similar to: Can't compute row means of two columns of a dataframe.

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

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 <-

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 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

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

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

Colleagues, I have started working with Amelia, with the aim of imputing missing data for time-series data. Although I have succeeded in getting Amelia to perform the imputation, I have not found any documentation describing how Amelia imputes time-series data. I have read the basic Amelia documentation, but it does not address how time-series data are imputed. The documentation describes

Colleagues, I have written an R function (see fully annotated code below), with which I want to process a dataframe within levels of the variable StepType. My program works, it processes the data within levels of StepType, but the usual headers that separate the output by levels of StepType are at the end of the listing rather than being used as separators, i.e. I get Regression results StepType

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

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

"Sexy code" may get a job done and demonstrate the code's knowledge of a programming language, but it often does this at the expense of clear, easy to document (i.e. annotate what the code does), easy to read, and easy to understand code. I fear that this is what this thread has developed "sexy" but not easily understandable code. While I send kudos to all of you, remember

R List, Please excuse this ultra-newbie post. I looked at this page but it's a bit beyond me. https://www2.kenyon.edu/Depts/Math/hartlaub/Math305%20Fall2011/R.htm I'm interested in R construct(s) to be entered at the command line that would output slope, y-intercept, and r-squared values read from a csv or other filename entered at the command line, and the same for standard deviation

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

Dear all; Thank you for your reply. David has explained an interesting method. David I have DEM file of the region and I have extracted the xyz data from that. Also I can extract bathymetry data as xyz file. I have calculated the storage (volume) of reservoir at the current elevation. But the method I have used to calculate the volume is different from your method. I have crop DEM by the reservoir