# [R] applying a function recursively

Marc Schwartz marc_schwartz at comcast.net
Wed Jun 11 18:23:27 CEST 2008

```on 06/11/2008 10:51 AM Georg Otto wrote:
> Hi,
>
> I have a question about applying a function recursively through a
> list. Suppose I have a list where the different elements have
> different levels of recursion:
>
>
>> test.list<-list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f"), "C"=c("g", "h", "i")),
> +                 "II"=list("A"=list("a"=c("a", "b", "c"), "b"=c("d", "e", "f"),
> +                             "c"=c("g", "h", "i")),
> +                   "B"=c("d", "e", "f"), "C"=c("g", "h", "i")))
>
>> test.list
> \$I
> \$I\$A
> [1] "a" "b" "c"
>
> \$I\$B
> [1] "d" "e" "f"
>
> \$I\$C
> [1] "g" "h" "i"
>
>
> \$II
> \$II\$A
> \$II\$A\$a
> [1] "a" "b" "c"
>
> \$II\$A\$b
> [1] "d" "e" "f"
>
> \$II\$A\$c
> [1] "g" "h" "i"
>
>
> \$II\$B
> [1] "d" "e" "f"
>
> \$II\$C
> [1] "g" "h" "i"
>
>
>
> I would like to apply a function recursively to that list, in a way
> that the function does someting with each vector (eg. rev()) and
> returns a list of modified vectors that has the same structure as the
> input list, in my example:
>
>
> \$I
> \$I\$A
> [1] "c" "b" "a"
>
> \$I\$B
> [1] "f" "e" "d"
>
> \$I\$C
> [1] "i" "h" "g"
>
>
> \$II
> \$II\$A
> \$II\$A\$a
> [1] "c" "b" "a"
>
> \$II\$A\$b
> [1] "f" "e" "d"
>
> \$II\$A\$c
> [1] "i" "h" "g"
>
>
> \$II\$B
> [1] "f" "e" "d"
>
> \$II\$C
> [1] "i" "h" "g"
>
>
>
> I understand that with a fixed number of recursion levels one can use
> lapply() in a nested way, but what if the numbers of recursion levels
> is not fixed or is different between the list elements as it is in my
> example?
>
> Any hint will be appreciated.
>
> Best,
>
> Georg

See ?rapply, which is a recursive version of lapply():

> rapply(test.list, rev, how = "list")
\$I
\$I\$A
[1] "c" "b" "a"

\$I\$B
[1] "f" "e" "d"

\$I\$C
[1] "i" "h" "g"

\$II
\$II\$A
\$II\$A\$a
[1] "c" "b" "a"

\$II\$A\$b
[1] "f" "e" "d"

\$II\$A\$c
[1] "i" "h" "g"

\$II\$B
[1] "f" "e" "d"

\$II\$C
[1] "i" "h" "g"

HTH,

Marc Schwartz

```