Re: Pattern problem: How to count from a long list of numbers all

*To*: mathgroup at smc.vnet.net*Subject*: [mg87421] Re: Pattern problem: How to count from a long list of numbers all*From*: "jwmerrill at gmail.com" <jwmerrill at gmail.com>*Date*: Thu, 10 Apr 2008 02:12:34 -0400 (EDT)*References*: <fti404$ok1$1@smc.vnet.net>

On Apr 9, 5:58 am, "Nasser Abbasi" <n... at 12000.org> wrote: > Hello; > > I think using Pattern is my weakest point in Mathematica. > > I have this list, say this: (it is all a list of integers, no real numbers). > > x = {1, 3, 3, 3, 2, 3, 3, 1, 3, 3} > > And I want to count how many say a 3 followed immediately by 3. So in the > above list, there will be 4 such occurrences. And if I want to count how > many 1 followed by a 3, there will be 2 such cases, etc... > > I tried Count[] but I do not know how to set the pattern for "3 followed by > a comma followed by 3" or just "3 followed immediately by 3". > > I tried few things, such as the following > > In[68]:= Count[x, {3, 3}_] > Out[68]= 0 > > Also tried Cases, but again, I am not to good with Patterns, so not sure how > to set this up at this moment. > > Any ideas will be appreciated. > > Nasser > I really need to sit down and study Patterns in Mathematica really well one > day :) I don't think you can solve this problem using only Count with a pattern. The pattern only gets to look at one element at a time. You can set things up using Partition, though. In[20]:= x = {1, 3, 3, 3, 2, 3, 3, 1, 3, 3} Out[20]= {1, 3, 3, 3, 2, 3, 3, 1, 3, 3} In[21]:= Partition[x, 2, 1] Out[21]= {{1, 3}, {3, 3}, {3, 3}, {3, 2}, {2, 3}, {3, 3}, {3, 1}, {1, 3}, {3, 3}} Now you can count how many times {3,3} occurs: In[32]:= Count[Partition[x, 2, 1], {3, 3}] Out[32]= 4 If you wanted to count how many times any element is repeated, you could do In[33]:= Count[Partition[x, 2, 1], {y_, y_}] Out[33]= 4 Regards, Jason