在Python中，相同的函数以相反的顺序给出不同的结果。为什么？

我正在使用以下代码：

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}

想法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-使用数据_string.value

为了运行一些测试，我使用了使用上述函数的代码：

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

结果如下：

   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.'])

如您所见，有时我会在从主机应用程序复制内容之前获取状态信息，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

我不熟悉使用

dict

返回信息，所以我想这可能是个问题，特别是当我试图返回两个变量时

有人能解释为什么会这样吗

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变量中，但我真的认为这是一个更优雅的解决方案-我错了吗？

是无序的（或者更好，它们的顺序是任意的）。对此，除了使用有序数据类型之外，您无法执行任何操作

例如，通过删除

集合

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，该代码返回一个值（第一个元素是“错误消息字典”的查找结果，第二个元素包含在

数据\u string.value

中）

您具体返回的是一个

集合

，该集合被定义为无序数据类型。也就是说，集合的元素可以以任何顺序返回。集合针对成员资格测试进行了优化（

如果集合中有x:

）。集合就像字典的键：它们可以按任意顺序迭代

我想对您来说更好的数据类型应该是元组：

return（a，b）

然后结果将始终以相同的顺序排列

请注意文字符号的差异：

• 字典有冒号来配对项。{'a'：'b'，'c'：'d'）
• 集合没有冒号，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']

---

它们的顺序是任意的

，但在你做某事时可能会改变。这种情况下通常要做的事情是返回一个元组。@MarkRansom:这就是当

集

构造函数被删除时发生的情况。我想你在我撰写注释时或之后用元组编辑了示例。@MarkRansom:可能是在你编写注释时nting；）忍者编辑FTW！元组通常用括号括起来，但正是逗号使其成为元组，因此

'a'，b'，c'，d'

也是元组。括号是可选的，但空元组除外，或者出于语法原因（例如，在没有paren的函数参数列表中，逗号将分隔参数）@Duncan:没错。一元素元组也需要逗号。可以说逗号创建元组，但它们不是语法的一部分。对于一元素元组，只需要逗号，不需要括号：

x=1，

将创建元组。