如何在 R 中解决这个问题：在 while (t <= cc[i]) { ... : 条件长度 > 1 并且只使用第一个元素

Question

I choose division 0 = d0 < d1 < . . . < dm =tau, and then, for t ∈ (dj−1, dj ] the intensity function λ(t) is equal to λj.

My data set

parlocation<-125.0949
parscale<-34.98857
tau<-365
trendlinier$coefficients[1]<-78.37888
trendlinier$coefficients[2]<--5.809776
lambdaj<-c(8.665351,178.934646,161.189187,57.680814,43.540869,102.405160,108.003538,157.690273,138.600075,269.314099,353.897407,197.891330)
deretbln2<-c(0,31,62,90,121,151,182,212,243,274,304,335,365)

and then, I make this function for λ(t) in R

lambda2 <- function(t){for (i in 1:12) {while (t <=deretbln2[i]) {return(lambdaj[i-1])}}}

I also make this function in R

FU0 <- function(t){pcauchy((tau-t)*trendlinier$coefficients[1]/((trendlinier$coefficients[2]*t)+trendlinier$coefficients[1]),location=parlocation,scale=parscale)}

intensityIBNR <- function(t){lambda2(t)*(1-FU0(t))}

when I run this code

c <- integrate(f=intensityIBNR, lower=0, upper=tau,
           subdivisions=200)

its result like this

There were 50 or more warnings (use warnings() to see the first 50)
> warnings()
Warning messages:
1: In while (t <= deretbln2[i]) { ... :
the condition has length > 1 and only the first element will be used

Please, help me to solve this problem..

Answer 1

a few remarks on the side:

1) the object trendlinier is non-existant in your MWE; hence trendlinier$coefficients[1] should terminate with an error. We need to instantiat that object first as

trendlinier <- list(coefficients = NULL)

2) I would recommend not using t or c as object names in R , since both are used for function names; rather use tt or cc (or better, variable names that have more meaning); Let's stay with tt and cc (and leave the lambda2 - function on the side), we'll obtain

parlocation<-125.0949
parscale<-34.98857
tau<-365
trendlinier$coefficients[1]<-78.37888
trendlinier$coefficients[2]<--5.809776
lambdaj<-c(8.665351,178.934646,161.189187,57.680814,43.540869,102.405160,108.003538,157.690273,138.600075,269.314099,353.897407,197.891330)
deretbln2<-c(0,31,62,90,121,151,182,212,243,274,304,335,365)

FU0 <- function(tt){pcauchy((tau-tt)*trendlinier$coefficients[1]/((trendlinier$coefficients[2]*tt)+trendlinier$coefficients[1]),location=parlocation,scale=parscale)}
intensityIBNR <- function(tt){lambda2(tt)*(1-FU0(tt))}
cc <- integrate(f=intensityIBNR, lower=0, upper=tau, subdivisions=200)

Now, note that R 's indices start at 1; that is regarding the for -loop in the lambda2 -function, for i=1 , there is no lambdaj[i-1] . You need to consider and implement different routines for i=1 and i>1 (or add a default value to lambdaj ; eg, lambdaj <- c(0, lambdaj) if 0 is a reasonble value in this case). (Another remark on the side: the statement while (expression) { return } is an unusual implementation of the if -clause). If you prefer not altering lambdaj you could use

lambda2 <- function(tt){for (i in 1:12) {if (i == 1) return(0) else if (tt <= deretbln2[i]) return(lambdaj[i-1])}}

PS: that is not the best way to typeset R -Code (cf., Hadley's style guide )