如何在python中将所有参数转换为字典 [英] How to convert all arguments to dictionary in python

问题描述

我想对函数func(*args, **kwargs)返回一个字典，其中包含我提供给它的所有参数.例如:

I would like to my function func(*args, **kwargs) return one dictionary which contains all arguments I gave to it. For example:

func(arg1, arg2, arg3=value3, arg4=value4)

应该返回一个这样的字典:

should return one dictionary like this:

{'arg1': value1, 'arg2': value2, 'arg3': value3, 'arg4': value4 }

推荐答案

您可以使用locals()或vars():

def func(arg1, arg2, arg3=3, arg4=4):
    print(locals())

func(1, 2)

# {'arg3': 3, 'arg4': 4, 'arg1': 1, 'arg2': 2}

但是，如果仅使用*args，则将无法使用locals()进行区分:

However if you only use *args you will not be able to differentiate them using locals():

def func(*args, **kwargs):
    print(locals())

func(1, 2, a=3, b=4)

# {'args': (1, 2), 'kwargs': {'a': 3, 'b': 4}}

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