It is not possible to sort a dict, only to get a representation of a dict that is sorted. Dicts are inherently orderless, but other types, such as lists and tuples, are not. So you need a sorted representation, which will be a list—probably a list of tuples.



In [8]:

    
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
#itemgetter(0) if you want sort by key
sorted_x = sorted(x.items(), key=operator.itemgetter(1),reverse=True) 
for k,v in sorted_x:
    print(k,v)



In [14]:

    
print(sorted_x)
print(type(sorted_x))
print(type(sorted_x[0]))









    



[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
<class 'list'>
<class 'tuple'>

Other Approach



In [2]:

    
#Less efficient
mydict = {'a':1,'b':3,'c':2}
sorted(mydict.items(), key=lambda x: x[1])









    Out[2]:





[('a', 1), ('c', 2), ('b', 3)]



In [10]:

    
#Less efficient
mydict = {'a':1,'b':3,'c':2}
sorted(mydict, key=mydict.get)









    Out[10]:





['a', 'c', 'b']



In [14]:

    
for w in sorted(mydict, key=mydict.get, reverse=True):
  print(w, mydict[w])



In [11]:

    
sorted(mydict, key=lambda key: mydict[key])









    Out[11]:





['a', 'c', 'b']



In [2]:

    
mydict = {'a':1,'b':3,'c':2}
sorted(mydict, key=lambda key: mydict[key],reverse=True)









    Out[2]:





['b', 'c', 'a']



In [6]:

    
mydict = {'a':1,'b':3,'c':2}
sorted_dict=sorted(mydict, key=lambda key: mydict[key],reverse=True)
for i in sorted_dict:
    print(i,mydict[i])