我有一个从数据库中两个字段读取的值字典：字符串字段和数字字段。字符串字段是唯一的，因此它是字典的关键字。
我可以按关键字排序，但如何根据值排序？
注意：我在这里读过堆栈溢出问题。如何根据字典的值对字典列表进行排序？并且可能会更改我的代码，使其具有字典列表，但由于我确实不需要字典列表，我想知道是否有更简单的解决方案来按升序或降序排序。
当前回答
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
其他回答
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；-）
我刚刚从Python for Everyone学习了相关技能。
您可以使用临时列表来帮助您对词典进行排序：
# Assume dictionary to be:
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# Create a temporary list
tmp = []
# Iterate through the dictionary and append each tuple into the temporary list
for key, value in d.items():
tmptuple = (value, key)
tmp.append(tmptuple)
# Sort the list in ascending order
tmp = sorted(tmp)
print (tmp)
如果要按降序排序列表，只需将原始排序行更改为：
tmp = sorted(tmp, reverse=True)
使用列表理解，一行是：
# Assuming the dictionary looks like
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# One-liner for sorting in ascending order
print (sorted([(v, k) for k, v in d.items()]))
# One-liner for sorting in descending order
print (sorted([(v, k) for k, v in d.items()], reverse=True))
样本输出：
# Ascending order
[(1.0, 'orange'), (500.1, 'apple'), (789.0, 'pineapple'), (1500.2, 'banana')]
# Descending order
[(1500.2, 'banana'), (789.0, 'pineapple'), (500.1, 'apple'), (1.0, 'orange')]
除了使用内置模块等，我尝试手动解决它。。。
首先，我制作了一个函数，其任务是返回dict的每个项的最小值：
def returnminDict(_dct):
dict_items = _dct.items()
list_items = list(dict_items)
init_items = list_items[0]
for i in range(len(list_items)):
if list_items[i][1] > init_items[1]:
continue
else:
init_items = list_items[i]
return init_items
第二，现在我们有一个函数，它返回一个具有最小值的项。然后我做了一个新的格言，并在格言上循环：
def SelectDictSort(_dct):
new_dict = {}
while _dct:
mindict = returnminDict(_dct)
new_dict.update(dict((mindict,)))
_dct.pop(mindict[0])
return new_dict
我尝试使用SelectDictSort（｛2:5，5:1，4:3，1:1，0:1，9:2，8:2｝）。它将返回：
{0: 1, 1: 1, 5: 1, 8: 2, 9: 2, 4: 3, 2: 5}
嗯……我不知道哪一个是正确的，但这是我尝试过的。。。
您还可以使用可以传递给参数键的自定义函数。
def dict_val(x):
return x[1]
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=dict_val)
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)]
