在Python 2.x中交叉引用两个列表

Question

我刚开始学Python（用的是2.6和2.7版本），我在docs.Python.org和这个网站上查过资料。我找到一个类似的问题：在Python中比较列表

...但是我想要对比两个大小不同、顺序也不确定的列表。下面是我想做的一个例子：

>>> mammals = ["gorilla","cat","rat","chimpanzee","dog","beaver"]
>>> apes =  ["orangutan","chimpanzee","human","gorilla"]

# magic happens here

>>> print result # order doesn't matter
['chimpanzee', 'gorilla']

希望能得到一个共同元素的结果。了解Python的人应该知道，这种简单的问题应该有一个简单又优雅的解决办法。

Answer 1

这对于内置的 set 类型来说非常合适。

首先把基础列表变成一个集合：

mammals = set(["gorilla","cat","rat","chimpanzee","dog","beaver"])

然后使用交集函数（被比较的列表不一定要是集合）

>>> mammals.intersection(["orangutan","chimpanzee","human","gorilla"])
set(['gorilla', 'chimpanzee'])

Answer 2

使用集合：

mammals = ["gorilla","cat","rat","chimpanzee","dog","beaver"]
apes =  ["orangutan","chimpanzee","human","gorilla"]
print set(mammals).intersection(apes)

打印输出

set(['gorilla', 'chimpanzee'])

Answer 3

list(set(mammals) & set(apes))

set(mammals) & set(apes)

如果你需要的结果是一个列表，但如果你可以接受把它保持为一个集合，那就直接这样做：