Dear Josh,
thanks for pointing this out - the idea behind writing this function is
plotting gradients on branches of phylogenetic trees - 'tree' refers to
a phylogenetic tree. It's easy to create a random phylogenetic tree in R:
library(ape)
library(plotrix)
rtree(15) -> tree
This gives you a tree with 15 taxa. You can plot it with plot() if you
want to take a look.
then the data - you can create a fake data set:
rnorm(15, mean = 0.5, sd = 0.15) -> data
for the data which the function needs, you also need:
ace(data, tree) -> results
data <- append(data,results$ace)
names(data) <- NULL
I also tried with the following updated code I still got the same error
message:
create.gradient <- function(i){
colorgrad01<-color.scale(seq(0,1,by=0.01), extremes=c("red","blue"))
tree$edge[i,1] -> x
tree$edge[i,2] -> y
print(x)
print(y)
data[x] -> z
data[y] -> z2
round(z, digits = 2) -> z
round(z2, digits = 2) -> z2
z*100 -> z
z2*100 -> z2
print(z)
print(z2)
colorgrad<-colorgrad01[z:z2]
colorgrad
}
lapply(tree$edge, create.gradient)
- Error in FUN(X[[26L]], ...) : subscript out of bounds
I hope this help and you can replicate the problem too.
Thanks!
Annemarie
Joshua Wiley wrote:
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$edge directly
to lapply, when it is also hardcoded into the function, but I do not
have a sense exactly for what you are doing and without data it is
hard to play around.
Cheers,
Josh
On Wed, Jul 6, 2011 at 12:31 PM, Annemarie Verkerk
<annemarie.verk...@mpi.nl> wrote:
Dear R-help subscribers,
I have a quite stupid question about using lapply. I have the following
function:
create.gradient <- function(i){
colorgrad01<-color.scale(seq(0,1,by=0.01), extremes=c("red","blue"))
tree1$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@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.
--
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@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.
