similar to: Trying to learn how to write an "advanced" function

Although I owe thanks to Ramus and Ivan, I still do not know how to write and "advanced" function. My most recent try (after looking at the material Ramus and Ivan set) still does not work. I am trying to run the lm function on two different formulae: 1) y~x, 2) y~x+z Any corrections would be appreciated! Thank you, John doit <- function(x){ ds <- deparse(substitute(x))

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

David, Again you have my thanks!. You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. I don't mind if I get these on the log scale; I can get the anti-log. Can you suggest how I can get the point estimate and its SE for "regional"? The predict function will give the point estimate, but not (to my knowledge)

I am trying to get the mean of a row of a data frame. My code follows: roop <- data.frame(x=c(1,2,3),y=c(4,5,2),z=c(0,9,4)) roop mean(roop[1,]) mean(roop[1,c("x","y","z")]) I get the following output: > roop x y z 1 1 4 0 2 2 5 9 3 3 2 4 > mean(roop[1,]) [1] NA Warning message: In mean.default(roop[1, ]) : argument is not numeric or logical:

David, predict.glm and se.fit were exactly what I was looking for. Many thanks! John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax)

David, Thank you for responding to my post. Please consider the following output (typeregional is a factor having two levels, "regional" vs. "general"): Call: glm(formula = events ~ type, family = poisson(link = log), data = data, offset = log(SS)) Deviance Residuals: Min 1Q Median 3Q Max -43.606 -17.295 -4.651 4.204 38.421 Coefficients:

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

> On Oct 22, 2017, at 5:01 PM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > David, > Again you have my thanks!. > You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. There needs to be a reference value for the contrast. Contrasts are differences. I gave you the choice of two references

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 have also had difficulty running R in RStudio. Has anyone else had problems? It will be a shame if we need to abandon R Studio. It is a very good IDE. John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene

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,

And to answer the dependency question. Neither is dependent on the other. But both can be complimentary. If you consider that SQL*may* be a route to accessing your data (if it's in a database). And R *may* be a route to analysis of the data. If the data is in a CSV file, Excel file, API etc. you don't need SQL. IF it is in a database, you might extract it to CSV etc. or you might

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

Hi All, I am trying to import a .sas7bdat file into R. I tried using Hmisc package's sasxport.get() function. temp <- sasxport.get("path\abcd.sas7bdat") I get an error that says "Error in lookup.xport(file) : file not in SAS transfer format" I am not familiar with SAS transfer format. Could somebody please clarify what is it that I am missing, Thanks for your

How would one generate data to be used in a simulation of a repeated measures ANOVA given a known (1) within-person correlation with known (2) mean and SD of data obtained at each of three times of observation? Thanks, John John Sorkin Chief Biostatistics and Informatics Univ. of Maryland School of Medicine Division of Gerontology and Geriatric Medicine JSorkin at grecc.umaryland.edu

??I would like to apply a function, fract, to the columns of a dataframe. I tried the following apply(data5NonEventEpochs,2,fract) but, no surprise it did not work as apply works on matrices not data frames. How can I apply a fuction to the columns of a data frame? (I can't covert data5NonEventsEpochs to a matrix as it contains character data). Thank you, John John David Sorkin M.D., Ph.D.

Can anyone suggest a good non-parametric test, and an R implementation of the test, that allows for two or more factors? Wilcoxon signed rank allows for only one. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of

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

Widows XP R 2.3.1 I have been trying to make a data structure that will contain both the coefficients from a linear regression along with column and row titles AND the call, i.e. myreg<-lm(y~x+y+z) whatIwant<-cbind(c(summary(myreg)$call,"",""),summary(myreg)$coefficients) Neither the statement above, nor any one of twenty variations I have tried work. I would appreciate

Colleagues, I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, Error in yfu n(x,10,20) : object 'x' not found. I hope someone can tell we how I can fix the problem