similar to: Collapse factor levels

Greetings, I have generated several ROC curves and would like to compare the AUCs. The data are cross sectional and the outcomes are binary. I am testing which of several models provide the best discrimination. Would it be most appropriate to report AUC with 95% CI's? I have been looking in to the "net reclassification improvement" (see below for reference) but thus far I can only

I was messing around with some data in R and SAS (the reason is unimportant) fitting a multiple linear regression and got a curious discrepancy. The data set is too big to post, but if someone wants it, I can send it. So, here are the (partial) results: From R: Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 61.11434 1.48065 41.275 < 2e-16 *** sexWomen

Good morning. Novice usR. Here. I am following this string, among many, learning as I go. Quick question please? I thought that perhaps ata.frame was part of the zoo pkg, b/c when I searched it came up in help? However, evidently not or I am not using it properly. Please advise, thank you. x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) x2 <- do.call(rbind,

It looks like you made a copy/paste error below. Your ata.frame should be data.frame. Kevin On 05/04/2018 08:18 AM, Bill Poling wrote: > Good morning. > > Novice usR. Here. > > I am following this string, among many, learning as I go. > > Quick question please? > > I thought that perhaps ata.frame was part of the zoo pkg, b/c when I > searched it came up in

Hi Kevin, There is probably a better way, but it can be done in two steps like this temp <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) temp <- lapply(names(temp), function(n, temp) { temp[[n]]$type <- n return(temp[[n]]) }, temp = temp) do.call(rbind, temp) On Wed, May 2, 2018 at 1:11 PM, Kevin E. Thorpe <kevin.thorpe at utoronto.ca> wrote: > I suspect

I tried to use write.foreign() to export to SAS this morning and got an error. When I looked at the code for writeForeignSAS() I saw this line: dfn < -df which I think should be dfn <- df So, I tried to run update.packages() to see if there was an updated version and got the following result. > update.packages(c("foreign")) Warning message: In list.files(lib) :

Is it an undocumented (at least I missed it if it's documented) feature of the reshape function to do numeric variables followed by character? I ask because that seems to be the case below. > str(rcw) 'data.frame': 23 obs. of 21 variables: $ ICU : int 1 18 17 9 22 19 6 16 25 26 ... $ Q6.RC.1 : chr "SM" "JF" "IW"

I suspect this is pretty easy, but I'm having trouble figuring it out. Basically, I have a list of data frames such as the following example: list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) I would like to turn this into data frame where the list elements are essentially rbind'ed together and the element name becomes a new variable. For example, I would like to turn the

Dear R users, I am not asking questions specifically on R, but I know there are many statistical experts here in the R community, so here it goes my questions: Freedman (1982) propose an approximation of sample size/power calculation based on log-rank test using the formula below (This is what nQuery does): (Z(1-?/side)+Z(power))^2*(hazard.ratio+1)^2 N =

Or add the type column first and then rbind: x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8)) x2 <- do.call(rbind, lapply(names(x), function(z) data.frame(type=z, dat[[z]]))) ---------------------------------------- David L Carlson Department of Anthropology Texas A&M University College Station, TX 77843-4352 -----Original Message----- From: R-help

Hi - I am conducting a meta-analysis and I have a matrix of f-statistics, Ns and dfs from a series of studies that tested for an interaction in a 2x2 anova. I'd like to test whether the 2x2 interaction is significant in the aggregate. Similarly, I have a matrix of chi-square statistics that I'd like to meta-analyze. How can I input these test statistics into the "metafor"

Dear R users, I recently downloaded the library lme4a by svn checkout svn://svn.r-forge.r-project.org/svnroot/lme4. I tried to install the library lme4a by copying the downloaded document in the location where all the R libraries are saved. When I try to load the library, I obtain the message library(lme4a) Error in library(lme4a) : 'lme4a' is not a valid installed package R

Hi, ESS replaces "_" by "<-". How can I switch off this feature? I need to be able to type the underscore Thanks Eryk -- Witold Eryk Wolski Heidmark str 5 D-28329 Bremen tel.: 04215261837

I have two matrices A and B > dim (A) [1] 30380   104 > dim(Bt) [1] 30380    63 I want to  combine both A and B to matrix C where >dim(C) [1] 30380   167 How do I do that ? Regards, Pankaj Barah Department of Biology, Norwegian University of Science & Technology (NTNU) Realfagbygget, N-7491 [[alternative HTML version deleted]]

Hello, TO update R under Ubuntu on my macchine, I downloaded R-2.13.1. When running ./configure, I got the following error message: checking for X... no configure: error: --with-x=yes (default) and X11 headers/libs are not available But I could use X11() function in R-2-10. So where does the problem come from? Cheers, Carol

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

Typo: dat[[z]] should be x[[z]]: x2 <- do.call(rbind, lapply(names(x), function(z) data.frame(type=z, x[[z]]))) x2 type x y 1 A 1 3 2 A 2 4 3 B 5 7 4 B 6 8 David C -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of David L Carlson Sent: Wednesday, May 2, 2018 3:51 PM To: William Dunlap <wdunlap at tibco.com>; Kevin E.

Dear All I can not download R for Linux. I do not know which file I should install? Best Regards, Soheila

Hey all, I learned that using the equals sign "=" to assign objects is generally OK, but will not work in some cases. As I always use "<-" for assignments, I have not encoutered any problems. Could somebody provide an example or explanation, why getting used to "=" is not a good idea? Or is it? Thanks ahead, Berry ------------------------------------- Berry

> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE))) > x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1) > str(x2) 'data.frame': 4 obs. of 3 variables: $ type: Factor w/ 2 levels "A","B": 1 1 2 2 $ x : int 1 2 5 6 $ y : int 3 4 7 8 Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe