根据多个条件将列表拆分为多个子列表的最快方法

Question

What is the fastest way to split a list into multiple sublists based on conditions?

One way to split a listOfObjects into sublists (three sublists for demonstration, but more are possible):

listOfObjects = [.......]
l1, l2, l3 = [], [], []
for l in listOfObjects:
    if l.someAttribute == "l1":
        l1.append(l)
    elif l.someAttribute == "l2":
        l2.append(l)
    else:
        l3.append(l)

This way does not seem pythonic at all and also takes quite some time. Are there faster approaches, eg using map ?

Similar question , but with only two conditions and no statement about speed.

Answer 1

You could collections.defaultdict here for mapping.

from collections import defaultdict

d = defaultdict(list)

for l in listOfObjects:
    d[l.someAttribute].append(l)

out = d.values() 
l1 , l2, l3 = d['l1'], d['l2'], d['l3']

d would be of the form.

{ 
  attr1 : [...],
  attr2 : [...],
  ...
  attrn : [...]
}

Answer 2

Omg, that similar question's answer is amazing. I haven't thought about that for splitting... Anyway, you can do something similar but it would be less readable:

for l in listOfObjects:
    (l3, l2, l1)[(l.someAttribute == "l1")*2 or l.someAttribute == "l2"].append(l)

This will work for any boolean conditions. or returns first truthy value (or False). True==1 , so we add *2 for the index that we want to be equal to 2.

But as I said, it's not really readable.

As for speed: or is short-circuiting, returns first truthy value, so the check of conditions should be similar to your approach. You might want to keep the lookup tuple defined outside of the loop.

But: unless you do something really specialized that requires every single thing to be optimized, it doesn't make a difference.

---

And more readable thing using dict because your conditions are based on equality (note: attribute you want also has to be hashable)

lookup = {"l1": l1, "l2": l2}
for l in listOfObjects:
    lookup.get(l.someAttribute, l3).append(l)

dict.get gets default value as second - so it's perfect for our else catchall.