IM not able to figure out algorithm to find the runs. Here is the code I have:
def ProgressCalc(items): counts = [items[0]] for i in range(1, len(items)-1): print "for loop", items[i], items[i + 1] if counts[- 1] < items[i]: counts += [items[i]] print "inside", items[i], items[i - 1], counts print counts counts = [items[0]] for i in range(1, len(items) - 1): print "for loop", items[i], items[i + 1] if counts[- 1] <= items[i] and items[i] < items[i + 1]: counts += [items[i]] print "inside", items[i], items[i - 1], counts elif counts[- 1] <= items[i] and items[i] > items[i + 1]: counts += [items[i]] i += 2 print counts ProgressCalc(items) ---------- Original Message ---------- From: Alan Gauld <alan.ga...@yahoo.co.uk> To: "monik...@netzero.net" <monik...@netzero.net> Cc: tutor@python.org Subject: Re: python programmin problem Date: Thu, 21 Jul 2016 00:11:24 +0100 On 20/07/16 22:11, monik...@netzero.net wrote: > ... if not in python, then in pseudo code. The first question to ask is can you do it without a computer? In other words given input [1,7,2,3,5,4,6] Can you first of all produce a list of all valid runs? [1,2,3,5,6] and [1,2,3,4,6] both have length 5 I also see shorter runs: [7] and [4,6] for example. Can you manually create a list of all valid runs? Once you can do that can you write a program to generate that as a list of lists in Python? If so then the answer to your question is a matter of finding the length of the longest valid run which should be fairly easy. -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos ____________________________________________________________ Affordable Wireless Plans Set up is easy. Get online in minutes. Starting at only $9.95 per month! www.netzero.net?refcd=nzmem0216 _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor