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")?
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.