创建单独变量字典的更简单方法?

问题:创建单独变量字典的更简单方法?

我希望能够以字符串的形式获取变量的名称,但是我不知道Python是否具有这么多的自省功能。就像是:

>>> print(my_var.__name__)
'my_var'

我想这样做是因为我想将一堆变量变成一个像这样的字典:

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

但是我想要比这更自动的东西。

Python具有locals()vars(),所以我猜有办法。

I would like to be able to get the name of a variable as a string but I don’t know if Python has that much introspection capabilities. Something like:

>>> print(my_var.__name__)
'my_var'

I want to do that because I have a bunch of variables I’d like to turn into a dictionary like :

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

But I’d like something more automatic than that.

Python have locals() and vars(), so I guess there is a way.


回答 0

您是否正在尝试这样做?

dict( (name,eval(name)) for name in ['some','list','of','vars'] )

>>> some= 1
>>> list= 2
>>> of= 3
>>> vars= 4
>>> dict( (name,eval(name)) for name in ['some','list','of','vars'] )
{'list': 2, 'some': 1, 'vars': 4, 'of': 3}

Are you trying to do this?

dict( (name,eval(name)) for name in ['some','list','of','vars'] )

Example

>>> some= 1
>>> list= 2
>>> of= 3
>>> vars= 4
>>> dict( (name,eval(name)) for name in ['some','list','of','vars'] )
{'list': 2, 'some': 1, 'vars': 4, 'of': 3}

回答 1

正如unwind所说,这实际上不是您在Python中所做的事情-变量实际上是到对象的名称映射。

但是,这是尝试的一种方法:

 >>> a = 1
 >>> for k, v in list(locals().iteritems()):
         if v is a:
             a_as_str = k
 >>> a_as_str
 a
 >>> type(a_as_str)
 'str'

As unwind said, this isn’t really something you do in Python – variables are actually name mappings to objects.

However, here’s one way to try and do it:

 >>> a = 1
 >>> for k, v in list(locals().iteritems()):
         if v is a:
             a_as_str = k
 >>> a_as_str
 a
 >>> type(a_as_str)
 'str'

回答 2

我已经很想这样做了。这种破解与rlotun的建议非常相似,但它是单行的,这对我很重要:

blah = 1
blah_name = [ k for k,v in locals().iteritems() if v is blah][0]

Python 3+

blah = 1
blah_name = [ k for k,v in locals().items() if v is blah][0]

I’ve wanted to do this quite a lot. This hack is very similar to rlotun’s suggestion, but it’s a one-liner, which is important to me:

blah = 1
blah_name = [ k for k,v in locals().iteritems() if v is blah][0]

Python 3+

blah = 1
blah_name = [ k for k,v in locals().items() if v is blah][0]

回答 3

这是一个hack。不适用于所有Python实现发行版(尤其是那些没有的发行版traceback.extract_stack)。

import traceback

def make_dict(*expr):
    (filename,line_number,function_name,text)=traceback.extract_stack()[-2]
    begin=text.find('make_dict(')+len('make_dict(')
    end=text.find(')',begin)
    text=[name.strip() for name in text[begin:end].split(',')]
    return dict(zip(text,expr))

bar=True
foo=False
print(make_dict(bar,foo))
# {'foo': False, 'bar': True}

请注意,此hack很脆弱:

make_dict(bar,
          foo)

(在2行上调用make_dict)将不起作用。

与其尝试根据 foo and bar来生成dict,不如使用字符串变量名 'foo'和来生成dict 'bar'

dict([(name,locals()[name]) for name in ('foo','bar')])

This is a hack. It will not work on all Python implementations distributions (in particular, those that do not have traceback.extract_stack.)

import traceback

def make_dict(*expr):
    (filename,line_number,function_name,text)=traceback.extract_stack()[-2]
    begin=text.find('make_dict(')+len('make_dict(')
    end=text.find(')',begin)
    text=[name.strip() for name in text[begin:end].split(',')]
    return dict(zip(text,expr))

bar=True
foo=False
print(make_dict(bar,foo))
# {'foo': False, 'bar': True}

Note that this hack is fragile:

make_dict(bar,
          foo)

(calling make_dict on 2 lines) will not work.

