similar to: scaling-centering a vector using an index

I have to solve a clustering problem. My first step is to determinate the number of clusters, that's why I 'm using the Calinski index ( [tr(b)/(k-1)]/[tr(w)/(k-1)] ) which i try to maximize to have the best number of clusters. A function is already implemented in R to calculate this index : clustIndex(cl,x, index="calinski") where cl is the result of a clustering method ,

Dear All: I would like to center the data in 'x' by 'group'. The following code scale the data and I have not been able to figure out how to change it so I get the centered data. x <- c(1, 2, 3, 4, 5, 6, 7, 8) group <- c(1,1,1,2,2,2,2,2) unsplit(lapply(split(x,group),scale),group) I would appreciate your help. Ashraf

https://bugs.freedesktop.org/show_bug.cgi?id=85352 Bug ID: 85352 Summary: [NVC0/Gallium] Scaling/centering issue in shadertoy shader Product: Mesa Version: git Hardware: Other OS: All Status: NEW Severity: normal Priority: medium Component: Drivers/DRI/nouveau

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()`

Hello, I would like to have any help with the function Kmeans of R.. I use this to do a classification of my data...I have chosen 12 classes but, I have always an error message: Error: empty cluster: try a better set of initial centers So, I don't understand the probleme with this function.. Thank you to help me!! All the Best Clothilde Clothilde Kussener CNRS - CEBC 79360 Villiers en bois

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

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

Hedderik van Rijn <hedderik@cmu.edu> writes: > If the second argument to unsplit is not a simple vector (but a "list > containing multiple lists"), the function seems to have some problems. > > Given a slight modification of the examples in help(split): > > > xg <- split(x,list(g1=g,g2=g)) > > unsplit(xg,list(g1=g,g2=g)) > [1] -0.7877109

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

Hello, I'm trying to get percentiles (PERCENTRANK for excel users) by factor in the following data.frame: myExample <- data.frame(Ret=seq(-2, 2.5, by=0.5),PE=seq(10,19),Sectors=rep(c("Financial","Industrial"),5)) myExample <- na.omit(myExample) Thanks to Patrick I I managed to put together the following lines which does it for the "Ret" column: myecdf

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

Hello all, I'm trying to transform data frames by grouping the rows by the values in a particular column, ordered by another column, then picking the first row in each group. I'd like to convert a data frame like this: x y z 1 10 20 1 11 19 2 12 18 4 13 17 into one with three rows, like this, where i've discarded one row: x y z 1 1 11 19 2 2 12 18 4 4 13 17 I've got a

I'm trying to kick off some of the rust and learn some of the R packages for Rasch modeling. When I tried using the eRm package, I get item difficulty estimates for all my items accept the first (in terms of order) item. #Begin code library(eRm) r.simulation <- sim.rasch(20,100) r.data <- r.simulation$items #eRm results erm.rasch <- RM(r.data) names(erm.rasch) erm.items <-

Dear list, I have two vectors: x <- c("one","two") y <- paste(rep(x,2),"blah") I want to replace all occurrences of each element of x in y with something else, so that y looks like this: y [1] "something else blah" "something else blah" "something else blah" [4] "something else blah" I can do this using a loop: for (

Hello, I discovered SAGA, an interesting free GIS, a few days ago and now, I would like to use it from within R 2.6.2 using the RSAGA package. I read the documentation for this package and thought that I understood it correctly for trying to call some of the SAGA modules. For getting the information on the usage of and arguments required by the SAGA command line "Import Binary Raw

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

Hi, I am running the non-parametric rasch model tests using eRm. I have a reasonably large dataset for this type of exercise (110 items, 248 persons). I run: > allb2=as.matrix(allb) > rsample <- rsampler(allb2, ctr) > t102<-NPtest(rsample, method="T10") #global test, subgroup inv and receive error message as follows. "Error in m[idx1, idx2] <- 1 : subscript out

Dear R Users, I am using the LLTM and the LRSM functions in the eRm package to do repeated measurements where there are 2 measurement points for a list of 10 items. I am trying to get ability estimates but am having trouble. I don't think that it is appropriate to use the pmat function since the person parameters are based on all 20 items. Rather, I think it would be more appropriate to