Re: Pattern problem: How to count from a long list of numbers all occurrences of 2 numbers next to each others?

*To*: mathgroup at smc.vnet.net*Subject*: [mg87403] Re: Pattern problem: How to count from a long list of numbers all occurrences of 2 numbers next to each others?*From*: "Nasser Abbasi" <nma at 12000.org>*Date*: Thu, 10 Apr 2008 02:09:12 -0400 (EDT)*Reply-to*: "Nasser Abbasi" <nma at 12000.org>

"Nasser Abbasi" <nma at 12000.org> wrote in message news:... > 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 :) Hello; After more studying this problem, I came up with this solution below. I use Partitions, then I use Count[]: x = {1, 3, 3, 3, 2, 3, 3, 1, 3, 3}; allPairs = Partition[x, 2, 1] Out[38]= {{1, 3}, {3, 3}, {3, 3}, {3, 2}, {2, 3}, {3, 3}, {3, 1}, {1, 3}, {3, 3}} Count[allPairs, {3, 3}] Out[39]= 4 Count[allPairs, {1, 3}] Out[40]= 2 I see what my problem was. Pattern applies to "one element" in the list. So in the list {a,b,c} It is not possible to make a pattern to find {a,b}? Since {b,c} are not one element in the list, it is 2 elements. So what I partitioned the list, now all adjacent pairs of numbers became individual elements in a new list (a list in a list) so a pattern would work on {b,c} now since {b,c} is now "one element" in a list. Nasser

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