Combining loops and lapply

Question

I am pretty new to R and really finding it powerful. I am trying to create a dataframe with multiple levels. I have 3 groups (of subjects) called "group1", "group2" and "group3" who have each 19 tests (names do not matter). In each of these 19 tests there are 3 components called b1, b2 and b3 out of 7 components that I am interested in.

What I tried so far: For each patient, the 19 tests form 19 columns with 7 rows (components) hence, the components I am interested in are 5,6,7 for each of the 19 tests:

sapply(x, '[', XXX, j)

My final code is:

for (j in 1:19) {
   x = lapply(c('group1', 'group2', 'group3'), function(i) {
      filenames = dir(file.path('c:/19/7', i),
                 pattern = 'sRL.txt', full.names = T)
      x = lapply(filenames, read.table, sep = '')
      bb1 = sapply(x, '[', 5, j)
      bb2 = sapply(x, '[', 6, j)
      gg = sapply(x, '[', 7, j)
      pcode = as.numeric(gsub('_.*', '', basename(filenames)))
      data.frame(group = i, tests= j, pcode, b1= bb1, b2= bb2, g = gg)
   })
}

Could you please advise how to create a nice dataframe that combine the components 5,6,7 of the 19 tests for each subject of each group?

Welcome to SO. Could you provide some sample data to run your code and easily understand the transformation from input to output? Have a look at [how to make a great reproducible example](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). — talat, Jun 15 '14 at 13:30

score 0 · Accepted Answer · answered Jun 15 '14 at 16:47

It would be better to provide sample data, but in your case I can see how it would be difficult as the organization of the files is the problem you are trying to deal with.

If I understand you correctly, each patient has their own file with all the tests. Each file has 19 columns (one column per test) and 7 rows (one row per test component). The files (patients) are organized into groups where each group is in a different directory.

As output you want rows 5:7 of each file, transposed so there is one column per component and 1 row for each test, for each patient, in each group.

I think this has a reasonable shot at doing what you want, although it's impossible to test without your complete dataset...

## not tested...
get.group <- function(g) {
  get.file <- function(x,g) {
    df <- read.table(x,sep="")
    data.frame(group=g, test=1:ncol(x), 
               pcode=gsub('_.*', '', basename(x)), t(x[5:7,]))
  }
  filenames <- dir(file.path('c:/19/7', g), pattern = 'sRL.txt', full.names = T)
  result    <- do.call(rbind,lapply(filenames,get.file,g))
}
final <- do.call(rbind,lapply(paste0("group",1:3),get.group))
colnames(final)[4:6] <- c("bb1","bb2","g")

So working from the outside in: for each group the function get.group(...) is called, with the results bound together row-wise. get.group(...) in turn grabs the file names for that group and calls get.file(...) for each file name, binding the results together row-wise. Finally, get.file(...) reads the file and creates a data frame with the group name, the patient code, the test number, and the results for the three components of all 19 tests.

Combining loops and lapply

1 Answers1