有一种简单的方法来整理字典。

根据你的问题，

解决方案是:

c={2:3, 1:89, 4:5, 3:0}
y=sorted(c.items())
print y

(其中c是你的字典名。)

这个程序给出如下输出:

[(1, 89), (2, 3), (3, 0), (4, 5)]

如你所愿。

另一个例子是:

d={"John":36,"Lucy":24,"Albert":32,"Peter":18,"Bill":41}
x=sorted(d.keys())
print x

给出输出:['Albert'， 'Bill'， 'John'， 'Lucy'， 'Peter']

y=sorted(d.values())
print y

给出输出:[18,24,32,36,41]

z=sorted(d.items())
print z

给出输出:

[('Albert', 32), ('Bill', 41), ('John', 36), ('Lucy', 24), ('Peter', 18)]

因此，通过将其更改为键、值和项，您可以像您想要的那样打印。希望这能有所帮助!

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)

最简单的解决方案是，你应该得到一个dict键排序顺序的列表，然后遍历dict。例如

a1 = {'a':1, 'b':13, 'd':4, 'c':2, 'e':30}
a1_sorted_keys = sorted(a1, key=a1.get, reverse=True)
for r in a1_sorted_keys:
    print r, a1[r]

以下是输出(按顺序排列)

e 30
b 13
d 4
c 2
a 1

Python字典是无序的。通常，这不是问题，因为最常见的用例是执行查找。

最简单的方法是创建一个集合。OrderedDict以排序顺序插入元素。

ordered_dict = collections.OrderedDict([(k, d[k]) for k in sorted(d.keys())])

如果需要迭代，就像上面其他人建议的那样，最简单的方法是迭代排序的键。的例子,

按键排序打印值:

# create the dict
d = {k1:v1, k2:v2,...}
# iterate by keys in sorted order
for k in sorted(d.keys()):
    value = d[k]
    # do something with k, value like print
    print k, value

获取按键排序的值列表:

values = [d[k] for k in sorted(d.keys())]

对于CPython/PyPy 3.6，以及任何Python 3.7或更高版本，这很容易做到:

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

2.7中两个方法的时间比较显示它们实际上是相同的:

>>> setup_string = "a = sorted(dict({2:3, 1:89, 4:5, 3:0}).items())"
>>> timeit.timeit(stmt="[(k, val) for k, val in a]", setup=setup_string, number=10000)
0.003599141953657181

>>> setup_string = "from collections import OrderedDict\n"
>>> setup_string += "a = OrderedDict({1:89, 2:3, 3:0, 4:5})\n"
>>> setup_string += "b = a.items()"
>>> timeit.timeit(stmt="[(k, val) for k, val in b]", setup=setup_string, number=10000)
0.003581275490432745