与汉克·盖伊的回答大致相同：

sorted([(value,key) for (key,value) in mydict.items()])

或者根据John Fouchy的建议进行略微优化：

sorted((value,key) for (key,value) in mydict.items())

months = {"January": 31, "February": 28, "March": 31, "April": 30, "May": 31,
          "June": 30, "July": 31, "August": 31, "September": 30, "October": 31,
          "November": 30, "December": 31}

def mykey(t):
    """ Customize your sorting logic using this function.  The parameter to
    this function is a tuple.  Comment/uncomment the return statements to test
    different logics.
    """
    return t[1]              # sort by number of days in the month
    #return t[1], t[0]       # sort by number of days, then by month name
    #return len(t[0])        # sort by length of month name
    #return t[0][-1]         # sort by last character of month name


# Since a dictionary can't be sorted by value, what you can do is to convert
# it into a list of tuples with tuple length 2.
# You can then do custom sorts by passing your own function to sorted().
months_as_list = sorted(months.items(), key=mykey, reverse=False)

for month in months_as_list:
    print month

为了完整起见，我发布了一个使用heapq的解决方案。注意，此方法适用于数值和非数值

>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> x_items = x.items()
>>> heapq.heapify(x_items)
>>> #To sort in reverse order
>>> heapq.nlargest(len(x_items),x_items, operator.itemgetter(1))
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> heapq.nsmallest(len(x_items),x_items, operator.itemgetter(1))
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]

这适用于3.1.x：

import operator
slovar_sorted=sorted(slovar.items(), key=operator.itemgetter(1), reverse=True)
print(slovar_sorted)

我有同样的问题，我这样解决了：

WantedOutput = sorted(MyDict, key=lambda x : MyDict[x]) 

（回答“不可能对字典进行排序”的人没有读到这个问题！事实上，“我可以对键进行排序，但如何根据值进行排序？”显然意味着他想要一个根据其值排序的键列表。）

请注意，顺序没有很好地定义（具有相同值的键在输出列表中的顺序是任意的）。

遍历dict并按其值降序排序：

$ python --version
Python 3.2.2

$ cat sort_dict_by_val_desc.py 
dictionary = dict(siis = 1, sana = 2, joka = 3, tuli = 4, aina = 5)
for word in sorted(dictionary, key=dictionary.get, reverse=True):
  print(word, dictionary[word])

$ python sort_dict_by_val_desc.py 
aina 5
tuli 4
joka 3
sana 2
siis 1