R error: Wrong length for a vector, should be 2

问题

Here is a gist of what I want to do:

I've got 2 data frames:
x (id is unique)

id          timestamp
282462839   2012-12-05 10:55:00
282462992   2012-12-05 12:08:00
282462740   2012-12-05 12:13:00
282462999   2012-12-05 12:48:00

y (id is not unique)

id          value1    value2
282462839   300       100
282462839   300       200
282462839   400       300
282462999   500       400
282462999   300       150

I also have a function myfunc(id,pvalue) that computes something and returns one of the value2 values depending on pvalue and other value1s (more complicated than just pvalue==value1)

I want to create a 3rd column for x that contains the corresponding computed myfunc(id,pvalue), where pvalue is an integer that is constant(say 20).

so in essence, I want to do this:

x$t20 <- myfunc(x$id,20)

I tried using lappy and sapply this way:

x$t20 <- sapply(as.vector(x$id),myfunc,pvalue=20)

I tried using lapply and without the as.vector as well, but I kept getting this error:

Error in .pointsToMatrix(p2) : Wrong length for a vector, should be 2

It works when I just give mean where it just replicates $id in $t20.

How do I do this?

EDIT 1: Here's a skeleton of myfunc:

myfunc <- function(xid,pvalue) {
  result <- subset(y,id==xid)
  retVal <- -1
  if(nrow(result) < 12){
    return(NaN)
  }
  for(i in (1:nrow(result))){
    #code to process result
  }
  return(retVal)
}

回答1:

It was very difficult to help without full code, but here are some tips. First you can obtain the logical vector of id's which should be processed, then use vectorized ifelse statment.

tmp <- table(y$id) >= 12
y$t20 <- ifelse(tmp[as.character(y$id)], your_new_func(), NaN)

来源：https://stackoverflow.com/questions/16932390/r-error-wrong-length-for-a-vector-should-be-2

标签

user-defined-functions

lapply

sapply