# [R] Break during the recursion?

Duncan Murdoch murdoch at stats.uwo.ca
Sun Jul 15 16:13:56 CEST 2007

```On 15/07/2007 10:06 AM, Atte Tenkanen wrote:
> Hi,
>
> Is it possible to break using if-condition during the recursive function?

You can do

if (condition) return(value)

>
> Here is a function which almost works. It is for inorder-tree-walk.
>
> iotw<-function(v,i,Stack,Indexes) # input: a vector and the first index (1), Stack=c(), Indexes=c().
> {
> 	print(Indexes)
> 	# if (sum(i)==0) break # Doesn't work...

if (sum(i)==0) return(NULL)

should work.

Duncan Murdoch

>
> 	if (is.na(v[i])==FALSE & is.null(unlist(v[i]))==FALSE)
> 		{Stack=c(i,Stack); i=2*i; iotw(v,i,Stack,Indexes)}
> 	Indexes=c(Indexes,Stack[1])
> 	Stack=pop.stack(Stack)\$vector
> 	Indexes=c(Indexes,Stack[1])
> 	i=2*Stack[1]+1
> 	Stack=pop.stack(Stack)\$vector
> 	iotw(v,i,Stack,Indexes)
> }
>
>
>> v=c(`-`,`+`,1,`^`,`^`,NA,NA,"X",3,"X",2)
>> Stack=c()
>> Indexes=c()
>
>> iotw(v,1,Stack,Indexes)
> NULL
> NULL
> NULL
> NULL
> NULL
> [1] 8 4
> [1] 8 4
> [1] 8 4 9 2
> [1] 8 4 9 2
> [1] 8 4 9 2
> [1]  8  4  9  2 10  5
> [1]  8  4  9  2 10  5
> [1]  8  4  9  2 10  5 11  1
> [1]  8  4  9  2 10  5 11  1
> [1]  8  4  9  2 10  5 11  1  3
> Error in if (is.na(v[i]) == FALSE & is.null(unlist(v[i])) == FALSE) { :
> 	argument is of length zero
>
> Regards,
>
> Atte Tenkanen
> University of Turku, Finland
>
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