similar to: More on ties with rank()

Hello All: I have a vector of data, z > z [1] 0.1 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.5 0.5 0.7 0.7 0.7 0.9 0.9 1.1 [20] 1.1 1.2 1.3 1.4 The first 4 elements have values of 0.1 followed 2 elements with values 0.2. When I invoke rank(z), I expected to get (1+2+3+4)/4 = 2.5 for the first 4 elements in the ranking and (5+6)/2 = 5.5 for elements 5 and 6. But what I do

Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get ?NA? in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached

I have a data frame (gom) or a matrix of trace metal data and some other observations from water column samples taken at sea (e.g., 19 samples (rows), 19 variables) I can calc. the rank individually from each column of the attached object. How can I create a matrix that contains the ranked data for each variable (either 1-19, ties=avg)? For example: >gom<-read.csv ("gomdata.csv")

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

Hello! I need to use Kruskal-Wallis test and post-hoc test (Dunn's test) for my data. But when I searched around, I only found this function: kruskal.test. But nothing for Dunn's test. So I started to write one myself. But I do not know how to count ties in the data frame. I can use for loops but it seems long and unnecessary since the rank function actually knows the ties. So

Hi, Is there a function like rank but that solves the ties by randomly assigning a value (doesn't average ranks of ties). This is what I actually need: I want to make NA all elements of each column in an array that are ranked in a position larger that rankmax for each column. # Say I've got an array b: b<-cbind(c(1:5,5:1),c(1,12,14,2,5,4:8)) #> b # [,1] [,2] #[1,] 1 1

On Tue, Oct 20, 2015 at 10:26 AM, Henric Winell <nilsson.henric at gmail.com> wrote: > Den 2015-10-09 kl. 12:14, skrev Martin Maechler: > I think so: the code above doesn't seem to do the right thing. Consider > the following example: > > > x <- c(1, 1, 2, 3) > > rank2(x, ties.method = "last") > [1] 1 2 4 3 > > That doesn't look right

Dear all, I have the following data frame: > dat V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 AAAACACCCACCCCCCCCCCCCCCCCCCCCCCCC 9.0 18 12.00 18.0 15.0 12.0 6.0 2 1 ACGATACGGCGACCACCGAGATCTACACTCTTCC 18.0 8 12.00 18.0 15.0 12.0 18.0 3 1 ACTACTGCTCCCCCCCCACTCCCCCCCCCCCCCC 15.0 8 12.00 12.0 18.0 12.0 12.0 4 1 ACTTATACGGCGACCACCGAGATCTACACTCTTT 15.0

Marius Hofert-4------------------------------ > Den 2015-10-09 kl. 12:14, skrev Martin Maechler: > I think so: the code above doesn't seem to do the right thing. Consider > the following example: > > > x <- c(1, 1, 2, 3) > > rank2(x, ties.method = "last") > [1] 1 2 4 3 > > That doesn't look right to me -- I had expected > > >

Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately

Hi, I ran into a problem where I actually need rank(, ties.method="last"). It would be great to have this feature in base and it's also simple to get (see below). Thanks & cheers, Marius rank2 <- function (x, na.last = TRUE, ties.method = c("average", "first", "last", # new "last" "random", "max",

Hi, I am very new to R and I've been trying to work through the R book to gain a better idea of the code (which is also completely new to me). Initially I imputed my data from a text file and that seemed to work ok, but I'm trying to examine linear relationships between gdist and gair, gdist and gsub, m6dist and m6air, etc. This didn't work and I think it might have something to do

This will start off sounding very easy, but I think it will be very complicated. Let's say that I have a matrix, which shows the number of apples that each person in a group has. OriginalMatrix<-matrix(c(2,3,5,4,6),nrow=5,ncol=1,byrow=T,dimnames=list(c("Bob","Frank","Joe","Jim","David"),c("Apples"))) Apples Bob 2

Hi, I would like to check which rows of 'types.prev' matrix pop up in 'types', following R in-built procedure. I tried 'match' function but it works in case of the one dimensional vectors. Will appreciate any suggestions. best, robert > types edate K [1,] 20060819 12.5 [2,] 20060819 17.5 [3,] 20060819 22.5 [4,] 20070217 12.5 [5,] 20060617 10.0

On 09/09/2013 05:18 AM, Star Tan wrote: > > At 2013-09-09 05:52:35,"Tobias Grosser" <tobias at grosser.es> wrote: > >> On 09/08/2013 08:03 PM, Star Tan wrote: >> Also, I wonder if your runs include the dependence analysis. If this is >> the case, the numbers are very good. Otherwise, 30% overhead seems still >> to be a little bit much. > I think

Hi all! I tried to perform Shapiro-Wilk test for my sample of 243 values. > Us [1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3 [13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8 [25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8 [37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8 [49]

Dear R users, I'm trying to perform flexible discriminant analysis (fda) with method bruto. I applied the fda function on my training data: bruto.fda <- fda (fda.formula,data=train.data) where fda.formula is: PRES ~ VA_D123 + VA_D124 + VA_D127 + VA_DARU + VA_DCAN + VA_DFON + VA_DLAP + VA_DRID + VA_DRIR + VA_VVEG + VA_WDIN + VA_DIF3 + VA_DIF4 + VA_DIF5 + VA_CAAC + VA_CABC +

Hello ALL: I ran with success the following commands in R getting a file saved ------------------------------------------------------------------------------------ postscript() postscript('~/data/st202/2003/lecture00/lecture00-graph-01.eps', horizontal = FALSE, height = 6, pointsize = 10) hist(trial.outcome.5, breaks = 5, main = '1000 Replications of 5 Trials of a

Hi All: Walter Moreira wrote a small extension module for using the R programming language from within Python. Tim Church's example at http://www.cmat.edu.uy/~walterm/rpy was so compelling, I could not resist installing it on my Linux Mandrake 8.1 box. But I ran into problems. I have installed on Mandrake 8.1 both python and R: R-base-1.4.1-1mdk.i586.rpm

At 2013-09-09 05:52:35,"Tobias Grosser" <tobias at grosser.es> wrote: >On 09/08/2013 08:03 PM, Star Tan wrote: >> Hello all, >> >> >> I have done some basic experiments about Polly canonicalization passes and I found the SCEV canonicalization has significant impact on both compile-time and execution-time performance. > >Interesting. > >>