Python：基于字典中存在的值对列表进行排序

Question

Consider list x = [obj1,obj2,obj3,obj4,obj5] they are grouped in a manner where

obj1.freq = obj2.freq = frequency1 
and 
obj3.freq = obj4.freq = obj5.freq = frequency2 

and I have a dict

y = {obj1 : 40, obj2 :50, obj3:60, obj4:10, obj5:70, obj6:30, obj7:20}

I have to SORT list x by considering obj of same frequency and sort based on the values of the obj present in dict and my final result should be

x = [obj2,obj1, obj5,obj3,obj4]

1st consider obj1 and obj2 only because they belong to same frequency and sort looking into their values in dict y . obj2 value is 50 and obj1 value is 40 .

so list x now will be sorted such that its 1st element will be obj2 followed by obj1

and i have to do the same for next set of objects belonging to same frequency and sort based on the value present in dict y How do i do it ?

Answer 1

Well, I guess you can do it in this way:

x = [obj1,obj2,obj3,obj4,obj5]
y = {obj1 : 40, obj2 :50, obj3:60, obj4:10, obj5:70, obj6:30, obj7:20}

sorted_list=sorted(lambda e: (e.freq, 0-y[e]), x)

Note, you need 0-y[e] if you want order from bigger values to smaller. Normally it can be reached by reverse=True , but in our case it will affect sorting by frequency as well

Answer 2

This code uses the tuple (frequency, value-in-y) as the sort key; the list is sorted in reverse order, so that the highest frequency comes first (was not specified in the question, if this is wrong, you can use -i.freq there); objects having frequencies get sorted by the second item in the tuple (the value from dictionary y , if any, or 0:

class Obj:
    def __init__(self, name, freq):
        self.freq = freq
        self.name = name

    def __repr__(self):
        return self.name


obj1 = Obj('obj1',42)
obj2 = Obj('obj2',42)
obj3 = Obj('obj3',6)
obj4 = Obj('obj4',6)
obj5 = Obj('obj5',6)
obj6 = Obj('obj6',332)
obj7 = Obj('obj7',123)


x = [obj2, obj1, obj5, obj3, obj4]
y = {obj1:40, obj2:50, obj3:60, obj4:10, obj5:70, obj6:30, obj7:20}

print(sorted(x, key=lambda i: (i.freq, y.get(i, 0)), reverse=True))

Prints

[obj2, obj1, obj5, obj3, obj4]