calendar.monthrange provides this information:
calendar.monthrange(year, month)
Returns weekday of first day of the month and number of days in month, for the specified year and month.
>>> import calendar
>>> calendar.monthrange(2002, 1)
(1, 31)
>>> calendar.monthrange(2008, 2) # leap years are handled correctly
(4, 29)
>>> calendar.monthrange(2100, 2) # years divisible by 100 but not 400 aren't leap years
(0, 28)
so:
calendar.monthrange(year, month)[1]
seems like the simplest way to go.
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
I don't know what's wrong about
A different option is to use
operator.itemgetter():I don't think this is any better, though. (It might be slightly faster if you don't count the time needed to instantiate the
itemgetterinstance, which can be reused if this operation is needed often.)