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R

ProvideSampleId <- function (pathexpfile) {
# Returns a "unique" sample ID when supplied
# with a path to an experimental file.
## Note: the sample ID must derive directly from the file or path.
sampleid <- sub("\\.[\\w]+$", "", basename(pathexpfile), perl = TRUE)
#### The code below is the old ProvideSampleId() function
# ### OBS! Only very rudimentary error-checking.
# ### If the filename is formatted as \w*-\w*-\w*, we use the middle segment,
# ### otherwise we use the whole string (excluding the extension)
# # Extract the name of the parent directory of the datafilename argument
# substrateid <- basename(dirname(fullpathwithfilename))
# # Extract the name of the method from the filename-part
# # First split the filename over all hyphens
# nameparts <- strsplit(basename(fullpathwithfilename), "-")[[1]]
# # If the number of nameparts exceed 3, save the whole filename
# # as methodid, otherwise use the middle part
# if (length(nameparts) > 3) {
# # We need to lose the file extension from the last namepart
# nameparts[length(nameparts)] <-
# strsplit(nameparts[length(nameparts)], "\\.")[[1]][1]
# methodid <- paste(nameparts, collapse = "-")
# } else {
# methodid <- nameparts[2]
# }
# # Make an informative sampleid
# sampleid <- paste(substrateid, methodid, sep = "-")
# #
# return(sampleid)
####
return(sampleid)
}