Question

I have a data set with the following row-naming scheme:

a.X.V
where:
a is a fixed-length core ID
X is a variable-length string that subsets a, which means I should keep X
V is a variable-length ID which specifies the individual elements of a.X to be averaged
. is one of {-,_}

What I am trying to do is take column averages of all the a.X's. A sample:

sampleList <- list("a.12.1"=c(1,2,3,4,5), "b.1.23"=c(3,4,1,4,5), "a.12.21"=c(5,7,2,8,9), "b.1.555"=c(6,8,9,0,6))
sampleList
$a.12.1
[1] 1 2 3 4 5

$b.1.23
[1] 3 4 1 4 5

$a.12.21
[1] 5 7 2 8 9

$b.1.555
[1] 6 8 9 0 6

Currently I am manually gsubbing out the .Vs to get a list of general :

sampleList <- t(as.data.frame(sampleList))
y <- rowNames(sampleList)
y <- gsub("(\\w\\.\\d+)\\.d+", "\\1", y)

Is there a faster way to do this?

This is one half of 2 issues I’ve encountered in a workflow. The other half was answered here.

Answer 1

You can use a vector of patterns to find the locations of the columns you want to group. I included a pattern I knew wouldn’t match anything in order to show that the solution is robust to that situation.

# A *named* vector of patterns you want to group by
patterns <- c(a.12="^a.12",b.12="^b.12",c.12="^c.12")
# Find the locations of those patterns in your list
inds <- lapply(patterns, grep, x=names(sampleList))
# Calculate the mean of each list element that matches the pattern
out <- lapply(inds, function(i) 
  if(l <- length(i)) Reduce("+",sampleList[i])/l else NULL)
# Set the names of the output
names(out) <- names(patterns)