Re: Re: Pattern Matching in Lists

*To*: mathgroup at smc.vnet.net*Subject*: [mg35597] Re: [mg35586] Re: Pattern Matching in Lists*From*: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>*Date*: Mon, 22 Jul 2002 02:10:53 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

I was too quick with sending my alleged solution , not noticing immediately that it fails in cases when w is of the form {0,...,1}. (It works however if we consider "cyclic" solutions, meaning that {0,1} is thought to contain one instance of {1,0}!) Since there seems no way of dealing with this problem without making the solutions slower, Alan's Count[Drop[w,-1]-Drop[w,1],1] is indeed (almost certainly) the fastest solution. (of course one can still make it faster by compiling: f = Compile[{{w, _Integer, 1}}, Count[Drop[w, -1] - Drop[w, 1], 1]] is about 6 times faster on my machine than the uncompiled version). Andrzej On Sunday, July 21, 2002, at 04:48 PM, Andrzej Kozlowski wrote: > Having quickly glanced through the avalanche of proposed solutions I > did not see following one.(if there was I apologize for claiming it as > my own): > > Count[w - RotateLeft[w], 1] > > According to my comparisons it is the fastest so far (the one that I > sent originally is one of the slowest) > > In[1]:= > w=Table[Random[Integer],{200000}]; > > In[2]:= > Count[Partition[w,2,1],{1,0}]//Timing > > Out[2]= > {0.86 Second,50131} > > In[3]:= > Count[Drop[w,-1]-Drop[w,1],1]//Timing > > Out[3]= > {0.34 Second,50131} > > In[5]:= > Count[w-RotateLeft[w],1]//Timing > > Out[5]= > {0.29 Second,50131} > Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ > > > On Sunday, July 21, 2002, at 02:01 PM, Allan Hayes wrote: > >> [second posting in view of reported technical problem] >> >> Anthony, >> Take >> w = Table[Random[Integer], {200000} >> >> My first thought was, and several posts used this, >> >> Count[Partition[w, 2,1],{1,0}]//Timing >> >> {3.24 Second,49851} >> >> Later it occured to me to use arithmetic, which turned out to be twice >> as >> fast: >> >> Count[ Drop[w,-1] - Drop[w,1],1]//Timing >> >> {1.49 Second,49851} >> >> This is close to Selwyn Hollis's code >> >> Count[Drop[w+2RotateRight[w],1],2]//Timing >> >> {1.6 Second,49851} >> >> -- >> Allan >> >> --------------------- >> Allan Hayes >> Mathematica Training and Consulting >> Leicester UK >> www.haystack.demon.co.uk >> hay at haystack.demon.co.uk >> Voice: +44 (0)116 271 4198 >> Fax: +44 (0)870 164 0565 >> >> >> "Anthony Mendes" <amendes at zeno.ucsd.edu> wrote in message >> news:ah5qce$59o$1 at smc.vnet.net... >>> Hello, >>> >>> Suppose w={1,1,1,0,0,1,0,1,0,0,1,0,0}. >>> >>> How can I count the number of occurrences of a 1 in w immediately >>> followed by a 0 in w? >>> >>> I have tried every incarnation of Count[] I can think of; for example, >>> >>> Count[w,{___,1,0,___}] >>> >>> does not seem to work. In general, how can I count the number of >>> occurrences of a 1 followed by a 0 in a list of 1's and 0's? Thank >>> you! >>> >>> >>> -- >>> Tony >>> _____________________ >>> amendes at math.ucsd.edu >>> >>> >> >> >> >> >> >> >> >> >