我希望能够以字符串的形式获取变量的名称，但我不知道Python是否有那么多的自省功能。喜欢的东西:
>>> print(my_var.__name__)
'my_var'
我想这样做是因为我有一堆变量，我想把它们转换成一个字典，比如:
bar = True
foo = False
>>> my_dict = dict(bar=bar, foo=foo)
>>> print my_dict
{'foo': False, 'bar': True}
但我想要更自动的。
Python有locals()和vars()，所以我猜有一种方法。
当前回答
I was working on a similar problem. @S.Lott said "If you have the list of variables, what's the point of "discovering" their names?" And my answer is just to see if it could be done and if for some reason you want to sort your variables by type into lists. So anyways, in my research I came came across this thread and my solution is a bit expanded and is based on @rlotun solution. One other thing, @unutbu said, "This idea has merit, but note that if two variable names reference the same value (e.g. True), then an unintended variable name might be returned." In this exercise that was true so I dealt with it by using a list comprehension similar to this for each possibility: isClass = [i for i in isClass if i != 'item']. Without it "item" would show up in each list.
__metaclass__ = type
from types import *
class Class_1: pass
class Class_2: pass
list_1 = [1, 2, 3]
list_2 = ['dog', 'cat', 'bird']
tuple_1 = ('one', 'two', 'three')
tuple_2 = (1000, 2000, 3000)
dict_1 = {'one': 1, 'two': 2, 'three': 3}
dict_2 = {'dog': 'collie', 'cat': 'calico', 'bird': 'robin'}
x = 23
y = 29
pie = 3.14159
eee = 2.71828
house = 'single story'
cabin = 'cozy'
isClass = []; isList = []; isTuple = []; isDict = []; isInt = []; isFloat = []; isString = []; other = []
mixedDataTypes = [Class_1, list_1, tuple_1, dict_1, x, pie, house, Class_2, list_2, tuple_2, dict_2, y, eee, cabin]
print '\nMIXED_DATA_TYPES total count:', len(mixedDataTypes)
for item in mixedDataTypes:
try:
# if isinstance(item, ClassType): # use this for old class types (before 3.0)
if isinstance(item, type):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isClass.append(mapping_as_str)
isClass = [i for i in isClass if i != 'item']
elif isinstance(item, ListType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isList.append(mapping_as_str)
isList = [i for i in isList if i != 'item']
elif isinstance(item, TupleType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isTuple.append(mapping_as_str)
isTuple = [i for i in isTuple if i != 'item']
elif isinstance(item, DictType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isDict.append(mapping_as_str)
isDict = [i for i in isDict if i != 'item']
elif isinstance(item, IntType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isInt.append(mapping_as_str)
isInt = [i for i in isInt if i != 'item']
elif isinstance(item, FloatType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isFloat.append(mapping_as_str)
isFloat = [i for i in isFloat if i != 'item']
elif isinstance(item, StringType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isString.append(mapping_as_str)
isString = [i for i in isString if i != 'item']
else:
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
other.append(mapping_as_str)
other = [i for i in other if i != 'item']
except (TypeError, AttributeError), e:
print e
print '\n isClass:', len(isClass), isClass
print ' isList:', len(isList), isList
print ' isTuple:', len(isTuple), isTuple
print ' isDict:', len(isDict), isDict
print ' isInt:', len(isInt), isInt
print ' isFloat:', len(isFloat), isFloat
print 'isString:', len(isString), isString
print ' other:', len(other), other
# my output and the output I wanted
'''
MIXED_DATA_TYPES total count: 14
isClass: 2 ['Class_1', 'Class_2']
isList: 2 ['list_1', 'list_2']
isTuple: 2 ['tuple_1', 'tuple_2']
isDict: 2 ['dict_1', 'dict_2']
isInt: 2 ['x', 'y']
isFloat: 2 ['pie', 'eee']
isString: 2 ['house', 'cabin']
other: 0 []
'''
其他回答
I was working on a similar problem. @S.Lott said "If you have the list of variables, what's the point of "discovering" their names?" And my answer is just to see if it could be done and if for some reason you want to sort your variables by type into lists. So anyways, in my research I came came across this thread and my solution is a bit expanded and is based on @rlotun solution. One other thing, @unutbu said, "This idea has merit, but note that if two variable names reference the same value (e.g. True), then an unintended variable name might be returned." In this exercise that was true so I dealt with it by using a list comprehension similar to this for each possibility: isClass = [i for i in isClass if i != 'item']. Without it "item" would show up in each list.
