R help - Feb 2010 - unexpected results with higher-order functions and lapply

I want to use lapply and a function returning a function in order to build a
list of functions.
> genr1 <- function(k) {function() {k}}
> l1 <- lapply(1:2,genr1)
> l1[[1]]()
[1] 2

This was unexpected. I had expected the answer to be 1, since that is the
value k should be bound to when genr1 is applied to the first element of
1:2.

By itself genr1 seems to work fine.
> genr1(5)()
[1] 5

I defined a slightly different higher-order function:
> genr2 <- function(k) {k;function() {k}}
> l2 <- lapply(1:2,genr2)
> l2[[1]]()
[1] 1

This gives the answer I expected.

Now I am confused. The function returned by genr2 is exactly the same
function that was being returned by genr1. Why should evaluating k make a
difference?

I am using R 2.9.2 on Ubuntu Linux.

Jyotirmoy Bhattacharya

	[[alternative HTML version deleted]]

Hi,

I believe it's lazy evaluation. See ?force

HTH,

baptiste

On 14 February 2010 20:32, Jyotirmoy Bhattacharya
<jyotirmoy at jyotirmoy.net> wrote:
> I want to use lapply and a function returning a function in order to build
a
> list of functions.
>
>> genr1 <- function(k) {function() {k}}
>> l1 <- lapply(1:2,genr1)
>> l1[[1]]()
> [1] 2
>
> This was unexpected. I had expected the answer to be 1, since that is the
> value k should be bound to when genr1 is applied to the first element of
> 1:2.
>
> By itself genr1 seems to work fine.
>> genr1(5)()
> [1] 5
>
> I defined a slightly different higher-order function:
>> genr2 <- function(k) {k;function() {k}}
>> l2 <- lapply(1:2,genr2)
>> l2[[1]]()
> [1] 1
>
> This gives the answer I expected.
>
> Now I am confused. The function returned by genr2 is exactly the same
> function that was being returned by genr1. Why should evaluating k make a
> difference?
>
> I am using R 2.9.2 on Ubuntu Linux.
>
> Jyotirmoy Bhattacharya
>
> ? ? ? ?[[alternative HTML version deleted]]
>
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