how to keep elements of a list based on another list

Question

I have two lists looking like:

list1 = [\'a\',\'a\',\'b\',\'b\',\'b\',\'c\',\'d\',\'e\',\'e\',\'g\',\'g\']

list2 = [\'a\',\'c\',\'z\',\'y\']

Answer 1

Using in operator, you can check whether an element is in a seqeunce.

>>> list2 = ['a','c','z','y']
>>> 'x' in list2
False
>>> 'y' in list2
True

Using list comprehension:

>>> list1 = ['a','a','b','b','b','c','d','e','e','g','g']
>>> list2 = ['a','c','z','y']
>>> [x for x in list1 if x in list2]
['a', 'a', 'c']

But x in list is not efficient. You'd better convert list2 to a set object.

>>> set2 = set(list2)
>>> [x for x in list1 if x in set2]
['a', 'a', 'c']

Answer 2

From Python 3 onwards use itertools.filterfalse

>>> import itertools
>>> list1 = ['a','a','b','b','b','c','d','e','e','g','g']
>>> list2 = ['a','c','z','y']
>>> list(itertools.filterfalse(lambda x:x not in list2,list1))
['a', 'a', 'c']

The list call is necessary as filterfalse returns an itertools object.

You can also use the filter function

>>> list(filter(lambda x: x in list2 , list1))
['a', 'a', 'c']

Answer 3

One alternative approach with numpy:

import numpy as np

np.asarray(list1)[np.in1d(list1, list2)].tolist()
#['a', 'a', 'c']