similar to: Extracting numeric part from a string

... Or if you just want to stick with basic regex's without extra packages: > x <- "\"cm_ffm\":\"563.77\"" > sub("[^[:digit:]]*([[:digit:]]*.?[[:digit:]]*).*","\\1",x) [1] "563.77" Cheers, Bert On Wed, Aug 2, 2017 at 5:16 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > >> On 3 Aug 2017, at 02:59,

> On 3 Aug 2017, at 02:59, Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > > 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

... and Marc's solution is **much** better than mine. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Wed, Aug 2, 2017 at 5:59 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ... Or if you just want to stick with

> On Aug 2, 2017, at 6:59 PM, Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > > 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

Hi there, I have to extract some relevant portion from a defined string, which is a mix of numeric and character. However this has following sequence: Some String - Some numerical - "c/C" (or "p/P") - then again some set of numbers. Examples of such string is "fdahsdfcha163517253c463278643" or "fdahsdfcha163517253C463278643" or

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>

Hello again, Let say I have following string: Vec <- c("sada", "asdsa", "sa") Now I want to make each element of this vector with equal length. Basically I want following vector: c("sada ", "asdsa", "sa ") Therefore we can get: > nchar(c("sada ", "asdsa", "sa ")) [1] 5 5 5 Is there any

Hello again, in the help page of grep() function, it is written that pattern: character string containing a regular expression (or character string for fixed = TRUE) to be matched in the given character vector. Coerced by as.character to a character string if possible. If a character vector of length 2 or more is supplied, the first element is used with a warning. Missing values are allowed

Hi, I'm trying to present a table of some experimental data, and I want to order the rows by the instance names. The issue I've got is that there are a variety of conventions for the instance names (e.g. competition01, competition13, small_1, big_20, med_9). What I want to be able to sort them first in category order so: competition < small < med < big, and then perform the

A very common use case for regmatches is to extract regex matches into a new column in a data.frame (or data.table, etc.) or otherwise use the extracted strings alongside the input. However, the default behavior is to drop empty matches, which results in mismatches in column length if reassignment is done without subsetting. For consistency with other R functions and compatibility with this use

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,

@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

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,

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

Buenas tarde a todo en s: Tenia la versión de R 3.6 y utilizaba la paquetería de pdftools para extraer información de archivos en pdf actualice la versión 3.6.1 y ya no reconoce la paquetería alguien que me pueda ayudar. Prácticamente no reconoce las funciones de pdftools library(pdftools) library(stringr)? library(NLP)? library(tm)? library(tesseract)? library(magick)?

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 =