Instead of trying to generate the dict out of the values foo and bar, it would be much more Pythonic to generate the dict out of the string variable names 'foo' and 'bar':

dict([(name,locals()[name]) for name in ('foo','bar')])

回答 4

在Python中这是不可能的,因为Python实际上没有“变量”。Python有名称,并且同一对象可以有多个名称。

This is not possible in Python, which really doesn’t have “variables”. Python has names, and there can be more than one name for the same object.


回答 5

我认为我的问题将有助于说明为什么这个问题很有用,并且可能会使您对如何回答这个问题有更多的了解。我编写了一个小函数来对代码中的各个变量进行快速内联头检查。基本上,它列出了变量名称,数据类型,大小和其他属性,因此我可以快速发现所犯的任何错误。代码很简单:

def details(val):
  vn = val.__name__                 #  If such a thing existed
  vs = str(val)
  print("The Value of "+ str(vn) + " is " + vs)
  print("The data type of " + vn + " is " + str(type(val)))

因此,如果您遇到一些复杂的字典/列表/元组情况,那么让解释器返回您分配的变量名称将非常有帮助。例如,这是一个奇怪的字典:

m = 'abracadabra'
mm=[]    
for n in m:
  mm.append(n)
mydic = {'first':(0,1,2,3,4,5,6),'second':mm,'third':np.arange(0.,10)}



details(mydic)

The Value of mydic is {'second': ['a', 'b', 'r', 'a', 'c', 'a', 'd', 'a', 'b', 'r', 'a'], 'third': array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.]), 'first': [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]}
The data type of mydic is <type 'dict'>

details(mydic['first'])
The Value of mydic['first'] is (0, 1, 2, 3, 4, 5, 6)]
The data type of mydic['first'] is <type 'list'>

details(mydic.keys())
The Value of mydic.keys() is ['second', 'third', 'first']
The data type of mydic.keys() is <type 'tuple'>

details(mydic['second'][0])
The Value of mydic['second'][0] is a
The data type of mydic['second'][0] is <type 'str'>

我不确定是否将其放在正确的位置,但是我认为这可能会有所帮助。我希望能。

I think my problem will help illustrate why this question is useful, and it may give a bit more insight into how to answer it. I wrote a small function to do a quick inline head check on various variables in my code. Basically, it lists the variable name, data type, size, and other attributes, so I can quickly catch any mistakes I’ve made. The code is simple:

def details(val):
  vn = val.__name__                 #  If such a thing existed
  vs = str(val)
  print("The Value of "+ str(vn) + " is " + vs)
  print("The data type of " + vn + " is " + str(type(val)))

So if you have some complicated dictionary / list / tuple situation, it would be quite helpful to have the interpreter return the variable name you assigned. For instance, here is a weird dictionary:

m = 'abracadabra'
mm=[]    
for n in m:
  mm.append(n)
mydic = {'first':(0,1,2,3,4,5,6),'second':mm,'third':np.arange(0.,10)}



details(mydic)

The Value of mydic is {'second': ['a', 'b', 'r', 'a', 'c', 'a', 'd', 'a', 'b', 'r', 'a'], 'third': array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.]), 'first': [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]}
The data type of mydic is <type 'dict'>

details(mydic['first'])
The Value of mydic['first'] is (0, 1, 2, 3, 4, 5, 6)]
The data type of mydic['first'] is <type 'list'>

details(mydic.keys())
The Value of mydic.keys() is ['second', 'third', 'first']
The data type of mydic.keys() is <type 'tuple'>

details(mydic['second'][0])
The Value of mydic['second'][0] is a
The data type of mydic['second'][0] is <type 'str'>

I’m not sure if I put this in the right place, but I thought it might help. I hope it does.


