Question

我试着稍微搜索一下这个问题但却找不到实际的答案。我试图实现一个函数（magic_debug），以便在另一个函数（somefunc）中调用时，它可以访问变量在somefunc中打印出来如下：

def magic_debug(s, *args, **kwargs):
    s2 = s.format(x=x,y=y,z=args[0])
    print(s2)


def somefunc():
    x = 123
    y = ['a', 'b']
    magic_debug('The value of x is {x}, and the list is {y} of len {z}', len(y))

somefunc()

预期输出 - ＆gt; x的值是123，列表是[＆＃39; a＆＃39;，＆＃39; b＆＃39;] len 2

Answer 1

这确实是一个常见问题，请尝试使用inspect。

def magic_debug(s, *args, **kwargs):
    import inspect
    parent_local_scope = inspect.currentframe().f_back.f_locals
    s2 = s.format(**parent_local_scope, z=args[0])
    print(s2)


def somefunc():
    x = 123
    y = ['a', 'b']
    magic_debug('The value of x is {x}, and the list is {y} of len {z}', len(y))

somefunc()

输出：

The value of x is 123, and the list is ['a', 'b'] of len 2

Answer 2

你的意思是？

def magic_debug(s, vars_dict):

    s2 = s.format(**vars_dict)
    print(s2)


def somefunc():
   x = 123         # variables are indent in python
   y = ['a', 'b']  # so they're in the function scope

                   # and so is this function that somefunc calls - 
   vars_dict = vars()
   vars_dict['z'] = len(y)
   magic_debug('The value of x is {x}, and the list is {y} of len {z}', vars_dict)

somefunc()

Answer 3

试试这个 - 你需要缩进你的功能

def magic_debug(s, *args, **kwargs):

    s2 = s.format(x=x,y=y,z=args[0])
    print(s2)


def somefunc():
   x = 123         # variables are indent in python
   y = ['a', 'b']  # so they're in the function scope

                   # and so is this function that somefunc calls - 
   magic_debug('The value of x is {x}, and the list is {y} of len {z}', len(y))

somefunc()

Answer 4

更改一些代码，如下所示：

def magic_debug(s, *args, **kwargs):
    s2 = s.format(x=args[1],y=kwargs.pop('y', ''),z=args[0])
    print(s2)


def somefunc():
    x = 123
    y = ['a', 'b']
    magic_debug('The value of x is {x}, and the list is {y} of len {z}', len(y), x, y=y)

somefunc()

您的输出将是完美的。你正在添加** Kargs，但你不使用它。上面的代码使用** Karg来存储数组。

输出：

The value of x is 123, and the list is ['a', 'b'] of len 2

编辑更少的参数：

def magic_debug(s, *args, **kwargs):
    s2 = s.format(x=args[1],y=args[0],z=len(args[0]))
    print(s2)


def somefunc():
    x = 123
    y = ['a', 'b']
    magic_debug('The value of x is {x}, and the list is {y} of len {z}', y, x)

somefunc()

Answer 5

如果您保留问题的精神，如果允许任何修改，您可以使用locals()将本地范围传递给magic_debug函数，即：

def magic_debug(s, *args, **kwargs):
    s2 = s.format(z=args[0], **kwargs)
    print(s2)

def somefunc():
    x = 123
    y = ['a', 'b']
    magic_debug('The value of x is {x}, and the list is {y} of len {z}', len(y), **locals())

somefunc()
# The value of x is 123, and the list is ['a', 'b'] of len 2

如果您被允许更改功能签名，您也可以通过locals()而不进行扩展。但是如果你不能改变被调试的函数，那么窥视前一帧是唯一的方法。 @Sraw已涵盖in his answer。

Python变量作用域的两个函数

5 个答案: