The sorted() function takes a key= parameter
newlist = sorted(list_to_be_sorted, key=lambda d: d['name'])
Alternatively, you can use operator.itemgetter instead of defining the function yourself
from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name'))
For completeness, add reverse=True to sort in descending order
newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)
Here's a generator that yields the chunks you want:
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
[40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
[50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
If you're using Python 2, you should use xrange() instead of range():
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in xrange(0, len(lst), n):
yield lst[i:i + n]
Also you can simply use list comprehension instead of writing a function, though it's a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:
[lst[i:i + n] for i in range(0, len(lst), n)]
Python 2 version:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
Best Solution
In modern versions of Python I'd suggest what SilentGhost posted (repeating here for clarity):
In an earlier version of this answer I had suggested this, which was necessary because SilentGhost's version didn't work in the version of Python (2.3) that was current at the time:
Now that version of Python is beyond obsolete, and SilentGhost's code works in all currently-maintained versions of Python, so there's no longer any reason to recommend the version I had originally posted.