similar to: [R] bug in unsplit()? (PR#1843)

Hedderik van Rijn <hedderik@cmu.edu> writes: > > Thanks for pointing this out. I think your code should work, although > > I'm slightly uneasy about actually modifying f, so how about > > Just curious, why are you uneasy about that? Does it have side effects? It is largely due superstition, but if you change f before calling split<-() you have split() and

even for people who didn't like the outcome of the US elections this year, it must have been a joy to see all the nice (and not so nice) graphs that were shown. As an exercise, I recreated the map shown on the NY-Times website [ http://www.nytimes.com/packages/khtml/2004/11/03/politics/ 20041103_px_ELECT_GRAPHIC.html ] If you're interested,

Hi, [some background] I have a dataset which describes a number of subjects doing a "scientific discovery". That is, they have to discover the rules underlying a particular domain. To do so, they have to set the levels of 5 variables which leads to a certain outcome. To identify what kind of "experiments" the subjects did, I want to combine these levels into one variable

I'm trying to get Sweave running for automatic report generation, and it seems to run fine when just using verbatim output. However, I've ran into a problem with xtable. I would like to print the following matrix using xtable: > dim(counts) [1] 19 15 All columns are filled with real/integer numbers > 0 and < 1000. Just typing xtable(counts) gives correct LaTeX output, but

Hello everyone, I use the split function splitting with the f function on a 3 columns and more than 100 000 rows data frame. Once it's split I have a list of data frames still with 3 columns and n rows. I manipulate those list elements and get a list of data frames still with 3 columns but less rows. So when I unsplit it, I get an error as I use the same factor function I used to split ( f in

R-devel, Below is a simple example calling split and unsplit on a numeric vector of length 2 where 'f' is c(1,NA). > unsplit(split(c(1,2), c(1,NA)), c(1,NA)) [1] 1 0 I noticed that the call to vector in unsplit gives us 0 as the 2nd element of the result. Is this the intended result, as opposed to NA? Thanks for your help, Jeff -- Jeff Enos Kane Capital Management jeff at

A couple of months ago, probably using an older version of R, R used to run the following code just fine: library("foreign") data.exp1 <- as.data.frame(read.spss("dataDef.sav")) Issuing the same commands now (after starting R using --vanilla), gives me the following behavior: > library("foreign") > x <- read.spss("dataDef.sav") Error

Hello, using the `unsplit()` function with tibbles currently leads to the following error: > mtcars_tb <- as_tibble(mtcars, rownames = NULL) > s <- split(mtcars_tb, mtcars_tb$gear) > unsplit(s, mtcars_tb$gear) Error: Must subset rows with a valid subscript vector. ? Logical subscripts must match the size of the indexed input. x Input has size 15 but subscript `rep(NA, len)` has

In data OME in MASS I would like to extract the first 5 observations per subject (=ID). So I do library(MASS) OMEsub <- split(OME, OME$ID) OMEsub <- lapply(OMEsub,function(x)x[1:5,]) unsplit(OMEsub, OME$ID) - which results in [[1]] [1] 1 1 1 1 1 [[2]] [1] 30 30 30 30 30 [[3]] [1] low low low low low Levels: N/A high low [[4]] [1] 35 35 40 40 45 [[5]] [1] coherent incoherent coherent

I get the sentiment, but this is really just bad coding (on my own part, I suspect), so we might as well just fix it... -pd > On 21 Nov 2020, at 17:42 , Marc Schwartz via R-devel <r-devel at r-project.org> wrote: > > >> On Nov 21, 2020, at 10:55 AM, Mario Annau <mario.annau at gmail.com> wrote: >> >> Hello, >> >> using the `unsplit()`

Perhaps this is the intended behavior, but I discovered that unsplit throws an error when it tries to set rownames of a variable that has no dimension. This occurs when unsplit is passed a list of data.frames that have only a single column. An example: df <- data.frame(letters[seq(25)]) fac <- rep(seq(5), 5) unsplit(split(df, fac), fac) For reference, I'm using R version 2.9.0

Hi everyone, I have already used split() and unsplit() in data frames without problems, but now I’m applying these functions to other data and when using unsplit() I have received the following message: Error in `row.names<-.data.frame`(`*tmp*`, value = c("1", "2", "3", "4", : duplicate ''row.names'' are not allowed In

Dear all, I've got a quite large dataframe (stor) with rows subject and rt (reaction time). I would like to split the reaction times per subject into 6 bins of equal size. Right now, I'm using the following code: bindata <- function(rt) { bindata <- rep(-1,length(rt)) binwidth <- length(rt)/6 bindata[order(rt)[(0*binwidth)+1:(1*binwidth)]] <- 1

Took me a while but I figured out how to put in common values of group means/counts, etc. to do the same thing as egen. lapply with split and then unsplit. Thomas Davidoff Assistant Professor Haas School of Business UC Berkeley Berkeley, CA 94720 phone: (510) 643-1425 fax: (510) 643-7357 davidoff@haas.berkeley.edu http://faculty.haas.berkeley.edu/davidoff [[alternative HTML

Dear R-bug-people I have encountered a problem with "unsplit", which I believe may be caused by a bug in the function. However, unexpericend with bug-reports I apologise if this is barely a user problem rather than a problem within R. The problem occurs if an object is split by several grouping factors with levels not occuring in the data, and using drop = TRUE. This may appear as

After some fiddling around with the tapply command, I discovered that the factors (the INDEX argument) given to tapply must be specified in fastest-cycling first order. The following code shows how I discovered my error: (R version 1.2.1) -o-o-o-o-o- x <- as.data.frame(list(data=c(-9,0,3,1,-9,1,0,-9,0,3,1,-9,1,0), subj=c(rep(1,7),rep(2,7)),

> On Nov 21, 2020, at 10:55 AM, Mario Annau <mario.annau at gmail.com> wrote: > > Hello, > > using the `unsplit()` function with tibbles currently leads to the > following error: > >> mtcars_tb <- as_tibble(mtcars, rownames = NULL) >> s <- split(mtcars_tb, mtcars_tb$gear) >> unsplit(s, mtcars_tb$gear) > Error: Must subset rows with a valid

(only of interest -- maybe! -- to those who followed this thread of a couple of weeks ago) Just for the heckuva it, I compared the timing of Deepayan's unsplit(x,f) solution to my as.vector(do.call(rbind, x)) approach to the query for a list of 3 vectors each of length 1000 (the original toy example was for a list of 3 vectors of length 5). Unsurprisingly, I think, because the unsplit()

Cool - thank you Peter! @Marc: This is really not a tidyverse vs base-R debate and I personally think that they should both work together for most parts. The common environment is still R. But just to give you the full picture I also filed a bug for tibbles (https://github.com/tidyverse/tibble/issues/829). With these two fixes I think that split/unsplit would work for tibbles and users (like me)

Hi, the function outer can not apply a constant function as in the last line of the following example: > xg <- 1:4 > yg <- 1:4 > fxyg <- outer(xg, yg, function(x,y) x*y) > fconstg <- outer(xg, yg, function(x,y) 1.0) Error in outer(xg, yg, function(x, y) 1) : dims [product 16] do not match the length of object [1] Of course there are simpler ways to construct a constant