我想知道python中是否有办法调用实例变量的名称?例如,如果我定义一个类
>>>class A(object):
... def get_instance_name(self):
... return # The name of the instance variable
>>>obj = A()
>>>obj.get_instance_name()
obj
>>>blah = A()
>>>blah.get_instance_name()
blah
答案 0 :(得分:3)
提出异常。它不仅是发出错误信号的合适方式,而且对调试也更有用。回溯包括执行方法调用的行,还包括其他行,行号,函数名等,这些对于调试而言只比变量名更有用。例如:
class A:
def do(self, x):
if x < 0:
raise ValueError("Negative x")
def wrong(a, x):
a.do(-x)
wrong(A(), 1)
如果未捕获到异常,则会提供与此类似的回溯:
Traceback (most recent call last):
File "...", line 1, in <module>
wrong(A(), 1)
File "...", line 7, in wrong
a.do(-x)
File "...", line 4, in do
raise ValueError("Negative x")
ValueError: Negative x
您也可以使用traceback
模块以编程方式获取此信息,即使没有例外(print_stack
和朋友)。
答案 1 :(得分:1)
globals()
返回一个代表模块命名空间的字典(命名空间不是这个字典,后者只代表它)
class A(object):
def get_instance_name(self):
for name,ob in globals().iteritems():
if ob is self:
return name
obj = A()
print obj.get_instance_name()
blah = A()
print blah.get_instance_name()
tu = (obj,blah)
print [x.get_instance_name() for x in tu]
结果
obj
blah
['obj', 'blah']
考虑到这些评论,我写了这个新代码:
class A(object):
def rondo(self,nameinst,namespace,li,s,seen):
for namea,a in namespace.iteritems():
if a is self:
li.append(nameinst+s+namea)
if namea=='__builtins__':
#this condition prevents the execution to go
# in the following section elif, so that self
# isn't searched among the cascading attributes
# of the builtin objects and the attributes.
# This is to avoid to explore all the big tree
# of builtin objects and their cascading attributes.
# It supposes that every builtin object has not
# received the instance, of which the names are
# searched, as a new attribute. This makes sense.
for bn,b in __builtins__.__dict__.iteritems():
if b is self:
li.append(nameinst+'-'+b)
elif hasattr(a,'__dict__') \
and not any(n+s+namea in seen for n in seen)\
and not any(n+s+namea in li for n in li):
seen.append(nameinst+s+namea)
self.rondo(nameinst+s+namea,a.__dict__,li,'.')
else:
seen.append(nameinst+s+namea)
def get_instance_name(self):
li = []
seen = []
self.rondo('',globals(),li,'')
return li if li else None
使用以下
bumbum = A()
blah = A()
print "bumbum's names:\n",bumbum.get_instance_name()
print "\nmap(lambda y:y.get_instance_name(), (bumbum,blah) :\n",map(lambda y:y.get_instance_name(), (bumbum,blah))
print "\n[y.get_instance_name() for y in (bumbum,blah)] :\n",[y.get_instance_name() for y in (bumbum,blah)]
结果是
bumbum's names:
['bumbum']
map(lambda y:y.get_instance_name(), (bumbum,blah) :
[['bumbum'], ['blah']]
[y.get_instance_name() for y in (bumbum,blah)] :
[['bumbum', 'y'], ['blah', 'y']]
第二个列表理解显示必须小心使用函数get_instance_name()。在列表comp中,标识符y
依次分配给(bumbum,blah)的每个元素,然后finction将其作为实例的名称找到它!
现在,情况更为复杂:
ahah = A() # ahah : first name for this instance
class B(object):
pass
bobo = B()
bobo.x = ahah # bobo.x : second name for ahah
jupiter = bobo.x # jupiter : third name for ahah
class C(object):
def __init__(self):
self.azerty = jupiter # fourth name for ahah
ccc = C()
kkk = ccc.azerty # kkk : fifth name for ahah
bobo.x.inxnum = 1005
bobo.x.inxwhat = kkk # bobo.x.inxwhat : fifth name for ahah
# Since bobo.x is instance ahah, this instruction also
# creates attribute inxwhat in ahah instance's __dict__ .
# Consequently, instance ahah having already 5 names,
# this instruction adds 5 additional names, each one
# ending with .inxwhat
# By the way, this kkk being ahah itself, it results that ahah
# is the value of its own attribute inxwhat.
print ahah.get_instance_name()
结果
['bobo.x', 'bobo.x.inxwhat',
'ahah', 'ahah.inxwhat',
'jupiter', 'jupiter.inxwhat',
'kkk', 'kkk.inxwhat',
'ccc.azerty', 'ccc.azerty.inxwhat']
我同意判断这个解决方案有点重,如果编码器认为他需要这么重的功能,可能是因为算法不是最优的。但我觉得有趣的是,虽然看起来不太明显,但可以用Python做到这一点。
顺便说一句,我说重,不是hacky,我发现它不是hacky。
答案 2 :(得分:0)
a1 = a2 = a3 = A()
A()实例的名称是什么?