similar to: merger two 3-d scatter plot

Hi, I recently ran into the problem that I needed a Siegel-Tukey test for equal variability based on ranks. Maybe there is a package that has it implemented, but I could not find it. So I programmed an R function to do it. The Siegel-Tukey test requires to recode the ranks so that they express variability rather than ascending order. This is essentially what the code further below does. After the

I'm sure this is a simple task, but how to do it has escaped me. I have imported data from two separate files (each file contains the results from an information retrieval algorithm) organized into a list. They are organized by File,Query, and Rank (in that order): [[1]] Doc Query Rank 5 1 1 9 1 2 7 1 3 5 2 1 7 2 2 9 2 3 [[2]]

Hi R-users I have a data set. There are 10 products and the numbers of people who ranked the products. The format of the data set is productID rank1 rank2 rank3 rank4 rank5 rank6 rank7 rank8 rank9 rank10 ------------------------------------------------------------------------------------------------------- 1 10 2 3 3 6 4 2 5

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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",

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 > > >

Hopefully this is an easy problem... I'm trying to add a partitioned rank column to a data frame where the rank is calculated separately across a partition by categories, the way you could easily do in SQL. I found this solution in the archives that looked like it might work: http://tolstoy.newcastle.edu.au/R/e11/help/10/09/8675.html The example has a data frame with several car companies,

Hi, I have two ranks of labels (strings) representing user preferences of colors. For instance, here is a simple example with 4 preferences for each user: > rank1 [1] "red" "blue" "green" "black" > rank2 [1] "white" "gray" "black" "blue" How can I compute Kendall's Tau for this scenario? Thanks in

------------------ >>>>> Henric Winell <[hidden email]> >>>>> on Wed, 21 Oct 2015 13:43:02 +0200 writes: > Den 2015-10-21 kl. 07:24, skrev Suharto Anggono Suharto Anggono via R-devel: >> Marius Hofert-4------------------------------ >>> Den 2015-10-09 kl. 12:14, skrev Martin Maechler: >>> I think so: the code above

Suppose I have two columns, x,y. I can use order(x,y) to calculate a permutation that puts them into increasing order of x, with ties broken by y. I'd like instead to calculate the rank of each pair under the same ordering, but the rank() function doesn't take multiple values as input. Is there a simple way to get what I want? E.g. > x <- c(1,2,3,4,1,2,3,4) > y <-

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

This is a general question about Design of experiments. If i am not allowed to post general questions like this here please accept my apologies and ignore the question. I did a DOE with six factors in eight runs. I know i cannot check for interactions using this design, but i tried the interaction plot and it showed me many interactions. After this I foldover the design and ran the 8 runs to learn