[BangPypers] golf problem
Gaurav Kalra
gvkalra at gmail.com
Mon Dec 26 08:39:15 CET 2011
From: steve <steve at lonetwin.net>
> I am sorry, I didn't quite understand this bit "it is necessary to generate a
> random list of 6 holes.", since the list below has 18 elements.
That means out of the 18 elements from the list, we need 6 random
elements but with one constraint.
> In any case, whether you need 6 or 18 -- the simplest way to do this would be:
>
> >>> l = []
> >>> for i in range(6):
> ... l.append(random.choice([3,4,5]))
random.choice([3,4,5]) may also possibly generate [3,3,4,3,4,3]
(http://docs.python.org/library/random.html#random.choice), which
violates our constraint. We need at least 1 occurrence of all the
three elements.
> ...
> >>> l
> [4, 5, 3, 4, 5, 5]
>
> # ...where the index is the hole number and the element is the par, or if you
> # need it in the format you showed:
>
> >>> l = []
> >>> for i in range(1, 19):
> ... l.append((i, random.choice([3,4,5])))
> ...
> >>> l
> [(1, 5), (2, 3), (3, 4), (4, 5), (5, 3), (6, 3), (7, 5), (8, 4), (9, 3),
> (10, 5), (11, 5), (12, 5), (13, 4), (14, 5), (15, 3), (16, 5), (17, 3), (18, 5)]
> >>>
You are getting it wrong. e.g there is no hole (4,5) in the sample data.
Every hole has a fixed par type and we can't randomly associate one
hole with a par type (as above).
--
Gaurav Kalra
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