回答 6

我根据这个问题的答案写了一个简洁实用的函数。我把它放在这里以防有用。

def what(obj, callingLocals=locals()):
    """
    quick function to print name of input and value. 
    If not for the default-Valued callingLocals, the function would always
    get the name as "obj", which is not what I want.    
    """
    for k, v in list(callingLocals.items()):
         if v is obj:
            name = k
    print(name, "=", obj)

用法:

>> a = 4
>> what(a)
a = 4
>>|

I wrote a neat little useful function based on the answer to this question. I’m putting it here in case it’s useful.

def what(obj, callingLocals=locals()):
    """
    quick function to print name of input and value. 
    If not for the default-Valued callingLocals, the function would always
    get the name as "obj", which is not what I want.    
    """
    for k, v in list(callingLocals.items()):
         if v is obj:
            name = k
    print(name, "=", obj)

usage:

>> a = 4
>> what(a)
a = 4
>>|

回答 7

我发现,如果您已经有一个特定的值列表,则使用@S描述的方式。Lotts是最好的。但是,下面描述的方法可以很好地获取整个代码中添加的所有变量和类,而无需提供变量名,尽管您可以根据需要指定它们。可以扩展代码以排除类。

import types
import math  # mainly showing that you could import what you will before d

# Everything after this counts
d = dict(globals())

def kv_test(k,v):
    return (k not in d and 
            k not in ['d','args'] and
            type(v) is not types.FunctionType)

def magic_print(*args):
    if len(args) == 0: 
        return {k:v for k,v in globals().iteritems() if kv_test(k,v)}
    else:
        return {k:v for k,v in magic_print().iteritems() if k in args}

if __name__ == '__main__':
    foo = 1
    bar = 2
    baz = 3
    print magic_print()
    print magic_print('foo')
    print magic_print('foo','bar')

输出:

{'baz': 3, 'foo': 1, 'bar': 2}
{'foo': 1}
{'foo': 1, 'bar': 2}

I find that if you already have a specific list of values, that the way described by @S. Lotts is the best; however, the way described below works well to get all variables and Classes added throughout the code WITHOUT the need to provide variable name though you can specify them if you want. Code can be extend to exclude Classes.

import types
import math  # mainly showing that you could import what you will before d

# Everything after this counts
d = dict(globals())

def kv_test(k,v):
    return (k not in d and 
            k not in ['d','args'] and
            type(v) is not types.FunctionType)

def magic_print(*args):
    if len(args) == 0: 
        return {k:v for k,v in globals().iteritems() if kv_test(k,v)}
    else:
        return {k:v for k,v in magic_print().iteritems() if k in args}

if __name__ == '__main__':
    foo = 1
    bar = 2
    baz = 3
    print magic_print()
    print magic_print('foo')
    print magic_print('foo','bar')

Output:

{'baz': 3, 'foo': 1, 'bar': 2}
{'foo': 1}
{'foo': 1, 'bar': 2}

回答 8

在python 3中这很容易

myVariable = 5
for v in locals():
  if id(v) == id("myVariable"):
    print(v, locals()[v])

这将打印:

myVariable 5

In python 3 this is easy

myVariable = 5
for v in locals():
  if id(v) == id("myVariable"):
    print(v, locals()[v])

this will print:

myVariable 5


回答 9

Python3。使用inspect捕获调用本地命名空间,然后使用此处介绍的想法。正如已经指出的,可以返回多个答案。

def varname(var):
  import inspect
  frame = inspect.currentframe()
  var_id = id(var)
  for name in frame.f_back.f_locals.keys():
    try:
      if id(eval(name)) == var_id:
        return(name)
    except:
      pass

Python3. Use inspect to capture the calling local namespace then use ideas presented here. Can return more than one answer as has been pointed out.

def varname(var):
  import inspect
  frame = inspect.currentframe()
  var_id = id(var)
  for name in frame.f_back.f_locals.keys():
    try:
      if id(eval(name)) == var_id:
        return(name)
    except:
      pass

