similar to: emacs and R

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

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 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 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,

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

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

R 2.2 on windows XP I have a dataset with multiple columns. Some of the columns represent independent variables, some represent dependent variables. I would like to run the same analyses on a fixed set of independent variables, changing only the dependent variable, e.g. y1-y2=x1+x2+x3 y3-y4=x1+x2+x3 y5-y6=x1+x2+x3, etc. I know I can write a function to perform the analyses, however in order to

R 2.7.2 Windows XP I am using apply to compute a series of Student's t-test from two matrices, sample1 and sample2. boo<-apply(sample1,1,t.test,sample2) I want to pick of the p-values from the tests, but can't seem to get it to work. I have tried several methods to get the values including: boo<-apply(sample1,1,t.test$t.test,sample2) boo<-apply(sample1,1,t.test,sample2)$t.test

R 2.2.0 windows XP How can I perform a regression analyses using a vector of means, a variance-covariance matrix? I looked at the help screen for lm and did not see any option for using the afore mentioned structures as input to lm. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC, University of Maryland School of Medicine Claude 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 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

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

Fellow Rers, Please forgive me if I have posted this to the wrong R list serve. Over the course of the years that I have used R and participated in this list server, I have noted a large number of questions and answers that direct people to specific packages. The multitude of packages is one of the great strengths of R. Unfortunately there is no (or at least I am not aware of) any single source

Peter You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a

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

Please forgive a statistics question. I know that a simple bivariate linear regression, y=f(x) or in R parlance lm(y~x) can be solved using the variance-covariance matrix: beta(x)=covariance(x,y)/variance(x). I also know that a linear regression with multiple independent variables, for example y=f(x,z) can also be solved using the variance-covariance matrix, but I don't know how to do this.

I think I am doing something wrong when I try to bootstrap R square obtained from lm. My code is included below. No matter how many times I run the simulation, I always get exactly the same result, the bias and std.error are always zero. I would think that these values should be non-zero. I would appreciate any suggestions as to what I am doing wrong, or perhaps what I fail to understand. R 2.1.0

R 2.3.1 Windows XP Question: How can I obtain the values of BIAS and STD. ERROR from a bootstrap. Background: I am running a bootstrap: result2<-boot(1:400,regSEvssample,R=5000) and obtain the following results: Bootstrap Statistics : original bias std. error t1* 1.876602 -0.0001368616 0.1630380 I would like to get the values of ORIGINAL and BIAS. I can get the value of

Given a square matrix of variables X, is there any way to compute a multivariable correlation among all the variables in the array? It is possible to calculate the correlation of all pairs of variables in the array, but I want to know the correlation of all the variables taken together, i.e. for the matrix X=x1 x2 x3 x4 x5 x6 x7 x8 x9 I don't want the pair-wise correlations

Fedora Linux 8 R 2.6.1 Intel CPU I am trying to get started with R under Linux. It has been suggested to me that I run R under EMACS. I have looked at the CRAN website, and several other places to try to find an RPM for EMACS and, if I need it for ESS. I have not succeeded in finding one. Can someone tell me where to go to get the RPMs? If anyone would have a suggestion for an editor other than