similar to: coef(summary) and plyr

I am trying to store regression objects in a data.table df <- data.frame(x = rnorm(20)) df[, "y"] <- with(df, x + 0.1 * x^2 + 0.2 * rnorm(20)) mydt <- data.table(mypower = c(1, 2), myreg = list(lm(y ~ x, data = df), lm(y ~ x + I(x^2), data = df))) mydt #?? mypower??? myreg #???? <num>?? <list> #1:?????? 1 <lm[12]> #2:?????? 2 <lm[12]> But mydt[1, 2]

This code gives unexpected result. library(data.table) library(lattice) set.seed(123) mydt <- data.table(date = seq.Date(as.IDate("2024-01-01"), by = 1, length.out = 50), xgroup = "A", x = runif(50, 0, 1)) mydt <- rbindlist(list(mydt, data.table(date = mydt$date, xgroup = "B", x = runif(50, 0, 3)))) mydt[, `:=`(xcumsum = cumsum(x)), by = .(xgroup)] mydt[,

Well, you may have good reasons to do things this way -- and you certainly do not have to explain them here. But you might wish to consider using R's poly() function and a basic nested list structure to do something quite similar that seems much simpler to me, anyway: x <- rnorm(20) df <- data.frame(x = x, y = x + .1*x^2 + rnorm(20, sd = .2)) result <- with(df,

Hello I have a rather simple code and for some reason it produces an error message. If someone can tell me why and how to fix it, I would be very greatful. Thank you in advance. ##### create data set.seed(10) n <- 200 # number of training points n.test <- 200 # number of test points p<-2 # dimension of input space z <-

I am calculating the mean of each column grouped by the variable 'id'. I do this using aggregate, data.table, and plyr. My aggregate results do not match the other two, and I am trying to figure out what is incorrect with my syntax. Any suggestions? Thanks. Here is the data. myData <- structure(list(var1 = c(31.59, 32.21, 31.78, 31.34, 31.61, 31.61, 30.59, 30.84, 30.98, 30.79, 30.79,

Thanks everyone for their responses. My data is organized in a data.table.? My goal is to perform analyses according to some groups.? The results of analysis are objects.? If these objects could be stored as elements of a data.table, this would help downstream summarizing of results. Let me try another example. carsdt <- setDT(copy(mtcars)) carsdt[, unique(cyl) |> length()] #[1] 3

Buen día estimados Estoy tratando de hacer un tibble con los resultados de un apply que se muestran en la consola que me da R, no estoy seguro si eso se pueda hacer, pero me gustaría organizar los resultados de esa manera. mi código es: data("mtcars") Mtcars_matriz <- as.matrix(mtcars) apply(Mtcars_matriz, MARGIN =2, FUN = shapiro.test) DF2 <- tibble(Variable = NA, W = NA, Pvalue =

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

List, Consider the following data. gender mygroup id 1 F A 1 2 F B 2 3 F B 2 4 F B 2 5 F C 2 6 F C 2 7 F C 2 8 F D 2 9 F D 2 10 F D 2 11 F D 2 12 F D 2 13 F D 2 14 M A 3 15 M A 3 16 M A 3 17

I have a data set with about 6 million rows and 50 columns. It is a mixture of dates, factors, and numerics. What I am trying to accomplish can be seen with the following simplified data, which is given as dput output below. > head(myData) mydate gender mygroup id 1 2012-03-25 F A 1 2 2005-05-23 F B 2 3 2005-09-08 F B 2 4 2005-12-07 F B 2

This is no doubt trivial but after searching the help files and the web, I cannot seem to find it. 1) How do I convert 'hgt' into 'HGT' in R? 2) How should I have used the help facilities to find this? At the end of the day, all I want to do is case insensitive string matching... i.e. 'if ("HGT" == 'hgt') print('this should be true')' I tried