Below are my codes:
library(fitdistrplus)
s <- c(11, 4, 2, 9, 3, 1, 2, 2, 3, 2, 2, 5, 8,3, 15, 3, 9, 22, 0, 4, 10, 1, 9, 10, 11,
2, 8, 2, 6, 0, 15, 0 , 2, 11, 0, 6, 3, 5, 0, 7, 6, 0, 7, 1, 0, 6, 4, 1, 3, 5,
2, 6, 0, 10, 6, 4, 1, 17, 0, 1, 0, 6, 6, 1, 5, 4, 8, 0, 1, 1, 5, 15, 14, 8, 1,
3, 2, 9, 4, 4, 1, 2, 18, 0, 0, 10, 5, 0, 5, 0, 1, 2, 0, 5, 1, 1, 2, 3, 7)
o <- fitdist(s, "gamma", method = "mle")
summary(o)
plot(o)
and the error says:
Error in fitdist(s, "gamma", method = "mle") : the function mle failed to estimate the parameters, with the error code 100
The Gamma distribution doesn't allow zero values (the likelihood will evaluate to zero, and the log-likelihood will be infinite, for a response of 0) unless the shape parameter is exactly 1.0 (i.e., an exponential distribution - see below) ... that's a statistical/mathematical problem, not a programming problem. You're going to have to find something sensible to do about the zero values. Depending on what makes sense for your application, you could (for example)
fitdist(s[s>0], ...)
)fitdist(replace(s,which(s==0),0.1),...)
which (if any) of these is best depends on your application.
@Sandipan Dey's first answer (leaving the zeros in the data set) appears to make sense, but in fact it gets stuck at the shape parameter equal to 1.
o <- fitdist(s, "exp", method = "mle")
gives the same answer as @Sandipan's code (except that it estimates rate=0.2161572, the inverse of the scale parameter=4.626262 that's estimated for the Gamma distribution - this is just a change in parameterization). If you choose to fit an exponential instead of a Gamma, that's fine - but you should do it on purpose, not by accident ...
To illustrate that the zeros-included fit may not be working as expected, I'll construct my own negative log-likelihood function and display the likelihood surface for each case.
mfun <- function(sh,sc,dd=s) {
-sum(dgamma(dd,shape=sh,scale=sc,log=TRUE))
}
library(emdbook) ## for curve3d() helper function
Zeros-included surface:
cc1 <- curve3d(mfun(x,y),
## set up "shape" limits" so we evaluate
## exactly shape=1.000 ...
xlim=c(0.55,3.55),
n=c(41,41),
ylim=c(2,5),
sys3d="none")
png("gammazero1.png")
with(cc1,image(x,y,z))
dev.off()
In this case the surface is only defined at shape=1 (i.e. an exponential distribution); the white regions represent infinite log-likelihoods. It's not that shape=1 is the best fit, it's that it's the only fit ...
Zeros-excluded surface:
cc2 <- curve3d(mfun(x,y,dd=s[s>0]),
## set up "shape" limits" so we evaluate
## exactly shape=1.000 ...
xlim=c(0.55,3.55),
n=c(41,41),
ylim=c(2,5),
sys3d="none")
png("gammazero2.png")
with(cc2,image(x,y,z))
with(cc2,contour(x,y,z,add=TRUE))
abline(v=1.0,lwd=2,lty=2)
dev.off()
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