Unpack multiple dictionaries as functional arguments without repeating double asterisk-CodePudding

I use API with long name of argument parameters. Consequently, I create following dictionaries for most common combinations of values which are then unpacked in function calls.

a_T = {'API parameter a': True}
a_F = {'API parameter a': False} 
b_100 = {'API parameter b': 100}
b_0 = {'API parameter b': 0}
hello = {'API parameter c': 'hello'}
bye = {'API parameter d': 'goodbye'}    

myf(**a_T, **bye)
myf(**b_0)
myf(**a_F, **b_100, **hello, **bye)

Is there any way to avoid repeat double asterisk? The code becomes quite unreadable with many of these strange characters.

Once could then add this unpacking utility to myf:

myf(a_T, bye)
myf(b_0)
myf(a_F, b_100, hello, bye)

CodePudding user response：

You can actually use | for Python 3.9 to combine all the dictionaries then send unpacked version.

def fun(**kwargs):
    print(kwargs)

>>> fun(**a_F| b_100| hello| bye)
{'API parameter a': False, 'API parameter b': 100, 'API parameter c': 'hello', 'API parameter d': 'goodbye'}

Or just use *args and pass multiple dictionaries:

def fun(*args):
    print(args)
    
>>> fun(a_F,b_100,hello,bye)
({'API parameter a': False}, {'API parameter b': 100}, {'API parameter c': 'hello'}, {'API parameter d': 'goodbye'})

Another solution is to use Python decorator and take care of ugly part inside the decorator function:

def decorator(fun):
    def wrapper(*args):
        from functools import reduce
        kwargs = reduce(lambda a, b: a | b, args)
        return fun(**kwargs)
    return wrapper
@decorator
def main_fun(**kwargs):
    print(kwargs)
    

>>> main_fun(a_F,b_100,hello,z)
{'API parameter a': False, 'API parameter b': 100, 'API parameter c': 'hello', 'parameter 4': 'goodbye'}

CodePudding user response：

To unpack a series of dicts, use dict.update, or a nested comprehension:

def myf(*dicts):
    merged = {k: v for d in dicts for k, v in d.items()}
    # do stuff to merged

def myf(*dicts):
    merged = {}
    for d in dicts:
        merged.update(d)