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, 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)

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

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,

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 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

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

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 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 <-

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

hi guys..please help me with this i am working on two data frames one goes like this: DF1 Sample_id RepairHours Denatured Dose ZeroMean FourtyFiveMean NinetyMean 1 SDM071 0 1 B 60.5 19.0 45.0 2 SDM071 1 1 B 46.0 23.0 42.5 3 SDM071 2 1 B 52.5 24.0 40.0 4 SDM071

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

Hello! I have a column in my data frame that I have to split: I have to distill the numbers from the text. Below is my example and my solution. x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1")) x library(stringr) out<-as.data.frame(str_split_fixed(x$x,"aaa",2)) out2<-as.data.frame(str_split_fixed(out$V2,"_bbb",2))

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