R中的deSolve的lsoda函数出错

Question

I'm trying to simulate the logistic population growth model (dp/dt = k p (1-P/K)) using the lsoda function of deSolve library. However, I keep getting an error on the defined parameters:

tiempo <- seq(0,10,0.5) #define time interval
ic2 <- 1 #define initial population density
parms <- c(K=100, k=2) #define the parameters of the model

log.gr <- function(t,x,k,K){ #define function.

x1 <- k * x[1] * (1-(x[1]/K))
list(c(x1))
}

log.gr.out <- lsoda(ic2, tiempo,log.gr,parms)

Error: Error in func(time, state, parms, ...) : argument "K" is missing, with no default

I already defined parameter K in the parms vector, so I don't know where the error is comming from. This is my first time using deSolve. I tried to look for a similar answer in the forums, but I wasn't successful. Any help is greatly appreciated.

Answer 1

It looks like the parameter variable ( parms ) is accessible via with() (see here ). Try this:

tiempo <- seq(0, 10, 0.5) #define time interval
ic2 <- 1 #define initial population density
parms <- c(K = 100, k=2) #define the parameters of the model
log.gr <- function(t, x, parms) {
        with(as.list(c(parms, x)), {
                x1 <- k * x[1] * (1-(x[1]/K))
                list(c(x1))
        })}
log.gr.out <- lsoda(ic2, tiempo,log.gr,parms)
log.gr.out # output
   time         1
1   0.0  1.000000
2   0.5  2.672371
3   1.0  6.945310
4   1.5 16.866424
5   2.0 35.546072
6   2.5 59.985918
7   3.0 80.295546
8   3.5 91.719949
9   4.0 96.785724
10  4.5 98.793065
11  5.0 99.552603
12  5.5 99.834928
13  6.0 99.939218
14  6.5 99.977638
15  7.0 99.991767
16  7.5 99.996957
17  8.0 99.998889
18  8.5 99.999599
19  9.0 99.999844
20  9.5 99.999940
21 10.0 99.999977