search for: 2.00000

Dear All, In R 3.6.3 (and earlier), method dispatch used to work for methods stored in local environments that are attached to the search path. For example: myfun <- function(y) { out <- list(y=y) class(out) <- "myclass" return(out) } print.myclass <- function(x, ...) print(formatC(x$y, format="f", digits=5)) myfun(1:4) # prints: [1]

hey what should I do to replace in a data frame NA?s with zeroes? Thanks in advance Juan Pablo

Dear Sir, I get an error message when I use polr() in MASS package. My data is "ord.dat". I made "y" a factor. y y1 y2 x lx 1 0 0 0 3.2e-02 -1.49485 2 0 0 0 3.2e-02 -1.49485 3 0 0 0 1.0e-01 -1.00000 4 0 0 0 1.0e-01 -1.00000 5 0 0 0 3.2e-01 -0.49485 6 0 0 0 3.2e-01 -0.49485 7 1 1 0 1.0e+00 0.00000 8 0 0 0 1.0e+00 0.00000 9 1 1 0

Dear R users, I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:  Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable my response variable rl <-

Dear list, I'm currently trying to use the rms package to get predicted ordinal responses from a conditional ratio model. As you will see below, my model seems to fit well to the data, however, I'm having trouble getting predicted mean (or fitted) ordinal response values using the predict function. I have a feeling I'm missing something simple, however I haven't been able to

So, trying to convert a very long, somewhat technical bit of lin alg MATLAB code to R. Most of it working, but raninto a stumbling block that is probaably simple enough for someone to explain. Basically, trying to 'line up' MATLAB results from an element-wise division of a matrix by a vector with R output. Here is a simplified version of the MATLAB code I'm translating: NN = [1,

Hi, is there a way I can calculate a summary statistics for a columns matrix let say we have this matrix x <- matrix(sample(1:8000),nrow=100) colnames(x)<- paste("Col",1:ncol(x),sep="") if I used summary summary(x) i get the output for each column but I need the output to be in matrix with rownames and all the columns beside it this how I want it

Hi all, I was conducting a meta-analysis of single proportions(i.e. without a control group) using the metafor package. When I performed a classic fail-safe N, I noticed that the result (the number of missing studies that would bring p-value to the alpha, to be exact)was different than that I got in Comprehensive Meta-Analysis Version 2.0. I wonder why R and CMA got different results. *Below is

Dear Wolfgang, I think this new behaviour is related to the following R 4.0.0 NEWS item: > S3 method lookup now by default skips the elements of the search path between the global and base environments. Your environment "myenv" is attached at position 2 of the search() path and thus now skipped in S3 method lookup. I have just noticed that

Dear Rusers, The default behavior in R when performing a regression model with missing values is to exclude any case that contains a missing value? How could i set the bahavior that R deal with missing values? e.g.: exclude cases listwise exclude cases pairwise replace with mean Thanks very much! -- Kind Regards, Zhi Jie,Zhang ,PHD Department of Epidemiology School of Public Health Fudan

Dear all, a program that worked well for weeks today gave me consistently the error message no lines available in input referring to the lines for (i in (0:(timeintervals-1))) { j=subjectquantity+6+i*(subjectquantity+7) print (j) results<-read.table(file, header=F, skip=j, nrows=subjectquantity) timeintervals have been specified as argument as 12 subjectquantity as 9 the file it should

Hi, I have generated a profile likelihood for a parameter (x) and am trying to get 95% confidence limits by calculating the two points where the log likelihood (LogL) is 2 units less than the maximum LogL. I would like to do this by linear interpolation and so I have been trying to use the function approxfun which allows me to get a function to calculate LogL for any value of x within

Hi, I am not sure if this is just me using R (R-2.3.1 and R-2.4.0) in the wrong way or if there is a more serious bug. I was having problems getting some calculations to add up so I ran the following tests: > (2.34567 - 2.00000) == 0.34567 <------- should be true [1] FALSE > (2.23-2.00) == 0.23 <------- should be true [1] FALSE > 4-2==2 [1] TRUE > (4-2)==2 [1] TRUE >

I would suggest to post this to the (recently created) R-sig-meta-analysis mailing list. See: https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis Best, Wolfgang >-----Original Message----- >From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Naike Wang >Sent: Monday, June 26, 2017 08:32 >To: R-help at r-project.org >Subject: [R] Classic fail-safe N >

> t(t(NN)/lambda) [,1] [,2] [,3] [1,] 0.5 0.6666667 0.75 [2,] 2.0 1.6666667 1.50 > R matrices are column-based. MATLAB matrices are row-based. > On Feb 27, 2024, at 14:54, Evan Cooch <evan.cooch at gmail.com> wrote: > > So, trying to convert a very long, somewhat technical bit of lin alg > MATLAB code to R. Most of it working, but raninto a stumbling block

I am playing with the a 1-way anova with and without the "-1" option. I have a simple cooked up example below but it behaves the same on a more complex real example. From what I can tell: 1) the estimated means of the different levels are correctly estimated either way (although reported as means with the -1 and as contrasts without the -1 as expected) 2) the residuals are

Why anything but sweep? The fundamental data type in Matlab is a matrix... they don't have vectors, they have Nx1 matrices and 1xM matrices. Vectors don't have any concept of "row" vs. "column". Straight division is always elementwise with recycling as needed, and matrices are really vectors in row-major order: 1 2 3 4 5 6 is really 1 4 2 5 3 6 and when you do

Hello, I have a table that was constructed in a wrong way (dput data on bottom - wrong data-frame): Local Mês Dia Colonia X6h X7h X8h X9h X10h X11h X12h X13h X14h X15h X16h X17h 1 Conceição Junho 1 3 2.16137 2.20412 2.08991 1.72428 1.69897 1.62325 1.44716 1.51851 1.43136 1.47712 1.51851 1.04139 2 Conceição Junho 2 3 2.46538 2.13672

Wolfgang, It is common to handle relative risk problems using Poisson regression. In your example you have 8 events out of 508 tries, and 0/500 in the second data set. > tdata <- data.frame(y=c(8,0), n=c(508,500), group=1:0) > fit <- glm(y ~ group + offset(log(n)), data=tdata, family=poisson) Because of the zero, the standard beta/se(beta) confidence intervals don't work.

Hello, everyone, I'm using R to deal with a REML problem. I found "lmer" is the right function for this. But I got stuck because I couldn't interpret the result. I'm attaching a short example of my executing log. Please have a look and give me some advice on it. Thanks a lot! Plot Block Treatment Data 1 1 2 7.8 2 1 1