I want to fit a function in the increase form of exponential decay (or asymptotic curve), such that:
Richness = C*(1-exp(k*Abundance)) # k < 0
I've read on this page about expn()
function, but simply can't find it (or a nls
package). All I found was a nlstools package, but it has no expn()
. I tried with the usual nls
and exp
function, but I only get increasing exponentials...
I want to fit the graph like below (drawn in Paint), and I don't know where the curve should stabilize (Richness = C). Thanks in advance.
This should get you started. Read the documentation on nls(...)
(type ?nls
at the command prompt). Also look up ?summary
and ? predict
.
set.seed(1) # so the example is reproduceable
df <- data.frame(Abundance=sort(sample(1:70,30)))
df$Richness <- with(df, 20*(1-exp(-0.03*Abundance))+rnorm(30))
fit <- nls(Richness ~ C*(1-exp(k*Abundance)),data=df,
algorithm="port",
start=c(C=10,k=-1),lower=c(C=0,k=-Inf), upper=c(C=Inf,k=0))
summary(fit)
# Formula: Richness ~ C * (1 - exp(k * Abundance))
#
# Parameters:
# Estimate Std. Error t value Pr(>|t|)
# C 20.004173 0.726344 27.54 < 2e-16 ***
# k -0.030183 0.002334 -12.93 2.5e-13 ***
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.7942 on 28 degrees of freedom
#
# Algorithm "port", convergence message: relative convergence (4)
df$pred <- predict(fit)
plot(df$Abundance,df$Richness)
lines(df$Abundance,df$pred, col="blue",lty=2)
Thanks, jlhoward. I've got to something similar after reading the link sent by shujaa.
R <- function(a, b, abT) a*(1 - exp(-b*abT))
form <- Richness ~ R(a,b,Abundance)
fit <- nls(form, data=d, start=list(a=20,b=0.01))
plot(d$Abundance,d$Richness, xlab="Abundance", ylab="Richness")
lines(d$Abundance, predict(fit,list(x=d$Abundance)))
I've found the initial values by trial and error, though. So your solution looks better :)
EDIT: The result:
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.