在 Python 中，如何定义一个函数包装器来验证具有特定名称的参数？

Question

我正在编写几个接受名为 policy 的参数的函数，该参数只允许具有特定值（即 'allow' 或 'deny'）。如果没有，我想提出一个ValueError。

为简洁起见，我想为此定义一个装饰器。到目前为止，我已经想出了以下内容：

def validate_policy(function):
    '''Wrapper which ensures that if the function accepts a 'policy' argument, that argument is either 'allow' or 'deny'.'''
    def wrapped_function(policy, *args, **kwargs):
        if policy not in ['allow', 'deny']:
            raise ValueError("The policy must be either 'allow' or 'deny'.")
        return function(policy, *args, **kwargs)
    return wrapped_function

问题是，这仅在 policy 是函数的第一个位置参数时有效。但是，我想允许 policy 出现在任何位置。

具体来说，这里有一些名为 make_decision 和 make_informed_decision 的（虚拟）函数，它们在不同位置接受参数 policy，以及一些测试用例：

import pytest

@validate_policy
def make_decision(policy):      # The 'policy' might be the first positional argument
    if policy == 'allow':
        print "Allowed."
    elif policy == 'deny':
        print "Denied."

@validate_policy
def make_informed_decision(data, policy):   # It also might be the second one
    if policy == 'allow':
        print "Based on the data {data} it is allowed.".format(data=data)
    elif policy == 'deny':
        print "Based on the data {data} it is denied.".format(data=data)


'''Tests'''
def test_make_decision_with_invalid_policy_as_positional_argument():
    with pytest.raises(ValueError):
        make_decision('foobar')

def test_make_decision_with_invalid_policy_as_keyword_argument():
    with pytest.raises(ValueError):
        make_decision(policy='foobar')

def test_make_informed_decision_with_invalid_policy_as_positional_argument():
    with pytest.raises(ValueError):
        make_informed_decision("allow", "foobar")

def test_make_informed_decision_with_invalid_policy_as_keyword_argument():
    with pytest.raises(ValueError):
        make_informed_decision(data="allow", policy="foobar")


if __name__ == "__main__":
    pytest.main([__file__])

目前除第三个测试外所有测试都通过了，因为第一个位置参数 'allow' 被解释为 policy 而不是应有的 data。

如何调整 validate_policy 装饰器，使所有测试都通过？

Answer 1

您可以使用 inspect 模块的 Signature.bind 功能：

import inspect

def validate_policy(function):
    '''Wrapper which ensures that if the function accepts a 'policy' argument, that argument is either 'allow' or 'deny'.'''
    signature= inspect.signature(function)
    def wrapped_function(*args, **kwargs):
        bound_args= signature.bind(*args, **kwargs)
        bound_args.apply_defaults()
        if bound_args.arguments.get('policy') not in ['allow', 'deny']:
            raise ValueError("The policy must be either 'allow' or 'deny'.")
        return function(*args, **kwargs)
    return wrapped_function

Answer 2

这是另一个使用 inspect.getcallargs 的解决方案：

def validate_policy(function):
    '''Wrapper which ensures that if the function accepts a 'policy' argument, that argument is either 'allow' or 'deny'.'''
    def wrapped_function(*args, **kwargs):
        call_args = inspect.getcallargs(function, *args, **kwargs)
        if 'policy' in call_args:
            if call_args['policy'] not in ['allow', 'deny']:
                raise ValueError("The policy must be either 'allow' or 'deny'.")
        return function(*args, **kwargs)
    return wrapped_function

它使所有测试都通过。

在 Python 中，如何定义一个函数包装器来验证具有特定名称的参数？

In Python, how to define a function wrapper which validates an argument with a certain name?

python

wrapper

python-decorators