我已经读了如何获得一个函数名作为字符串?。

如何对变量做同样的事情呢?与函数相反,Python变量没有__name__属性。

换句话说,如果我有一个变量,比如:

foo = dict()
foo['bar'] = 2

我正在寻找一个函数/属性,例如retrieve_name(),以便从这个列表中创建一个DataFrame in Pandas,其中列名由实际字典的名称给出:

# List of dictionaries for my DataFrame
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [retrieve_name(d) for d in list_of_dicts] 

当前回答

您可以尝试以下方法来检索您定义的函数的名称(但不适用于内置函数):

import re
def retrieve_name(func):
    return re.match("<function\s+(\w+)\s+at.*", str(func)).group(1)

def foo(x):
    return x**2

print(retrieve_name(foo))
# foo

其他回答

如果有两个值相同的变量,前面的一些情况会失败。所以提醒它很方便:

定义函数:

# Variable to string of variable name

def var_name(variable,i=0):

  results = []
  for name in globals():   
     if eval(name) == variable:
       results.append(name)

  if len(results) > 1:
    print('Warning:' )
    print('   var_name() has found',len(results), 'possible outcomes.')
    print('   Please choose the suitable parameter "i". Where "i" is the index')
    print('   that matches your choice from the list below.')
    print('  ',results) ; print('')

  return results[i]

Use:

var_1 = 10
var_name(var_1) # Output will be "var_1"

如果你有两个相同值的变量,比如var_1 = 8和var_2 = 8,那么就会出现一个警告。

var_1 = 8
var_2 = 8
var_name(var_2)  # Output will be "var_1" too but Warning will appear
from __future__ import annotations
import inspect
import pandas as pd


# 변수 이름 가져오기
def getVariableName(variable: str) -> (str | Exception):
    callers_local_vars = inspect.currentframe().f_back.f_locals.items()

    df = pd.DataFrame(callers_local_vars)
    df.columns = ['함수명', '값']

    try:
        return df[(df['값'] == variable)].iloc[0]['함수명']
    except Exception as e:
        return e


test = 'Hello World!'

if __name__ == '__main__':
    print(getVariableName(test))

我不相信这是可能的。考虑下面的例子:

>>> a = []
>>> b = a
>>> id(a)
140031712435664
>>> id(b)
140031712435664

a和b指向同一个对象,但是对象不知道哪些变量指向它。

如何对变量做同样的事情呢?与函数相反,Python变量没有__name__属性。

问题出现的原因是您对术语、语义或两者都感到困惑。

"variables" don't belong in the same category as "functions". A "variable" is not a thing that takes up space in memory while the code is running. It is just a name that exists in your source code - so that when you're writing the code, you can explain which thing you're talking about. Python uses names in the source code to refer to (i.e., give a name to) values. (In many languages, a variable is more like a name for a specific location in memory where the value will be stored. But Python's names actually name the thing in question.)

In Python, a function is a value. (In some languages, this is not the case; although there are bytes of memory used to represent the actual executable code, it isn't a discrete chunk of memory that your program logic gets to interact with directly.) In Python, every value is an object, meaning that you can assign names to it freely, pass it as an argument, return it from a function, etc. (In many languages, this is not the case.) Objects in Python have attributes, which are the things you access using the . syntax. Functions in Python have a __name__ attribute, which is assigned when the function is created. Specifically, when a def statement is executed (in most languages, creation of a function works quite differently), the name that appears after def is used as a value for the __name__ attribute, and also, independently, as a variable name that will get the function object assigned to it.

但大多数对象都没有这样的属性。

换句话说,如果我有一个变量,比如:

That's the thing: you don't "have" the variable in the sense that you're thinking of. You have the object that is named by that variable. Anything else depends on the information incidentally being stored in some other object - such as the locals() of the enclosing function. But it would be better to store the information yourself. Instead of relying on a variable name to carry information for you, explicitly build the mapping between the string name you want to use for the object, and the object itself.

这是另一种基于输入变量内容的方法:

(它返回第一个与输入变量匹配的变量名,否则为None。可以修改它,以获得所有与输入变量具有相同内容的变量名)

def retrieve_name(x, Vars=vars()):
    for k in Vars:
        if isinstance(x, type(Vars[k])):
            if x is Vars[k]:
                return k
    return None