Extract list of attributes from list of objects in python

Question

I have an uniform list of objects in python:

class myClass(object):
    def __init__(self, attr):
        self.attr = attr
        self.other = None

objs = [myClass (i) for i in range(10)]

Now I want to extract a list with some attribute of that class (let's say attr), in order to pass it to some function (for plotting that data for example)

What is the pythonic way of doing the following:

attr = [o.attr for o in objs]

Maybe derive list and add a method to it, so I can use some idiom like objs.getattribute("attr")?

Answer 1

attrs = [o.attr for o in objs] was the right code for making a list like the one you describe. Don't try to subclass list for this. Is there something you did not like about that snippet?

Answer 2

You can also write:

attr=(o.attr for o in objsm)

This way you get a generator that conserves memory. For more benefits look at Generator Expressions.

Answer 3

Another way could be to map getattr() on objs list.

# for a generator
from itertools import repeat
attrs = map(getattr, objs, repeat('attr'))
# or even without itertools
attrs = map(getattr, objs, ['attr']*len(objs))

# for a list
attrs = list(map(getattr, objs, repeat('attr')))

A nice thing about getattr is that a default value can be returned if no such attribute exist. For example, the below line returns 'NA' if attr is not defined in an object.

attrs = [getattr(o, 'attr', 'NA') for o in objs]

If multiple attributes need to be fetched, operator.attrgetter could be useful.

from operator import attrgetter
attrs = list(map(attrgetter('attr', 'other'), objs))

N.B. itertools and operator are in the standard library.