快捷方式:

list(set(a).intersection(set(b)))

还有一个找到共同价值观的方法:

a = [1, 2, 3, 4, 5]
b = [9, 8, 7, 6, 5]
matches = [i for i in a if i in b]

这不是最有效的方法，但到目前为止最明显的方法是:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a) & set(b)
{5}

如果顺序很重要，你可以用这样的列表推导式来做:

>>> [i for i, j in zip(a, b) if i == j]
[5]

(仅适用于同等大小的列表，这意味着顺序重要性)。

>>> s = ['a','b','c']   
>>> f = ['a','b','d','c']  
>>> ss= set(s)  
>>> fs =set(f)  
>>> print ss.intersection(fs)   
   **set(['a', 'c', 'b'])**  
>>> print ss.union(fs)        
   **set(['a', 'c', 'b', 'd'])**  
>>> print ss.union(fs)  - ss.intersection(fs)   
   **set(['d'])**

you can | for set union and & for set intersection.
for example:

    set1={1,2,3}
    set2={3,4,5}
    print(set1&set2)
    output=3

    set1={1,2,3}
    set2={3,4,5}
    print(set1|set2)
    output=1,2,3,4,5

curly braces in the answer.

最简单的方法是使用set:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a) & set(b)
set([5])