similar to: Extract part of a numer

Hola a tod en s ?alquien sabria como convertir estas frases con expresiones regulares? 1.3ptd -> 1.3 ptd 1.3ptdm -> 1.3 ptdm 4.4ptdm23j -> 4.4 ptdm 23j 7.716s -> 7.7 16s 1.4hola -> 1.4 hola 1.4hola.hola -> 1.4 hola.hola 5.5v6 -> 5.5 v6 5.5v6sdp -> 5.5 v6 sdp 5.5v10sdp -> 5.5 v10 sdp de forma que esta frase "hola 1.3ptd 1.3ptdm 4.4ptdm23j 7.716s 1.4hola pepe

Hi, I have a vector of strings like: c("a1b1","a2b2","a1b2") which I want to spilt into two parts like: c("a1","a2","a2") and c("b1","b2,"b2"). So there is always a first part with a+number and a second part with b+number. Unfortunately there is no separator I could use to directly split the vectors.. Any idea

Hola esta es una solución library(data.table) library(stringr) dt <- data.table( V1a = sample(c("1","0"), 10, TRUE) , V1b = sample(c("1","0"), 10, TRUE) , V2a = sample(c("1","0"), 10, TRUE) , V2b = sample(c("1","0"), 10, TRUE) , V3a =

Hola, Vaya desastre al mandar la cabecera de la base de datos. Al enviarlo, se veía bien, pero está visto que por el camino la cosa se torció. Siguiendo el consejo de Javier, he subido un pequeño fragmento al DropBox. He quitado algunas especies para no hacerlo muy largo. Espero que ésta vez no haya problemas. https://www.dropbox.com/s/q7zla50oq7owg8k/CPUE.csv?dl=0 Un saludo y gracias Juan

Hola Javier, Intenta los siguientes cambios g4 <- function(codigo_llega, n_caracteres){ codigo_llega <- as.character(codigo_llega) if(n_caracteres == 6) res <- substr(codigo_llega, start=0, stop=4) else res <- codigo_llega res } g4 <- Vectorize(g4) x <- c('Jorge Velez','Javier Marcuzzi','Daniel Merino','Susana deus Alvarez', 'Carlos

> -----Original Message----- > From: Richard Lyman [mailto:pchammer@dynx.net] > Sent: Wednesday, December 20, 2006 4:29 PM > To: Asterisk Users Mailing List - Non-Commercial Discussion > Subject: Re: [asterisk-users] Re: Match a Numer - then continue with, > dialplan > > > Douglas Garstang wrote: > >> -----Original Message----- > >> From: David

> -----Original Message----- > From: Richard Lyman [mailto:pchammer@dynx.net] > Sent: Wednesday, December 20, 2006 4:29 PM > To: Asterisk Users Mailing List - Non-Commercial Discussion > Subject: Re: [asterisk-users] Re: Match a Numer - then continue with, > dialplan > > > Douglas Garstang wrote: > >> -----Original Message----- > >> From: David

I seriously doubt he'd know how to get on the 'Internets' -----Original Message----- From: Doug Crompton [mailto:doug@crompton.com] Sent: Wed 12/20/2006 8:07 PM To: Asterisk Users Mailing List - Non-Commercial Discussion Cc: Subject: RE: [asterisk-users] Re: Match a Numer - then continue with, dialplan On Wed, 20 Dec 2006, Michael Collins wrote: > After listing all of that,

Hi, I would like to replace the last tree characters of the values of a certain column in a dataframe. This replacement should only take place if the last three characters correspond to the value "/:/" and they should be replaced with ""(blank) I cannot perform a simple gsub because the characters /:/ might also be present somewhere else in the string values and then they

I think you're making it far too difficult. What I do is something like this: [outgoing] include => internal include => longdistance ;Always include internal first, as matches from the first include ;will be used first. This allows you to make sure your internal ;extensions don't go out your trunks. [longdistance] ignorepat => 9; include => default; already included from

> -----Original Message----- > From: Tony Mountifield [mailto:tony@softins.clara.co.uk] > Sent: Wednesday, December 20, 2006 2:41 PM > To: asterisk-users@lists.digium.com > Subject: [asterisk-users] Re: Match a Numer - then continue with, > dialplan > > > In article > <645FEC31A18FE54A8721500CDD55A7B6035D0C6C@mail.oneeighty.com>, > Douglas Garstang

> -----Original Message----- > From: David Gomillion [mailto:dgomillion@eyecarenow.com] > Sent: Wednesday, December 20, 2006 10:27 AM > To: asterisk-users@lists.digium.com > Subject: [asterisk-users] Re: Match a Numer - then continue with, > dialplan > > > I think you're making it far too difficult. > > What I do is something like this: > > [outgoing]

That's basically what I did 1. Get text lines using readLines 2. use tryCatch to parse each line using read.csv(text=...) 3. in the catch, use?gregexpr to find any quotes not adjacent to a comma (gregexpr("[^,]\"[^,]",...) 4. escape any quotes found by adding a second quote (using str_sub from stringr) 6. parse the patched text using read.csv(text=...) 7. write out the parsed

Anyone know if there's a way to match a dialplan extension, execute some code, say set a variable, and then continue with the dialplan? I want to set a variable when the dialplan flows beyond a certain context. This would be a great feature. Doug.

?s 06:47 de 08/04/2024, Dave Dixon escreveu: > Greetings, > > I have a csv file of 76 fields and about 4 million records. I know that > some of the records have errors - unmatched quotes, specifically. > Reading the file with readLines and parsing the lines with read.csv(text > = ...) is really slow. I know that the first 2459465 records are good. > So I try this: >

Hola a todos, Gracias por las respuestas. Los nombres ciéntificos son los nombres de las columnas de un csv. Usando vuestros comentarios y sugerencias, estoy tratando de "automatizarlo", ya que tengo unas 40 especies. De momento sin mucho éxito por que no consigo dar con la forma de que me abrevie las especies correctamente y mezclo campos La aproximación de Jorge me valdría (y he

e.g. a= a b c d e 1 1 1 3 1 1 2 1 2 3 4 5 3 1 3 3 8 3 4 1 4 3 3 5 5 1 1 3 1 1 I want to delete colume a and colume c, because they have the same values in every row, then ,I want to get this data.frame . b= b d e 1 1 1 1 2 2 4 5 3 3 8 3 4 4 3 5 5 1 1 1 the following is my code but it's wrong. rm(list=ls()) a=c(1,1,1,1,1); b=c(1,2,3,4,1); c=c(3,3,3,3,3);

I am a new R-language user. I have set up a data frame "mydata",one of the colume of which is "skill". Now I want to select the observations whose "skill" value is equal to 1,by what command can I get it?

Dear all, I want to import just part of an excel data file into R. I would like to have the data imported without rownames or colume names. I used read.delim("clipboard", header=F). Somehow even though I added the argument "header=F", I still have the row names V1, V2, ..., Does anyone know how to fix this? Thanks very much in advance. Hannah [[alternative

Dear all, I hope to remove a whole column from a data frame or matrix (> 2000 columns). All value in the column are same. The first thing is to select those columns. For instance, I hope to remove the V3~6 column, for all the value in those colume is zero. V3 V4 V5 V6 V7 V8 V9 V10 1 0 0 0 0 0.000 0.000 0.000 0.000 2 0 0 0 0 0.000 0.000 0.000 0.000 3 0 0 0