Storing Objects in a List in Python

Question

If an object exists with variable .X

randomData.X

is the created object. If multiple objects are stored in a list and can be accessed via

randomList[3].X

Is there a way to pull all values of X from the list without looping through every object in the list as below:

for x i range(0,10)
randomList[x].X

No. There isn't. You'll have to loop one way or anther, list object do no support vectorized attribute access, or any vectorized operations. Although in this case (in most cases), you should just loop over the list directly: for item in randomList: item.X — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jun 13, 2019 at 22:40
Are you asking if you can pull fewer than the whole lists worth of Xs, or how to pull the Xs efficiently? — ShadowRanger
– ShadowRanger, Commented Jun 13, 2019 at 22:46

WiseDev · Accepted Answer · 2019-06-13 22:45:35Z

3

You are probably looking for a list comprehension.

[obj.X for obj in randomList]

This produces a list with all properties X of every object in your list of objects.

Keep in mind that you can't get away from looping over the list. This is just syntactic sugar for the same loop as before.

edited Jun 13, 2019 at 22:45

answered Jun 13, 2019 at 22:43

WiseDev

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3 Comments

Mark Over a year ago

I suspect this is what the OP is after, but probably worth mentioning that it still loops over every item — it's just a more succinct way of doing it.

WiseDev Over a year ago

Edited my answer

Alexander Ridgers Over a year ago

Thanks very much. Maybe storing lots of data in this object format is not the most efficient as I will need to do lots of searches within the data.

ShadowRanger · Accepted Answer · 2019-06-13 23:00:01Z

Just in case you're looking for maximum efficiency on larger lists, an alternative to the list comprehension in this case is using map + operator.attrgetter. You can either loop over the map directly:

from operator import attrgetter

for X in map(attrgetter('X'), randomList):

which involves no temporary list (map lazily pulls items on demand in Python 3), or if you really need the list, just wrap in the list constructor or use list unpacking to run it out eagerly:

Xs = list(map(attrgetter('X'), randomList))
# or
Xs = [*map(attrgetter('X'), randomList)]

For small input lists, this will be slower than the list comprehension (it has a slightly higher setup overhead), but for medium to large inputs, it will be faster (the per-item overhead is slightly lower, as it involves no per-item bytecode execution).

To be clear, it's still going to have to loop over the list. There is no magic way to get the attributes of every item in a list without looping over it; you could go to extreme lengths to make views of the list that seamlessly read the attribute from the underlying list, but if you accessed every element of that view it would be equivalent to the loop in work required.

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