Thank you all.
Problem solved.
Regards,
Phil
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h
Michael originally suggested ?outer. I think that was enough in this
case (no need for mv or sapply):
wlpk3 <- outer(k3,wl,'+')
ln.phiDIC <- log(k1)+k2/wlpk3
phiDIC<- t(exp(ln.phiDIC))
colnames(phiDIC)<- stations
str(phiDIC)
Cheers
On Thu, Apr 12, 2012 at 8:58 PM, R. Michael Weylandt
wrote:
>
Strange, that isn't the error I get:
> mouter(wl, k1, k2, k3, FUN = function(w, k1, k2, k3) k1 *exp(k2 / (w + k3)))
Error in FUN(X, Y, ...) : argument "k2" is missing, with no default
Still, it's a problem with my mouter() function which was only tested
on binary operators (and then only really
Hi and thank you for your time.
I got this error when trying your function.
mouter(wl, k1, k2, k3, FUN = function(w, k1, k2, k3) k1 *exp(k2 / (w + k3)))
"Error in k3/(w + k3) : 'k3' is missing"
Regards,
Phil
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Perhaps ?outer -- well, not outer directly, but a multivariate outer
-- I keep this one around for personal use:
`mouter` <- function(..., FUN = "*"){
dotArgs <- list(...)
FUN <- match.fun(FUN)
if(length(dotArgs) == 1L)
return(unlist(dotArgs))
if (length(dotArgs) == 2L)
Hi every one. I have a exponential function (3 fitting parameters) that I
would like to use to produce data (6 series) without having to use a loop.
Here
wl = seq(300,500,1)
k1 = c(1.2e-6, 4.9e-6, 9.6e-6, 2.7e-10, 6.7e-8, 7.44e-6)
k2 = c(726, 352, 128, 5232, 1538, 128)
k3 = c(-176, -224, -257, 88
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