similar to: Correlation with small N

Dear all, Sorry to post my query once again in the list, since I did not get attention from anyone in my previous mail to this list. Now I make it simple here that please give me a code for find out the columns of a dataframe whose correlation coefficient is below a pre-determined threshold. (For detailed query please see my previous message to this list, pasted hereunder) Thanks and regards,

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Is there a library which is capable of identifying distinct clusters of size n from a series of XY coordinates? Failing this, I'd like to be able to to something like: Using a sliding window of size n along the x-axis I'd like to determine the distance between the center of the points in the window and the closest point outside the window. I

Hi, I have a data.frame with 294 columns and 211 rows. I am calculating correlations between all pairs of columns (excluding column 1) and based on these correlation values I delete one column from any pair that shows a R^2 greater than a cuttoff value. (Rather than directly delete the column all I do is store the column number, and do the deletion later) The code I am using is: ndesc

Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm)

Have got any R-proffessional a starting point for me how i can write me a function which multiply every column with every other column in the data.frame - indenpendent from the dim's . Thanks in advance regards,Christian -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Hi, everyone. I know it may be a basic statistical question. But I can't find a good answer. I have a question raised by one of the reviewers. Factor A expression was strongly correlated with B expression (chi-square) in this series. Prior reports by the same authors showed that B expression strongly correlated with survival (Log-rank). Please provide an explanation why then were the results

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 <-

Under the hood, sapply() is also a loop (at the interpreted level). As is lapply(), etc. -- Bert On Sun, Oct 15, 2023 at 2:34?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > That's very helpful and instructive, thank you! > > Jason Stout, MD, MHS > Box 102359-DUMC > Durham, NC 27710 > FAX 919-681-7494 > ________________________________ > From: John

If one makes the reasonable assumption that Pct is much larger than Cutoff, sorting Cutoff is the expensive part e.g O(nlog2(n) for Quicksort (n = length Cutoff). I believe looping is O(n^2). Jeff's approach using findInterval may be faster. Of course implementation details matter. -- Bert On Mon, Oct 16, 2023 at 4:41?AM Leonard Mada <leo.mada at syonic.eu> wrote: > > Dear

Dear List,   I am new to ROC analysis and the package ROCR. I want to compute the confidence intervals of sensitivity and specificity for a given cutoff value. I have used the following to calculate sensitivity and specificity:   data(ROCR.simple) pred <- prediction(ROCR.simple$predictions, ROCR.simple$labels)   se.sp <- function (cutoff, performance) {     sens <-

Dear Jason, The code could look something like: dummyData = data.frame(Tract=seq(1, 10, by=1), ?? ?Pct = c(0.05,0.03,0.01,0.12,0.21,0.04,0.07,0.09,0.06,0.03), ?? ?Totpop = c(4000,3500,4500,4100,3900,4250,5100,4700,4950,4800)) # Define the cutoffs # - allow for duplicate entries; by = 0.03; # by = 0.01; cutoffs <- seq(0, 0.20, by = by) # Create a new column with cutoffs dummyData$Cutoff

Hello, some misc. cleanup patches for speexdsp, nothing big I'm not sure about how to submit patches, so this is a test balloon :) ultimately, I'd like to fix the FIXED_POINT issue, see http://lists.xiph.org/pipermail/speex-dev/2013-December/008465.html currently, I think the only way to find out how speexdsp has been compiled is to resample some bytes and observe the output; which is

That's very helpful and instructive, thank you! Jason Stout, MD, MHS Box 102359-DUMC Durham, NC 27710 FAX 919-681-7494 ________________________________ From: John Fox <jfox at mcmaster.ca> Sent: Saturday, October 14, 2023 10:13 AM To: Jason Stout, M.D. <jason.stout at duke.edu> Cc: r-help at r-project.org <r-help at r-project.org> Subject: Re: [R] Create new data frame with

Well, here's one way to do it: (dat is your example data frame) Cutoff <- seq(0, .15, .01) Pop <- with(dat, sapply(Cutoff, \(p)sum(Totpop[Pct >= p]))) I think there must be a more efficient way to do it with cumsum(), though. Cheers, Bert On Sat, Oct 14, 2023 at 12:53?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > This seems like it should be simple but I

Dear R helpers: I wonder how to pass more than one argument to the function called by lapply. For example, #R code below --------------------------- indf <- data.frame(id=I(c('a','b')),y=c(1,10)) #I want to add an addition argument cutoff into the function called by lapply. outside.fun <- function(indf, cutoff) { unlist(lapply(split(indf, indf[,'id']),

Dear Jason, I do not think that the solution based on aggregate offered by GPT was correct. That quasi-solution only aggregates for every individual level. As I understand, you want the cumulative sum. The idea was proposed by Bert; you need only to sort first based on the cutoff (e.g. using an ordered factor). And then only extract the last value for each level. If Pct is unique, than you

Dear all (copying the package author), I have a question on the vlmc package. I am trying to model a time series, where each element can take one of 11 values (the result of some clustering). When I run the following command (synthetic data to facilitate self-contained example) I get the following warning: ("alphabet with >1-letter strings; trying to abbreviate") +++ START+++ >

I found that predict.gnls failed with a wierd error message about a non-numeric argument to a binary vector in one of three nearly identical uses. Error in Inh/Ki : non-numeric argument to binary operator (Inh and Ki are arguments to the function used in the formula for the object whose predictions were requested). It turns out that the problem is in getCovariateFormula(). The final line in

I have a question about "deparse" function in R What is the reason that "deparse" use an argument like "width.cutoff" ? Why the maximum cutoff is 500? I was manipulating an R formula and used "deparse". Since the length of user's formula was greater then 500, my code didnt work. thanks Johan johan Faux <johanfaux@yahoo.com> wrote: I have a

This seems like it should be simple but I can't get it to work properly. I'm starting with a data frame like this: Tract Pct Totpop 1 0.05 4000 2 0.03 3500 3 0.01 4500 4 0.12 4100 5 0.21 3900 6 0.04 4250 7 0.07 5100 8 0.09