我遇到了一个无法解决的问题。我想使用
我成功地使用
如果我们看一下,我们有一个适当的数据集:
我看起来像是
我们可以看到非线性拟合在单个信号上效果很好。现在,如果我想对带有随机效应的数据集进行回归分析,则语法应为:
这里提供的函数中的猫:
所以,
nlme
或nlmODE
来执行具有随机效应的非线性回归,以固定系数的二阶微分方程的解作为模型(阻尼振荡器)。我成功地使用
nlme
进行简单模型的建立,但似乎使用deSolve
生成微分方程的解会导致问题。以下是一个示例和我面临的问题。
数据和函数
这是使用deSolve
生成微分方程解的函数:library(deSolve)
ODE2_nls <- function(t, y, parms) {
S1 <- y[1]
dS1 <- y[2]
dS2 <- dS1
dS1 <- - parms["esp2omega"]*dS1 - parms["omega2"]*S1 + parms["omega2"]*parms["yeq"]
res <- c(dS2,dS1)
list(res)}
solution_analy_ODE2 = function(omega2,esp2omega,time,y0,v0,yeq){
parms <- c(esp2omega = esp2omega,
omega2 = omega2,
yeq = yeq)
xstart = c(S1 = y0, dS1 = v0)
out <- lsoda(xstart, time, ODE2_nls, parms)
return(out[,2])
}
我可以为给定的周期和阻尼系数生成解决方案,例如这里的周期为20,阻尼系数略微为0.2:
# small example:
time <- 1:100
period <- 20 # period of oscillation
amort_factor <- 0.2
omega <- 2*pi/period # agular frequency
oscil <- solution_analy_ODE2(omega^2,amort_factor*2*omega,time,1,0,0)
plot(time,oscil)
现在,我生成了一个包含10个个体的面板,并采用随机起始相位(即不同的起始位置和速度)。目标是进行非线性回归,其中起始值会产生随机效应。library(data.table)
# generate panel
Npoint <- 100 # number of time poitns
Nindiv <- 10 # number of individuals
period <- 20 # period of oscillation
amort_factor <- 0.2
omega <- 2*pi/period # agular frequency
# random phase
phase <- sample(seq(0,2*pi,0.01),Nindiv)
# simu data:
data_simu <- data.table(time = rep(1:Npoint,Nindiv), ID = rep(1:Nindiv,each = Npoint))
# signal generation
data_simu[,signal := solution_analy_ODE2(omega2 = omega^2,
esp2omega = 2*0.2*omega,
time = time,
y0 = sin(phase[.GRP]),
v0 = omega*cos(phase[.GRP]),
yeq = 0)+
rnorm(.N,0,0.02),by = ID]
如果我们看一下,我们有一个适当的数据集:
library(ggplot2)
ggplot(data_simu,aes(time,signal,color = ID))+
geom_line()+
facet_wrap(~ID)
问题
使用nlme
在简单的例子(非线性函数不使用deSolve)上使用类似语法使用nlme
,我尝试了以下操作:
fit <- nlme(model = signal ~ solution_analy_ODE2(esp2omega,omega2,time,y0,v0,yeq),
data = data_simu,
fixed = esp2omega + omega2 + y0 + v0 + yeq ~ 1,
random = y0 ~ 1 ,
groups = ~ ID,
start = c(esp2omega = 0.08,
omega2 = 0.04,
yeq = 0,
y0 = 1,
v0 = 0))
我得到:
在 checkFunc(Func2, times, y, rho) 中出错:func() 返回的导数数量(2)必须等于初始条件向量的长度(2000)
回溯信息:
12. stop(paste("The number of derivatives returned by func() (", length(tmp[[1]]), ") must equal the length of the initial conditions vector (", length(y), ")", sep = ""))
11. checkFunc(Func2, times, y, rho)
10. lsoda(xstart, time, ODE2_nls, parms)
9. solution_analy_ODE2(omega2, esp2omega, time, y0, v0, yeq)
.
.
