假设函数a_method的定义如下

def a_method(arg1, arg2):
    pass

从a_method本身开始,我怎么能得到参数名-例如,作为字符串的元组,如("arg1", "arg2")?


当前回答

在decorator方法中,你可以这样列出原始方法的参数:

import inspect, itertools 

def my_decorator():   
        def decorator(f):
            def wrapper(*args, **kwargs):
                # if you want arguments names as a list:
                args_name = inspect.getargspec(f)[0]
                print(args_name)

                # if you want names and values as a dictionary:
                args_dict = dict(itertools.izip(args_name, args))
                print(args_dict)

                # if you want values as a list:
                args_values = args_dict.values()
                print(args_values)

如果**狼对你来说很重要,那就有点复杂了:

def wrapper(*args, **kwargs):
    args_name = list(OrderedDict.fromkeys(inspect.getargspec(f)[0] + kwargs.keys()))
    args_dict = OrderedDict(list(itertools.izip(args_name, args)) + list(kwargs.iteritems()))
    args_values = args_dict.values()

例子:

@my_decorator()
def my_function(x, y, z=3):
    pass


my_function(1, y=2, z=3, w=0)
# prints:
# ['x', 'y', 'z', 'w']
# {'y': 2, 'x': 1, 'z': 3, 'w': 0}
# [1, 2, 3, 0]

其他回答

下面是另一种不使用任何模块获得函数参数的方法。

def get_parameters(func):
    keys = func.__code__.co_varnames[:func.__code__.co_argcount][::-1]
    sorter = {j: i for i, j in enumerate(keys[::-1])} 
    values = func.__defaults__[::-1]
    kwargs = {i: j for i, j in zip(keys, values)}
    sorted_args = tuple(
        sorted([i for i in keys if i not in kwargs], key=sorter.get)
    )
    sorted_kwargs = {
        i: kwargs[i] for i in sorted(kwargs.keys(), key=sorter.get)
    }   
    return sorted_args, sorted_kwargs


def f(a, b, c="hello", d="world"): var = a
    

print(get_parameters(f))

输出:

(('a', 'b'), {'c': 'hello', 'd': 'world'})

从python 3.0开始,简单易读的答案:

import inspect


args_names = inspect.signature(function).parameters.keys()
args_dict = {
    **dict(zip(args_names, args)),
    **kwargs,
}


在python 3中,下面是将*args和**kwargs放入dict(对于python < 3.6使用OrderedDict来维护dict顺序):

from functools import wraps

def display_param(func):
    @wraps(func)
    def wrapper(*args, **kwargs):

        param = inspect.signature(func).parameters
        all_param = {
            k: args[n] if n < len(args) else v.default
            for n, (k, v) in enumerate(param.items()) if k != 'kwargs'
        }
        all_param .update(kwargs)
        print(all_param)

        return func(**all_param)
    return wrapper

操作一些函数的参数名称的最简单方法:

parameters_list = list(inspect.signature(self.YOUR_FUNCTION).parameters))

结果:

['YOUR_FUNCTION_parameter_name_0', 'YOUR_FUNCTION_parameter_name_1', ...]

这样做会更容易,因为你得到了具体的一个:

parameters_list = list(inspect.signature(self.YOUR_FUNCTION).parameters)[0]

结果:

'YOUR_FUNCTION_parameter_name_0'

这里有一些东西,我认为会为你想要的工作,使用装饰。

class LogWrappedFunction(object):
    def __init__(self, function):
        self.function = function

    def logAndCall(self, *arguments, **namedArguments):
        print "Calling %s with arguments %s and named arguments %s" %\
                      (self.function.func_name, arguments, namedArguments)
        self.function.__call__(*arguments, **namedArguments)

def logwrap(function):
    return LogWrappedFunction(function).logAndCall

@logwrap
def doSomething(spam, eggs, foo, bar):
    print "Doing something totally awesome with %s and %s." % (spam, eggs)


doSomething("beans","rice", foo="wiggity", bar="wack")

运行它,它将产生以下输出:

C:\scripts>python decoratorExample.py
Calling doSomething with arguments ('beans', 'rice') and named arguments {'foo':
 'wiggity', 'bar': 'wack'}
Doing something totally awesome with beans and rice.