回答 10

这是我创建的用于读取变量名称的函数。它更通用,可以在不同的应用程序中使用:

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]

要在指定问题中使用它:

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

Here’s the function I created to read the variable names. It’s more general and can be used in different applications:

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]

To use it in the specified question:

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

回答 11

在阅读线程时,我看到了很多摩擦。给出错误答案很容易,然后让某人给出正确答案。无论如何,这就是我所发现的。

来自:[effbot.org](http://effbot.org/zone/python-objects.htm#names

名称有些不同-它们实际上不是对象的属性,并且对象本身不知道它叫什么。

一个对象可以具有任意数量的名称,也可以完全没有名称。

名称存在于命名空间中(例如模块命名空间,实例命名空间,函数的本地命名空间)。

注意:它说对象本身不知道它叫什么,所以这就是线索。Python对象不是自引用的。然后说,名称存在于命名空间中。我们在TCL / TK中有这个。所以也许我的回答会有所帮助(但确实有帮助)

    jj = 123
    打印eval(“'” + str(id(jj))+“'”)
    打印目录()

166707048
['__builtins __','__ doc __','__ file __','__ name __','__ package __','jj']

因此,列表末尾有“ jj”。

将代码重写为:

    jj = 123
    打印eval(“'” + str(id(jj))+“'”)
    对于dir()中的x:
        列印编号(eval(x))

161922920
['__builtins __','__ doc __','__ file __','__ name __','__ package __','jj']
3077447796
136515736
3077408320
3077656800
136515736
161922920

代码ID的这个讨厌的部分是变量/对象/您调用它的名字。

就是这样。当我们直接查找’jj’的内存地址时,就像在全局命名空间中查找字典时一样。我确定您可以创建一个函数来执行此操作。只要记住您的变量/对象/ wypci位于哪个命名空间即可。

QED。

In reading the thread, I saw an awful lot of friction. It’s easy enough to give a bad answer, then let someone give the correct answer. Anyway, here is what I found.

From: [effbot.org] (http://effbot.org/zone/python-objects.htm#names)

The names are a bit different — they’re not really properties of the object, and the object itself doesn’t know what it’s called.

An object can have any number of names, or no name at all.

Names live in namespaces (such as a module namespace, an instance namespace, a function’s local namespace).

Note: that it says the object itself doesn’t know what it’s called, so that was the clue. Python objects are not self-referential. Then it says, Names live in namespaces. We have this in TCL/TK. So maybe my answer will help (but it did help me)


    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.


回答 12

也许我想得太多了,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'

Maybe I’m overthinking this but..

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'

回答 13

import re
import traceback

pattren = re.compile(r'[\W+\w+]*get_variable_name\((\w+)\)')
def get_variable_name(x):
    return pattren.match( traceback.extract_stack(limit=2)[0][3]) .group(1)

a = 1
b = a
c = b
print get_variable_name(a)
print get_variable_name(b)
print get_variable_name(c)
import re
import traceback

pattren = re.compile(r'[\W+\w+]*get_variable_name\((\w+)\)')
def get_variable_name(x):
    return pattren.match( traceback.extract_stack(limit=2)[0][3]) .group(1)

a = 1
b = a
c = b
print get_variable_name(a)
print get_variable_name(b)
print get_variable_name(c)

回答 14

我向pypi上传了一个解决方案。它是一个模块,定义了C#的等效nameof功能。

它通过字节码指令对其所调用的帧进行迭代,以获取传递给它的变量/属性的名称。该名称在找到.argreprLOAD指令下列函数的名称。

I uploaded a solution to pypi. It’s a module defining an equivalent of C#’s nameof function.

It iterates through bytecode instructions for the frame its called in, getting the names of variables/attributes passed to it. The names are found in the .argrepr of LOAD instructions following the function’s name.


