[R] Proba( Ut+2=1 / ((Ut+1==1) && (Ut==1))) ?

[Tony Plate]

table() can return all the n-gram statistics, e.g.:

 > v <- sample(c(-1,1), 1000, rep=TRUE)
 > table("v_{t-2}"=v[-seq(to=length(v), len=2)], 
"v_{t-1}"=v[-c(1,length(v))], "v_t"=v[-(1:2)])
, , v_t = -1

        v_{t-1}
v_{t-2}  -1   1
      -1 136 134
      1  131 112

, , v_t = 1

        v_{t-1}
v_{t-2}  -1   1
      -1 131 113
      1  115 126

 >

This says that there were 136 cases in which a -1 followed two -1's (and 
126 cases in which a 1 followed to 1's).

If you're really only interested in particular contexts, you can do 
something like:

 > table(v[-seq(to=length(v), len=2)]==1 & v[-c(1,length(v))]==1 & 
v[-(1:2)]==1)

FALSE  TRUE
   872   126
 > table(v[-seq(to=length(v), len=2)]==-1 & v[-c(1,length(v))]==-1 & 
v[-(1:2)]==-1)

FALSE  TRUE
   862   136

or

 > sum(v[-seq(to=length(v), len=2)]==-1 & v[-c(1,length(v))]==-1 & 
v[-(1:2)]==-1)
[1] 136
 >
vincent wrote:
> Dear all,
> 
> First I apologize if my question is quite simple,
> but i'm very newbie with R.
> 
> I have vectors of the form v = c(1,1,-1,-1,-1,1,1,1,1,-1,1)
> (longer than this one of course).
> The elements are only +1 or -1.
> 
> I would like to calculate :
> - the frequencies of -1 occurences after 2 consecutives -1
> - the frequencies of +1 occurences after 2 consecutives +1
> 
> It looks probably something like :
> Proba( Ut+2=1 / ((Ut+1==1) && (Ut==1)))
> 
> could someone please give me a little hint about how
> i should/could begin to proceed ?
> 
> Thanks
> (Thanks also to the R creators/contributors, this soft
> seems really great !)
> 
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