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)]

在最近的Python2.7中，我们有了新的OrderedDict类型，它可以记住添加项目的顺序。

>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}

>>> for k, v in d.items():
...     print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1

>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}

要从原始字典创建新的有序字典，请按值排序：

>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))

OrderedDict的行为类似于普通dict：

>>> for k, v in d_sorted_by_value.items():
...     print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4

>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])

这是代码：

import operator
origin_list = [
    {"name": "foo", "rank": 0, "rofl": 20000},
    {"name": "Silly", "rank": 15, "rofl": 1000},
    {"name": "Baa", "rank": 300, "rofl": 20},
    {"name": "Zoo", "rank": 10, "rofl": 200},
    {"name": "Penguin", "rank": -1, "rofl": 10000}
]
print ">> Original >>"
for foo in origin_list:
    print foo

print "\n>> Rofl sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rofl")):
    print foo

print "\n>> Rank sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rank")):
    print foo

以下是结果：

起初的

{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}

Rofl

{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}

等级

{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}

在Python 2.7中，只需执行以下操作：

from collections import OrderedDict
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}

# dictionary sorted by key
OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])

# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

复制粘贴自：http://docs.python.org/dev/library/collections.html#ordereddict-示例和食谱

享受；-）

简单如：sorted（dict1，key=dict1.get）

嗯，实际上可以进行“按字典值排序”。最近，我不得不在代码高尔夫（堆栈溢出问题代码高尔夫：单词频率表）中这样做。简言之，问题是这样的：给定一个文本，计算每个单词遇到的频率，并显示一个按频率递减排序的顶级单词列表。

如果您构建一个字典，将单词作为关键字，将每个单词的出现次数作为值，则此处简化为：

from collections import defaultdict
d = defaultdict(int)
for w in text.split():
    d[w] += 1

然后，您可以得到一个单词列表，按使用频率排序，排序（d，key=d.get）-排序将使用单词出现的次数作为排序关键字，在字典关键字上迭代。

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

我写这篇详细的解释是为了说明人们通常所说的“我可以很容易地按关键字对字典进行排序，但我如何按值排序”——我认为最初的文章正试图解决这一问题。解决方案是根据这些值列出一些键，如上所示。