similar to: permutation test for PLS/PLSDA

Hi, I am quite new to R and trying to analyze the following data. I have 28 controls and 25 patients. I measured X values of 4 different locations (A,B,C,D) in the brain image of each subject. And X ranges from 0 to 1. I think "control or patient" is a between subject factor and location is a within subject factor. So, controls: 28 patients: 25 (unbalanced data set) respone measure:

Hi all, I'm trying to do model reduction for logistic regression. I have 13 predictor (4 continuous variables and 9 binary variables). Using subject matter knowledge, I selected 4 important variables. Regarding the rest 9 variables, I tried to perform data reduction by principal component analysis (PCA). However, 8 of 9 variables were binary and only one continuous. I transformed the data by

Hi, I'm trying to calculate 95% confidence interval of C statistic of logistic regression model using rcorr.cens in rms package. I wrote a brief function for this purpose as the followings; CstatisticCI <- function(x) # x is object of rcorr.cens. { se <- x["S.D."]/sqrt(x["n"]) Low95 <- x["C Index"] - 1.96*se Upper95 <- x["C

Hello, I would like to do a partial least square discriminant analysis (PLSDA) in R using the package "ropls" Which is in R available via the R command : source("https://bioconductor.org/biocLite.R") When I try to do a PLSDA using my own data. The impact of two genders (AP,C) on 5 compounds measured in persons (samples) should be illustrated. When I try to do a PLSDA I get

Hello, I would like to do a partial least square discriminant analysis (PLSDA) in R using the package "ropls" Which is in R available via the R command : source("https://bioconductor.org/biocLite.R") I try to do a PLSDA to illustrate the impact of two genders (AP,C) on 5 compounds measured in persons (samples) should be illustrated. When I try to do a PLSDA I get the warning

Hello, I would like to perform a Partial least square discriminate analysis (PLSDA) in R. To do this I use the package mixOmics. I could perform the PLSDA in R. however I would also like to perform a leave-one-out cross validation in order to assess the performance of my model. My supervisor told me that I should focus on the R2/Q2 ratios. However when I read the instruction for running the

Hi. I was used to draw 2D plot to show distribution of my samples. It was very powerful in small samples. Recently , I handle a lot of type of samples as more 15. so, I want to use more dimention. I was used that I conduct plsda function and draw biplot, such as, p1=plsda(mw, sp) biplot(p1) How can I draw 3D graph with my data?? I don't know exactly numerical value each spot. please,

Hi guRus, My discipline (experimental psychology) is gradually moving away from Null Hypothesis Testing and towards measures of evidence. One measure of evidence that has been popular of late is the likelihood ratio. Glover & Dixon (2005) demonstrate the calculation of the likelihood ratio from ANOVA tables, but I'm also interested in non-parametric statistics and wonder if anyone has any

Hi, I have optimized the shrinkage parameter (GCV)for ridge and got my r2 value is 70% . to check the sensitivity of the result, I did permutation test. I permuted the response vector and run for 1000 times and draw a distribution. But now, I get r2 values highest 98% and some of them more than 70 %. Is it expected from such type of test? *I was under impression that, r2 with real data set

Hi, I am currently attempting to write a small program for a randomization test (based on rank/combination) for matched pairs. If you will please allow me to introduce you to some background information regarding the test prior to my question at hand, or you may skip down to the bold portion for my issue. There are two sample sizes; the data, as I am sure you guessed, is matched into pairs and

I all, is there a way with R to perform an exact permutation test to replace the wilcoxon test to compare paired series and/or to perform pairwise multiple comparisons for related series after a Friedman test ? Thanks Gilles

R users, Is it possible to do 'exact' permutation tests in R? If I run a perm.test or permutation.test.discrete on a data set that has, for example, 720 possible permutations, will that test permute ALL of those possible permutations, or will it just permute 720 random permutations (thus allowing some possibilities to repeat and some not to be permuted at all)? I'm a little confused

I am looking for an algorithm (written in R (preferably) or C, but even pseudo-code in a text book maybe fine) to determine the sign of a permutation. What is that? Well, a permutation is either even or odd, the sign is +1 or -1, respectively, see, e.g., http://en.wikipedia.org/wiki/Signature_of_a_permutation which also says >> In practice, in order to determine whether a given

Hi, 1/ I wonder why a anova.lme on a single lme object does not print the sum of squares (as expected from the help: "a data frame with the sums of squares, numerator degrees of freedom, denominator degrees of freedom, F-values, and P-values"). Example: > fm2 <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1) > anova(fm2) numDF denDF F-value p-value

I saw the manual of this function but not sure what to do. I have a array contain 7 numbers and want to choose 4 to do permutation test. But using this function with parameters as onet.permutation(scores,4), it returns 0. Instead, with no parameter, onet.permutation(), it returns something but every time it's different. Please tell me how am I supposed to do it. Thanks a lot -- View this

Insightful are pleased to announce we are now taking bookings for our latest course on "Bootstrap methods and permutation tests" in the UK on 23rd - 24th October. This course will focus in particular on two resampling methods, bootstrapping and permutation tests, which have been applied successfully to areas of statistical modelling where "traditional" standard errors,

Dear R-list members, This is in fact a question about statistics, not directly about the R software. The coin package, for conditional inference through permutation methods, has as it main function the function independence_test. One of its parameters is teststat, about which the package documentation says: teststat: "a character, the type of test statistic to be applied: either a

Hello, Like David said, what you are trying to do with sort() can be done with order() in a much easier way. First, your code x <- sort(c("bc","ac","dd"), index.return=TRUE) Now, with function order() i <- order(c("bc", "ac", "dd")) y <- c("D","E", "F")[i] y #[1] "E" "D"

I am perfectly aware that this question is not an R question, at least not yet, but I have not succeeded in finding what I want in other ways, so ... What I am looking for are two algorithms, preferabley in Pascal, but other languages may do. For (a) systematic (complete) permutations for grouped data with unequal group sizes, and (b) random permutations for the same kind of data. I know

> On May 22, 2018, at 10:57 PM, John <miaojpm at gmail.com> wrote: > > Thanks, David. > I got the answer from the web. > Is there any easy way to permute a set (e.g., a set of characters) by the permutation it returns? Thanks, > > > > x <- c(10,7,4,3,8,2) > > sort(x, index.return=TRUE) > $x > [1] 2 3 4 7 8 10 > > $ix > [1] 6 4 3 2