Aggregate nested lists without nested loops in Python 3.x

Question

I'm about to work my way into python's 3.5 lambda notation and I'm wondering wether nested loops can simply be replaced with a lambda one-liner. e.g.: I have this simple dummy class hierarchy:

class Resource:
    def __init__(self, name="foo"):
        self.name = name

class Course:
    def __init__(self):
        self.resources = list()

class College:
    def __init__(self):
        self.courses = list()

I have an instance of Collegewith multiple Courses and Resources as my starting point.

college = College()

Now if I want a listof all the Resources in my College I could easily do this with 2 for-loops:

all_resources = list()
for course in  college.courses:
    for resource in course.resources:
        all_resources.append(resource)

This is indeed very simple but I wondered whether I could also achieve this by doing something like this:

all_resources = list(map(lambda r: r, [c.resources for c in college.courses]))

But unfortunately this gives me a listof lists and not a listof Resources, what I wanted to achieve. Is lambda suitable for something like that? What would be the most pythonic way for a operation like this?

holdenweb · Accepted Answer · 2016-05-02 18:12:46Z

6

First of all, please stop creating empty lists by calling list() — it's both simpler and quicker to use a literal empty list [] (cheeck this with timeit if you don't believe me).

Secondly, there is absolutely no need to use lambda to create the list you want. A simple list comprehension will do the job:

all_resources = [r for r in course.resources for course in college.courses]

Don't overthink it. Keep it as simple as you can.

answered May 2, 2016 at 18:12

holdenweb

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1 Comment

MartinD Over a year ago

I think these loops are the wrong way round? It should be: all_resources = [r for course in college.courses for r in course.resources]

holdenweb · Accepted Answer · 2016-05-02 19:41:51Z

2

You shouldn't be using lambdas for this. You should be list comprehensions.

all_resources = [resource for resource for course in college.courses in course.resources]

Lambdas would be completely unnecessary here. If you absolutely wanted to, you could do something like:

all_resources = []
[(lambda c: [all_resources.append(r) for r in c])(x) for x in college.courses)]

But I'd recommend the first one. An alternative giving a different order would be

all_resources = [resource for resource in course.resources for course in college.courses]

edited May 2, 2016 at 19:41

holdenweb

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answered May 2, 2016 at 18:19

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