similar to: How to convert time to days

Hola a todos: Tengo un problema con las fechas, básicamente necesito una diferencia en días y siempre se sale en segundos. Salvo un ejemplo copiado de un libro, pero yo necesito procesar los datos propios. Les paso el código en R (copiar y pegar, son solo dos días que se comparan) y al final como me salen los resultados porque es medio complicada mi redacción. ¿Alguna idea? fechas1

Dear all, I apologize if my question is quite simple. I have a dataset (20 columns & 1000 rows) which some of columns have the same value and the others have different values. Here are some piece of my dataset: obj <- cbind(c(1,1,1,4,0,0,1,4,-1), c(0,1,1,4,1,0,1,4,-1), c(1,1,1,4,2,0,1,4,-1), c(1,1,1,4,3,0,1,4,-1), c(1,1,1,4,6,0,1,5,-1),

Dear R-helpers, I have dataset (data.frame) like below, x1 x2 x3 x4 x5 x6 x7 x8 x9 ... x1200 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 ... How can I change automatically 0=no and 1=yes. Thank you very much in advance. Kindly regards, Muhammad

Dear all, I have a dataset train <- cbind(c(0,2,2,1,0), c(8,9,4,0,2), 6:10, c(-1, 1, 1, -1, 1)) test <- cbind(1:5, c(0,1,5,1,3), c(1,1,2,0,3) ,c(1, 1, -1, 1, 1)) I want to find which rows of train and test it different in its last column (column 4). The solution must be something like train [,1] [,2] [,3] [,4] [1,] 0 8 6 -1 [3,] 2 4 8 1 [4,] 1 0 9 -1

Hi, Excuse me for this simple question. How to convert as.data.frame to as.character? ?data.frame > L3 <- LETTERS[1:3] > L10 <- LETTERS[1:10] > d <- data.frame(cbind(x=c("XYZ"), y=L10), fac=sample(L3, 10, repl=TRUE)) > d x y fac 1 XYZ A A 2 XYZ B A 3 XYZ C A 4 XYZ D A 5 XYZ E B 6 XYZ F C 7 XYZ G A 8 XYZ H C 9 XYZ I B 10 XYZ

Thank you very much for your useful suggestions. These are exactly what I was looking for. foo <- list(foo1, foo2, foo3) lapply(foo, function(x) matrix(unlist(x), nrow = length(x), byrow = TRUE)) or lapply(foo, function(x) do.call('rbind', x)) Best, Muhammad Subianto On 4/11/06, Muhammad Subianto <msubianto at gmail.com> wrote: > Dear all, > I have a result my experiment

Dear R-helper, Is there possible to make this array: > a <- array(1:12, c(4, 3)) > a [,1] [,2] [,3] [1,] 1 5 9 [2,] 2 6 10 [3,] 3 7 11 [4,] 4 8 12 > like: c(1,5,9) c(2,6,10) c(3,7,11) c(4,8,12) Thank you very much in advance. Regards, Muhammad Subianto

Dear all, Are there any functions to convert decimals to fractions in R? I have the result: > summary(as.factor(complete.ID)) 0 0.0133333333333333 0.04 2256 488 230 0.0666666666666667 0.0933333333333333 0.106666666666667 2342 310 726 0.133333333333333

Dear all, #Last week, I asked about merge x and y as list. #Now I have a dataset with list of list like: x <- list(list(matrix(1:20, 5, 4),matrix(1:20, 5, 4)), list(matrix(1:20, 5, 4),matrix(1:20, 5, 4))) y <- list(list(c(1, -1, -1, 1, 1),c(1, 1, -1, -1, -1)), list(c(1, 1, 1, 1, 1),c(1, 1, -1, 1, -1))) x y #I need merge x and y, I have tried with list.uni <-

Dear all, I have a dataset. I want to make barplot from this data. Zero1 <- " V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1

Hi All, I've a couple of questions i've been struggling with using the time features, can anyone help? sample data Timestamp user_id 27/05/08 22:57 763830873067 27/05/08 23:00 763830873067 27/05/08 23:01 763830873067 27/05/08 23:01 763830873067 05/06/08 11:34 763830873067 29/05/08 23:08 765253440317 29/05/08 23:06 765253440317 29/05/08 22:52 765253440317

