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

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;-)

其他回答

这是一个在d.values()和d.keys()上使用zip的解决方案。此链接(在Dictionary视图对象上)下面的几行是:

这允许使用zip()创建(value,key)对:pairs=zip(d.values(),d.keys())。

因此,我们可以执行以下操作:

d = {'key1': 874.7, 'key2': 5, 'key3': 8.1}

d_sorted = sorted(zip(d.values(), d.keys()))

print d_sorted 
# prints: [(5, 'key2'), (8.1, 'key3'), (874.7, 'key1')]

您可以使用:

sorted(d.items(), key=lambda x: x[1])

这将根据字典中每个条目的值从最小到最大对字典进行排序。

要按降序排序,只需添加reverse=True:

sorted(d.items(), key=lambda x: x[1], reverse=True)

输入:

d = {'one':1,'three':3,'five':5,'two':2,'four':4}
a = sorted(d.items(), key=lambda x: x[1])    
print(a)

输出:

[('one', 1), ('two', 2), ('three', 3), ('four', 4), ('five', 5)]

我有同样的问题,我这样解决了:

WantedOutput = sorted(MyDict, key=lambda x : MyDict[x]) 

(回答“不可能对字典进行排序”的人没有读到这个问题!事实上,“我可以对键进行排序,但如何根据值进行排序?”显然意味着他想要一个根据其值排序的键列表。)

请注意,顺序没有很好地定义(具有相同值的键在输出列表中的顺序是任意的)。

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
from django.utils.datastructures import SortedDict

def sortedDictByKey(self,data):
    """Sorted dictionary order by key"""
    sortedDict = SortedDict()
    if data:
        if isinstance(data, dict):
            sortedKey = sorted(data.keys())
            for k in sortedKey:
                sortedDict[k] = data[k]
    return sortedDict