python：比较两个列表并按顺序返回匹配项

Question

我有两个不等长的列表，我想按照第一个列表的顺序比较和拉出匹配的值，所以在这个例子中。

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

预期产量：

['a','d']

我是这样做的，但我忘记了套装不保留订单！ 如何编辑下面的代码以保留订单。

 set(a).intersection(b)

链接到此如何比较python中的两个列表并返回匹配项

Answer 1

将b转换为一个集合，然后遍历a项目并检查它们是否存在于该集合中：

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

Answer 2

你需要使用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)