Hello,
Sorry, disregard my previous e-mail.
Instead of your FPG function try
FPG <- function(pH, model) {
? coef(model)[1] + coef(model)[2]*pH + coef(model)[3]*pH^2
}
FPG(aedf$pH, m)
fitted(m)
Hope this helps,
Rui Barradas
?s 12:30 de 02/04/20, Rui Barradas escreveu:
Hello,
Tip: in a formula, pH^2 = pH*pH is an interaction.
pH^2 = pH*pH = pH + pH + pH:pH = pH
Try I(pH^2)
Hope this helps,
Rui Barradas
?s 12:07 de 02/04/20, Troels Ring escreveu:
Dear friends - I'm on Win10 with R 6.3.1 and have a very simple
problem with
nls which apparently gives a decent fit to the parable below, even
without
starting values. But when I then think I know the meaning of the three
parameters a, b, and d it goes very wrong. I guess I am again
overlooking
something easy but cannot spot it.
BW
Troels Ring,
Aalborg, Denmark
aedf <- structure(list(Flux = c(-0.141256, -0.154709, -0.215247,
-0.302691,
???????????? -0.32287, -0.511211, -0.605381, -0.813901, -1.11659,
-1.76906
), pH = c(7.06273, 7.11182, 7.16182, 7.18818, 7.21455, 7.23818,
?????????? 7.26273, 7.28455, 7.31182, 7.34364)), class = "data.frame",
row.names = c(NA,
-10L))
m <- with(aedf,nls(Flux~a + b*pH + d*pH^2))
with(aedf,plot(pH,Flux))
with(aedf,lines(pH,fitted(m),lty=1,col="red",lwd=3))
m
# a??????? b??????? d
# -1630.70?? 457.67?? -32.11
fitted(m)
# 1] -0.22087053 -0.09989926 -0.13579690 -0.21936385 -0.34761742
-0.50048799
# [7] -0.69729602 -0.90471194 -1.20692552 -1.61994657
FPG <- function(pH) -1630.70 + 457.67*pH -32.11*pH^2
FPG(aedf$pH)
# [1] -0.016359649? 0.107602395? 0.074773375 -0.007166685 -0.133786467
# [6] -0.285187665 -0.480463769 -0.686513537 -0.987013685 -1.398026917
# So why aren't fitted(m) and FPG(aedf$pH) not closer ("equal")?
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