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Estimate sex ratio according to constant incubation temperature The data.frame has the temperatures or durations in columns and the quantiles in rows. Note that incubation duration is a very bad proxy for sex ratio. See Georges, A., Limpus, C. J. & Stoutjesdijk, R. 1994. Hatchling sex in the marine turtle _Caretta caretta_ is determined by proportion of development at a temperature, not daily duration of exposure. J. Exp. Zool., 270, 432-444. If replicate.CI is 0 or NULL, point estimate for maximum likelihood is returned.
# S3 method for tsd
predict(
object,
temperatures = NULL,
durations = NULL,
SD.temperatures = NULL,
SD.durations = NULL,
resultmcmc = NULL,
chain = 1,
replicate.CI = 10000,
probs = c(0.025, 0.5, 0.975),
...
)
A result file generated by tsd
A vector of temperatures
A vector of durations
SD of temperatures
SD of durations
A result of tsd_MHmcmc()
What chain to be used is resultmcmc is provided
Number of replicates to estimate CI
The quantiles to be returned, default=c(0.025, 0.5, 0.975)
Not used
A data.frame with informations about sex-ratio
predict.tsd Estimate sex ratio according to constant incubation temperature
Other Functions for temperature-dependent sex determination:
DatabaseTSD.version()
,
DatabaseTSD
,
P_TRT()
,
TSP.list
,
plot.tsd()
,
stages
,
tsd_MHmcmc_p()
,
tsd_MHmcmc()
,
tsd()
# NOT RUN {
library(embryogrowth)
m <- c(10, 14, 7, 4, 3, 0, 0)
f <- c(0, 1, 2, 4, 15, 10, 13)
t <- c(25, 26, 27, 28, 29, 30, 31)
result <- tsd(males=m, females=f, temperatures=t)
plot(result)
predict(result, temperatures=c(25, 31), replicate.CI = 10000)
predict(result, temperatures=c(25, 31), SD.temperatures = c(1, 2), replicate.CI = 10000)
d <- c(72, 70, 65, 63, 62, 60, 59)
result <- tsd(males=m, females=f, durations=d)
predict(result, durations=c(67, 68), replicate.CI = 10000)
# }
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