement, but that does not work.

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
c = [x for x in a if x in b]

This is the current solution I have, 'I also tried to alter the c list comprehension to:

c = [x for x in a if x in b and x not in c]

But this did not work. Is this not possible using list comprehension? I am aware that I can do this using sets quite easily. I am just practicing the use of list comprehension.

Answer 1

Depending on your reasons for using a list for c, you could consider using the built-in set structure to do intersection operations and which guarantees uniqueness of elements. For instance, set(a) will produce a set containing the unique elements in a .

c = [x for x in a if x in b] does not work since the duplicate elements in a are still contained in b and therefore not excluded by your if statement. ( 1 is duplicated in a , but both elements will be contained in c by your first definition, since 1 is in b ).

EDIT: if you want to simply modify your list comprehension but continue using it, you could do something like: c = [x for x in set(a) if x in b]

Answer 2

If you want a new list of all elements that appear in either of the 2 original lists, you could use the set class to achieve that. Maybe like this:

>>> a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
>>> b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

>>> list(sorted(set(a) | set(b)))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 21, 34, 55, 89]

What this does is building 2 sets from the lists, then finding the union of all the elements and then convert the result back to a list.

This approach will be faster than checking elem for elem in a if elem in b for large lists, because membership tests are O(1) for sets but up to O(n) for lists.

Answer 3

I will present a slightly different solution: The idea would be to first count the frequency of numbers using Counter and then perform the list comprehension using the keys. Since Counter returns a dictionary, you will not have duplicate keys.

from collections import Counter

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

f1 = Counter(a)
f2 = Counter(b)

c = [x for x in f1.keys() if x in f2.keys()]
# [1, 2, 3, 5, 8, 13]