Pass self automatically in lambda function?-CodePudding

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()

CodePudding user response：

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')}