可以用循环或应用或向量化来完成它。

 > x（1，2，3，4，5）
> r<  -  NA 
>对于（n in 2：length（x））r [n] > （r）
 [1] NA 1 1 1 NA 
> 
> r<  -  NA 
> lapply（2：length（x），function（n）r [n]  [1] 1 

 [[2]] 
 [1] 1 

 [[3]] 
 [1] 1 

 [[4]] 
 [1] NA 

> （r）
 [1]不适用1 1 1不适用

> r<  -  NA 
> （x（x），x（1）），其中x（x，y） （r）
 [1] NA 1 1 1 NA 
 

矢量化是最高效的代码难以破译

 > x<  -  runif（50000）
> 
> r<  -  NA 
> system.time（for（n in 2：length（x））r [n]  8.55 0.01 8.58 
> 
> r<  -  NA 
> system.time（lapply（2：length（x），function（n）r [n] 用户系统消耗
 11.36 0.00 11.39 
> 
> r<  -  NA 
>系统时间（r 用户系统消耗
 0.01 0.00 0.01 
 

I have a vector containing simple time series data (extracted from a deSolve matrix), which for testing purposes can be:

x <- c(1, 2, 3, 4, 5)

and would like to apply the nonlinear filter

x[n]*x[n]-x[n-1]*x[n+1]

to all elements of the vector except the first and last elements because the filter can't be applied to these two elements (e.g., when the x[n-1] term meets the first element or the x[n+1] term meets the last element). Therein lies my problem.

Things I've tried: 1) The filter() command expects a linear filter (i.e., without multiplication of filter coefficients). 2) lapply() requires that the function applies to all elements of the list.

Is a loop the only alternative?

Thanks for your help, Carey

Can do it with loop or apply or vectorized.

> x <- c(1, 2, 3, 4, 5)
> r <- NA
> for (n in 2:length(x)) r[n] <- x[n]*x[n]-x[n-1]*x[n+1]
> (r)
[1] NA  1  1  1 NA
> 
> r <- NA
> lapply(2:length(x),function(n) r[n] <<- x[n]*x[n]-x[n-1]*x[n+1])
[[1]]
[1] 1

[[2]]
[1] 1

[[3]]
[1] 1

[[4]]
[1] NA

> (r)
[1] NA  1  1  1 NA

> r <- NA
> r <- x^2 - c(NA,x[1:(length(x)-1)]) * c(x[2:length(x)],NA)
> (r)
[1] NA  1  1  1 NA

vectorization is the most efficient but code is harder to decipher

> x <- runif(50000)
> 
> r <- NA
> system.time(for (n in 2:length(x)) r[n] <- x[n]*x[n]-x[n-1]*x[n+1])
   user  system elapsed 
   8.55    0.01    8.58 
> 
> r <- NA
> system.time(lapply(2:length(x),function(n) r[n] <<- x[n]*x[n]-x[n-1]*x[n+1]))
   user  system elapsed 
  11.36    0.00   11.39 
> 
> r <- NA
> system.time(r <- x^2 - c(NA,x[1:(length(x)-1)]) * c(x[2:length(x)],NA))
   user  system elapsed 
   0.01    0.00    0.01