Custom Types in help shows parser name

[afikbar]

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Example Code

import typer
from typing_extensions import Annotated


class CustomClass:
    def __init__(self, value: str):
        self.value = value

    def __str__(self):
        return f"<CustomClass: value={self.value}>"


def parse_custom_class(value: str):
    return CustomClass(value * 2)


def main(
    custom_arg: Annotated[CustomClass, typer.Argument(parser=parse_custom_class)],
    custom_opt: Annotated[CustomClass, typer.Option(parser=parse_custom_class)] = "Foo",
):
    print(f"custom_arg is {custom_arg}")
    print(f"--custom-opt is {custom_opt}")


if __name__ == "__main__":
    typer.run(main)

Description

I've been trying to use typer with custom types, recently a version was released with such a support (thx!).

However, when specifying a parser for a typer.Option(...), the "type" description in --help of the cli shows the parser function name as the argument type.

For example, using the example provided in docs:

Usage: my-cli main [OPTIONS] CUSTOM_ARG

Arguments:
  CUSTOM_ARG  [required]

Options:
  --custom-opt PARSE_CUSTOM_CLASS
                                  [default: Foo]
  --help                          Show this message and exit.

(Note PARSE_CUSTOM_CLASS)

Operating System

macOS

Operating System Details

No response

Typer Version

0.9.0

Python Version

3.9.13

Additional Context

No response

[danmysak]

One way to handle this:

from typing import Callable


def name_function[T: Callable](function: T, name: str) -> T:
    """Set the name of a function."""
    function.__name__ = name
    return function


def named_function[T: Callable](name: str) -> Callable[[T], T]:
    """Decorator to set the name of a function."""
    return lambda function: name_function(function, name)


@named_function('custom type name to display or an empty string')
def parse_custom_class(value: str) -> CustomClass:
    return CustomClass(value * 2)