search for: rank2

Dear R groups: I have the data as follows, I want to plot the "Rank1 ~ obs30*Cases" and "Rank2 ~ obs30*Cases" on the same plot as one 3-D scatter plot, how to do that? Any help is highly appreciated. ID obs30 Cases RANK1 RANK2 1 0.03175 63 82 81 2 0.00000 34 1 34 3 0.00000 36 2 41 4 0.00000 54 3 26 5 0.00000 22 4 42 6 0.00746 134 39 32 7 0.00000 2 5 53 8 0.01190 168 46 31 9 0.03012 166...

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", "min")) { nas <- is.na(x) nm <- names(x) ties.method <- match.arg(ties.method) if (is.facto...

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

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 > > > rev(sort.list(x, decreasing = TRUE)) > [1] 2 1 3 4 > Indeed, well spotted, that seems to be correct. > > Henric Winell > ------------------------------...

...1 2 7 1 3 5 2 1 7 2 2 9 2 3 [[2]] Doc Query Rank 4 1 1 5 1 2 9 1 3 8 2 1 5 2 2 7 2 3 I need to rearrange the data so that it is sorted by Query and Document, with columns for rank1 and rank2 (from files 1 and 2, respectively). For example: [[1]] Doc Query Rank1 Rank1 4 1 NA 1 5 1 1 2 7 1 3 NA 9 1 2 3 5 2 1 2 7 2 2 3 8 2 NA 1 9 2 3 NA My goal is to perfor...

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 24 6 8 7 3...

...---------------- >>> 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 >>> >>> > rev(sort.list(x, decreasing = TRUE)) >>> [1] 2 1 3 4 >>> >> >> I...

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,

...ue, 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 to me -- I had expected > > > rev(sort.list(x, decreasing = TRUE)) > [1] 2 1 3 4 > Indeed, well spotted, that seems to be correct. > > Henric Winell >

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 advance, -- S?rgio Nunes

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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 <-

...frame by the combined rank of Res1 & Res2, but > > where weigths are assigned to the importeance av each column. Lets > assume > > that Res1 is twice as important and lower values rank better. > > > > MyRanks<-data.frame > (Rank1=rank(MyFrame[,"Res1"]),Rank2=rank(MyFrame[,"Res2"]),CombR=2*rank(MyFrame[,"Res1"])+rank(MyFrame[,"Res2"]), > row.names=c("G1","G2","G3","G4","G5")) > > > > Rank1 Rank2 CombR > > G1 2 1 5 > > G2 5 5...