similar to: Script file bug

Please always reply-all so the mailing list can record the answer along with your question. I am cc'ing this time. Thanks for the unusual case. -- Sent from my phone. Please excuse my brevity. On March 1, 2018 12:46:53 AM PST, zn l <esetlzn at gmail.com> wrote: >It is my mistake. I find that I input # character in the Chinese. The >R.app won?t display any error messages. >

Hello, I'm having problems working with date values in POSIXct format. Here is what I got (eg.lig attached): x <- read.table("eg.txt", sep = ',', col.names=c("ok","time","secs","lig")) # it gives time as factor z <- cbind(x,colsplit(x$time, split="\\s", names=c("date", "clock")))

Hi all, I can't get the functionality of the package parallel to work. Specifically, makeCluster() hangs when I run it. I first noticed the problem when trying to run Rstan with multiple cores and the traced it back to the core package parallel. The following results in R hanging after the call to makeCluster. library(parallel) # Calculate the number of cores no_cores <-

Dear Henrik, thank you, for the quick reply. Bizarrely enough, the problem vanished when I woke the computer from sleep (I had previously replicated the problem after several restarts of both R and the MacOS). I will follow-up if I can again replicate the problem. Florian On 2/10/18 4:39 PM, Henrik Bengtsson wrote: > A few quick comments: > > * You mention R --vanilla, but make sure

Dear All, Suppose I have a dataframe like this with many thousands rows all with different names: data.frame(gene=c("a","b","c","d","c","d","c","f"),value=c(20,300,48,55,9,2,100,200)), I want to set column "gene" as row.names, but there are duplicates (c, d), which I want to transform into this as row names:

On Wed, 28 Feb 2018, Stephen HonKit Wong wrote: > Dear All, > Suppose I have a dataframe like this with many thousands rows all with > different names: > data.frame(gene=c("a","b","c","d","c","d","c","f"),value=c(20,300,48,55,9,2,100,200)), > > I want to set column "gene" as row.names, but

Hi... I have a dataframe with n columns and n rows. I need to find how many rows contains zero raw read count across all column. Thanks -- *Yogesh Gupta* *Postdoctoral Researcher* *Department of Biological Science* *Seoul National University* *Seoul, South Korea* [[alternative HTML version deleted]]

Dear member, When I run a code on a computer B from a computer A through shh using for example: system("ssh login at IPadress \"R -e 'print(1)'\"") [Note that I don't need indicate password because ~/.ssh/authorized_keys is used] I get a warning: During startup - Warning message: Setting LC_CTYPE failed, using "C" The code is working (it prints 1)

Hi With the example, na.locf seems to be the easiest way. > library(zoo) > na.locf(df1) ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 4 b aa TRUE 5 5 b ab FALSE 5 Cheers Petr > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff > Newmiller > Sent: Monday, January

Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote: >I just came

Upon closer examination I see that you are not using the split version of df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ]

You can enforce these assumptions by sorting on multiple columns, which leads to na.locf(df1[ order(df1$ID,df1$Value), ]) On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > Yes, you are right if the IDs are always sequentially-adjacent and the > first non-NA value appears in the first record for each ID. > -- > Sent from my phone. Please

Are they the same with .RData being the newer format?? Thanks, ...Tao

Hej, i need the which() funktion to find the positions of an entry in a matrix. the entries i'm looking for are : seq(begin,end,0.01) and there are no empty spaces i'm searching in the right range. so i was looking for the results R can find and i recieved this answer. for (l in

Dear all, I have an R script that uses Rcpp, and I have been trying to parallelize it using mclapply (I tried with the multicore and the parallel library) Sometimes (not always, interestingly), the CPU use for each core drops, usually so that the total over all cores reaches 100%, i.e., as fast as if using just one single core fully. I tried my code directly from within emacs, and also using a

Hi, I'm trying to user lmer function from lmerTest package because, if I understood correectly, it allows to make better inference than lmer method from lme4 package. However, whatever I do I keep getting this error: Error in lme4::lFormula(formula = mark ~ ssCount + sTime+ : rank of X = 1660 < ncol(X) = 1895 any ideas what could be a problem? thanks, Srecko [[alternative HTML

Hello r-help, I've been banging my head against the computer in an attempt to learn how to divide my matrix into segments by rows. I want to be able to return each segment as a newly named object. I've tried looking at the apply functions and creating a for loop but brain no work. Here's the basic starting objects that I believe would be needed to separate the matrix. mat <-

Dear R users, I am running simulations (1000), and in my simulation I am looking at specific sums. For example, if the sum is >=4 then count this, if say <3, then don't count, if the sum=3, then generate a random number from uniform distribution, if this number is say less than 0.5, then count this sum, if greater than 0.5, then don't count. I am having trouble with introducing this

Hi all: My OS is 32bit winxp,but I wanna install 64bit R2.14.1. >From the following website,it says "You can also go back and add 64-bit components to a 32-bit install, or vice versa" http://cran.r-project.org/bin/windows/rw-FAQ.html#Can-both-32_002d-and-64_002dbit-R-be-installed-on-the-same-machine_003f Does it mean that I can install and run 64bit R2.14.1 under 32bit

In the following data.frame there are 6 columns, but 7 are written to the CSV file. install.packages("pmlr") library(pmlr) data(enzymes) write.table(enzymes, sep=",", eol="\n",file="albert.csv")