python：比较两个列表并按顺序返回匹配项

Question

I have two lists of unequal length and I would like to compare and pull matched values in the order from the first list, so a in this example.

a = ['a','s','d','f']
b = ['e','d','y','a','t','v']

expected output:

['a','d']

I was doing it like this but I forgot that set doesnt retain the order! how can I edit my code below to retain the order.

 set(a).intersection(b)

Linked to this How can I compare two lists in python and return matches

Answer 1

Convert b to a set and then loop over a 's items and check if they exist in that set:

>>> s = set(b)
>>> [x for x in a if x in s]
['a', 'd']

Answer 2

you need to use set:

>>> a = ['a','s','d','f']
>>> b = ['e','d','y','a','t','v']
>>> sorted(set(a) & set(b), key=a.index) # here sorting is done on the index of a
['a', 'd']

Answer 3

a = ['a','s','d','f']
b = ['e','d','y','a','t','v']
st_b = set(b)
print([ele for ele in a if ele in st_b])
['a', 'd']

Answer 4

a = ['a','s','d','f']
b = ['e','d','y','a','t','v']
matches=[]
for item_a in a:
    for item_b in b:
        if item_a == item_b:
            matches.append(item_a)
print(matches)