similar to: Help with Regular expression

Hello again, I am having a problem on Regular expression. Let say I have following code: > gsub("[',]", "", "'asd'f") [1] "asdf" This is perfect. However I am having problem if I include "" (i.e. the double quote) in the first argument as the pattern search: > gsub("[',"]", "",

Dear R-er, I would like format integer number as characters with leading 0 for a fixed width, for example: 1 shoud be "01" 2 shoud be "02" 20 should be "20" Now I use: x <- c(1, 2, 20) gsub(" ", "0", format(x, width=2)) But I suspect more elegant way could be done directly with format options, but I don't find. Thanks a lot Marc

Dear R People: Could someone recommend a good reference on regular expressions, please? Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com

On Sun, Jun 04, 2017 at 08:57:44AM -0400, Michael Butler wrote: > It seems that {rpc.}lockd no longer runs after the ino64 changes on any > of my systems after a full rebuild of src and ports. No log entries > offer any insight as to why :-( > > imb I don't tend to use NFS on my systems that are running head, so I haven't had occasion to test this as stated. However, I

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

But they row.names() cannot give me the IDs On Wednesday, November 1, 2017 9:45 AM, David Wolfskill <r at catwhisker.org> wrote: On Wed, Nov 01, 2017 at 04:13:42PM +0000, Elahe chalabi via R-help wrote: > Hi all, > I have two data frames that one of them does not have the column ID: > > > str(data) > 'data.frame': 499 obs. of 608 variables:

Hello again, Let say I have following vector: Vec <- c("0.0365780769", "(1.09738648244378)", "(0.812507787221523)", "0.5778069963", "(0.452456601362355)", "-1.8900812605", "-1.8716093762", "0.0055217041", "-0.4769192333", "-2.4133018880") Now I want to convert this vector to numeric vector. I

I am acquiring some sampled data that is time-stamped (with a POSIXct). Some of the data is in the form of "counters" -- that is, what is interesting isn't value of a given counter at a given time, but the change in the counter from one sample to a later one. As the counters are only incremented, they would be perceived to be monotonically increasing -- ideally. Unfortunately, the

Thanks for your pointer. Is there any way in R how to replace " ' " with " /' " programmatically? My actual string is quite lengthy, so changing it manually may not be possible. I am aware of gsub() function, however not sure I can apply it directly on my original string. Regards, On Tue, Jul 18, 2017 at 10:27 PM, John McKown <john.archie.mckown at gmail.com>

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

Hi, I was under the impression a successful make world was updating include files in /usr/include/netinet too. When I have a good make world, the files in /usr/include have new time stamps, but the ones in netinet have not! Am I missing something here? My current stable doesn't compille properly it fails in kdump on a missing ioctlcmd_t typedef. Peter

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Hello. I once read somewhere that it's possible to limit SSH pubkeys to 'tunnel-only'. I can't seem to find any information about this in any of the usual places. I'm going to be deploying a few servers in a couple of days and I'd like them to log to a central server over an SSH tunnel (using syslog-ng) however I'd like to prevent actual logins (hence

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =