Question

我目前正在研究用Python 2.6.6编写的程序。它使用的字典如下：

{ 'somekeystring': someobject("a name", 1, 3),
'anotherkey': someobject("another name", 2, 2),
'keythree': someobject("third name", 3, 1) }

该对象具有以下属性：

name
startOrder
stopOrder

我想要完成的是将字典排序。一次通过someobject.startOrder，一次通过someobject.stopOrder。

我试过了

sortedByStartOrder = sorted(mydict.iteritems(), key=lambda x: x[1].startOrder)

但这似乎不起作用。无论我在上面的示例中使用startOrder还是stopOrder，列表项都按相同的顺序排序。

任何提示？

Answer 1

这个例子似乎对我有用：

class P:
    def __init__(self, x):
        self.x = x

d = { 'how': P(3), 'hi': P(2), 'you': P(5), 'are': P(4) }
print list(d.iteritems())
print sorted(d.iteritems(), key=lambda x: x[1].x)

产生

>> [('how', <__main__.P instance at 0x7f92028e52d8>), ('you', <__main__.P instance at 0x7f92028e5368>), ('hi', <__main__.P instance at 0x7f92028e5320>), ('are', <__main__.P instance at 0x7f92028e53b0>)]
>> [('hi', <__main__.P instance at 0x7fc210e6c320>), ('how', <__main__.P instance at 0x7fc210e6c2d8>), ('are', <__main__.P instance at 0x7fc210e6c3b0>), ('you', <__main__.P instance at 0x7fc210e6c368>)]

我猜这个问题不在排序本身;在您尝试排序的结构中可能存在问题。

Answer 2

非常感谢您的所有投入！我写了一个像Odexios＆＃39;这样的简单例子。

与之完美配合

sortedByStartOrder = sorted(mydict.iteritems(), key=lambda x: x[1].startOrder)

线索是：

我猜这个问题不在排序本身;可能有在你尝试排序的结构中出现了问题。

我查看了代码并找到了一个代码段，我将两个变量都设置为0.

Answer 3

您可以定义sorted内置于关键字cmp内置def sort_by_start(obj1, obj2): # this is just an example using '>' and '<' as comparison operators # you can compare 1 and 2 using whatever methods you wish; just # return a negative number if 1 is less than 2, 0 if they are equal # or a positive number if 1 is greater than 2. if obj1.startOrder < obj2.startOrder: return -1 if obj1.startOrder > obj2.startOrder: return 1 else: return 0 sorted(mydict.iteritems(), cmp=sort_by_start, key=lambda x: x[1]) def sort_by_stop(obj1, obj2): # let's make this more specific: imagine `stopOrder` is a string, # but we need to remove a prefix before we sort them alphabetically common_prefix = 'id-' # or whatever comp1 = obj1.stopOrder.lstrip(common_prefix) comp2 = obj2.stopOrder.lstrip(common_prefix) if comp1 < comp2: return -1 if comp1 > comp2: return 1 else: return 0 sorted(mydict.iteritems(), cmp=sort_by_stop, key=lambda x: x[1])的{{3}}。

    HTTP Status 500 - Internal Server Error

type Exception report

messageInternal Server Error

descriptionThe server encountered an internal error that prevented it from fulfilling this request.

exception

javax.servlet.ServletException: 
            SELECT name, city, state FROM customer
        : La tabla/vista 'CUSTOMER' no existe.
root cause

java.sql.SQLSyntaxErrorException: La tabla/vista 'CUSTOMER' no existe.
root cause

org.apache.derby.client.am.SqlException: La tabla/vista 'CUSTOMER' no existe.
note The full stack traces of the exception and its root causes are available in the GlassFish Server Open Source Edition 4.1 logs.

按照作为实例属性的值对字典进行排序

3 个答案: