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

?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 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,

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

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

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

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

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

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

On 2/23/24 16:28, Sorkin, John wrote: > David, > My apologies regarding the format of my email. I am replying using my > iPhone, and I can?t find a way to switch from what I suspect is html > to txt format. > The link you sent told me that R tools allows compilation of code. It's specifically designed to provide the code tools missing in Windows that would other wise have

An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are

I am using read.csv to read a CSV file (produced by saving an Excel file as a CSV file). The columns containing dates are being read as factors. Because of this, I can not compute follow-up time, i.e. Followup<-postDate-preDate. I would appreciate any suggestion that would help me read the dates as dates and thus allow me to calculate follow-up time. Thanks John John Sorkin M.D., Ph.D. Chief,

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John

I want to place four plots on a page, and I would like to have all four plots share a common title. I have tried the following code, but the title is centered over the fourth graph and not centered across all four plots. Does anyone have any suggestions? R 2.1.1 windows xp oldpar<-par(mfcol =c(1,4),ask=TRUE) plot(p,varp) plot(p,SEp) plot(p,CVp) plot(p,ppval) title(paste("P and 95%CI

I am trying to model data in which subjects are followed through time to determine if they fall, or do not fall. Some of the subjects fall once, some fall several times. Follow-up time varies from subject to subject. I know how to model time to the first fall (e.g. Cox Proportional Hazards, Kaplan-Meir analyses, etc.) but I am not sure how I can model the data if I include the data for those

I am having trouble selecting rows of a dataframe that will be included in a regression. I am trying to select those rows for which the variable Meno equals PRE. I have used the code below: difffitPre<-lm(data[,"diff"]~data[,"Age"]+data[,"Race"],data=data[data[,"Meno"]=="PRE",]) summary(difffitPre) The output from the summary indicates that

I have written an R function that produces multiple graphs. I use par(ask=TRUE) to allow for the inspection of each graph before the next graph is drawn. I am looking for a way to recall all graphs drawn in an R session, and a method that can be used to print all the graphs at one time. I know that I could simply print each graph after I inspect the graph, but this gets tiresome if one's

Water engineer here. The standard approach is to 1) get the storage vs. elevation data from the designers of the reservoir or, barring that, 2) get the bathymetry data from USBR or state DWR, or, if available, get the DEM data from USGS if the survey was done before the reservoir was built or 3) get a boat+sonar with GPS? +lots of time and survey the bottom elevation yourself. Put the xyz