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I have a 'named expression' like expr <- expression(rep(1,d)) and would like to replace the argument d with say 5 without actually evaluating the expression. So I try substitute(expr, list(d=5)) in which case R simply returns expr which when I 'evaluate' it gives eval(expr) Error in rep.default(1, d) : invalid number of copies in rep() I've looked at ?substitute and

Consider the following example: # substitute a with b in the indicated function. Seems to work. > z <- substitute( function()a+1, list(a=quote(b)) ) > z function() b + 1 # z is an object of class call so use eval # to turn it into an object of class expression; however, # when z is evaluated, the variable a returns. > eval(z) function()a+1 Why did a suddenly reappear again

Hallo I'm having trouble figuring out how to evaluate an expression when one of the variables in the expression is defined separately as a sub expression. Here's a simplified example mat <- expression(0, f1*s1*g1) # vector of formulae g1 <- expression(1/Tm) # expansion of the definition of g1 vals <- data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for

Hello Folks, This is the long awaited (like 2 weeks) 0.3.12 release of Mongrel. This release has received heavier testing than previous releases and supports a whole raft of improvements to existing functionality plus some new stuff. For those not clued in, Mongrel is a web server written in (mostly) Ruby. Check the funny dogs and read the docs about it at http://mongrel.rubyforge.org/. The