CODEWITHSUNDEEP
pythonpython-3.5
Dancer PhD
Dancer PhD

Reputation: 307

TypeError: group_by_retval() missing 1 required keyword-only argument: 'grouper_func'

guys below is my code for dictionary generating:

import collections
from typing import Callable

def group_by_retval(*args, grouper_func: Callable[[], None]):
    my_list = []
    for elem in args:
        my_tr = grouper_func(elem)
        my_list.append([my_tr,elem])
    my_tuple = tuple(tuple(elem) for elem in my_list)
    my_diction = collections.defaultdict(list)
    for k,v in my_tuple:
        my_diction[k].append(v)
    return my_diction

I have read from python official documentation that I can announce callable function after the *args. as above. However, when I run simple example like below I got an error. Can someone please help me to solve this issue. Thank you for reading:

l = ["ab", 12, "cd", "d", 3]


print(group_by_retval(l, lambda x: isinstance(x, str)))

Upvotes: 0

Views: 633

Answers (1)

kvorobiev
kvorobiev

Reputation: 5070

As mentioned in comments, after *arg (arbitrary arguments list) you could use only keyword arguments:

group_by_retval(l, grouper_func=lambda x: isinstance(x, str))

to do it without providing a name you could swap arguments:

def group_by_retval(grouper_func: Callable[[], None], *args):
...

print(group_by_retval(lambda x: isinstance(x, str), l))

Output:

defaultdict(<class 'list'>, {False: [['ab', 12, 'cd', 'd', 3]]})

But in fact, you just do not need to use arbitrary arguments list, because your first argument is a list. Right solution

def group_by_retval(args, grouper_func: Callable[[], None]):
...
print(group_by_retval(l, lambda x: isinstance(x, str)))

Output:

defaultdict(<class 'list'>, {False: [12, 3], True: ['ab', 'cd', 'd']})

Upvotes: 1

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