Python dict.get(k)返回none,即使键存在
也许我对python字典的理解不太好。但问题是 有没有发生过这样的情况,一对Python dict.get(k)返回none,即使键存在,python,python-3.x,dictionary,Python,Python 3.x,Dictionary,也许我对python字典的理解不太好。但问题是 有没有发生过这样的情况,一对{yelk:shell}存在在字典中说鸡蛋,但是鸡蛋.get(yelk)可以返回无 因此,在一个大型代码中,我对一个字典执行多个get操作,在某些迭代之后,我观察到这种情况 >>> for key, value in nodehashes.items(): ... print(key, nodehashes.get(key), value) ...................
{yelk:shell}
存在在字典中说鸡蛋,但是鸡蛋.get(yelk)
可以返回无
因此,在一个大型代码中,我对一个字典执行多个get
操作,在某些迭代之后,我观察到这种情况
>>> 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
............................
............................
如果在可变对象上有一个自定义的\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
方法,则可以复制该方法:
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
A类:
定义散列(自我):
返回散列(self.a)
>>>a1=A()
>>>a2=A()
>>>a1.a=1
>>>a2.a=2
>>>d={a1:1,a2:2}
>>>a1.a=3
>>>d.项目()
口述项目([(1),(2)])
>>>d.get(a1)
没有一个
您可以看到d.items()
仍然可以访问这两个A
对象,但是get
无法再找到它,因为hash
值已更改。这是并发代码吗?有可能还有另一个线程/。。。这可以在迭代时修改节点hash
?不,它不是并发的。类节点
是否定义了\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu,classs节点同时定义为'def uuu eq uu(self,other):if(self.x==other.x)和(self.y==other.y):return True-else:return False-def uu hash uu(self):return hash(tuple([self.x,self.y])以及任何与等式不一致的hash方法。在散列数据结构中考虑<代码>浮点('n')的行为。但是我怀疑你的猜测是正确的。@juanpa不一致是一个问题,但在你的情况下,代码将是完全确定的,它显然不在这里。嗯,这实际上解释了这一点,所以我必须考虑另一种方法来识别节点,而不是散列方法。@JaswantP你可以使用id(node)
作为dict中的键。@JaswantP好的,是的,id将是相同的。不过,我忘了,如果定义的是\uuuuuueq\uuuuuu
而不是\uuuuuuuuu散列
,那么\uuuuuuuu散列
会隐式设置为无
,这将在尝试对其进行散列时引发类型错误。因此,您可以使用\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
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