similar to: applying ifelse to dataframe

Dear R-help community, If I have a data.frame df as follows: > df x1 x2 x3 x4 x5 x6 1 5 5 1 1 2 1 2 5 5 5 5 1 5 3 1 5 5 5 5 5 4 5 5 1 4 5 5 5 5 1 5 2 4 1 6 5 1 5 4 5 1 7 5 1 5 4 4 5 8 5 1 1 1 1 5 9 1 5 1 1 2 5 10 5 1 5 4 5 5 11 1 5 5 2 1 1 12 5 5 5 4 4 1 13 1 5 1 4 4 1 14 1 1 5 4 5 5 15 1 5 5 4

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

Hi, i'm trying to simplify some R code but i got stucked in this: test<-data.frame(cbind(id,x1,x2,x3,x4,x5,x6,x7)) test > test id x1 x2 x3 x4 x5 x6 x7 1 1 36 26 21 32 31 27 31 2 2 45 21 46 50 22 36 29 3 3 49 47 35 44 33 31 46 4 4 42 32 38 28 39 45 32 5 5 29 42 39 48 25 35 34 6 6 39 31 30 37 46 43 44 7 7 41 40 25 23 42 40 24 8 8 27 29 47 34 26 38 28 9 9 25 35 29 36

Hi, You might need to check library(boot).? I have never used that before.? So, I can't comment much.? It is better to post on R-help list.? I had seen your postings on Nabble in the past.? Unfortunately those postings were not accepted in R-help.? You have to directly post at ? r-help at r-project.org after registering at: https://stat.ethz.ch/mailman/listinfo/r-help ?

Hola, me gustar?a hacer algo como en el siguiente ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5',

Yo copio y pego este código y me sale correctamente. Se me ocurre que pueda deberse a la versión de R ¿cuál usas? El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11',

Dear All, I have a question, suppose X is a dataframe, with column names as "x1", "x2", "x3", ..... And I would like to use the i-th column by X[,'xi']. But it seems the single quote and double quote are different. So if I run X[, names(X)[i]], it has some error. Please use the below example code X = matrix(rnorm(50),ncol = 5) X = data.frame(X)

Hi everyone, Let me have a dataframe named ?mydata? and created as below, *> n=c(5,5,5,5) #number of trils > x1=c(2,3,1,3) ) #number of successes > x2=c(5,5,5,5) #number of successes > x3=c(0,0,0,0) #number of successes > x4=c(5,0,5,0) #number of successes > mydata=data.frame(n,x1,x2,x3,x4) > mydata* n x1 x2 x3 x4 1 5 2 5 0 5 2 5 3 5 0 0 3 5 1 5 0 5 4 5 3 5 0

This might be a trivial question, but I would appreciate if anybody could suggest an elegant way of plotting a function such as the following (a simple distribution function): F(x) = 0 if x<=0 =(x^2)/2 if 0<x<=1 =2x-((x^2)/2)-1 if 1<x<=2 =1 if x>2 This is just an example. In this case it is a continuous function. But how to do it in general in an elegant way.

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Hola: Como dice Carlos, algo así, por ejemplo: transforma <- function(df) sapply(df, function(x) ifelse(x%in%c("x1","x2","x3"), "prueba1",ifelse(x%in%c("x4","x5","x6"),"prueba2",x))) > transforma(df1)       col1  [1,] "prueba1"  [2,] "prueba1"  [3,] "x11"  [4,]

Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: > dd [1] "lm(y ~ 1,data=Cement)" "lm(y ~ X,data=Cement)" "lm(y ~ X1,data=Cement)" [4] "lm(y ~ X2,data=Cement)" "lm(y ~

I posted this one as an R bug (https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14767), but Prof. Ripley says I'm premature, and I should raise the question here. Here's the behavior I assert is a bug: The output from delete.response on a terms object alters the formula by removing the dependent variable. It removes the response from the "variables" attribute and it changes

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Dear r users, I have 4 variables x1,x2,x3,x4 and each one has two levels, for example Y and N. For x1: prob(Y)=0.6, prob(N)=0.4; For x2: prob(Y)=0.5, prob(N)=0.5; For x3: prob(Y)=0.8, prob(N)=0.2; For x4: prob(Y)=0.9, prob(N)=0.1; Therefore, the sample space for (x1, x2, x3, x4)={YYYY, YYYN, YYNY,......} (16 possible combination) and the corresponding probabilities are {(0.6)(0.5)(0.8)(0.9),

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20

Dear R-users I have a probelm running stepAIC in R1.7.1 I wrote a program which used stepAIC as a part of it, and it worked fine while I was using the previous version of R1.7.0. However, I found the program did not work any more. Now, R produces a message which tells "Error in as.data.frame.default(data) : can't coerce function into a data.frame" every time I run the part of

Hello I have data like this x1 x2 x3 x4 x5 I want to create a matrix similar to a correlation matrix, but with the difference between the two values, like this x1 x2 x3 x4 x5 x1 x2-x1 x3-x1 x4-x1 x5-x1 x2 x3-x2 x4-x2 x5-x2 x3 x4-x3 x5-x3 x4 x5-x4 x5 Then I

There's something peculiar that I do not understand here. However, did you realize that the thing you are assigning into parts of `a' is NULL? Check you're my.test.boot.ci.1: It's NULL. Be that as it may, I get: > a <- data.frame(matrix(1:4, nrow=2), X3=NA, X4=NA) > a X1 X2 X3 X4 1 1 3 NA NA 2 2 4 NA NA > a[a$X1 == 1,]$X3 <- NULL > a X1 X2 X3 X4 1 1

Dear all, I have been trying to investigate the behaviour of different weights in weighted regression for a dataset with lots of missing data. As a start I simulated some data using the following: library(MASS) N <- 200 sigma <- matrix(c(1, .5, .5, 1), nrow = 2) sim.set <- as.data.frame(mvrnorm(N, c(0, 0), sigma)) colnames(sim.set) <- c('x1', 'x2') # x1 & x2 are