它不会返回变量的名称，但您可以轻松地从全局变量创建字典。

class CustomDict(dict):
    def __add__(self, other):
        return CustomDict({**self, **other})

class GlobalBase(type):
    def __getattr__(cls, key):
        return CustomDict({key: globals()[key]})

    def __getitem__(cls, keys):
        return CustomDict({key: globals()[key] for key in keys})

class G(metaclass=GlobalBase):
    pass

x, y, z = 0, 1, 2

print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}

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 []
'''

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。

我上传了一个解决方案到pypi。它是一个模块，定义了c#的nameof函数的等价物。

它遍历被调用帧的字节码指令，获取传递给它的变量/属性的名称。这些名称可以在函数名后面的LOAD指令的.argrepr中找到。

下面是我创建的读取变量名的函数。它更通用，可以用于不同的应用:

def get_variable_name(*variable):
    '''gets string of variable name
    inputs
        variable (str)
    returns
        string
    '''
    if len(variable) != 1:
        raise Exception('len of variables inputed must be 1')
    try:
        return [k for k, v in locals().items() if v is variable[0]][0]
    except:
        return [k for k, v in globals().items() if v is variable[0]][0]

在特定的问题中使用:

>>> foo = False
>>> bar = True
>>> my_dict = {get_variable_name(foo):foo, 
               get_variable_name(bar):bar}
>>> my_dict
{'bar': True, 'foo': False}

大多数对象没有__name__属性。(类、函数和模块可以;还有其他内置类型吗?)

除了print("my_var")，你还期望print(my_var.__name__)有什么?你能直接使用字符串吗?

你可以"slice" a dict:

def dict_slice(D, keys, default=None):
  return dict((k, D.get(k, default)) for k in keys)

print dict_slice(locals(), ["foo", "bar"])
# or use set literal syntax if you have a recent enough version:
print dict_slice(locals(), {"foo", "bar"})

另外:

throw = object()  # sentinel
def dict_slice(D, keys, default=throw):
  def get(k):
    v = D.get(k, throw)
    if v is not throw:
      return v
    if default is throw:
      raise KeyError(k)
    return default
  return dict((k, get(k)) for k in keys)