similar to: Inaccuracy in seq() function (PR#7503)

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 =

I have a dataset of the form below, consisting of one unique ID per row, followed by a series of visit dates. At each visit there are values for 3 dichotomous variables. Of the 8 different possible combinations of the three variables, 4 are "abnormal" and the remaining 4 are "normal". Everyone starts out abnormal, and then either continues to be abnormal at subsequent visits,

Suppose I have two data frames A and B A has three variables and B also has three variables. I would like to merge these two database but the requirement to merge is that the value of the second column in database A is less than the value of the second column in database B. Is there a R code to do this? Thanks Dataframe A: V1a V2a V3a 1 2 3 5

Hi, First let me thank you for the incredible help and resource that this forum is. I am trying to compare the repeated measurement of more than 100 analytes that have been take in 70 subjects at 2 time points adjusted for the time difference of sample times(TimeDifferenceDays), therefore I wanted to do it with a function that allows me to do all at once. (131 is the column difference that

Hola: Muchas gracias por responder. Lo pruebo. Saludos. On Fri, 27 Jan 2023 01:40:48 +0100 Carlos Ortega <cof en qualityexcellence.es> wrote: > Hola, > > Otra alternativa... > > #-------------------- > > library(data.table) > > library(tidytable) > > library(stringi) > > > > df <- data.frame( V1a =

Hola, Carlos: Gracias, funciona también a la perfección y muy ingeniosa la solución. Disculpa si no te he respondido antes, pero hasta ahora no he podido privarlo. Gracias por la ayuda y saludos. On Fri, 27 Jan 2023 01:40:48 +0100 Carlos Ortega <cof en qualityexcellence.es> wrote: > Hola, > > Otra alternativa... > > #-------------------- > > library(data.table)

Hola a todos: Se me ha planteado un problema que no está ligado a ningún problema concreto. Es más teórico. Supongamos que tenemos tres variables: V1 <- c (47, 71, 41, 23, 83, 152, 82, 8, 160, 18) V2a <- c (NA, 36, 15, 5, 56, 18, NA, 5, NA, 5) V2b <- c (37, NA, 15, NA, NA, NA, 90, NA, 161, NA) Supongamos que tengo la expresión (que no puedo asignarlo a

No tuve tiempo de mirarlo, pero, ¿es coherente lo que dice? El vie, 11 ago 2023 a las 21:02, Griera-yandex (<griera en yandex.com>) escribió: > Muchas gracias, Manuel: > > Que bueno! No se me había ocurrido lo de GPT! > > Lo pruebo. > > Saludos. > > On Fri, 11 Aug 2023 18:15:18 +0200 > Manuel Mendoza <mmendoza en fulbrightmail.org> wrote: > > >

Esta es la respuesta que te da ChatGPT-4: Entiendo tu pregunta y, aunque no hay una función nativa en R que te permita hacer exactamente lo que estás pidiendo, puedes lograr el mismo resultado utilizando una función. Una función te permitiría encapsular la lógica de la expresión que quieres reutilizar y luego llamar a esa función donde sea necesario. He aquí cómo podrías hacerlo: V1 <-

Muchas gracias, Manuel: Que bueno! No se me había ocurrido lo de GPT! Lo pruebo. Saludos. On Fri, 11 Aug 2023 18:15:18 +0200 Manuel Mendoza <mmendoza en fulbrightmail.org> wrote: > Esta es la respuesta que te da ChatGPT-4: > > Entiendo tu pregunta y, aunque no hay una función nativa en R que te > permita hacer exactamente lo que estás pidiendo, puedes lograr el mismo >

A ver... con que xfunc() esté preparada para tomar un parámetro de tipo "carácter" y evaluarlo, claro que se puede hacer... Si el problema lo tienes en evaluar la expresión, la función "eval()" te lo hace. Si no te he entendido bien, explícate más ? Saludos Isidro -----Mensaje original----- De: R-help-es <r-help-es-bounces en r-project.org> En nombre de Griera Enviado

Sipura has implemented auto-answer in version 0.9.5 of the SPA-841 firmware. However, it is implemented via the Call-Info header, which Asterisk stable doesn't currently support. The attached patch implments a quick hack to support the Call-Info header from the Dial() application by way of setting the CALL_INFO variable. For example, the following macro can be used to dial up a single

Hello to all,I have a data file as Class V1 V2A -2.0 0.0A 0.9 0.7B 0.1 0.6C 4.1 0.4C 1.0 1.9B 1.1 0.5 I am plotting this data in R as V1 verses V2> temp<-read.table('temp.dat', header=T)> attach(temp)> plot (V1,V2, col='red')> text(x=V1, y=V2, labels=Class, pos=4) But I want to change the  'plotting symbol'  by the 'Class of  the row' (which is

Dear all, I a bit unsure, whether this qualifies as a bug, but it is definitly a strange behaviour. That why I wanted to discuss it. With the following function, I want to test for evenly space numbers, starting from anywhere. .is_continous_evenly_spaced <- function(n){ if(length(n) < 2) return(FALSE) n <- n[order(n)] n <- n - min(n) step <- n[2] - n[1] test <-

On two laptops running 32-bit kubuntu, I have found that svd(), invoked within R 2.9.1 as supplied with the current ubuntu package, returns very incorrect results when presented with complex-valued input. One of the laptops is a Dell D620, the other a MacBook Pro. I've also verified the problem on a 32-bit desktop. On these same systems, R compiled from source provides apparently

Agreed that's it's rounding error, and all.equal would be the way to go. I wouldn't call it a bug, it's simply part of working with floating point numbers, any language has the same issue. And while we're at it, I think the function can be a lot shorter: .is_continous_evenly_spaced <- function(n){ length(n)>1 && isTRUE(all.equal(n[order(n)], seq(from=min(n),

I noticed the following peculiarity with `serialize()' when `ascii = TRUE' is used. In today's (svn r37299) R-devel, I get > set.seed(10) > x <- rnorm(10) > > a <- serialize(x, con = NULL, ascii = TRUE) > b <- unserialize(a) > > identical(x, b) ## FALSE [1] FALSE > x - b [1] -3.469447e-18 2.775558e-17 -4.440892e-16 0.000000e+00

El vie., 31 ago. 2018 a las 16:00, Mark van der Loo (<mark.vanderloo at gmail.com>) escribi?: > > how about > > is_evenly_spaced <- function(x,...) all.equal(diff(sort(x)),...) This doesn't work, because 1. all.equal does *not* return FALSE. Use of isTRUE or identical(., TRUE) is required if you want a boolean. 2. all.equal compares two objects, not elements in a vector.

Bugs item #7503, was opened at 2006-12-30 19:28 You can respond by visiting: http://rubyforge.org/tracker/?func=detail&atid=218&aid=7503&group_id=35 Category: Incorrect behavior Group: None Status: Open Resolution: None Priority: 3 Submitted By: Alex Fenton (brokentoy) Assigned to: Alex Fenton (brokentoy) Summary: Segfaults on exception, related to sizer Initial Comment: In

Hi, Cosmetic. Starting R with e.g. --max-ppsize=-10 produces the following warning: WARNING: '-max-ppsize' value is negative: ignored The name of the option displayed in the warning is incorrect. Could that be fixed? See src/main/CommandLineArgs.c (there are 3 places in that file where the name of this option needs to be adjusted). This is with current R-alpha. Thanks! H. --