I like Don's answer which is clean and clear. A frequently useful
alternative to lapply() is sapply().
Taking the sapply() route also avoids the need for do.call(). So a modified
version of Don's code would be:
## example data
output <- list(1:5, 1:7, 1:4)
maxlen <- max( sapply(output, length) )
outputmat <- sapply(output, function(x, maxl) c(x, rep(NA,
maxl-length(x))), maxl=maxlen)
write.csv(outputmat, na='')
On Thu, Mar 29, 2018 at 2:18 AM, MacQueen, Don <macque...@llnl.gov> wrote:
> Perhaps this toy example will help:
>
> ## example data
> output <- list(1:5, 1:7, 1:4)
>
> lens <- lapply(output, length)
> maxlen <- max(unlist(lens))
> outputmod <- lapply(output, function(x, maxl) c(x, rep(NA,
> maxl-length(x))), maxl=maxlen)
> outputmat <- do.call(cbind, outputmod)
> write.csv(outputmat, na='')
>
> The idea is to pad the shorter vectors with NA (missing) before converting
> to a matrix structure.
>
> I don't really need to know where the data came from, or that it's ncdf
> data, or how many months or years, etc. But I do need to know the structure
> of your "output" list. I'm assuming each element is a simple vector of
> numbers, and that the vectors' lengths are not all the same. If that's
> correct, then my example may be what you need.
>
> This uses only base R methods, which I generally prefer. And no doubt it
> can be done more cleverly, or in a way that needs fewer intermediate
> variables ... but I don't really care.
>
> -Don
>
> --
> Don MacQueen
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
> Lab cell 925-724-7509
>
>
> On 3/28/18, 9:32 AM, "R-help on behalf of orlin mitov via R-help" <
> r-help-boun...@r-project.org on behalf of r-help@r-project.org> wrote:
>
> Hello,
> I have no previous experience with R, but had to learn on the fly in
> the past couple of weeks. Basically, what I am trying to do is read a
> certain variable from a series of files and save it as csv-table. The
> variable has an hourly value for each month in a year for the past 20 years
> and has to be read for different geographical locations. So there will be
> 12 files per year (1 for each month) and the values for the variable from
> each file will be 696 to 744 (depending on how many days x 24 hours there
> were in the month).What I achieved is to to read the values from all 12
> files stored in directory with a function and add them as vectors to a
> lapply-list:
>
>
>
> Myfunction <- function(filename) {
> nc <- nc_open(filename)
> lon <- ncvar_get(nc, "lon")
> lat <- ncvar_get(nc, "lat")
> RW <- ncvar_get(nc, "X")
> HW <- ncvar_get(nc, "Y")
> pt.geo <- c(8.6810 , 50.1143)
> dist <- sqrt( (lon - pt.geo[1])^2 + (lat - pt.geo[2])^2 )
> ind <- which(dist==min(dist, na.rm=TRUE),arr.ind=TRUE)
> sis <- ncvar_get(nc, "SIS", start=c(ind[1],ind[2],1), count=c(1,1,-1))
> vec <- c(sis)
> }
>
> filenames <- list.files(path = "C:/Users/Desktop/VD/Solardaten/NC",
> pattern = "nc", full.names = TRUE)
> output <- lapply(filenames, Myfunction)
>
>
>
> And here start my problems with saving "output" as a csv table. Output
> would contain 12 vectors of different lenght.I want to have them as 12
> columns (1x per month) in Excel and each column should have as many
> row-entries as there are values for this month.Whatever I tried with
> write.table I was not able to achieve this (tried converting the output to
> a matrix, also no successes).Please help! Or should I be trying to have the
> 12 elements as data frames and not vectors?
> This is how I want the table for each year to look - 12 columns and
> all the respective values in the rows (column names I can add by myself):
> Best regardsOrlin
>
>
>
> ______________________________________________
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