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Dear Wiza[R]ds, I have a data.frame header that looks like this: v2FfaPre15 v2FfaPre10 v2FfaPre5 v2Ffa2 v2Ffa3 v2Ffa4 I need it to look like this, 15 10 5 2 3 4 i.e., with v2FfaPre and v2Ffa stripped off Any suggestions, Thanks in advance! -- Oscar Oscar A. Linares, MD Translational Medicine Unit LaPlaisance Bay, Bolles Harbor Monroe, Michigan 48161

??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.

Dear Rexperts, I am trying to add a '+' identifying the mean in a boxplot using the following sizelist <- split(size, grp) centers <- boxplot(sizelist, style.bxp = "att", medpch = "o", ylab = "Prostate Volume (cm3)") points(centers, unlist(lapply(sizelist, mean)), pch = "+") But, I get error Error in xy.coords(x, y) :

Dear Wiza[R]ds, I have the following data in a data.frame. I need to do an anova with multiple comparison but I don't know how to factor the groups for analysis. There are 3 groups, 1,2 and 3 labelled in column 1. Help appreciated with thanks in advance. Group SI Sif SG Io I2 lol2 1 9.08e-05 9.08e+00 0.060842287 1.798556446 32.574500

Dear Wiz[R]ds, I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun, and others. We are almost there. For whatever reason, I can't change the symbol from a circle to a triangle in the upright posture plots. Any ideas? I have included the problem in full. # tritiated (3H)-Norepinephrine(NE) disappearance from plasma # concentrations supine and upright # supine datasu <-

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,

Windows 10 64-bit, R-Studio, R version 3.4.3 Several questions relating to groupedData: (1) I am trying to create a groupedData object that can be used to run an analysis that I have been able to urn using lmer. When I include the interaction terms in the groupedData opbject I get an error message stating that + is not meaningful for factors. How do I include factors in my model? See code and

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, 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)

Windows 7 (local computer) Windows server (server I am trying to reach) I need to read a file whose windows path is of the form \\Theserver\mydirectory\data.csv You will note that as per windows standards the server name is preceded by two backslashes. I am not sure how to specify this in R. One usually needs to specify an escape characters in a path new which would suggest my path should

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:

> 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 am interested in finding an R version of SAS "Proc Varclus". SAS's Proc Varclus implements an oblique cluster analysis based on principal components. How can I find out if R has a package that runs the same algorithm implemented in SAS "Proc Varclus"? Thank you, Mary Helen Black __________________________ Mary Helen Black, M.S. Keck School of Medicine of USC

Yes, keep the $(lsb-release -cs) there. This allows you to change the version of the OS without having to change anything else (for instance we are migrating all of our docker to 24.04 LTS right now, so we only have to modify the FROM entry and all of our Dockerfile will still work). From: Gerber, Lauren J <lauren.gerber at helsinki.fi> Date: Thursday, July 11, 2024 at 2:21?AM To:

Jeff, Perhaps I was sloppy with my notation: I want groups >=0 <10 >=10 <20 >=20<30 ...... >=90 <100 In any event, my question remains, why did the four different versions of cut give me the same results? I hope someone can explain to me the function of include.lowest and right in the call to cut. As demonstrated in my example below, the parameters do not seem to alter

How can I get the p values from a glm ? I want to get the p values so I can add them to a custom report fitwean<- glm(data[,"JWean"]~data[,"Group"],data=data,family=binomial(link ="logit")) summary(fitwean) # This lists the coefficeints, SEs, z and p values, but I can't isolate the pvalues. names(summary(fitwean)) # I see the coefficients,

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

Dear John You are only allowed to have the primary covariate on the left hand side of the vertical bar. Other covariates go in inner or outer. Michael On 22/03/2018 12:59, Sorkin, John wrote: > Windows 10 64-bit, R-Studio, R version 3.4.3 > > > Several questions relating to groupedData: > > (1) I am trying to create a groupedData object that can be used to run an analysis

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,

Dear Rxperts, Running the following code: ======================================================= twlo=10; twhi=20; wt=154; vd=0.5; cl=0.046; tau=6; t=3; F=1; wtkg <- wt/2.2 # convert lbs to kg vd.pt <- wtkg * vd # compute weight-based vd (L) cl.pt <- wtkg * cl # compute CL (L/hr) k <- cl.pt/vd.pt # compute k (hr^-1) cave <-