用nlme和lsoda拟合一阶方程

Question

我试图使用nlme和拟合一阶微分模型lsoda。这是基本思想：我首先定义允许生成微分方程解的函数：

library(deSolve)

ODE1 <- function(time, x, parms) {with(as.list(c(parms, x)), {
  import <- excfunc(time)
  dS <- import*k/tau - (S-yo)/tau 
  res <- c(dS)
  list(res)})}


solution_ODE1 = function(tau1,k1,yo1,excitation,time){
  excfunc <- approxfun(time, excitation, rule = 2)
  parms  <- c(tau = tau1, k = k1, yo = yo1, excfunc = excfunc)
  xstart = c(S = yo1)
  out <-  lsoda(xstart, time, ODE1, parms)
  return(out[,2])
}

然后，根据两个ID的公式生成数据：

time <- 0:49
excitation <- c(rep(0,10),rep(1,10),rep(0,10),rep(1,10),rep(0,10))
simu_data <- data.frame(signal = c(solution_ODE1(3,2,0.1,excitation,time)+rnorm(length(time),0,0.1),
                                   solution_ODE1(3.2,1.5,0.3,excitation,time)+rnorm(length(time),0,0.1)),
                        time = rep(time,2),
                        excitation = rep(excitation,2),
                        ID = rep(c("A","B"),each = length(time)))

这是它的样子：

library(ggplot2)
ggplot(simu_data)+
  geom_point(aes(time,signal,color = "signal"),size = 2)+
  geom_line(aes(time,excitation,color = "excitation"))+
  facet_wrap(~ID)

然后，我尝试使用nlme进行调整：

fit1 <- nlme(signal ~ solution_ODE1(damping,gain,eq,excitation,time),
             data = simu_data,
             fixed = damping + gain + eq ~1,
             random =  damping   ~ 1 ,
             groups = ~ ID,
             start = c(damping = 5, gain = 1,eq = 0))

我得到这个错误，但我没有得到：

eval（替代（expr），数据，enclos = parent.frame（））中的错误：找不到对象'k'

该traceback显示的错误来自ODE1模型，生成值时，其工作原理。

16.    eval(substitute(expr), data, enclos = parent.frame()) 
15.    eval(substitute(expr), data, enclos = parent.frame()) 
14.    with.default(as.list(c(parms, x)), {
    import <- excfunc(time)
    dS <- import * k/tau - (S - yo)/tau
    res <- c(dS) ... 
13.    with(as.list(c(parms, x)), {
    import <- excfunc(time)
    dS <- import * k/tau - (S - yo)/tau
    res <- c(dS) ... 
12.    func(time, state, parms, ...) 
11.    Func2(times[1], y) 
10.    eval(Func2(times[1], y), rho) 
9.    checkFunc(Func2, times, y, rho) 
8.    lsoda(xstart, time, ODE1, parms) 
7.    solution_ODE1(damping, gain, eq, excitation, time) 
6.    eval(model, data.frame(data, pars)) 
5.    eval(model, data.frame(data, pars)) 
4.    eval(modelExpression[[2]], envir = nlEnv) 
3.    eval(modelExpression[[2]], envir = nlEnv) 
2.    nlme.formula(signal ~ solution_ODE1(damping, gain, eq, excitation, 
    time), data = simu_data, fixed = damping + gain + eq ~ 1, 
    random = damping ~ 1, groups = ~ID, start = c(damping = 5, 
        gain = 1, eq = 0)) 
1.    nlme(signal ~ solution_ODE1(damping, gain, eq, excitation, time), 
    data = simu_data, fixed = damping + gain + eq ~ 1, random = damping ~ 
        1, groups = ~ID, start = c(damping = 5, gain = 1, eq = 0)) 

有谁知道我应该如何进行？

---

编辑

我尝试根据mikeck的建议进行修改：

ODE1 <- function(time, x, parms) {
  import <- parms$excfunc(time)
  dS <- import*parms$k/parms$tau - (x["S"]-parms$yo)/parms$tau 
  res <- c(dS)
  list(res)}

生成数据可以正常工作。但是nlme现在使用给出：

checkFunc（Func2，times，y，rho）中的错误：func（）返回的导数数（0）必须等于初始条件向量的长度（100）

具有以下回溯：

> traceback()
10: stop(paste("The number of derivatives returned by func() (", 
        length(tmp[[1]]), ") must equal the length of the initial conditions vector (", 
        length(y), ")", sep = ""))
9: checkFunc(Func2, times, y, rho)
8: lsoda(xstart, time, ODE1, parms) at #5
7: solution_ODE1(damping, gain, eq, excitation, time)
6: eval(model, data.frame(data, pars))
5: eval(model, data.frame(data, pars))
4: eval(modelExpression[[2]], envir = nlEnv)
3: eval(modelExpression[[2]], envir = nlEnv)
2: nlme.formula(signal ~ solution_ODE1(damping, gain, eq, excitation, 
       time), data = simu_data, fixed = damping + gain + eq ~ 1, 
       random = damping ~ 1, groups = ~ID, start = c(damping = 5, 
           gain = 1, eq = 0))
1: nlme(signal ~ solution_ODE1(damping, gain, eq, excitation, time), 
       data = simu_data, fixed = damping + gain + eq ~ 1, random = damping ~ 
           1, groups = ~ID, start = c(damping = 5, gain = 1, eq = 0))

Answer 1

在您的示例中，您的times向量并不是单调运行的。我认为这与lsoda. 时间在这里运作的方式的背景/意义是什么？将随机效应模型与两组进行拟合并没有多大意义。您是否想将同一条曲线拟合到两个独立的时间序列？

这是一个经过一些调整的精简示例（并非所有内容都可以折叠为数字向量而不丢失必要的结构）：

library(deSolve)
ODE1 <- function(time, x, parms) {
    with(as.list(parms), {
        import <- excfunc(time)
        dS <- import*k/tau - (x-yo)/tau 
        res <- c(dS)
        list(res)
    })
}
solution_ODE1 = function(tau1,k1,yo1,excitation,time){
    excfunc <- approxfun(time, excitation, rule = 2)
    parms  <- list(tau = tau1, k = k1, yo = yo1, excfunc = excfunc)
    xstart = yo1
    out <-  lsoda(xstart, time, ODE1, parms)
    return(out[,2])
}
time <- 0:49
excitation <- c(rep(0,10),rep(1,10),rep(0,10),rep(1,10),rep(0,10))
simu_data <- data.frame(time = rep(time,2),
                        excitation = rep(excitation,2))
svec <- c(damping = 3, gain = 1.75, eq = 0.2)

这有效：

with(c(simu_data, as.list(svec)),
     solution_ODE1(damping,gain,eq,excitation[1:50],time[1:50]))

但如果我们再添加一个步骤（以便时间重置为 0），则会失败：

with(c(simu_data, as.list(svec)),
     solution_ODE1(damping,gain,eq,excitation[1:51],time[1:51]))

lsoda(xstart, time, ODE1, parms) 中的错误：在采取任何集成步骤之前检测到非法输入 - 请参阅书面消息