How can I pass method as function parameter without 'lambda' keyword?

Question

Can I somehow refer to a method without using the lambda keyword?

Say we have following example code:

class AbstractDummy: def size(self): raise NotImplementedError class Dummy1(AbstractDummy): def size(self): return 10 class Dummy2(AbstractDummy): def size(self): return 20

If I have my example objects:

dummies1 = [Dummy1(), Dummy1(), Dummy1()] dummies2 = [Dummy2(), Dummy2()]

Then if I want to map them, and I can do that with extracted function parameter to save me some characters:

f = lambda x : x.size() map(f, dummies1) map(f, dummies2)

Question here: can I somehow avoid this temporary f and/or lambda keyword?

To make a small comparison, in Java it would be possible to refer to AbstractDummy::size and so the invocation would look a bit like print(map(AbstractDummy::size, dummies1).

If you want this to work correctly, the lambda is the simplest way here. You cannot use AbstractDummy.size because that will always call the abstract-classes size method (because in Python, .size is just a regular function here). So the lambda is a perfectly sensible solution. Alternatively, you could use operator.methodcaller — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jan 13, 2020 at 22:28
Though if you are only assigning the lambda expression to a name, just use a def statement. — chepner
– chepner, Commented Jan 13, 2020 at 22:29
Note, idiomatic Python almost always prefers a list comprehension/ generator expression over an equivalent map. Guido actually wanted to remove map/filter/reduce and lambda from Python 3! The only times I use map are usually reserved for mapping simple built-in functions, so for number in map(int, user_input) for example... — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jan 13, 2020 at 22:35
Also note, in Python, methods only exist when they are called from an instance. Before that, they are simply regular functions existing in some class namespace. Note, a new method object is created on each invocation,so some_instance.some_method is some_instance.some_method will be false! — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jan 13, 2020 at 22:38

chepner · Accepted Answer · 2020-01-13 22:24:54Z

The operator module provides methodcaller for this.

from operator import methodcaller f = methodcaller('size') results1 = [f(x) for x in dummies1] results2 = [f(x) for x in dummies2]

though [x.size() for x in ...] is simpler, as in C_Z_'s answer. methodcaller is useful for when you need a function as a function argument, for example

# Sort some_list_of_objects on return value of each object's `a` method. sorted_list = sorted(some_list_of_objects, key=methodcaller('a'))

C_Z_ · Accepted Answer · 2020-01-13 22:25:01Z

In this case you would probably want to use a list comprehension

[x.size() for x in dummies1]

[x.size() for x in dummies2]

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