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question about getting things out of an lapply

Annemarie Verkerk 
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

Thread (5 messages)

Annemarie Verkerk question about getting things out of an lapply Jul 6 Joshua Wiley question about getting things out of an lapply Jul 6 Annemarie Verkerk question about getting things out of an lapply Jul 7 Joshua Wiley question about getting things out of an lapply Jul 7 Annemarie Verkerk question about getting things out of an lapply Jul 10