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
也许我想多了，但是…
str_l = next((k for k,v in locals().items() if id(l) == id(v)))
>>> bar = True
>>> foo = False
>>> my_dict=dict(bar=bar, foo=foo)
>>> next((k for k,v in locals().items() if id(bar) == id(v)))
'bar'
>>> next((k for k,v in locals().items() if id(foo) == id(v)))
'foo'
>>> next((k for k,v in locals().items() if id(my_dict) == id(v)))
'my_dict'
其他回答
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)
在python 2.7及更新版本中，还有字典理解，这使得它更短一些。如果可能的话，我会使用getattr代替eval (eval是邪恶的)，就像在顶部的答案。Self可以是任何有你正在看的上下文的对象。它可以是一个对象或locals=locals()等。
{name: getattr(self, name) for name in ['some', 'vars', 'here]}
使用python-varname你可以很容易地做到这一点:
PIP安装python-varname
from varname import Wrapper
foo = Wrapper(True)
bar = Wrapper(False)
your_dict = {val.name: val.value for val in (foo, bar)}
print(your_dict)
# {'foo': True, 'bar': False}
声明:我是python-varname库的作者。
在阅读这篇文章时，我看到了很多可怕的摩擦。给予很容易 一个糟糕的答案，然后让别人给出正确的答案。总之，这是我的发现。
来自:[effbot.org] (http://effbot.org/zone/python-objects.htm#names)
名称有点不同——它们不是对象的真正属性，对象本身也不知道它叫什么。
对象可以有任意数量的名称，也可以没有名称。
名称存在于名称空间中(例如模块名称空间、实例名称空间、函数的本地名称空间)。
注意:它说对象本身不知道它叫什么，所以这是线索。Python对象不是自引用的。然后它说，名称存在于名称空间中。我们在TCL/TK中有这个。所以也许我的回答会有所帮助(但它确实帮助了我)
jj = 123
print eval("'" + str(id(jj)) + "'")
print dir()
166707048 ['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj'] So there is 'jj' at the end of the list. Rewrite the code as: jj = 123 print eval("'" + str(id(jj)) + "'") for x in dir(): print id(eval(x)) 161922920 ['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj'] 3077447796 136515736 3077408320 3077656800 136515736 161922920 This nasty bit of code id's the name of variable/object/whatever-you-pedantics-call-it. So, there it is. The memory address of 'jj' is the same when we look for it directly, as when we do the dictionary look up in global name space. I'm sure you can make a function to do this. Just remember which namespace your variable/object/wypci is in. QED.
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 []
'''
