similar to: icon for an R package

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

(1) I know that \n when used in cat, e.g. cat("\n") produces a line feed (i.e. skips to the next line). Is there any escape sequence that will go to the top of the next page? (2) I know that control L will clear the console. Is there an equivalent function or other means that can be used in R code to clear the console? Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics

I am looking for a package that will allow me to choose R (a real number) that minimizes the correlation of y and x^R, i.e. find R such that corr(y,x^R) is minimized. Any suggestions for packages I might look at would be helpful. Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA

??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.

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 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 am trying to learn more about how to write functions. I would like to pass a data frame (or matrix) and depending on the parameters passed to the function work with a given column of the data frame or matrix. My function, learnfn is given below as are two calls to the function. The first call is an attempt to print the x column from the data frame, the second call is an attempt to print the y

I would like to select rows if a row contains any one of several values. I can do the selection as follows: result[,"Subject"]=="JEFF" | result[,"Subject"]=="BG" But this is very unwieldily if one wishes to select many, many rows as one has to continuously repeat the source: result[,"Subject"]=="JEFF" |

I would like to plot 3-dimensional data on a two-dimensional scatter-plot. Is there a way I can automatically modify the plot symbol (e.g. changing size or color) to indicate the value of a third variable? E.g. How can I plot weight vs. age and indicate the value of muscle mass for each value weight-age pair by making the plot point proportional to the subject's muscle mass? Thanks, John

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

R 2.6 Windows XP I need to select from the integers 1,2,3,4,5 with some pre-determined probability, e.g. probability of selecting 5 80%, probability of selecting 1 or 2 or 3 or 4 20%. Any suggestions for how I might accomplish this? I need to do it very efficiently as I will be doing it 500,000 times. Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of

windows XP R 2.10 As pointed out by Prof. Venables and Ripley (MASS 4th edition, p275), the results obtained from lme using method="ML" and method="REML" are often different, especially for small datasets. Is there any way to determine which method is preferable for a given set of data? Thanks, john John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics

I am trying to plot where data points from a give subject are connected by a line. Each subject is represented by a single row of data. Each subject can have up to five observations. The first five columns of mydata give the time of observation, columns 6-10 give the values at each time point. Some subjects have all data, some are missing values. The code I wrote to draw the plot is listed below.

Windows XP R 2.3.1 I have a funciton fit1<-lm(y~x+z) Is there a function that will produce a 3-dimensional plot of y,x,z? I looked at the help files, but did not find a clean answer to my question. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street

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

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)

Is there an R function that can be used to calculate a sample size for a linear regression: i.e. Given a linear regression: y=f(x,z) i.e. fit1<-lm(y~x+z) is there a function which can be passed coef(z), SE(z), and other parameters to determine a sample size based on z, SEz and other information from the regression? Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics

I am trying to drop rows of a dataframe based on values of the column PID, but my strategy is not working. I hope someoen can tell me what I am doing incorrectly. # Values of PID column > jdata[,"PID"] [1] 16608 16613 16355 16378 16371 16280 16211 16169 16025 11595 15883 15682 15617 15615 15212 14862 16539 [18] 12063 16755 16720 16400 16257 16209 16200 16144 11598 13594 15419 15589

Forgive my resending this post. To data I have received only one response (thank you Bert Gunter), and I still do not have an answer to my question. Respectfully, John Windows XP R 2.12.1 contrast package. I am trying to understand how to create contrasts for a model that contatains an interaction. I can get contrasts to work for a model without interaction, but not after adding the

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: