CODEWITHSUNDEEP
python
Adam Kotwasinski
Adam Kotwasinski

Reputation: 4574

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

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).

Upvotes: 1

Views: 239

Answers (2)

C_Z_
C_Z_

Reputation: 7806

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

[x.size() for x in dummies1]

[x.size() for x in dummies2]

Upvotes: 4

chepner
chepner

Reputation: 531948

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'))

Upvotes: 4

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