Pass self automatically in lambda function?

Question

Is there a way to automatically pass self to a lambda function in a class? I know I can pass self in the __init__ function but then I'm redefining token_functions for every instance of Parser. token_functions never changes so this seems quite inefficient. Is there a better way to do this?

class Parser:
    token_functions = {r'\n': lambda: Parser.new_line(self)}

    def __init__(self):
        self.line_number = 0
        Parser.token_functions[r'\n']()
    
    def new_line(self):
        self.line_number += 1
        print(self.line_number)
    
Parser()

It would help if you showed more context of how you intend to use the Parser class and redefine token_functions for each instance. — mkrieger1
– mkrieger1, Commented Dec 17, 2022 at 17:06
@YevhenKuzmovych Yeah but you can access it in any function__init__, new_line, Or, lambda. — Sharim09
– Sharim09, Commented Dec 17, 2022 at 17:13
Probably user gets an error because the self is not defined Yet. — Sharim09
– Sharim09, Commented Dec 17, 2022 at 17:14

chepner · Accepted Answer · 2022-12-17 17:24:54Z

1

No. The function created by the lambda expression is not a class attribute, so the descriptor protocol is not triggered.

You could call its __get__ method directly, e.g. Parser.token_functions[r'\n'].__get__(self, type(self))(), but you probably don't want to be dealing with the low-level machinery directly.

Just define the function to accept an object with a new_line method as an explicit argument, and pass self when the time comes.

class Parser:
    token_functions = {r'\n': lambda obj: obj.new_line()}

    def __init__(self):
        self.line_number = 0
        Parser.token_functions[r'\n'](self)
    
    def new_line(self):
        self.line_number += 1
        print(self.line_number)

The operator module provides a methodcaller function to replace this use of a lambda expression.

token_functions = {'\n': operator.methodcaller('new_line')}

answered Dec 17, 2022 at 17:24

chepner

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2 Comments

user2820068 Over a year ago

This works. Just out of curiosity. Is this the same as token_functions = {r'\n': lambda obj: Parser.new_line(obj)}? Both seem to work.

chepner Over a year ago

More or less. The difference come up if you decide to subclass Parser, in which case self.new_line() would a subclass's override of new_line be used.

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