Dear useR, I have a dataset like this below, > prevRND.dat <- read.table("C:\\workdir\\prevRND.txt", + header=FALSE, # No header. + col.names = c("X","Y","Z"), + sep = ",") > prevRND.dat X Y Z 1 A A 0.950933 2 A B 0.143600 3 A C 0.956133 4 B A

I have a question, which very easy to solve, but I can't find a solution. I want to convert a data frame to matrix. Here my toy example: > L3 <- c(1:3) > L10 <- c(1:6) > d <- data.frame(cbind(x=c(10,20), y=L10), fac=sample(L3, + 6, repl=TRUE)) > d x y fac 1 10 1 1 2 20 2 1 3 10 3 1 4 20 4 3 5 10 5 2 6 20 6 2 > is.data.frame(d) [1] TRUE > sapply(d,

Dear R-helpers, First I apologize if my question is quite simple I have a large datasets which more 100 variables. For a research I need to change all name of variables with add one or more letters on each variables. For example, > data(Pima.tr) > Pima.tr[1:5,] npreg glu bp skin bmi ped age type 1 5 86 68 28 30.2 0.364 24 No 2 7 195 70 33 25.1 0.163 55 Yes 3 5

Dear all R-users, (My apologies if this subject is wrong) I have dataset: mydat <- as.data.frame( matrix(c(14,0,1,0,1,1, 25,1,1,0,1,1, 5,0,0,1,1,0, 31,1,1,1,1,1, 10,0,0,0,0,1), nrow=5,ncol=6,byrow=TRUE)) dimnames(mydat)[[2]]

Dear R-list, I have a dataset, say (the real dataset is 20 columns,110200 rows). > my.reducedID V1 V2 V3 V4 V5 V6 V7 V8 V9 [1,] 1 0 0 1 14 3 1 0 2 [2,] 2 0 0 1 14 3 1 0 2 [3,] 0 1 0 1 14 2 1 0 2 [4,] 0 0 1 1 14 3 1 0 2 [5,] 0 1 1 0 14 2 1 0 2 [6,] 0 0 0 1 14 3 1 0 2 [7,] 0 0 0 1 0 3 1 0 2 [8,] 0

Dear R users, My situation: (1) I have limited workspace for my work harddisk (about 10 GiB). (2) I have a lot of data files in R workspace (*.RData) which most of them > 200 MiB. For some reason I zip some of them, for instance "filename.RData (250 MiB)" to "filename.zip (3MiB)". In this work I have a lot of more space of my harddisk. Normally, If I want to use

Dear all, I have a result my experiment like this below (here my toy example): foo1 <- list() foo1[[1]] <- c(10, 20, 30) foo1[[2]] <- c(11, 21, 31) foo2 <- list() foo2[[1]] <- c(100, 200, 300) foo2[[2]] <- c(110, 210, 310) foo3 <- list() foo3[[1]] <- c(1000, 2000, 3000) foo3[[2]] <- c(1100, 2100, 3100) list(foo1,foo2,foo3) The result: > list(foo1,foo2,foo3) [[1]]

Hello, I wanted to compute the time differenze between to times: first =as.POSIXct( "2012-06-15 16:32:39.0025 CEST") second = as.POSIXct("2012-06-15 16:32:39.0086 CEST") second - first The result is Time difference of 0.006099939 secs instead of just 0.0061 secs So R adds aditional numbers after the result. I know I could round it in this case. But I am working with a

Dear all, I have a vector generated using the function strptime: > my.dt [1] "2004-04-19 08:35:00 W. Europe Daylight Time" "2004-04-19 09:35:00 W. Europe Daylight Time" "2004-04-19 11:35:00 W. Europe Daylight Time" [4] "2004-04-19 13:35:00 W. Europe Daylight Time" "2004-04-20 07:50:00 W. Europe Daylight Time" > class(my.dt) [1]