similar to: creating a matrix subset based on a threshold cutoff

Hi, i'd like to plot some data that I have with the value on the x axis and freq on the y axis. So, I need to calculate the freq a value is seen within my data vector for example, say i have a vector of data data=c(1,1,1,4,5,5,6) I want values<-c(1,4,5,6) freq<-c(3,1,2,1) in order to enable me to plot this. Sorry, i'm new to R. What is standard

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

Hello, I am trying to use R2WinBUGS to conduct a mixed treatment comparison (MTC) analysis. On the surface, it seems to me that I am following the correct steps: (1) reading the data into R, (2) specifying initial values for the parameters in the model and (3) fitting the model to the data using the bugs() function in R2WinBUGS. However, I get the error message

I am new to this thread. At the risk of presenting something that has been shown before, below I demonstrate how a column in a data frame can be dropped using a wild card, i.e. a column whose name starts with "th" using nothing more than base r functions and base R syntax. While additions to R such as tidyverse can be very helpful, many things that they do can be accomplished simply

My guess: It I clear from the link that they can use the R logo for commercial purposes. The issue is what to do about the "appropriate credit" and "link to the license." How would I do that on a hoodie? Would they need a web address or something? -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of John Fox Sent: Tuesday, March 21, 2023

Hello Avi, while something like d$something <- ... may seem like you're directly modifying the data it does not actually do so. Most R objects try to be immutable, that is, the object may not change after creation. This guarantees that if you have a binding for same object the object won't change sneakily. There is a data structure that is in fact mutable which are environments. For

On 4/6/23 03:49, Steven Yen wrote: > The RStudio list generally does not respond to free version users. I was hoping someone one this (R) list would be kind enough to help me. I don't think that is true. It is perhaps true that you cannot get personalized help from employed staff, but you can certainly submit to the Q&A forum. -- David > > Steven from iPhone > >>

Richard, A slight addition to your code shows an important aspect of R, local vs. global variables: x <- 137 f <- function () { a <- x x <- 42 b <- x list(a=a, b=b) } f() print(x) ________________________________________ From: R-help <r-help-bounces at r-project.org> on behalf of Richard O'Keefe <raoknz at gmail.com> Sent: Sunday,

This is large data set Spencer. What about when the dates change as below: COL1 COL2 Jan-14 1-Aug-16 Feb-14 1-Aug-16 Mar-14 1-Aug-16 Apr-14 1-Aug-16 May-14 1-Aug-16 Jun-14 1-Aug-16 Jul-14 1-Aug-16 Aug-14 1-Aug-16 Sep-14 1-Aug-16 Oct-14 1-Aug-16 Nov-14 1-Aug-16 Dec-14 1-Aug-16 Jan-15 1-Aug-16 Feb-15 1-Aug-16 Mar-15 1-Aug-16 Apr-15 1-Aug-16 May-15 1-Aug-16 Jun-15 1-Aug-16

Richard, I sent my prior email too quickly: A slight addition to your code shows an important aspect of R, local vs. global variables: x <- 137 f <- function () { a <- x x <- 42 b <- x list(a=a, b=b) } f() print(x) When run the program produces the following: > x <- 137 > f <- function () { + a <- x + x <- 42 +

I may be missing something but using the plain old c() combine function seems to work fine: df <- data.frame(left = 1:5, right = 6:10) df.combined <- data.frame(comb = c(df$left, df$right)) df left right 1 1 6 2 2 7 3 3 8 4 4 9 5 5 10 df.combined comb 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 -----Original

On 2017-08-22 1:30 PM, Patrick Casimir wrote: > Dear R Fellows, > > > I Have a dataset( data1) with 2 columns of date showing a class of factor. How to convert them to date? Then compare them, keep the greater date only in a new column. Using as.Date to change the class to Date but the data becomes NA. ????? When I specified a format with the second date, I got the desired

hi Jay, Consult a local statistician. Statistics is not you think is (namely simple computations, R and probably plotting..). regards, vito Jay Zola <jayjay.1988 at hotmail.nl> ha scritto: > Hello, > > > I am medical student, writing a meta-analysis on complication and > reoperation rates after the five most common treatments of distal > radius fractures. I have

Hello, I am medical student, writing a meta-analysis on complication and reoperation rates after the five most common treatments of distal radius fractures. I have been busy with the statistics for months by my self, but find it quite hard since our classes were very basic. Now I want to compare the treatment modalities to see if there are significant differences. Using R I was able to

Again, John, we are comparing different designs in languages that are often decades old and partially retrofitted selectively over the years. Is it poor form to use global variables? Many think so. Discussions have been had on how to use variables hidden in various ways that are not global, such as within a package. But note R still has global assignment operators like <<- and its partner

?library Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Oct 23, 2017 at 11:36 AM, Rene J Suarez-Soto < rene.j.suarez at gmail.com> wrote: > I installed R 3.4.2 (Windows) and noticed that 30 folders are under my

If we use the ROCR package to find the accuracy of a classifier pred <- prediction(svm.pred, testset[,2]) perf.acc <- performance(pred,"acc") Do we?find the maximum accuracy?as follows?(is there a simplier way?): > max(perf.acc at x.values[[1]]) Then to find the cutoff point that maximizes the accuracy?do we do the following?(is there a simpler way): > cutoff.list <-

Fixed 'maxiter' in the help file. Thanks. Please give the original source of that dataset. That dataset is a tiny sample of GUSTO-I and not large enough to fit this model very reliably. A nomogram using the full dataset (not publicly available to my knowledge) is already available in http://biostat.mc.vanderbilt.edu/tmp/bbr.pdf Use lrm, not lrm.fit for this. Adding maxit=20 will

I installed R 3.4.2 (Windows) and noticed that 30 folders are under my R_HOME/library folder. I assume all of these are R base packages. Is this correct? Where can I see a list of current R base packages? Also; are R base packages also in CRAN? and are these packages updated only when a new version of R is released or can the be updated at a different time? Thanks [[alternative HTML version

Hi, You were on the right track using stack(), but you just pass the entire data frame as a single object, not the separate columns: > stack(NamesWide) ? values ? ind 1 ? ?Tom Name1 2 ? Dick Name1 3 ?Larry Name2 4 ?Curly Name2 Note that stack also returns the index (second column of 'ind' values), which tells you which column in the source data frame the stacked values originated