from operator import itemgetter
# if you would like to play with multiple dictionaries then here you go:
# Three dictionaries that are composed of first name and last name.
user = [
    {'fname': 'Mo', 'lname': 'Mahjoub'},
    {'fname': 'Abdo', 'lname': 'Al-hebashi'},
    {'fname': 'Ali', 'lname': 'Muhammad'}
]
#  This loop will sort by the first and the last names.
# notice that in a dictionary order doesn't matter. So it could put the first name first or the last name first. 
for k in sorted (user, key=itemgetter ('fname', 'lname')):
    print (k)

# This one will sort by the first name only.
for x in sorted (user, key=itemgetter ('fname')):
    print (x)

Python字典在Python 3.6之前是无序的。在Python 3.6的CPython实现中，字典保持插入顺序。 从Python 3.7开始，这将成为一种语言特性。

在Python 3.6的更新日志(https://docs.python.org/3.6/whatsnew/3.6.html#whatsnew36-compactdict):

考虑了这个新实现的保序方面 一个实现细节，不应该依赖(这可能 将来会有变化，但希望有这个新词典 在语言中实现了几个版本，然后才更改 语言规范要求所有当前的语义保持有序 以及未来的Python实现;这也有助于保存 向后兼容该语言的旧版本 随机迭代顺序仍然有效，例如Python 3.5)。

Python 3.7文档(https://docs.python.org/3.7/tutorial/datastructures.html#dictionaries):

在字典上执行list(d)将返回所有使用的键的列表 在字典中，按插入顺序(如果你想排序，只需使用 排序(d))。

因此，与以前的版本不同，您可以在Python 3.6/3.7之后对字典进行排序。如果你想对包含子字典在内的嵌套字典进行排序，你可以这样做:

test_dict = {'a': 1, 'c': 3, 'b': {'b2': 2, 'b1': 1}}

def dict_reorder(item):
    return {k: dict_reoder(v) if isinstance(v, dict) else v for k, v in sorted(item.items())}

reordered_dict = dict_reorder(test_dict)

https://gist.github.com/ligyxy/f60f0374defc383aa098d44cfbd318eb

简单:

d = {2:3, 1:89, 4:5, 3:0}
sd = sorted(d.items())

for k,v in sd:
    print k, v

输出:

1 89
2 3
3 0
4 5

我认为最简单的事情是按键对字典进行排序，并将排序的键:值对保存在一个新的字典中。

dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2} 
dict2 = {}                  # create an empty dict to store the sorted values
for key in sorted(dict1.keys()):
    if not key in dict2:    # Depending on the goal, this line may not be neccessary
        dict2[key] = dict1[key]

更清楚地说:

dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2} 
dict2 = {}                  # create an empty dict to store the sorted     values
for key in sorted(dict1.keys()):
    if not key in dict2:    # Depending on the goal, this line may not be  neccessary
        value = dict1[key]
        dict2[key] = value

以下是建议解决方案的性能:

from collections import OrderedDict
from sortedcontainers import SortedDict
import json

keys = np.random.rand(100000)
vals = np.random.rand(100000)

d = dict(zip(keys, vals))

timeit SortedDict(d)
#45.8 ms ± 780 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit sorted(d.items())
#91.9 ms ± 707 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit OrderedDict(sorted(d.items(), key=lambda x: x[0]))
#93.7 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit dict(sorted(dic.items()))
#113 ms ± 824 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit OrderedDict(sorted(dic.items()))
#122 ms ± 2.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

timeit json.dumps(d, sort_keys=True)
#259 ms ± 9.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

如我们所见，格兰特·詹克斯的解决方案是目前为止最快的。

字典本身并没有这样的排序条目，如果你想按某种顺序打印它们，这里有一些例子:

在Python 2.4及以上版本中:

mydict = {'carl':40,
          'alan':2,
          'bob':1,
          'danny':3}

for key in sorted(mydict):
    print "%s: %s" % (key, mydict[key])

给:

alan: 2
bob: 1
carl: 40
danny: 3

(Python低于2.4:)

keylist = mydict.keys()
keylist.sort()
for key in keylist:
    print "%s: %s" % (key, mydict[key])

来源:http://www.saltycrane.com/blog/2007/09/how-to-sort-python-dictionary-by-keys/