我看起来像是
nlme
试图将起始条件的向量传递给solution_analy_ODE2
,并导致lasoda
中的checkFunc
出现错误。
我尝试使用nlsList
:
test <- nlsList(model = signal ~ solution_analy_ODE2(omega2,esp2omega,time,y0,v0,yeq) | ID,
data = data_simu,
start = list(esp2omega = 0.08, omega2 = 0.04,yeq = 0,
y0 = 1,v0 = 0),
control = list(maxiter=150, warnOnly=T,minFactor = 1e-10),
na.action = na.fail, pool = TRUE)
head(test)
Call:
Model: signal ~ solution_analy_ODE2(omega2, esp2omega, time, y0, v0, yeq) | ID
Data: data_simu
Coefficients:
esp2omega omega2 yeq y0 v0
1 0.1190764 0.09696076 0.0007577956 -0.1049423 0.30234654
2 0.1238936 0.09827158 -0.0003463023 0.9837386 0.04773775
3 0.1280399 0.09853310 -0.0004908579 0.6051663 0.25216134
4 0.1254053 0.09917855 0.0001922963 -0.5484005 -0.25972829
5 0.1249473 0.09884761 0.0017730823 0.7041049 0.22066652
6 0.1275408 0.09966155 -0.0017522320 0.8349450 0.17596648
我们可以看到非线性拟合在单个信号上效果很好。现在,如果我想对带有随机效应的数据集进行回归分析,则语法应为:
fit <- nlme(test,
random = y0 ~ 1 ,
groups = ~ ID,
start = c(esp2omega = 0.08,
omega2 = 0.04,
yeq = 0,
y0 = 1,
v0 = 0))
但我收到了完全相同的错误信息。
然后我尝试使用 nlmODE
,按照Bne Bolker在我几年前提出的类似问题中的评论所说。
使用nlmODE
library(nlmeODE)
datas_grouped <- groupedData( signal ~ time | ID, data = data_simu,
labels = list (x = "time", y = "signal"),
units = list(x ="arbitrary", y = "arbitrary"))
modelODE <- list( DiffEq = list(dS2dt = ~ S1,
dS1dt = ~ -esp2omega*S1 - omega2*S2 + omega2*yeq),
ObsEq = list(yc = ~ S2),
States = c("S1","S2"),
Parms = c("esp2omega","omega2","yeq","ID"),
Init = c(y0 = 0,v0 = 0))
resnlmeode = nlmeODE(modelODE, datas_grouped)
assign("resnlmeode", resnlmeode, envir = .GlobalEnv)
#Fitting with nlme the resulting function
model <- nlme(signal ~ resnlmeode(esp2omega,omega2,yeq,time,ID),
data = datas_grouped,
fixed = esp2omega + omega2 + yeq + y0 + v0 ~ 1,
random = y0 + v0 ~1,
start = c(esp2omega = 0.08,
omega2 = 0.04,
yeq = 0,
y0 = 0,
v0 = 0)) #
我遇到了这个错误:
在resnlmeode(esp2omega, omega2, yeq, time, ID)中出现错误:找不到对象“yhat”
在这里,我不明白错误的来源,也不知道如何解决它。
问题
- 你能重现这个问题吗?
- 有人有想法用
nlme
或者nlmODE
解决这个问题吗? - 如果没有,是否有其他包可以解决这个问题?我看到了
nlmixr
(https://cran.r-project.org/web/packages/nlmixr/index.html),但是我不熟悉它,安装很麻烦,而且最近从 CRAN 中移除了。
修改
@tpetzoldt提出了一种很好的调试nlme
行为的方法,让我大吃一惊。这里是一个带有非线性函数的工作示例,我生成了五个具有随机参数变化的个体:
reg_fun = function(time,b,A,y0){
cat("time : ",length(time)," b :",length(b)," A : ",length(A)," y0: ",length(y0),"\n")
out <- A*exp(-b*time)+(y0-1)
cat("out : ",length(out),"\n")
tmp <- cbind(b,A,y0,time,out)
cat(apply(tmp,1,function(x) paste(paste(x,collapse = " "),"\n")),"\n")
return(out)
}
time <- 0:10*10
ramdom_y0 <- sample(seq(0,1,0.01),10)
Nid <- 5
data_simu <-
data.table(time = rep(time,Nid),
ID = rep(LETTERS[1:Nid],each = length(time)) )[,signal := reg_fun(time,0.02,2,ramdom_y0[.GRP]) + rnorm(.N,0,0.1),by = ID]
这里提供的函数中的猫:
time : 11 b : 1 A : 1 y0: 1
out : 11
0.02 2 0.64 0 1.64
0.02 2 0.64 10 1.27746150615596
0.02 2 0.64 20 0.980640092071279
0.02 2 0.64 30 0.737623272188053
0.02 2 0.64 40 0.538657928234443
0.02 2 0.64 50 0.375758882342885
0.02 2 0.64 60 0.242388423824404
0.02 2 0.64 70 0.133193927883213
0.02 2 0.64 80 0.0437930359893108
0.02 2 0.64 90 -0.0294022235568269
0.02 2 0.64 100 -0.0893294335267746
.
.
.
现在我使用 nlme
:
nlme(model = signal ~ reg_fun(time,b,A,y0),
data = data_simu,
fixed = b + A + y0 ~ 1,
random = y0 ~ 1 ,
groups = ~ ID,
start = c(b = 0.03, A = 1,y0 = 0))
我得到:
time : 55 b : 55 A : 55 y0: 55
out : 55
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
time : 55 b : 55 A : 55 y0: 55
out : 55
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
0.03 1 0 0 0
0.03 1 0 10 -0.259181779318282
0.03 1 0 20 -0.451188363905974
0.03 1 0 30 -0.593430340259401
0.03 1 0 40 -0.698805788087798
0.03 1 0 50 -0.77686983985157
0.03 1 0 60 -0.834701111778413
0.03 1 0 70 -0.877543571747018
0.03 1 0 80 -0.909282046710588
0.03 1 0 90 -0.93279448726025
0.03 1 0 100 -0.950212931632136
...
所以,
nlme
将时间向量绑定 5 次(个体数),并将其传递给函数,参数重复相同的次数。这显然与 lsoda
和我的函数工作方式不兼容。
with
是否是原因,但是很容易摆脱它。只需直接访问状态变量和参数,例如S1 <- x[1]
或S1 <- x["S1"]
,然后再用parms
。代码会稍微不那么易读,这就是为什么大多数文档和许多人更喜欢使用with(as.list())
结构的原因。 - tpetzoldt