similar to: Convert character date time to R date-time variable.

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 am reading a string that has the following form: "2020-08-26_05:15:01" I want to convert the string to a date-time variable. I tired: x <- "2007-02-01_10:10:30" x <- as.POSIXct(x,tz=Sys.timezone()) x but this did not work; the time portion was ignored. I suspect the problem is the _ between the date and time, but I don't know how to account for this character in

?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

I need to write code that will give me the previous value of from a data.frame. I have written the following code using the shift function from data.table . It does not work. I hope someone can help me correct the code. ########################### # Try to understand shift # ########################### if(!require(data.table)) install.packages("data.table") library(data.table) # Create

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,

Dear R help folks, First my apologizes for sending several related questions to the list server. I am trying to learn how to manipulate data in R . . . and am having difficulty getting my program to work. I greatly appreciate the help and support list member give! I am trying to write a program that will run through a data frame organized by ID and for the first line of each new group of data

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

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

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

Rui: "f these two, diff is faster. But of all the solutions posted so far, Ben Bolker's is the fastest." But the explicit version of diff is still considerably faster: > D <- c(rep(1,10),rep(2,6),rep(3,2)) > microbenchmark(c(1L,diff(D)), times = 1000L) Unit: microseconds expr min lq mean median uq max neval c(1L, diff(D)) 3.075 3.198 3.34396

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

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 <-

I am trying to use ggplot2 to create a figure with multiple lines, one line for each value of the variable Day. Each group of data for Day requires seven lines. The dataframe has data for 4 days and thus 4*7=28 lines. I can create a plot, but the plot only contains dots. The dots for each day should be connected each day's data by a different line. There should be a total of four lines on the

Dear Askay, I believe my grey hair allows me to help answer your question. SQL, and its progenitor SEQUEL, were developed specifically to manipulate relational databases. It was developed in the early 1970s (equivalent to the historical bronze age) when the concept of a relational database (see https://en.wikipedia.org/wiki/Relational_database) and Codd's 12-rules were being developed (see

I am trying to use the aggregate function to run a function, catsbydat2, that produces the mean, minimum, maximum, and number of observations of the values in a dataframe, inJan2Test, by levels of the dataframe variable MyDay. The output should be in the form of a dataframe. #my code: # This function should process a data frame and return a data frame # containing the mean, minimum, maximum, and

?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 >