list列表

 

1. 判断list包含关系
   

   a = [1, 2, 3, 4, 5]
   b = [3, 4, 5, 3]
   if [v for v in b if v not in a]:
       print('a doesn\'t comprise b')
   else:
       print('a comprise b')

   采用列表解析式
   
   

   a = [1, 2, 3, 4, 5]
   b = [3, 4, 5, 3]
   
   if set(b)<=set(a):
       print('a embody b')
   else:
       print('a doesn\'t embody b')

   采用集合,集合可以直接比较
   
   

   a = [1, 2, 3, 4, 5]
   b = [3, 4, 5, 3]
   
   tmp=a[:]
   for v in b:
       if v in tmp:
           tmp.remove(v)
       else:
           print('a doesn\'t embody b')
           break
   else:
       print('a embody b')
   del tmp

   上述方法考虑了重复元素的情况