On 10/14/2015 05:00 AM, r-help-requ...@r-project.org wrote:
I am trying to fit this data to a weibull distribution:
My y variable is:1 1 1 4 7 20 7 14 19 15 18 3 4 1 3 1 1 1
1 1 1 1 1 1
and x variable is:1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
19 20 21 22 23 24
One could always use existing R functions that fit a Weibull regression, instead of
reinventing the wheel.
library(survival)
y <- scan()
1 1 1 4 7 20 7 14 19 15 18 3
4 1 3 1 1 1
1 1 1 1 1 1
wfit <- survreg(Surv(1:24) ~ 1, weights=y, dist="weibull")
wfit
Call:
survreg(formula = Surv(1:24) ~ 1, weights = y, dist = "weibull")
Coefficients:
(Intercept)
2.352121
Scale= 0.4130924
Loglik(model)= -351.4 Loglik(intercept only)= -351.4
n= 24
zz <- seq(0, 25, length=100)
plot(zz, dsurvreg(zz, 2.352121, 0.4130924), col=2, type='l', ylim=c(0, .15),
xlab="Value", ylab="Density")
points(1:24, y/sum(y))
-----
There are a half dozen ways to parameterize a Weibull distribution; the location-scale
form used by survreg is one of the less common. See help(survreg) for more information --
look at the example near the bottom of the page.
Terry Therneau
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