similar to: about reading files in order

Hi R users, I have a question about opening the txt files and putting them into a matrix. The txt files are in the folder01, while they have the name file.1.txt, file.2.txt, file.3.txt, etc. There are about 200 such text files. Each txt file contains one value inside. When I tried to use the code below, I found that the txt files are not in order, from 1, 2, 3, to 200. Rather, they are in the

You can use: > files <- list.files(path = "folder01") > files <- gtools::mixedsort(files) to order the files in a "human-friendly" order rather than lexicographic order (which sort() provides). FYI 1; it's preferred to use file.path("folder01", list[i]) rather than paste('folder01',lists[i],sep='/'). FYI 2; if you use

Thanks for that answer. I was not aware of gtools::mixedsort <https://www.rdocumentation.org/packages/gtools/versions/3.5.0/topics/mixedsort> function. Jean On Thu, Jun 29, 2017 at 2:47 PM, Henrik Bengtsson < henrik.bengtsson at gmail.com> wrote: > You can use: > > > files <- list.files(path = "folder01") > > files <- gtools::mixedsort(files) >

Hi, I'm having an error trying to read the contens of a DFS Share The mount is: sudo mount -t cifs -o username=mxxxxx ,uid=mxxxxx,gid=mxxxxx //cluster_share/share /mnt The share is mounted without errors, but when I try to list the share i have this: ls -la /mnt/folder ls: cannot open directory /mnt/folder: Operation now in progress and the syslog have the error reported

TCL IS BETTER THAN RUBY. THIS MESSAGE WAS BROUGHT TO YOU BY IRC.GANGNET.RU #TCLMAFIA SHOUTZ TO RYAN ETERNAL PRESIDENT OF GNAA, MALCOM, FLUNITRAZEPAM, DUTCHMINATI, CRACK, SNITCH -- Posted via http://www.ruby-forum.com/. -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To unsubscribe from this group and stop receiving emails from

TCL IS BETTER THAN RUBY. THIS MESSAGE WAS BROUGHT TO YOU BY IRC.GANGNET.RU #TCLMAFIA SHOUTZ TO RYAN ETERNAL PRESIDENT OF GNAA, MALCOM, FLUNITRAZEPAM, DUTCHMINATI, CRACK, SNITCH -- Posted via http://www.ruby-forum.com/. -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To unsubscribe from this group and stop receiving emails from

TCL IS BETTER THAN RUBY. THIS MESSAGE WAS BROUGHT TO YOU BY IRC.GANGNET.RU #TCLMAFIA SHOUTZ TO RYAN ETERNAL PRESIDENT OF GNAA, MALCOM, FLUNITRAZEPAM, DUTCHMINATI, CRACK, SNITCH -- Posted via http://www.ruby-forum.com/. -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To unsubscribe from this group and stop receiving emails from

Hi All Can you help me understand the meaning of the below error message then I will try to figure out what is going wrong: *Input Code:* > bodlane2 <- lapply( + lapply(split(lctolc2, lctolc2$Item), function(x) graph.data.frame(x[, 2:3])), + function(x) lapply( + V(x)[degree(x, mode = "in") == 0], + function(s) all_simple_paths(x, from = s, +

On 11/1/19 5:24 PM, Anshul Saravgi wrote: > Thanks Tomas for identifying the bug. > > *How to resolve this bug? Can you please help me here?* I would recommend you first try to create a complete reproducible example (the code below is just a snippet). Then try to minimize it - make it do as little as possible but still trigger the bug. Then try to reduce the number of packages used.

hello, This rbindlist(list(list(a=NA,b=NA),list(a=20,b=FALSE))) returns a b 1: NA NA 2: TRUE FALSE as per the documentation ?rbindlist is there a way to specify the column class of 'a' to be numeric? In actual usage, i wont be able to re-order the 2nd list entry to be the first. I did try

This is very nice to learn about, Denis, but it seems only fair to point out that the result of rbindlist is not a data frame. You can convert it to a data frame easily, but the copy and indexing semantics of data tables are quite different than data tables, which could be a real headache for someone not prepared for those differences. (To learn more, read the data tables vignette.) Tibbles (as

On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote: > I suspect this is pretty easy, but I'm having trouble figuring it out. > Basically, I have a list of data frames such as the following example: > > list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) > > I would like to turn this into data frame where the list elements are > essentially rbind'ed together and

In the lower right panel of R-studio interface, there is the "Export" button. I saved as PDF from there directly, rather than using functions On Sat, Aug 5, 2017 at 6:18 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > > > On 6 Aug 2017, at 03:01, lily li <chocold12 at gmail.com> wrote: > > > > I am using the plot() function, but have a problem.

If layer$z is a matrix and you want to reverse the order of the rows, you can do: n <- nrow(layer$z) layer$z <- layer$z[ n:1, ] HTH, Eric On Tue, Jan 16, 2018 at 8:43 AM, lily li <chocold12 at gmail.com> wrote: > Sorry for the emails, I just wanted to have an example. > layer$z > > 1 1 3 4 6 2 > 2 3 4 1 2 9 > 1 4 5 2 1 8 > > How to convert

Hi Bert, I think you are correct that I can use levelplot, but I have a question about converting data. For example, the statement: levelplot(Z~X*Y), Z is row-wise from the lower left corner to the upper right corner. My dataset just have gridded Z data as a txt file (or can be called matrix?), how to convert them to the vector in order for levelplot to use? Thanks. On Mon, Jan 15, 2018 at 6:04

Sorry for the emails, I just wanted to have an example. layer$z 1 1 3 4 6 2 2 3 4 1 2 9 1 4 5 2 1 8 How to convert the matrix to layer$z = c(1, 4, 5, 2, 1, 8, 2, 3, 4, 1, 2, 9, 1, 1, 3, 4, 6, 2)? I think this vector is the order that levelplot can use. Thanks again. On Mon, Jan 15, 2018 at 10:58 PM, lily li <chocold12 at gmail.com> wrote: > Hi Bert, > > I think

Dear Group: I have 72 files (.txt). Each file has 2 columns and column 1 is always identical for all 70 files. Each file has 90,799 rows and is standard across all files. I want to create a matrix 40(rows) x 70 columns. I tried : temp = list.files(pattern="*.txt") named.list <- lapply(temp, read.delim) library(data.table) files.matrix <-rbindlist(named.list) >

> On 6 Aug 2017, at 03:47, lily li <chocold12 at gmail.com> wrote: > > In the lower right panel of R-studio interface, there is the "Export" button. I saved as PDF from there directly, rather than using functions > > On Sat, Aug 5, 2017 at 6:18 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > > > On 6 Aug 2017, at 03:01, lily li <chocold12 at

Hi all, I ran the code: > s <- stack(replicate(2, raster(matrix(runif(100), 10)))) > xy <- data.frame(coordinates(sampleRandom(s, 10, sp=TRUE)), + z1=runif(10), z2=runif(10)) > levelplot(s, margin=FALSE, at=seq(0, 1, 0.05)) + + layer(sp.points(xy, pch=ifelse(pts$z1 < 0.5, 2, 3), cex=2, col=1), columns=1) + + layer(sp.points(xy, pch=ifelse(pts$z2 < 0.5, 2,

Hi all, I'm trying to add single points with known coordinates to a level plot, but could not find the proper answer. I got to know that layer() function is good for this, but I don't know which package is related to this function. The source is here: https://stackoverflow.com/questions/28597149/add-xy-points-to-raster-map-generated-by-levelplot but my question is a little different as I