# [R] Elementary sapply question

Liaw, Andy andy_liaw at merck.com
Mon Jun 21 20:49:48 CEST 2004

```At least two ways:

1. Use extra argument in the function being sapply()'ed; e.g.,

> f <- function(x, y) x*x + y*y
> x <- 3:5
> sapply(x, f, 3)
[1] 18 25 34

[See the "..." argument in ?sapply.]

2. More generally, if both x and y are vectors (of the same length), then
you can use mapply(); e.g.,

> x <- 1:3
> y <- 3:5
> pyth <- function(x, y) x*x + y*y
> mapply(pyth, x, y)
[1] 10 20 34

HTH,
Andy

> From: Ajay Shah
>
> elementary question?
>
> Here is what I know. The following two programs generate the
>
> --------------------------------+-----------------------------
> -----------
>        Loops version            |          sapply version
> --------------------------------+-----------------------------
> -----------
>                                 |
> f <- function(x) {              |       f <- function(x) {
>   return(x*x)                   |         return(x*x)
> }                               |       }
> values = c(2,4,8)               |       values = c(2,4,8)
> for (i in 1:3) {                |
> }                               |
>
> and this is cool!
>
> My problem is this. Suppose I have:
>      pythagorean <- function(x, y) {
>        return(x*x + y*y)
>      }
>
> then how do I utilise sapply to replace
>      fixed.x = 3
>      y.values = c(3,4,5)
>      for (i in 1:3) {
>      }
>
> ?
>
> I have read the sapply docs, and don't know how to tell him that the
> list values that he'll iterate over "fit in" as y.values[i].
>
> --
> Ajay Shah                                                   Consultant
> ajayshah at mayin.org                      Department of Economic Affairs
> http://www.mayin.org/ajayshah           Ministry of Finance, New Delhi
>
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