similar to: Resumen de R-help-es, Vol 167, Envío 10

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 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 =

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

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,

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 >

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: > > >

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

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

Hola, No hace falta (en este caso) capturarlo de la consola. El resultado de la función apply se puede capturar y procesar. > data("mtcars") > # Mtcars_matriz <- as.matrix(mtcars) > res_out <- apply(mtcars, MARGIN =2, FUN = shapiro.test) > > > res_df <- as.data.frame(unlist(res_out)) > res_df$vars <- rownames(res_df) > rownames(res_df) <- NULL

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

Full_Name: Vlad Stolin Version: R 2.0.0 OS: Windows 2000 Submission from: (NULL) (204.128.232.211) When generating the sequence using seq() function with non-integer numbers result is somewhat unpredictable. Example: > v1<-seq(1.60,1.90,.05) > v2<-c(1.60,1.65,1.70,1.75,1.80,1.85,1.90) > v1-v2 [1] 0.000000e+00 2.220446e-16 2.220446e-16 0.000000e+00 0.000000e+00 0.000000e+00

Hi everybody, using bwplot for producing panel boxplot with 3 dimensions i want to add a mark on each boxplot representing one individual (on all its dimensions) till now, i didn't succeed getting the desired solution I want as well to keep the median symbols as a line Many thanks for your help christophe here is the tested code: ######################## library(lattice) ex <-

Dear R-community, Using bwplot, how can I put the whiskers at percentile 5 and percentile 95, in place of the default position coef=1.5?? Using panel=panel.bwstrip, whiskerpos=0.05, from the package agsemisc gives satisfaction, but changes the appearance of my boxplot and works with an old version of R, what I don’t want, and I didn’t find the option in box.umbrella parameters Many thanks

Hello I need to use this command: cbind(ftable(xtabs(cnt~geo+slp+con+hey,data=dt3)) hey is in count of success /failure value but cbind gives failure/success counts. I want to change the order of this cbind as success /failure counts. for instance: I want cbind to give counts as 32-4552 rather than 4552-32 what should I do ? thanks in advance Ahmet TURKEY

Hi, my data set is data.frame(id, yr, y, l, e, k). I would like to estimate Lee and Schmidts (1993, OUP) model in R. My colleague wrote SAS code as follows: ** procedures for creating dummy variables are omitted ** ** di# and dt# are dummy variables for industry and time ** data a2; merge a1 a2 a; by id yr; proc sysnlin maxit=100 outest=beta2; endogenous y; exogenous l e k

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

Hi, my data set is data.frame(id, yr, y, l, e, k). I would like to estimate Lee and Schmidts (1993, OUP) model in R. My colleague wrote SAS code as follows: ** procedures for creating dummy variables are omitted ** ** di# and dt# are dummy variables for industry and time ** data a2; merge a1 a2 a; by id yr; proc sysnlin maxit=100 outest=beta2; endogenous y; exogenous l e k

Hello, I have error on compiling time I tried on HP-UX 11.00 system with gcc version 3.1 to compile samba source Samba latest version source was dowloaded from www.us1.samba.org/samba/ftp I attached log file with this error thank you. begin 600 samba_make.log.doc M(R!M86ME#0HO=7-R+VQO8V%L+W-A;6)A+V)I;B(@+410241$25(](B]U<W(O M;&]C86PO<V%M8F$O=F%R+VQO8VMS(B