即使存在键，Python dict.get(k)也不会返回任何值 [英] Python dict.get(k) returns none even though key exists

问题描述

也许我对python字典的理解不好.但这就是问题所在.

Perhaps my understanding of python's dictionary is not good. But here's the problem.

是否曾经在字典中出现 {yolk:shell} 对 ，说的是 eggs ，却是 eggs.get(卵黄)是否可以返回 None ?

Does it ever happen that a {yolk: shell} pair exists in the dictionary say eggs, but a eggs.get(yolk) can return None?

因此，在大型代码中，我对字典执行了多个 get 操作，并且在进行某些迭代之后，我观察到了这种情况.

So, in a large code, I do multiple get operations for a dictionary, and after certain iterations, I observe this situation.

   >>> for key, value in nodehashes.items():
   ...    print(key, nodehashes.get(key), value)
   ............................
   ...........................
   <Graph.Node object at 0x00000264128C4DA0> 3309678211443697093 3309678211443697093
   <Graph.Node object at 0x00000264128C4DD8> 3554035049990170053 3554035049990170053
   <Graph.Node object at 0x00000264128C4E10> None -7182124040890112571  # Look at this!!
   <Graph.Node object at 0x00000264128C4E48> 3268020121048950213 3268020121048950213
   <Graph.Node object at 0x00000264128C4E80> -1243862058694105659 -1243862058694105659
   ............................
   ............................

乍一看，看起来像在代码中的某个地方，键被删除了，但是 nodehashes.items()如何返回正确的键值对?我席卷了整个区域，我根本没有弹出任何项目.这怎么会发生?

At first sight, It looks like somewhere in the code, the key is deleted, but then how does nodehashes.items() return the correct key-value pair? I swept the entire region, I am not popping an item at all. How can this happen?

我知道我自己不发布示例是错误的，但是我真的不知道从哪里开始查找代码，这些节点在开头是散列的，并且只能通过 get进行访问.令人惊讶的是，甚至PyCharm的调试器也显示了键值对的存在.但是 get 返回None.因此，如果以前有人遇到过这种情况，我将不知所措.

I know it's wrong on my part not to post an example, but I really don't know where to start looking in the code, The Nodes are hashed in the beginning and they are only accessed with get. Surprisingly, even PyCharm's debugger shows the key-value pair to exist. But the get returns None. So if anyone else has hit upon this before, I am all ears.

def __eq__(self, other): 
    if (self.x == other.x) and (self.y == other.y): 
        return True 
    else: 
        return False 

def __hash__(self): 
    return hash(tuple([self.x, self.y]))

推荐答案

如果对可变对象具有自定义的 __ hash __ 方法，则可以重现:

You can reproduce that if you have a custom __hash__ method on mutable objects:

class A:
    def __hash__(self):
        return hash(self.a)

>>> a1 = A()
>>> a2 = A()
>>> a1.a = 1
>>> a2.a = 2
>>> d = {a1: 1, a2: 2}
>>> a1.a = 3
>>> d.items()
dict_items([(<__main__.A object at 0x7f1762a8b668>, 1), (<__main__.A object at 0x7f17623d76d8>, 2)])
>>> d.get(a1)
None

您会看到 d.items()仍然可以访问两个 A 对象，但是 get 找不到了，因为 hash 值已更改.

You can see that d.items() still has access to both A objects, but get can't find it anymore, because the hash value has changed.

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