Hi
not sure if this is what you want. It does not do fuzzy matching but make
a exact evaluation equal row sums of arrays.
rle(do.call("c",lapply(lapply(l, rowSums), function(x) paste(x,
collapse=""))))
Maybe something similar can be done without conversion to character.
Regards
Petr
[EMAIL PROTECTED] napsal dne 24.10.2008 12:34:15:
> Dear friends
> Hope you all are fine. Suppose we have a list of arrays.
> a1=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a1=array(a1,dim=c(4,4));
a2=c(4,4,4,
> 4,0,4,4,4,0,3,3,0,0,0,0,0); a2=array(a2,dim=c(4,4));
> a3=c(4,4,4,4,0,3,3,4,0,4,4,0,0,0,0,0); a3=array(a3,dim=c(4,4));
a4=c(4,4,4,4,
> 4,0,3,3,3,3,0,4,4,4,0,0,0,0,0,0); a4=array(a4,dim=c(5,4));
a5=c(4,4,4,4,4,0,4,
> 4,4,4,0,3,3,3,0,0,1,1,0,0); a5=array(a5,dim=c(5,4));
a6=c(4,4,4,4,4,0,1,1,1,1,
> 0,4,4,4,0,0,3,3,0,0); a6=array(a6,dim=c(5,4));
a7=c(1,1,1,1,1,0,4,4,4,4,0,3,3,
> 3,0,0,4,4,0,0); a7=array(a7,dim=c(5,4));
a8=c(4,4,4,4,4,0,3,3,3,3,0,1,1,1,0,0,
> 4,4,0,0); a8=array(a8,dim=c(5,4));
> l=list(a1,a2,a3,a4,a5,a6,a7,a8);
>
> x <- sapply(1:length(l), function(x) {
> sum(sapply(l, function(y) {
> if ( nrow(l[[x]]) != nrow(y) | ncol(l[[x]]) != ncol(y) ) FALSE
> else sum(y != l[[x]]) == 0
> }))
> } ); l; x
>
> Using the above function, we are able to get frequency of each most
repeated
> similar components of the list. For example, [[1]] and [[2]] are most
repeated
> similar out of all. But if we consider the "combinations" at each row of
each
> array. Then [[3]] will be included with [[1]] and [[2]]. Also [[5]],
[[6]] and
> [[8]] will be similar. How can we modify the above function to get the
desired
> most repeated sequence in this case? Any help in this regard is needed.
>
> best regards
> M.Azam
>
>
>
>
> [[alternative HTML version deleted]]
>
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