How to define property with lambda expression in Python

Question

I am trying to write Python properties with using less code, I'd like to define getter and setter functions with lambdas.

So, I try like this:

class Text(object):

    content = property(lambda self: self._content,
                       lambda self,content: self._content = content)

    def __init__(self, content):
        self._content = content
        pass

But unfortunately, I get error on second lambda expression (on setter), because you can't define lambda with assignment, right?

So is there some other way to writing a property (preferably inline) that would take less code. Private attribute _content is of type string, is there a way to assign a value to string without = operator.

Why are you even defining a property in the first place if you're not doing anything with the value? — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Apr 22, 2017 at 15:00
Haha, I just prefer concept of encapsulation from OOP :P. Maybe it's not that important in this example, but it might make more sense for some larger classes. How come you know I don't need this for some larger class, and this is just an example? :P — Scarass
– Scarass, Commented Apr 22, 2017 at 15:03
Because if this was a larger class you wouldn't care about saving two lines of code. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Apr 22, 2017 at 15:04
Of course I would, it actually takes 4 lines to write getter and setter, plus an empty line between them. — Scarass
– Scarass, Commented Apr 22, 2017 at 15:14

RaminNietzsche · Accepted Answer · 2017-04-22 15:06:12Z

1

You can do that like this:

content = property(lambda self: self._content,
                   lambda self,content: setattr(self, '_content' ,content))

answered Apr 22, 2017 at 15:06

RaminNietzsche

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