[R] counting strings of identical values in a matrix

[Marc Schwartz]

Moshe,

Gabor posted that same solution shortly after my reply on Thursday.

I had one of those "head banging" episodes shortly thereafter...  :-)

Regards,

Marc

On Mon, 2007-11-19 at 17:46 -0800, Moshe Olshansky wrote:
> How about adding an artificial last row containing no
> 1's (say a row of zeros)?
> 
> --- Marc Schwartz <marc_schwartz at comcast.net> wrote:
> 
> > 
> > On Thu, 2007-11-15 at 17:53 +0100, A M Lavezzi
> > wrote:
> > > thank you.
> > > I did not think about the case of overlapping of 
> > > 1's from the end of one column to the start of the
> > next,
> > > this would actually be a problem
> > > 
> > > In the simulations I am running each column 
> > > corresponds to the path followed by an agent 
> > > across states of a stochastic process,
> > > so I would like to avoid mixing up two different 
> > > paths (I made a mistake when I mentioned the
> > possibility of turning my matrix
> > > into a vector, sorry about that).
> > > 
> > > can I kindly ask again your help on this?
> > > 
> > > please excuse me.
> > > 
> > > Mario
> > 
> > <snip>
> > 
> > Not a problem.  After sending my follow up, I
> > suspected that you might
> > need a more general approach. This sort of ends up
> > being a combination
> > of the first two, in order to keep each column
> > sequence intact:
> > 
> > 
> > res <- do.call(cbind, apply(prova, 2, 
> >                             function(x)
> > do.call(rbind, rle(x))))
> > 
> > > res
> >         [3,] [5,] [7,]   [4,] [7,]   [4,] [8,]  
> > [2,] [3,]  
> > lengths    2    2    2 2    3    3 2    3    4 1   
> > 1    1 6
> > values     3    1    3 1    3    1 3    3    1 3   
> > 3    1 3
> > 
> > 
> > 
> > > table(res["lengths", res["values", ] == 1])
> > 
> > 1 2 3 4 
> > 1 2 1 1 
> > 
> > 
> > I think that should do it, but you might want to
> > test it on a known set
> > of data.
> > 
> > HTH,
> > 
> > Marc
> > 
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> > 
>