Hi,
Is it possible to break using if-condition during the recursive function?
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 (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
Oh. I forgot one extra-function:
pop.stack<-function(v){
if(length(v)==0){x=NA}
if(length(v)==1){x=v[1]; v=c()}
if(length(v)>1){x=v[1]; v=v[2:length(v)]}
return(list(vector=v,x=x))
}
Atte
> Hi,
>
> Is it possible to break using if-condition during the recursive
> function?
> 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 (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
>
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 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.