__metaclass__ = type
from types import *
class Class_1: pass
class Class_2: pass
list_1 = [1, 2, 3]
list_2 = ['dog', 'cat', 'bird']
tuple_1 = ('one', 'two', 'three')
tuple_2 = (1000, 2000, 3000)
dict_1 = {'one': 1, 'two': 2, 'three': 3}
dict_2 = {'dog': 'collie', 'cat': 'calico', 'bird': 'robin'}
x = 23
y = 29
pie = 3.14159
eee = 2.71828
house = 'single story'
cabin = 'cozy'
isClass = []; isList = []; isTuple = []; isDict = []; isInt = []; isFloat = []; isString = []; other = []
mixedDataTypes = [Class_1, list_1, tuple_1, dict_1, x, pie, house, Class_2, list_2, tuple_2, dict_2, y, eee, cabin]
print '\nMIXED_DATA_TYPES total count:', len(mixedDataTypes)
for item in mixedDataTypes:
try:
# if isinstance(item, ClassType): # use this for old class types (before 3.0)
if isinstance(item, type):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isClass.append(mapping_as_str)
isClass = [i for i in isClass if i != 'item']
elif isinstance(item, ListType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isList.append(mapping_as_str)
isList = [i for i in isList if i != 'item']
elif isinstance(item, TupleType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isTuple.append(mapping_as_str)
isTuple = [i for i in isTuple if i != 'item']
elif isinstance(item, DictType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isDict.append(mapping_as_str)
isDict = [i for i in isDict if i != 'item']
elif isinstance(item, IntType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isInt.append(mapping_as_str)
isInt = [i for i in isInt if i != 'item']
elif isinstance(item, FloatType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isFloat.append(mapping_as_str)
isFloat = [i for i in isFloat if i != 'item']
elif isinstance(item, StringType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isString.append(mapping_as_str)
isString = [i for i in isString if i != 'item']
else:
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
other.append(mapping_as_str)
other = [i for i in other if i != 'item']
except (TypeError, AttributeError), e:
print e
print '\n isClass:', len(isClass), isClass
print ' isList:', len(isList), isList
print ' isTuple:', len(isTuple), isTuple
print ' isDict:', len(isDict), isDict
print ' isInt:', len(isInt), isInt
print ' isFloat:', len(isFloat), isFloat
print 'isString:', len(isString), isString
print ' other:', len(other), other
# my output and the output I wanted
'''
MIXED_DATA_TYPES total count: 14
isClass: 2 ['Class_1', 'Class_2']
isList: 2 ['list_1', 'list_2']
isTuple: 2 ['tuple_1', 'tuple_2']
isDict: 2 ['dict_1', 'dict_2']
isInt: 2 ['x', 'y']
isFloat: 2 ['pie', 'eee']
isString: 2 ['house', 'cabin']
other: 0 []
'''
我写了包巫术健壮地做这种魔法。你可以这样写:
from sorcery import dict_of
my_dict = dict_of(foo, bar)
好吧，几天前我遇到了同样的需求，必须获得一个指向对象本身的变量名。
为什么这么有必要呢?
In short I was building a plug-in for Maya. The core plug-in was built using C++ but the GUI is drawn through Python(as its not processor intensive). Since I, as yet, don't know how to return multiple values from the plug-in except the default MStatus, therefore to update a dictionary in Python I had to pass the the name of the variable, pointing to the object implementing the GUI and which contained the dictionary itself, to the plug-in and then use the MGlobal::executePythonCommand() to update the dictionary from the global scope of Maya.
为了做到这一点，我所做的是:
import time
class foo(bar):
def __init__(self):
super(foo, self).__init__()
self.time = time.time() #almost guaranteed to be unique on a single computer
def name(self):
g = globals()
for x in g:
if isinstance(g[x], type(self)):
if g[x].time == self.time:
return x
#or you could:
#return filter(None,[x if g[x].time == self.time else None for x in g if isinstance(g[x], type(self))])
#and return all keys pointing to object itself
我知道这不是一个完美的解决方案，在全局许多键可以指向同一个对象，例如:
a = foo()
b = a
b.name()
>>>b
or
>>>a
而且这种方法不是线程安全的。如果我错了，请指正。
至少这种方法解决了我的问题，它在全局作用域中获取指向对象本身的任何变量的名称，并将其作为参数传递给插件，供它在内部使用。
我在int(原始整数类)上尝试了这一点，但问题是这些原始类不会被绕过(请纠正使用的技术术语，如果它是错误的)。你可以重新实现int，然后执行int = foo，但a = 3永远不会是foo的对象，而是原语的对象。为了克服这个问题，你必须使用a = foo(3)来让a.name()工作。
应该得到列表然后返回
def get_var_name(**kwargs):
"""get variable name
get_var_name(var = var)
Returns:
[str] -- var name
"""
return list(kwargs.keys())[0]
这在Python中是不可能的，因为Python真的没有“变量”。Python有名称，同一个对象可以有多个名称。
