R help - Jul 2011 - question about getting things out of an lapply

Dear R-help subscribers,

I have a quite stupid question about using lapply. I have the following 
function:

create.gradient <- function(i){
colorgrad01<-color.scale(seq(0,1,by=0.01),
extremes=c("red","blue"))
tree1$edge[i,1] -> x
tree1$edge[i,2] -> y
print(x)
print(y)
all2[x] -> z
all2[y] -> z2
round(z, digits = 2) -> z
round(z2, digits = 2) -> z2
z*100 -> z
z2*100 -> z2
print(z)
print(z2)
colorgrad<-colorgrad01[z:z2]
colorgrad
}

Basically, I want to pick a partial gradient out of a bigger gradient 
(colorgrad01) for values that are on row i, from a matrix called tree1.

when I use lapply:

lapply(tree1$edge, create.gradient)

I get the following error message:

Error in FUN(X[[27L]], ...) : subscript out of bounds

I'm not sure what's wrong: it could be either fact that
'colorgrad' is a
character string; i.e. consisting of multiple characters and not just 
one, or because 'i' doesn't come back in the object
'colorgrad' that it
has to return. Or it could be something else entirely...

In any case, what I prefer as output is a vector with all the different 
'colorgrad's it generates with each run.

Thanks a lot for any help you might be able to offer!
Annemarie

-- 
Annemarie Verkerk, MA
Evolutionary Processes in Language and Culture (PhD student)
Max Planck Institute for Psycholinguistics
P.O. Box 310, 6500AH Nijmegen, The Netherlands
+31 (0)24 3521 185
http://www.mpi.nl/research/research-projects/evolutionary-processes

Dear Annemarie,

Can you replicate the problem using a madeup dataset or one of the
ones built into R?  It strikes me as odd to pass tree1$edge directly
to lapply, when it is also hardcoded into the function, but I do not
have a sense exactly for what you are doing and without data it is
hard to play around.

Cheers,

Josh

On Wed, Jul 6, 2011 at 12:31 PM, Annemarie Verkerk
<annemarie.verkerk at mpi.nl> wrote:
> Dear R-help subscribers,
>
> I have a quite stupid question about using lapply. I have the following
> function:
>
> create.gradient <- function(i){
> colorgrad01<-color.scale(seq(0,1,by=0.01),
extremes=c("red","blue"))
> tree1$edge[i,1] -> x
this works, but it would typically be written:

x <- tree1$edge[i, 1]

flipping back and forth can be a smidge (about 5 pinches under an
iota) confusing.
> tree1$edge[i,2] -> y
> print(x)
> print(y)
> all2[x] -> z
> all2[y] -> z2
> round(z, digits = 2) -> z
> round(z2, digits = 2) -> z2
> z*100 -> z
> z2*100 -> z2
> print(z)
> print(z2)
> colorgrad<-colorgrad01[z:z2]
> colorgrad
> }
>
> Basically, I want to pick a partial gradient out of a bigger gradient
> (colorgrad01) for values that are on row i, from a matrix called tree1.
>
> when I use lapply:
>
> lapply(tree1$edge, create.gradient)
>
> I get the following error message:
>
> Error in FUN(X[[27L]], ...) : subscript out of bounds
>
> I'm not sure what's wrong: it could be either fact that
'colorgrad' is a
> character string; i.e. consisting of multiple characters and not just one,
> or because 'i' doesn't come back in the object
'colorgrad' that it has to
> return. Or it could be something else entirely...
>
> In any case, what I prefer as output is a vector with all the different
> 'colorgrad's it generates with each run.
>
> Thanks a lot for any help you might be able to offer!
> Annemarie
>
> --
> Annemarie Verkerk, MA
> Evolutionary Processes in Language and Culture (PhD student)
> Max Planck Institute for Psycholinguistics
> P.O. Box 310, 6500AH Nijmegen, The Netherlands
> +31 (0)24 3521 185
> http://www.mpi.nl/research/research-projects/evolutionary-processes
>
> ______________________________________________
> R-help at r-project.org 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.
>


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
https://joshuawiley.com/