I need to do something that is functionally equivalent to this:
for foo in foos:
bar = foo.get_bar()
# Do something with bar
My first i
I think the cleanest way is to forgo the for
-loop completely and use a list comprehension:
bars = [foo.get_bar() for foo in foos]
This answer is close to some other answers but differs in that the list is actually returned and not used in the subsequent loop. Depending on what you do with bars
, you may be able to use a list comprehension there to.
I don't think the map
function with lambdas
are particulary readable, besides the overhead for map
is considerable.
You want operator.methodcaller(). Or, of course, a list or generator comprehension.
Either with lambda
:
for bar in map(lambda foo: foo.get_bar(), foos):
Or simply with instance method reference on your instance's class:
for bar in map(Foo.get_bar, foos):
As this was added from a comment, I would like to note that this requires the items of foos
to be instances of Foo
(i.e. all(isinstance(foo, Foo) for foo in foos)
must be true) and not only as the other options do instances of classes with a get_bar
method. This alone might be reason enough to not include it here.
Or with methodcaller
:
import operator
get_bar = operator.methodcaller('get_bar')
for bar in map(get_bar, foos):
Or with a generator expression:
for bar in (foo.get_bar() for foo in foos):
You can use lambda
for bar in map(lambda foo: foo.get_bar(), foos):
# Do something with bar
This modified code will work:
for bar in map(lambda f: f.get_bar(), foos):
# Do something with bar
You provide lambda function here. Simply providing get_bar
doesn't work because it is accessible only through an instance of class (f.get_bar()
), never by itself.