回答 15

我编写了包裹法术来稳健地执行这种魔术。你可以写:

from sorcery import dict_of

my_dict = dict_of(foo, bar)

I wrote the package sorcery to do this kind of magic robustly. You can write:

from sorcery import dict_of

my_dict = dict_of(foo, bar)

回答 16

大多数对象没有__name__属性。(类,函数和模块可以使用;是否有内置类型可以包含一个?)

你会期望还有什么print(my_var.__name__)比其他print("my_var")?您可以简单地直接使用字符串吗?

您可以“切片”一个字典:

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"})

或者:

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)

Most objects don’t have a __name__ attribute. (Classes, functions, and modules do; any more builtin types that have one?)

What else would you expect for print(my_var.__name__) other than print("my_var")? Can you simply use the string directly?

You could “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"})

Alternatively:

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)

回答 17

好吧,几天前我遇到了同样的需求不得不获得一个指向对象本身的变量

为何如此必要呢?

简而言之,我正在为Maya构建一个插件。核心插件是使用C ++构建的,但GUI是通过Python绘制的(因为它不占用大量处理器)。由于我到目前为止还不知道如何return从插件中选择多个值(默认值除外)MStatus,因此要在Python中更新字典,我必须传递变量名,指向实现GUI的对象以及哪个将字典本身包含到插件中,然后使用MGlobal::executePythonCommand()来从Maya的全局范围更新字典。

要做到这一点,我所做的就是:

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

我知道这不是完美的解决方案,因为globals许多键可能指向同一个对象,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,请纠正我。

至少这种方法解决了我的问题,方法是获取全局范围内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,以供内部使用。

我在int(原始整数类)上尝试过此方法,但问题是这些原始类没有被绕过(如果错误,请更正所使用的技术术语)。您可以重新实现int,然后再做,int = foo但是a = 3绝对不能成为foo原始对象的对象。要克服,你必须a = foo(3)得到a.name()工作。

Well, I encountered the very same need a few days ago and had to get a variable’s name which was pointing to the object itself.

And why was it so necessary?

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.

To do that what I did was something like:

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

I know that it is not the perfect solution in in the globals many keys could be pointing to the same object e.g.:

a = foo()
b = a
b.name()
>>>b
or
>>>a

and that the approach isn’t thread-safe. Correct me if I am wrong.

At least this approach solved my problem by getting the name of any variable in the global scope which pointed to the object itself and pass it over to the plug-in, as argument, for it use internally.

I tried this on int (the primitive integer class) but the problem is that these primitive classes don’t get bypassed (please correct the technical terminology used if its wrong). You could re-implement int and then do int = foo but a = 3 will never be an object of foo but of the primitive. To overcome that you have to a = foo(3) to get a.name() to work.


回答 18

在python 2.7和更高版本中,还具有字典理解功能,这使其变得更短了。如果可能的话,我将使用getattr代替eval(eval是邪恶的),就像在最高答案中一样。自我可以是具有您所要查看的上下文的任何对象。它可以是一个对象或locals = locals()等。

{name: getattr(self, name) for name in ['some', 'vars', 'here]}

With python 2.7 and newer there is also dictionary comprehension which makes it a bit shorter. If possible I would use getattr instead eval (eval is evil) like in the top answer. Self can be any object which has the context your a looking at. It can be an object or locals=locals() etc.

{name: getattr(self, name) for name in ['some', 'vars', 'here]}

回答 19

我正在研究类似的问题。@ S.Lott说:“如果有变量列表,那么“发现”它们的名字有什么意义?” 我的答案只是看它是否可以完成,以及是否出于某种原因要按类型将变量排序到列表中。所以无论如何,在我的研究中,我遇到了这个线程,并且我的解决方案有所扩展,并且基于@rlotun解决方案。@unutbu说了另一件事,“这种想法是有好处的,但是请注意,如果两个变量名引用相同的值(例如True),则可能会返回意外的变量名。” 在这个练习中,所以我处理它通过使用类似这样的可能性,每一个列表理解这是真的:isClass = [i for i in isClass if i != 'item']。没有它,“项目”将显示在每个列表中。

