Python: sort list based on value present in dictionary -

consider list x = [obj1,obj2,obj3,obj4,obj5] grouped in manner

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

and have dict

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

i have sort list x considering obj of same frequency , sort based on values of obj present in dict , final result should

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

1st consider obj1 , obj2 because belong same frequency , sort looking values in dict y. obj2 value 50 , obj1 value 40.

so list x sorted such 1st element obj2 followed obj1

and have same next set of objects belonging same frequency , sort based on value present in dict y how do ?

this code uses tuple (frequency, value-in-y) sort key; list sorted in reverse order, highest frequency comes first (was not specified in question, if wrong, can use -i.freq there); objects having frequencies sorted second item in tuple (the value 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]