在Optim中约束单个功能参数-TERR

如何解决在Optim中约束单个功能参数-TERR

最终转到另一个代码。似乎工作得更好，但需要弄清楚两件事...

如何遍历col.Name输入中的所有唯一值。
防止Hyp.b在optim（）和剩余计算中大于1.00。

Hyp.q.forward.fun = function( time,Hyp.qi,Hyp.b,Hyp.Di){ 
  Hyp.q.theo = Hyp.qi*(1 + Hyp.b*Hyp.Di*time)^(-1/Hyp.b)
    return(Hyp.q.theo)}

residfun = function(x,x.days,y.prod){
  Hyp.qi = x[1]
  Hyp.b  = x[2]
  Hyp.Di = x[3]
  q.theo = 
  (365.25/12)*Hyp.q.forward.fun(time=x.days,Hyp.qi=Hyp.qi,Hyp.b=Hyp.b,Hyp.Di=Hyp.Di)
    residual = sqrt(sum((q.theo-y.prod)^2))
  return(residual)}

min.data = 3

result = 
    data.frame(time.days =  numeric(0),time.months = numeric(0),production = numeric(0),theo = numeric(0))
Hyp.qi       =as.numeric(NA)
Hyp.b        =as.numeric(NA)
Hyp.Di.daily =as.numeric(NA)
Hyp.Di.annual=as.numeric(NA)

ok = length(col.Production)>0
if(ok) ok = length(col.Date) == length(col.Production)

if(ok){
  u.order = order(col.Date)
  
  col.Date       = col.Date[u.order]
  col.Production = col.Production[u.order]
  
  t.days = as.numeric(difftime(col.Date,min(col.Date),units="days"))
  t.months = round(t.days*12/365.25)
  
  u0 = which.max(col.Production)
  u = u0:length(col.Production)
  
  x.months = t.months[u]
  x.days = t.days[u]
  y.prod = col.Production[u]
  x.date = col.Date[u]
  Well = col.Name[u]
  
  x0 = c(y.prod[1]*12/365.25,1.00,0.005)
  
  optim.result = 
    optim(par = x0,fn=residfun,x.days=x.days,y.prod=y.prod)
  
  Hyp.qi = optim.result$par[1]*(365.25/12) 
  Hyp.b  = optim.result$par[2]
  Hyp.Di.daily  = optim.result$par[3]
  Hyp.Di.annual = optim.result$par[3]*365.25 
  
  result = 
    data.frame(Well=Well,time.days = x.days,time.months = x.months,production = y.prod,qi= Hyp.qi,b = Hyp.b,did=Hyp.Di.daily,date=x.date)
}