我想使用numcaseswk0优化R中的值（optim）。但是，该值不是ODE的参数-只是初始值。
下面显示的代码尝试执行此操作，但是优化过程始终失败，并且仅产生用户提供的上限。我怀疑这可能是由于numcaseswk0不是ODE的参数之一。如果有人可以指出如何解决此问题，我将感到非常高兴。谢谢。
library(deSolve)
### ODE FUNCTION
HAVODE <- function(t, states, parameters){
with(as.list(c(states,parameters)),
{
N <- S + L + Z + I + R
dS <- -beta * S * (I/N)
dL <- beta * S * (I/N) - (1/durL)*L
dI <- (1/durL)*L +(1/durRel)*R - (1/durI)*I
dZ <- (1-propRelapse)*(1/durI)*I
dR <- propRelapse*(1/durI)*I - (1/durRel)*R
return(list(c(dS, dL, dI, dZ, dR)))
})
}
### COST FUNCTION
calib_function <- function(x, parameters,observed.){
## Variable to be optimized
numcaseswk0 <- x
initpop = parameters[1]
durL = parameters[2]
durI = parameters[3]
fracImmune = parameters[4]
durRel = parameters[5]
propRelapse = parameters[6]
probdetec = parameters[7]
beta = parameters[8]
## Starting values for states
S. = (1-fracImmune)*initpop
L. = numcaseswk0 # *** I want this value to be optimized
I. = 0
Z. = fracImmune*initpop
R. = 0
states = c(S=S., L= L. , I=I., Z=Z., R=R.)
## Parameters to be fed into ODE solver
parameters1 = c(durL = durL, durI = durI,durRel = durRel, propRelapse = propRelapse, beta = beta )
tspan = seq(0, length(observed.)+10);
# Run the ODE solver
result <- data.frame(ode(y = states, times = tspan, func = HAVODE, parms = parameters1))
# Calculating model response (number of detected incident cases)
IncDetec <- probdetec *((1/durL)*result[, 3] + (1/durRel)*result[, 6])
model_response <- IncDetec[-1][1:length(observed.)] # exclude initial week
# Calculate negative log likelihood of model responses
NLLK <- -sum(dpois(x = floor(model_response), lambda = observed., log = TRUE ))
if (NLLK == Inf){
NLLK = 999999 # if NLLK is infinity, replace by a large number
}
return(NLLK)
}
## vector of starting values
x0 <- 2
## set lower and upper bounds for these variables
upper <- 10
lower <- 1
## Call the cost function with optim
calib_parameters <- c(135722, 9.2088, 2.6047, 0.47, 3.930, 7.21, 0.094, 0.517)
optimization_results <- optim(par=x0, lower = lower, upper = upper, method = 'Brent', fn = calib_function, parameters = calib_parameters, observed. = abs(rnorm(50, mean=6, sd=3)))
运行上面的代码将给出：
> optimization_results
$par
[1] 1.000001
$value
[1] 113463174144
$counts
function gradient
NA NA
$convergence
[1] 0
$message
NULL
optim产生的估计值是所提供的下限（lower=1）的值。您可能还会注意到，没有功能评估。为什么优化不适用于numcaseswk0？
答案 0 :(得分：0)
正如Ben已经指出的那样，该代码示例不可复制。如果我从http://desolve.r-forge.r-project.org/user2014/examples/FME/fit_twocomp.svg
摘录的一个教程示例中发布代码段，也许对您有帮助该示例使用软件包FME，该软件包包装了optim（和其他优化程序）并提供了一些其他支持。
### ============================================================================
### code snippet from the useR!2014 (Los Angeles) tutorial
###
### Copyright tpetzoldt, license: GPL >= 2.0
### more see:
### http://desolve.r-forge.r-project.org/user2014/examples/FME/fit_twocomp.svg
### ============================================================================
library(deSolve)
library(FME)
## A two compartment pharmacokinetic model
twocomp <- function (time, y, parms, ...) {
with(as.list(c(parms, y)), {
dCL <- kFL * CF - kLF * CL - ke * CL # concentration in liver
dCF <- kLF * CL - kFL * CF # concentration in fat
list(c(dCL, dCF))
})
}
parms <- c(ke = 0.2, kFL = 0.1, kLF = 0.05)
times <- seq(0, 40, length=200)
y0 <- c(CL = 1, CF = 0)
out <- ode(y0, times, twocomp, parms)
## -----------------------------------------------------------------------------
## data in database format
## -----------------------------------------------------------------------------
dat2 <- data.frame(
label = rep(c("CL", "CF"), each=8), # must be the first column
time = rep(seq(0, 28, 4), 2),
value = c(1.31, 0.61, 0.49, 0.41, 0.20, 0.12, 0.16, 0.21,
0.001, 0.041, 0.050, 0.039, 0.031, 0.025, 0.017, 0.012)
)
## -----------------------------------------------------------------------------
## fit parameters and initial values
## -----------------------------------------------------------------------------
parms <- c(CL = 1.0, CF = 0.0, ke = 0.2, kFL = 0.1, kLF = 0.05)
cost <- function(p, data, ...) {
yy <- p[c("CL", "CF")] # initial values
pp <- p[c("ke", "kFL", "kLF")] # start parameters
out <- ode(yy, times, twocomp, pp)
modCost(out, data, y="value", ...)
