similar to: Splitting the left and right hand terms of a formula

Hi All, I sent the following message to R-help on July 14th, 2006: Let's say I have the following formula: a.formula <- x ~ y + z I want to extract the left and right-hand sides of the function so that I have two character vectors like the ones you would create using the following assignments: left.hand.side <- "x" right.hand.side <- c("y", "z") One

Hello all, Consider the following problem: There is a matrix of probabilities: > set.seed(1) > probs <- array(abs(rnorm(25, sd = 0.33)), dim = c(5,5), dimnames = list(1:5, letters[1:5])) > probs a b c d e 1 0.21 0.27 0.50 0.0148 0.303 2 0.06 0.16 0.13 0.0053 0.258 3 0.28 0.24 0.21 0.3115 0.025 4 0.53 0.19 0.73 0.2710 0.656 5 0.11 0.10 0.37 0.1960

Hello, I'm developing a GUI in R that will be used to monitor financial portfolio performance. The GUI will be distributed as an R package. So far, I've decided to use the "cairoDevice", "RGtk2", "gWidgets", and "gWidgetsRGtk2" packages to develop the GUI. I am trying to decide what would be the best way to structure the GUI would be. I've

Dear All, trying to write a data.frame, containing Surv objects to a csv-file I get "Error in dimnames(X) <- list(dn[[1L]], unlist(collabs, use.names = FALSE)) : length of 'dimnames' [2] not equal to array extent". See example below. May be, I overlooked something, but I expected that also data.frames containing Surv objects may be written to csv files. Is there a

Hi list, In a function I am writing, I need to extract the dimension names of an array. I know this can be acheived easily using dimnames() but my problem is that I want my function to be robust when the number of dimensions varies. Consider the following case: foo <- array(data = rnorm(32), dim = c(4,4,2), dimnames=list(letters[1:4], LETTERS[1:4], letters[5:6])) # What I want is to extract

Hello R Gurus, I have a matrix for which I am doing a heatmap using heatmap.2. I want to put the rownames on the lefthand side instead of the right side of the heatmap. how can i put the rownames on left hand side: I have already tried axis but could not make it work ccc<-structure(c(1, 0.283300333755851, 0.237863231117007, 0.0148696794159253, -0.0780756406815149, -0.106161465097659,

summary: The function 'by' inconsistently strips class from the data to which it is applied. quick reason: tapply strips class when simplify is set to TRUE (the default) due to the class stripping behaviour of unlist. quick answer: This can be fixed by invoking tapply with simplify=FALSE, or changing tapply to use do.call(c instead of unlist executable example:

hello, i have been trying to convert my data frames to matrices in the hopes of speeding up some of my more complicated scripts. to assist with this, i am trying to create a "matrix column operator" like $: "%$%" = function(data,field) { as.numeric(data[,grep(field,unlist(dimnames(data)[2]))]) } the idea here is that you can use a matrix like a dataframe:

On Fri, Jan 27, 2017 at 12:34 AM, Martin Maechler <maechler at stat.math.ethz.ch> wrote: > > > On Jan 26, 2017 07:50, "William Dunlap via R-devel" <r-devel at r-project.org> > > wrote: > > > It would be cool if the default for tapply's init.value could be > > FUN(X[0]), so it would be 0 for FUN=sum or FUN=length, TRUE for >

I work with a list of crypto assets daily closing prices in a xts class. Here is a limited example: asset.xts.lst <- list(BTCUSDT = structure(c(26759.63, 26862, 26852.48, 27154.15, 27973.45), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200, 1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"), class = c("xts", "zoo")), ETHUSDT

Hello, Does anyone know of a function that will determine whether or not a formula object has a left hand side? I.e., can differentiate between y ~ x + z and ~ x + z Perhaps I'm overlooking the obvious... Thanks!

I think that the "dimname names" of tables and arrays could make aperm() and apply() (and probably some other functions) easier to use. (dimname names are, for example, created by table() ) The use would be something like: -- x <-table( from=sample(3,100,rep=T), to=sample(5,100,rep=T)) trans <- x / apply(x,"from",sum) y <- aperm( trans,

>>>>> Hilmar Berger >>>>> on Sat, 14 Sep 2019 13:31:27 +0200 writes: > Dear all, > I did some more tests regarding the == operator in Ops.data.frame (see > below).? All tests done in R 3.6.1 (x86_64-w64-mingw32). > I find that errors are thrown also when comparing a zero length > data.frame to atomic objects with length>0

Arnaud, Short answer may be that the tibble data structure will not be supporting row names and you may want to simply save those names in an additional column or externally. My first thought was to simply save the names you need and then put them back on the tibble. In your code, something like this: save.names <- names(my.ret.lst) result.tib <- as_tibble_col(unlist(my.ret.lst),

I have two tables: table_1: tab1_id, other_id table_2: id, tab2_id, something_else I would like to specify a has_one mapping from table_1 to table_2 but specifying that the join is based on (table_2.tab2_id = table_1.other_id). I am half way there with: class Table1 < ActiveRecord::Base set_table_name "table_1" set_primary_key "tab1_id" has_one :Table2, :class_name

The lattice package uses special logic to allow for multiple left-hand-side variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to do this outside of lattice? I'm trying to implement a data summarization function that logically takes multiple dependent variables. The usual invocation of model.frame( ) causes R to try to do arithmetic addition to create a single dependent

Hello, I would like to create a function that gets passed a class name and then calls setClass, and a few other functions, inside. I have done this in the past with setmethod, creating accessors for all slots in a set of S4 classes. But setClass is choking when my function is called isnide a package, telling about an error in exists(cname, where). I assume this to be a problem with the

Hello all Is there a way reducing the number of characters in a list so that just the left n numbers of characters is given? For example, If I have a list, listnames, which consists of 4 strings of 6 characters; >listnames [1] "item12" "item34" "item56" "item78" Is there a way to reduce this so only the 5 characters on the lefthand side are

Sorry, I can't reproduce the example below even on the same machine. However, the following example produces the same error as NULL values in prior examples: > setClass("FOOCLASS", +????????? representation("list") + ) > ma = new("FOOCLASS", list(M=matrix(rnorm(300), 30,10))) > isS4(ma) [1] TRUE > data.frame(a=1:3) == ma Error in

Hello, I am trying to convert BY to a data frame, consider the following example: exampleDF<-data.frame(a=c(1,2),b=c(10,20),name=c("first","second")) exampleBY<-by(exampleDF,with(exampleDF,paste(a,b,sep="_")),               function(x) {                 data.frame(                     name=as.character(x$name),                     a=x$a,