If the value of list 1 is the index of list 2, how can I sort list 1 according to the list 2 value
You can transform the items in a using a key.  That key is a function of each element of a.
Try this:
a = sorted(a, key=lambda i: b[i])
Note that if any value in a is outside the range of b, this would fail and raise an IndexError: list index out of range.
Based on your description, however, you want the list to be sorted in reverse order, so:
a = sorted(a, key=lambda i: b[i], reverse=True)
Suppose there is a list a[i] which stores index value v, v is the index value of  another list b[v]. I want to according to the values of list b to sort the list a.
For example
a=[0,2,3,1]  b=[7,10,8,6] 
I want the list a become a=[1,2,0,3], is there some concise way to sort list a?