How to call same method for a list of objects?

Question

Suppose code like this:

class Base:     def start(self):         pass     def stop(self)         pass  class A(Base):     def start(self):         ... do something for A     def stop(self)         .... do something for A  class B(Base):     def start(self):      def stop(self):  a1 = A(); a2 = A() b1 = B(); b2 = B()  all = [a1, b1, b2, a2,.....] 

Now I want to call methods start and stop (maybe also others) for each object in the list all. Is there any elegant way for doing this except of writing a bunch of functions like

def start_all(all):     for item in all:         item.start()  def stop_all(all): 

Answer 1

This will work

all = [a1, b1, b2, a2,.....]  map(lambda x: x.start(),all)     

simple example

all = ["MILK","BREAD","EGGS"] map(lambda x:x.lower(),all) >>>['milk','bread','eggs'] 

and in python3

all = ["MILK","BREAD","EGGS"] list(map(lambda x:x.lower(),all)) >>>['milk','bread','eggs'] 

Answer 2

It seems like there would be a more Pythonic way of doing this, but I haven't found it yet.

I use "map" sometimes if I'm calling the same function (not a method) on a bunch of objects:

map(do_something, a_list_of_objects) 

This replaces a bunch of code that looks like this:

 do_something(a)  do_something(b)  do_something(c)  ... 

But can also be achieved with a pedestrian "for" loop:

  for obj in a_list_of_objects:        do_something(obj) 

The downside is that a) you're creating a list as a return value from "map" that's just being throw out and b) it might be more confusing that just the simple loop variant.

You could also use a list comprehension, but that's a bit abusive as well (once again, creating a throw-away list):

  [ do_something(x) for x in a_list_of_objects ] 

For methods, I suppose either of these would work (with the same reservations):

map(lambda x: x.method_call(), a_list_of_objects) 

or

[ x.method_call() for x in a_list_of_objects ] 

So, in reality, I think the pedestrian (yet effective) "for" loop is probably your best bet.