在Python中，相同的函数以相反的顺序给出不同的结果。为什么？

Question

我用的是这个代码：

def copy_part_of_space(row,column,lenght):
    #Copy String to Presentation Space (15)
    #Prerequisite Connect Presentation Space
    #Prerequisite function: connect_pcomm(presentation_space)    
    function_number = c_int(8)
    data_string = create_string_buffer(lenght*2*2) #number of unicode char *2*2
    lenght = c_int(lenght)
    ps_position = c_int(((row - 1) * 80)+ column)
    foo = hllapi(byref(function_number), data_string, byref(lenght), byref(ps_position))
    data_string.value
    return {{
        0 : 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.',
        1 : 'Your program is not connected to a host session.',
        4 : 'The host presentation space contents were copied. The connected host     presentation space was waiting for host response.',
        5 : 'The host presentation space was copied. The keyboard was locked.',
        9 : 'A system error was encountered.',
        'x' : 'Undocumented error found. Run in circles.',
        }.get(foo, 'x'),data_string.value}

其思想是从终端复制一些信息；函数需要返回状态信息(使用字典和0、1、4、5、9、x参数)和复制的信息--使用data_string.value。

为了运行一些测试，我使用了以下代码，该代码使用了上面的函数：

for a in range(15,22):
    print copy_part_of_space(a,7,8)

这是结果：

   set(['The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.', '36343581'])
   set(['36343663', 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.'])
   set(['The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.', '36343708'])
   set(['36344673', 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.'])
   set(['36344740', 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.'])
   set(['The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.', '36344758'])
   set(['36344869', 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.'])

正如您所看到的，有时我会在从主机应用程序复制的内容之前获得状态信息，比如第一行。

但有时我会得到在状态信息之前复制的信息，比如第二行。

我是不熟悉使用dict返回信息，所以我想这可能是一个问题，特别是当我试图返回两个变量时。

有人能解释一下为什么会发生这种事吗？

我知道我可以简单地使用dict并在传递返回之前将返回信息保存到一个变量中，但是我真的认为这是一个更优雅的解决方案--我错了吗？

Answer 1

set是无序的(更好的是，它们的顺序是任意的)。除了使用有序的数据类型之外，您什么也做不了。

例如，通过删除set构造函数{...}

return {
    0 : 'The host presentation space contents were copied to the application program. The target presentation space was active, and the keyboard was unlocked.',
    1 : 'Your program is not connected to a host session.',
    4 : 'The host presentation space contents were copied. The connected host     presentation space was waiting for host response.',
    5 : 'The host presentation space was copied. The keyboard was locked.',
    9 : 'A system error was encountered.',
    'x' : 'Undocumented error found. Run in circles.',
    }.get(foo, 'x'), data_string.value

现在，这段代码返回一个元组 (第一个元素是来自“错误消息字典”的查找结果，第二个元素包含在data_string.value中)。

Answer 2

您将具体返回一个set，它被定义为一个无序的数据类型。也就是说，集合的元素可以按任何顺序返回。集合是为成员资格测试(if x in set:)优化的。集合就像字典中的键:它们可以按任何顺序迭代。

我认为更好的数据类型应该是元组：return (a, b)

那么，结果将始终是相同的顺序。

注意文字表示法的差异：

• 字典使用冒号对项。：{'a': 'b', 'c': 'd')。
• Set没有冒号，只是任意排序的项：{'a', 'b', 'c', 'd'}。
• 元组使用括号：('a', 'b', 'c', 'd')
• 列表使用方括号，并且是可变的：['a', 'b', 'c', 'd']