similar to: read columns of quoted numbers as factors

Hi, I'm new to R and I'm stuck trying to import some data from a .dat file I've been given. The tricky bit for me is that the data has both variable values and labels? The data looks like this, Id=1 time=2011-03-27 19:23:40 start=1.4018 end=1.4017 Id=2 time=2011-03-27 19:23:40 start=1.8046 end=1.8047 Id=1 time=2011-03-27 19:23:50 start=1.4017 end=1.4018 Id=2

What is a good way to enter a very long model formula. For example: y ~ Input.2 + Input.3 + ... + Input.1000 (assuming the corresponding dataframe has many other columns). Is there a way to convert a character string to a formula? Are there command line expansions in R besides the simple '.'? Thanks. [[alternative HTML version deleted]]

Suppose df is a dataframe with one named row of numeric observations. I want to coerce df into a named vector. as.vector does not work as I expected: as.vector(df) returns the original dataframe, while as.vector(df,mode="numeric") returns an unnamed vector of NAs. This works: > v <- as.numeric(as.matrix(df)); names(v) <- names(df); I just wanted check if there

I have some square matrices with na values in corresponding rows and columns. M<-matrix(1:2,10,10) M[6,1:2]<-NA M[10,9]<-NA M<-as.matrix(as.dist(M)) print (M) 1 2 3 4 5 6 7 8 9 10 1 0 2 1 2 1 NA 1 2 1 2 2 2 0 1 2 1 NA 1 2 1 2 3 1 1 0 2 1 2 1 2 1 2 4 2 2 2 0 1 2 1 2 1 2 5 1 1 1 1 0 2 1 2 1 2 6 NA NA 2 2 2 0 1 2 1 2 7 1 1 1 1 1 1 0 2 1 2 8

In that case identical should be FALSE but it is TRUE identical(a1, a2) ## [1] TRUE On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > > They differ in whether the row names are "automatic": > > > .row_names_info(a1) > [1] -3 > > .row_names_info(a2) > [1] 3 > > Best, > -Deepayan > > On Tue, 14 Nov

Also why should that difference result in different behavior? On Tue, Nov 14, 2023 at 9:38?AM Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > In that case identical should be FALSE but it is TRUE > > identical(a1, a2) > ## [1] TRUE > > > On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar > <deepayan.sarkar at gmail.com> wrote: > > > >

On Tue, 14 Nov 2023 at 09:41, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > Also why should that difference result in different behavior? That's justifiable, I think; consider: > d1 = data.frame(a = 1:4) > d2 = d3 = data.frame(b = 1:2) > row.names(d3) = c("a", "b") > data.frame(d1, d2) a b 1 1 1 2 2 2 3 3 1 4 4 2 > data.frame(d1,

They differ in whether the row names are "automatic": > .row_names_info(a1) [1] -3 > .row_names_info(a2) [1] 3 Best, -Deepayan On Tue, 14 Nov 2023 at 08:23, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > What is going on here? In the lines ending in #### the inputs and outputs > are identical yet one gives a warning and the other does not. > >

Hallo, I would like to use to read.xls function from the gdata package to read data from Microsoft Excel files but I experienced a problem: For example I used the following code: testfile<-read.xls("/home/.../wsjecon0603.xls", #file path header=F, dec=",", na.strings="n.a.", skip=5, sheet=2,

What is going on here? In the lines ending in #### the inputs and outputs are identical yet one gives a warning and the other does not. a1 <- `rownames<-`(anscombe[1:3, ], NULL) a2 <- anscombe[1:3, ] ix <- 5:8 # input arguments to #### are identical in both cases identical(stack(a1[ix]), stack(a2[ix])) ## [1] TRUE identical(a1[-ix], a2[-ix]) ## [1] TRUE res1 <-

Hello all, I'm trying to use zoo.read but can't figure out how to deal with the time format. (example below) would be nice if someone could help. best regards, Immanuel --------------------------- L <- "Date,Time,Open,High,Low,Close,Up,Down 05.02.2001,00:30,421.20,421.20,421.20,421.20,11,0 05.02.2001,01:30,421.20,421.40,421.20,421.40,7,0

The idea is that one wants to write the line of code below in a general way which works the same whether you specify ix as one column or multiple columns but the naming entirely changes when you do this and BOD[, 1] and transform(BOD, X=..., Y=...) or other hard coding solutions still require writing multiple cases. ix <- 1:2 transform(BOD, X = BOD[ix] * seq(6)) On Tue, Jul 24, 2018 at

there are really two related problems here I have a 2D matrix A <- matrix(1:100,nrow=20,ncol =5) S <- matrix(1:10,nrow=2,ncol =5) #I want to subtract S from A. so that S would be subtracted from the first 2 rows of #A, then the next two rows and so on. #I have a the same problem with a 3D array # where I want to subtract Q for every layer (1-10) in Z # I thought I solved this one

Hello, We've released the int64 package to CRAN a few days ago. The package provides S4 classes "int64" and "uint64" that represent signed and unsigned 64 bit integer vectors. One further development of the package is to facilitate reading 64 bit integer data from csv, etc ... files. I have this function that wraps a call to read.csv to: - read the "int64"

Hello, I have a file that looks like this: Date,Hour,DA_DMD,DMD,DA_RTP,RTP,, 1/1/2006,1,3393.9,3412,76.65,105.04,, 1/1/2006,2,3173.3,3202,69.20,67.67,, 1/1/2006,3,3040.0,3051,69.20,77.67,, 1/1/2006,4,2998.2,2979,67.32,69.10,, 1/1/2006,5,3005.8,2958,65.20,68.34,, where the ',' is the separator and I tried to read it into R, but... > y <- read.csv("Data/Data_tmp.csv",

Note the inconsistency in the names in these two examples. X.Time in the first case and Time.1 in the second case. > transform(BOD, X = BOD[1:2] * seq(6)) Time demand X.Time X.demand 1 1 8.3 1 8.3 2 2 10.3 4 20.6 3 3 19.0 9 57.0 4 4 16.0 16 64.0 5 5 15.6 25 78.0 6 7 19.8 42 118.8 >

Hi R-helpers I'm looking for a vectorised function which does missing value replacement as in last observation carried forward in the zoo package but instead of a locf, I would like the locf function to add +1 to each time a missing value occurred. See below for an example. > require(zoo) > x <- 5:15 > x[4:7] <- NA > coredata(na.locf(zoo(x))) [1] 5 6 7 7 7 7 7 12 13

*Dear R users, Ive just started using the ff package. There is a csv file (~4Gb) with 7 columns and 6e+7 rows. I want to read only column from the file, skipping the first 100 rows. Below Ive provided different outcomes, which will clarify my problem * > sessionInfo() R version 2.14.2 (2012-02-29) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: ... attached base packages: [1] tools

I am new to Using R for data analysis. I have an incomplete time series dataset that is in daily format. I want to extract only Friday data from it. However, there are two problems with it. First, if Friday data is missing in that week, I need to extract the data of the day prior to that Friday (e.g. Thursday). Second, sometimes there are duplicate Friday data (say Friday morning and

Dear all: I have two larges files with 2000 columns. For each file I am performing a loop to extract the "i"th element of each file and create a data frame with both "i"th elements in order to perform further analysis. I am not extracting all the "i"th elements but only certain which I am indicating on a vector called "d". See an example of my code below