similar to: Convert string to date time

Colleagues, I have a matrix of character data that represents date and time. The format of each element of the matrix is "2020-09-17_00:00:00" How can I convert the elements into a valid R date-time constant? Thank you, John John David Sorkin M.D., Ph.D. Professor of Medicine, University of Maryland School of Medicine; Associate Director for Biostatistics and Informatics,

I am an old, long time SAS programmer. I need to produce R code that processes a dataframe in a manner that is equivalent to that produced by using a by statement in SAS and an if first.day statement and a retain statement: I want to take data (olddata) that looks like this ID Day 1 1 1 1 1 2 1 2 1 3 1 3 1 4 1 4 1 5 1 5 2 5 2 5 2 5 2 6 2 6 2 6 3 10 3 10 and make it look like this: (withing each

I have a data frame with three columns, TotalInches, Low20, High20. For each row of the dataset, I am trying to compute the mean of Low20 and High20. xxxz <- structure(list(TotalInches = c(58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76), Low20 = c(84, 87, 90, 93, 96, 99, 102, 106, 109, 112, 116, 119, 122,

I am trying to use ggplot to plot the data, and R code, below. The dates (jdate) are printing as Mar 01, Mar 15, etc. I want to have the date printed as MMM DD YYYY (or any other way that will show month, date, and year, e.g. mm/dd/yy). How can I accomplish this? yyy <- structure(list( jdate = structure(c(19052, 19053, 19054, 19055, 19058, 19059, 19060, 19061, 19062,

?s 16:21 de 07/12/2023, Sorkin, John escreveu: > Colleagues, > > I have a matrix of character data that represents date and time. The format of each element of the matrix is > "2020-09-17_00:00:00" > How can I convert the elements into a valid R date-time constant? > > Thank you, > John > > > > John David Sorkin M.D., Ph.D. > Professor of

Check out the dplyr package, specifically the mutate function. # Create new column based on existing column value df <- df %>% mutate(FirstDay = if(ID = 2, 5)) df Repeat as needed to capture all of the day/firstday combinations you want to account for. Like everything else in R, there are probably at least a dozen other ways to do this, between base R and all of the library packages

Look at the lubridate package in R. Regards, Tim -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Sorkin, John Sent: Thursday, December 7, 2023 11:22 AM To: r-help at r-project.org (r-help at r-project.org) <r-help at r-project.org> Subject: [R] Convert character date time to R date-time variable. [External Email] Colleagues, I have a matrix of

I am running a zero inflated regression using the zeroinfl function similar to the model below: fm_zinb2 <- zeroinfl(art ~ . | ., data = bioChemists, dist = "poisson") summary(fm_zinb2) I have three questions: 1) How can I obtain a value for the parameter pie, which is the fraction of the population that is in the zero inflated model vs the fraction in the count model? 2) For

On 12/7/23 08:21, Sorkin, John wrote: > Colleagues, > > I have a matrix of character data that represents date and time. The format of each element of the matrix is > "2020-09-17_00:00:00" > How can I convert the elements into a valid R date-time constant? You will not be able to store these datetime values in an R matrix, at least as class POSIXct. You could with class

Colleagues I want to convert a 10x2 array: # create a 10x2 matrix. datavals <- matrix(nrow=10,ncol=2) datavals[,] <- rep(c(1,2),10)+c(rnorm(10),rnorm(10)) datavals into a 10x3 array, ThreeDArray, dim(10,2,10). The values storede in ThreeDArray's first dimensions will be the data stored in datavalues. ThreeDArray[i,,] <- datavals[i,] The values storede in ThreeDArray's second

Use apply(), not by(). xxxz$av20 <- apply(xxxz[,c("Low20","High20")],1, mean) -- Bert On Sat, Jun 8, 2024 at 10:38?AM Sorkin, John <jsorkin at som.umaryland.edu> wrote: > I have a data frame with three columns, TotalInches, Low20, High20. For > each row of the dataset, I am trying to compute the mean of Low20 and > High20. > > xxxz <-

?s 16:30 de 27/11/2024, Sorkin, John escreveu: > I am an old, long time SAS programmer. I need to produce R code that processes a dataframe in a manner that is equivalent to that produced by using a by statement in SAS and an if first.day statement and a retain statement: > > I want to take data (olddata) that looks like this > ID Day > 1 1 > 1 1 > 1 2 > 1 2 > 1 3 >

Are you referring to the zeroinfl() function in the countreg package? If so, I think predict(fm_zinb2, type = "zero", newdata = some.new.data) will give you pi for each combination of covariate values that you provide in some.new.data where pi is the probability to observe a zero from the point mass component. As to your second question, I'm not sure that's possible, for any

On 11/27/24 08:30, Sorkin, John wrote: > I am an old, long time SAS programmer. I need to produce R code that processes a dataframe in a manner that is equivalent to that produced by using a by statement in SAS and an if first.day statement and a retain statement: > > I want to take data (olddata) that looks like this > ID Day > 1 1 > 1 1 > 1 2 > 1 2 > 1 3 > 1 3 >

John, Maybe you can clarify what you want the output to look like. It took me a while to realize what you may want as it is NOT properly described as wanting rowsums. There is a standard function called rowMeans() that probably does what you want if you want the mean of all rows as in: > rowMeans(xxxz) [1] 84.33333 87.00000 89.66667 92.33333 95.00000 97.66667 100.33333 103.66667

Aside from the fact that the original question might well be a class exercise (or homework), the question is unanswerable given the data given by the original poster. One needs to know the dimensions of the reservoir, above and below the current waterline. Are the sides, above and below the waterline smooth? Is the region currently above the waterline that can store water a mirror image of the

Was wondering when this would be suggested. But the question was about getting the final dataframe... newdta <- olddta newdta$FirstDay <- ave(newdata$date, newdata$ID, FUN = \(x) x[1L]) On November 27, 2024 11:13:49 AM PST, Rui Barradas <ruipbarradas at sapo.pt> wrote: >?s 16:30 de 27/11/2024, Sorkin, John escreveu: >> I am an old, long time SAS programmer. I need to

David, I greatly appreciate the explanation you gave regarding R tools providing tools available in Linux distros, but not found in Windows. (I am using a windows system). Does this mean that Linux users don't need to use R tools when they want to compile R code? Additionally, thank you for the information about what I should read. I will look at the material again, and hopefully things the

?s 13:27 de 07/04/2024, javad bayat escreveu: > Dear all; > I have a question about the water level of a reservoir, when the volume > changed or doubled. > There is a DEM file with the highest elevation 1267 m. The lowest elevation > is 1230 m. The current volume of the reservoir is 7,000,000 m3 at 1240 m. > Now I want to know what would be the water level if the volume rises to

John, Your reaction was what my original reaction was until I realized I had to find out what a DEM file was and that contains enough of the kind of depth-dimension data you describe albeit what may be a very irregular cross section to calculate for areas and thence volumes. If I read it correctly, this can be a very real-world problem worthy of a solution, such as in places like California