如何根据字典的键进行排序?
示例输入:
{2:3, 1:89, 4:5, 3:0}
期望的输出:
{1:89, 2:3, 3:0, 4:5}
如何根据字典的键进行排序?
示例输入:
{2:3, 1:89, 4:5, 3:0}
期望的输出:
{1:89, 2:3, 3:0, 4:5}
Python字典是无序的。通常,这不是问题,因为最常见的用例是查找。
实现你想要的最简单的方法是创建一个collections.OrderedDict
并按排序顺序插入元素。
ordered_dict = collections.OrderedDict([(k, d[k]) for k in sorted(d.keys())])
# 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
按键排序,获取值列表:
values = [d[k] for k in sorted(d.keys())]
for k,value in sorted(d.items()):
更好:避免在循环中再次通过键访问字典。 - Jean-François Fabre此函数将对任何字典进行递归排序,即如果字典中的任何值也是一个字典,则它也将按其键进行排序。 如果您在运行CPython 3.6或更高版本,则可以简单修改使用dict而不是OrderedDict。
from collections import OrderedDict
def sort_dict(d):
items = [[k, v] for k, v in sorted(d.items(), key=lambda x: x[0])]
for item in items:
if isinstance(item[1], dict):
item[1] = sort_dict(item[1])
return OrderedDict(items)
#return dict(items)
最简单的解决方案是获取一个键按排序顺序排列的字典列表,然后遍历该字典。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]
e 30
b 13
d 4
c 2
a 1
大家把事情复杂化了...其实很简单
from pprint import pprint
Dict={'B':1,'A':2,'C':3}
pprint(Dict)
{'A':2,'B':1,'C':3}
dictionary = {1:[2],2:[],5:[4,5],4:[5],3:[1]}
temp=sorted(dictionary)
sorted_dict = dict([(k,dictionary[k]) for i,k in enumerate(temp)])
sorted_dict:
{1: [2], 2: [], 3: [1], 4: [5], 5: [4, 5]}
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)
或者使用 pandas
,
演示:
>>> d={'B':1,'A':2,'C':3}
>>> df=pd.DataFrame(d,index=[0]).sort_index(axis=1)
A B C
0 2 1 3
>>> df.to_dict('int')[0]
{'A': 2, 'B': 1, 'C': 3}
>>>
请参见:
>>> 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
l = dict.keys()
l2 = l
l2.append(0)
l3 = []
for repeater in range(0, len(l)):
smallnum = float("inf")
for listitem in l2:
if listitem < smallnum:
smallnum = listitem
l2.remove(smallnum)
l3.append(smallnum)
l3.remove(0)
l = l3
for listitem in l:
print(listitem)
TreeMap
(https://docs.oracle.com/javase/8/docs/api/java/util/TreeMap.html),它的行为完全符合OP的要求。 - Nayuki