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Yes -- this seems to be the case; the following example works as expected. Thank you! R = matrix(rnorm(9),3,3) R[lower.tri(R)] = 0 R.inv = solve(R) Dmat = t(R) %*% R dvec = c(0,5,0) Amat...
The strptime object is a structured list object with 9 components. (Type "names(z)" to see the names of the individual components). You may want to convert the strptime object to a POSIXct object using Z = as.POSIXct(z) This...
It seems that you have a nonlinear constraint (rather than box constraints). Provided your constraint is reasonably well behaved (differentiable) you could still use optim on the Lagrangian, and then run an "outer" optimisation to find the langrangian multiplier and...
Hi Duncan Thanks for the reply. Unfortunately, the Excel process still persists indefinitely even after removing the handle E and issuing the gc() command -- even after closing R, too. For the moment, I have found a way round this by...
Duncan -- thanks for your reply. As you point out -- this is more a "nice-to-have" and should not really be prioritised in any way. I currently run a lengthy statistical process in which intermediate results are continually output to...
On the subject of the LAPACK error -- I also had this problem with the R (D)COM package. It appears that there is a problem accessing the newer LAPACK routines rather than the original LINPACK routines which were built into...
Try specifying quote=NULL as an argument to read.table. It could be that one of your fields has a quote symbol in it. -----Original Message----- From: Ajay Shah [mailto:ajayshah at mayin.org] Sent: 05 July 2004 11:15...
A bit cumbersome, but (somewhat) vectorised. dist = apply(B, 1, FUN = function(x, M) {rowSums(sweep(M, 2, x, "-")^2) }, A) m = row(t(dist))[t(dist==apply(dist,1,min))] -----Original Message----- From: Roger Bivand [mailto:Roger.Bivand at...
As previous replies have suggested -- you could use an "array" E.g. > X = array(0, dim=c(100,12,12)) Then X[i, , ] returns the i-th 12 by 12 matrix "slice" in this array, e.g. > X[1, ,] [,1...
A starting point might be the string splitting function strsplit For example, > X = c("1,4,5" "1,2,5" "5,1,2") > strsplit(X) [[1]] [1] "1" "4" "5" [[2]] [1] "1" "2" "5" [[3]] [1] "5" "1" "2...
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