顺序相同直接用“==”进行比较即可
list1 = ["one","two","three"]
list2 = ["one","two","three"]
list1 == list2 
“==”只有成员、成员位置都相同时才返回True，但有时候我们希望只要成员相同、即使成员位置不同也能返回True。
列表本身有sort()内置方法，可对自身成员进行排序；注意sort()方法对自身造成改变。
list1 = ["one","two","three"]
list2 = ["one","three","two"]
list1.sort() == list2.sort()
print(list1) 
上一小节介绍的sort()方法会对列表成员进行重排，但有时候我们并不希望列表本身被改动。
我们可以用一下变量将原先的列表保存起来，但更好的做法是使用sorted()方法，sorted()不改变列表原本顺序而是新生成一个排序后的列表并返回。
list1 = ["one","two","three"]
list2 = ["one","three","two"]
sorted(list1) == sorted(list2)
print(list1)
sorted(list1) 
直接用列表本身进行包含类比较，只能用遍历的方法这是比较麻烦的，使用set()转成集合进行包含比较就简单多了。
list1 = ["one","two","three"]
list2 = ["one","three","two","four"]
set(list1).issubset(set(list2))
set(list2).issuperset(set(list1)) 
2.2 获取两个列表相同成员（交集）
list1 = ["one","two","three","five"]
list2 = ["one","three","two","four"]
set(list1).intersection(set(list2)) 
2.3 获取两个列表不同成员
list1 = ["one","two","three","five"]
list2 = ["one","three","two","four"]
set(list1).symmetric_difference(set(list2)) 
2.4 获取一个列表中不是另一个列表成员的成员（差集）
list1 = ["one","two","three","five"]
list2 = ["one","three","two","four"]
set(list1).difference(set(list2))
set(list2).difference(set(list1)) 
2.5 获取两个列表所有成员（并集）
list1 = ["one","two","three","five"]
list2 = ["one","three","two","four"]
set(list1).union(set(list2)) 
参考：
https://stackoverflow.com/questions/9623114/check-if-two-unordered-lists-are-equal
https://stackoverflow.com/questions/3847386/testing-if-a-list-contains-another-list-with-python
http://www.runoob.com/python3/python3-set.html
转载于:https://www.cnblogs.com/lsdb/p/10275657.html