有没有办法检查测试为相等（==）的Python对象的（不同）内部结构？

发布于 2024-08-09 06:54:30 字数 518 浏览 9 评论 0原文

昨天我问（“在 Python 2.5 下行为不同的表面上相等的集合列表的情况（我认为……）”）为什么列表 W 构造如下：

r_dim_1_based = range( 1, dim + 1)
set_dim_1_based =  set( r_dim_1_based)

def listW_fill_func( val):
    if (val == 0):
        return set_dim_1_based
    else:
        return set( [val])

W = [ listW_fill_func( A[cid])  
            for cid in r_ncells ]

没有按照我的预期运行。特别是，它的行为不像其他列表那样显示与它相等（another_list == W --> True）。

是否有一个实用程序、技巧、内置的东西可以向我展示这些不同的内部结构？可能会产生类似 C 的对象声明，以便我立即看到我在一种情况下处理指针（列表 W）而在其他情况下处理值？

原文

Yesterday I asked ("A case of outwardly equal lists of sets behaving differently under Python 2.5 (I think …)") why list W constructed as follows:

r_dim_1_based = range( 1, dim + 1)
set_dim_1_based =  set( r_dim_1_based)

def listW_fill_func( val):
    if (val == 0):
        return set_dim_1_based
    else:
        return set( [val])

W = [ listW_fill_func( A[cid])  
            for cid in r_ncells ]

didn't behave as I expected. In particular, it did not behave like other lists that showed equality with it (another_list == W --> True).

Is there a utility, trick, builtin, whatever that would have shown these differing internal structures to me? Something that would have produced perhaps a C-like declaration of the objects so that I would have seen at once that I was dealing with pointers in one case (list W) and values in the others?

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生死何惧 2024-08-16 06:54:30

您在每种情况下都处理引用（更类似于指针而不是值）。您当然可以内省对象对您内心内容的引用 - 例如，如果您有一个列表并想要检查是否有任何项目是相同的引用，

if len(thelist) != len(set(id(x) for x in thelist)): ...

请注意我们在这里讨论的是引用 - 因此，两个相同的引用None 或对 int 值 17 的两个相同引用也会触发相同的警报。当然，您可以继续反省以删除这种情况，例如，如果您认为对同一不可变对象的多个引用很好，则在第一遍中从列表中消除不可变对象 - 例如：

immutyps = int, long, float, tuple, frozenset, str, unicode
mutables = [x for x in thelist if not isinstance(x, immutyps)]
if len(mutables) != len(set(id(x) for x in mutables)):
  cryhavocandletloosethedogsofwar()

但我会质疑投资回报率如此深刻的反省策略！

You're dealing with references in each case (more similar to pointers than to values). You can surely introspect your objects' references to your heart's contents -- for example, if you have a list and want to check if any items are identical references,

if len(thelist) != len(set(id(x) for x in thelist)): ...

DO note that we're talking about references here -- so, two identical references to None, or two identical references to the int value 17, would also trigger the same alarm. Of course you can keep introspecting to remove that case, eliminating immutables from the list in a first pass, for example, if you think that multiple references to the same immutable are fine -- e.g.:

immutyps = int, long, float, tuple, frozenset, str, unicode
mutables = [x for x in thelist if not isinstance(x, immutyps)]
if len(mutables) != len(set(id(x) for x in mutables)):
  cryhavocandletloosethedogsofwar()

but I would question the return-on-investment of such a deep introspection strategy!

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