I have a list of integers and I want to create a new list with all elements smaller than a given limit.
a=range(15) #example list
limit=9 #example limit
My approach to solve this problem was
[i for i in a if i < limit]
To me the beginning 'i for i in' looks pretty verbose. Is there a better implementation in Python?
You can use filter() (at least in the Python 2.x series... I think it might have been removed in 3.0)
newlist = filter(lambda item: item < limit, oldlist)
The first argument can be any callable (its result will be coerced to boolean, so it's best to use a callable that returns boolean anyway), and the second argument can be any sequence.
You could use filter
>>> filter(lambda i: i < limit, a)
[0, 1, 2, 3, 4, 5, 6, 7, 8]
But list comprehensions are the preferred way to do it
Here is what python docs has to say about this:
List comprehensions provide a concise way to create lists without resorting to use of map(), filter() and/or lambda. The resulting list definition tends often to be clearer than lists built using those constructs.
This is about the best you can do. You may be able to do better with filter, but I wouldn't recommend it. Bear in mind that the list comprehension reads almost like english: "i for i in a if i < limit". This makes it much easier to read and understand, if a little on the verbose side.
The nicer versions require boilerplate code, so the list comprehension is as nice as you can get.
This would be one different way to do it:
from operator import ge
from functools import partial
filter(partial(ge, limit), a)
(But if you were to use filter, Nadia's way would be the obvious way to do it)