On Mon, 02 May 2011 10:33:31 -0700, Raymond Hettinger wrote: > I think it is time to give some visibility to some of the instructive > and very cool recipes in ActiveState's python cookbook. [...] > What are your favorites?
I'm not sure if favourite is the right word, but I'm amazed by this one: McCarthy's "amb" (ambiguous) operator. http://code.activestate.com/recipes/577368 Here's how I might use it to solve the problem if making change. In Australian currency, I have 5, 10, 20, 50 cent and $1 and $2 coins. Ignoring the dollar coins, how can I make up change for any multiple of five cents up to a dollar? Suppose I have more 5 cent coins that I can deal with, and I want to make sure I hand out at least three of them. Here's how to make 45 cents worth of change: >>> amb = Amb() >>> a = amb(range(3, 21)) # number of 5 cent coins >>> b = amb(range(11)) # number of 10 cent coins >>> c = amb(range(6)) # number of 20 cent coins >>> d = amb(range(3)) # number of 50 cent coins >>> for _ in amb(lambda a,b,c,d: 5*a+10*b+20*c+50*d == 45): ... print a, b, c, d ... 3 1 1 0 3 3 0 0 5 0 1 0 5 2 0 0 7 1 0 0 9 0 0 0 Suppose I have some words, and want to put them together so that there are a certain number of vowels: >>> amb = Amb() >>> a = amb(['quick', 'slow', 'hungry', 'wise-old']) >>> b = amb(['fox', 'hare', 'turtle', 'kangaroo']) >>> c = amb(['lazy', 'stupid', 'sleepy', 'confused']) >>> d = amb(['dog', 'aardvark', 'sloth', 'wombat']) >>> >>> def test(a, b, c, d): ... s = "The %s brown %s jumped over the %s %s." % (a, b, c, d) ... num_vowels = sum(s.count(c) for c in 'aeiou') ... return num_vowels in (12, 18, 19) ... >>> for _ in amb(test): ... print a, b, c, d ... quick fox lazy sloth quick fox lazy dog quick kangaroo stupid aardvark [...more solutions cut for brevity] hungry kangaroo confused aardvark As written, amb is just a brute-force solver using more magic than is good for any code, but it's fun to play with. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
