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
也许这很有用:
def Retriever(bar):
return (list(globals().keys()))[list(map(lambda x: id(x), list(globals().values()))).index(id(bar))]
该函数遍历全局作用域的值的ID列表(可以编辑命名空间)，根据其ID找到想要/需要的var或函数的索引，然后根据获得的索引从全局名称列表中返回名称。
其他回答
如果目标是帮助您跟踪变量，您可以编写一个简单的函数来标记变量并返回其值和类型。例如，假设i_f=3.01，并将其舍入为一个名为i_n的整数以用于代码，然后需要一个字符串i_s用于报告。
def whatis(string, x):
print(string+' value=',repr(x),type(x))
return string+' value='+repr(x)+repr(type(x))
i_f=3.01
i_n=int(i_f)
i_s=str(i_n)
i_l=[i_f, i_n, i_s]
i_u=(i_f, i_n, i_s)
## make report that identifies all types
report='\n'+20*'#'+'\nThis is the report:\n'
report+= whatis('i_f ',i_f)+'\n'
report+=whatis('i_n ',i_n)+'\n'
report+=whatis('i_s ',i_s)+'\n'
report+=whatis('i_l ',i_l)+'\n'
report+=whatis('i_u ',i_u)+'\n'
print(report)
这将在每次调用时打印到窗口中用于调试，并为书面报告生成一个字符串。唯一的缺点是每次调用函数时都必须输入两次变量。
I am a Python newbie and found this very useful way to log my efforts as I program and try to cope with all the objects in Python. One flaw is that whatis() fails if it calls a function described outside the procedure where it is used. For example, int(i_f) was a valid function call only because the int function is known to Python. You could call whatis() using int(i_f**2), but if for some strange reason you choose to define a function called int_squared it must be declared inside the procedure where whatis() is used.
在Python中，def和class关键字将特定的名称绑定到它们定义的对象(函数或类)。类似地，模块通过在文件系统中调用某个特定的名称来命名。在这三种情况下，都有一种明显的方法可以将“规范”名称分配给有问题的对象。
然而，对于其他类型的对象，这样的规范名称可能根本不存在。例如，考虑列表中的元素。列表中的元素不是单独命名的，在程序中引用它们的唯一方法完全可能是在包含它们的列表上使用列表索引。如果将这样一个对象列表传递到函数中，则不可能为值分配有意义的标识符。
Python不会将赋值对象左边的名称保存到赋值对象中，因为:
这需要在多个冲突的对象中找出哪个名称是“规范的”， 对于那些从未赋值给显式变量名的对象来说，这是没有意义的， 这样效率会非常低， 实际上没有其他现存的语言能做到这一点。
例如，使用lambda定义的函数将始终具有"name" <lambda>，而不是特定的函数名。
最好的方法是简单地要求调用者传入一个(可选的)名称列表。如果输入“…”，“…”'太麻烦了，你可以接受一个包含逗号分隔的名字列表的字符串(像namedtuple那样)。
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.
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
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