}
## The default Marq optimizer fails here, so we use another, e.g. Port
fit6 <- modFit(f = cost, p = parms, data=dat2, weight="std",
lower=rep(0, 5), upper=c(2,2,1,1,1), method="Port")
summary(fit6)
y0 <- coef(fit6)[c("CL", "CF")]
pp <- coef(fit6)[c("ke", "kFL", "kLF")]
out6 <- ode(y0, times, twocomp, pp)
plot(out, out6, obs=dat2)
答案 1 :(得分：0)
我认为您的模型规格有问题。优化程序可以正常运行（从技术角度而言），并且最佳参数为为零。
请查看评论，更改内容：
以下内容可能仍然不是您想要的，但希望可以帮助使其正常运行。祝你好运！
library(deSolve)
### ODE FUNCTION
HAVODE <- function(t, states, parameters){
with(as.list(c(states,parameters)),
{
N <- S + L + Z + I + R
#cat("N=", N, "\n")
dS <- -beta * S * (I/N)
dL <- beta * S * (I/N) - (1/durL)*L
dI <- (1/durL)*L +(1/durRel)*R - (1/durI)*I
dZ <- (1-propRelapse)*(1/durI)*I
dR <- propRelapse*(1/durI)*I - (1/durRel)*R
return(list(c(dS, dL, dI, dZ, dR)))
})
}
### COST FUNCTION
calib_function <- function(x, parameters,observed.){
## Variable to be optimized
#cat("x=", x, "\n")
numcaseswk0 <- x
initpop = parameters[1]
durL = parameters[2]
durI = parameters[3]
fracImmune = parameters[4]
durRel = parameters[5]
propRelapse = parameters[6]
probdetec = parameters[7]
beta = parameters[8]
## Starting values for states
S. = (1-fracImmune)*initpop
L. = numcaseswk0 # *** I want this value to be optimized
I. = 0
Z. = fracImmune*initpop
R. = 0
states = c(S=S., L= L. , I=I., Z=Z., R=R.)
#cat(states, "\n")
## Parameters to be fed into ODE solver
parameters1 = c(durL = durL, durI = durI,durRel = durRel, propRelapse = propRelapse, beta = beta )
tspan = seq(0, length(observed.)+10);
# Run the ODE solver
result <- data.frame(ode(y = states, times = tspan, func = HAVODE, parms = parameters1))
# Calculating model response (number of detected incident cases)
IncDetec <- probdetec *((1/durL)*result[, 3] + (1/durRel)*result[, 6])
model_response <- IncDetec[-1][1:length(observed.)] # exclude initial week
# Calculate negative log likelihood of model responses
NLLK <- -sum(dpois(x = floor(model_response), lambda = observed., log = TRUE ))
## tpe: set it to a really large value
if (!is.finite(NLLK)){
NLLK = 0.1 * .Machine$double.xmax # if NLLK is infinity, replace by a large number
}
## tpe: re-scale return value to a numerically feasible range
return(NLLK * 1e-10)
}
## vector of starting values
x0 <- 2
## set lower and upper bounds for these variables
upper <- 10
lower <- 0
## Call the cost function with optim
calib_parameters <- c(135722, 9.2088, 2.6047, 0.47, 3.930, 7.21, 0.094, 0.517)
## tpe: reproducible comparison data
set.seed(42)
observed <- abs(rnorm(50, mean=6, sd=3))
## test manually
calib_function(1, calib_parameters, observed)
calib_function(0, calib_parameters, observed)
calib_function(10, calib_parameters, observed)
## tpe: we see that zero *is* the best among these
## optimize automatically
optimization_results <- optim(par=x0, lower = lower, upper = upper,
method = 'L-BFGS-B', fn = calib_function,
parameters = calib_parameters,
observed. = observed,
control=list(trace=TRUE))
optimization_results
## tpe: optimized par is again zero, that confirms the manual test