__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 []
'''

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 []
'''

回答 20

你可以使用easydict

>>> from easydict import EasyDict as edict
>>> d = edict({'foo':3, 'bar':{'x':1, 'y':2}})
>>> d.foo
3
>>> d.bar.x
1
>>> d = edict(foo=3)
>>> d.foo
3

另一个例子:

>>> d = EasyDict(log=False)
>>> d.debug = True
>>> d.items()
[('debug', True), ('log', False)]

you can use easydict

>>> from easydict import EasyDict as edict
>>> d = edict({'foo':3, 'bar':{'x':1, 'y':2}})
>>> d.foo
3
>>> d.bar.x
1
>>> d = edict(foo=3)
>>> d.foo
3

another example:

>>> d = EasyDict(log=False)
>>> d.debug = True
>>> d.items()
[('debug', True), ('log', False)]

回答 21

在python3上,此函数将在堆栈中获得最外部的名称:

import inspect


def retrieve_name(var):
        """
        Gets the name of var. Does it from the out most frame inner-wards.
        :param var: variable to get name from.
        :return: string
        """
        for fi in reversed(inspect.stack()):
            names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
            if len(names) > 0:
                return names[0]

它在代码的任何地方都很有用。遍历反向堆栈以查找第一个匹配项。

On python3, this function will get the outer most name in the stack:

import inspect


def retrieve_name(var):
        """
        Gets the name of var. Does it from the out most frame inner-wards.
        :param var: variable to get name from.
        :return: string
        """
        for fi in reversed(inspect.stack()):
            names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
            if len(names) > 0:
                return names[0]

It is useful anywhere on the code. Traverses the reversed stack looking for the first match.


回答 22

尽管这可能是一个糟糕的主意,但它与rlotun的答案是一致的,但是它会更频繁地返回正确的结果。

import inspect
def getVarName(getvar):
  frame = inspect.currentframe()
  callerLocals = frame.f_back.f_locals
  for k, v in list(callerLocals.items()):
    if v is getvar():
      callerLocals.pop(k)
      try:
        getvar()
        callerLocals[k] = v
      except NameError:
        callerLocals[k] = v
        del frame
        return k
  del frame

您这样称呼它:

bar = True
foo = False
bean = False
fooName = getVarName(lambda: foo)
print(fooName) # prints "foo"

While this is probably an awful idea, it is along the same lines as rlotun’s answer but it’ll return the correct result more often.

import inspect
def getVarName(getvar):
  frame = inspect.currentframe()
  callerLocals = frame.f_back.f_locals
  for k, v in list(callerLocals.items()):
    if v is getvar():
      callerLocals.pop(k)
      try:
        getvar()
        callerLocals[k] = v
      except NameError:
        callerLocals[k] = v
        del frame
        return k
  del frame

You call it like this:

bar = True
foo = False
bean = False
fooName = getVarName(lambda: foo)
print(fooName) # prints "foo"

回答 23

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]

should get list then return

def get_var_name(**kwargs):
    """get variable name
        get_var_name(var = var)
    Returns:
        [str] -- var name
    """
    return list(kwargs.keys())[0]

回答 24

它不会返回变量的名称,但是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}

It will not return the name of variable but you can create dictionary from global variable easily.

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}

回答 25

有了python-varname您,您可以轻松做到:

pip install 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}

免责声明:我是该python-varname库的作者。

With python-varname you can easily do it:

pip install 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}

Disclaimer: I’m the author of that python-varname library.


回答 26

>>> a = 1
>>> b = 1
>>> id(a)
34120408
>>> id(b)
34120408
>>> a is b
True
>>> id(a) == id(b)
True

这样就可以为“ a”或“ b”获取变量名。

>>> a = 1
>>> b = 1
>>> id(a)
34120408
>>> id(b)
34120408
>>> a is b
True
>>> id(a) == id(b)
True

this way get varname for a maybe ‘a’ or ‘b’.