similar to: A better way to Rank Data that considers "ties"

Hi, Here i have a matrix like this, OLDMatrix <- X1 X2 X3 ----- ------ ------ 22 24 23 25 27 27 10 13 15 the thing is, im running two function(SUM,COUNT) to get output in another matrix called NEWMatrix NEWMatrix <- c("SUM",colSums(OLDMatrix )) NEWMatrix <- c("COUNT",colSums(!is.na(OLDMatrix

Hello, I am sure I am not the only person with this problem. I have a list with n elements, each consisting of a single column matrix with different row lengths. Each row has a name ranging from A to E. Here is an example: alph[[1]] A 1 B 2 C 3 D 4 alph[[2]] A 1 C 3 D 4 alph[[3]] A 1 D 4 E 5 I would like to create a matrix from the elements in the list with n columns such that the row names

Let's say that I have a bunch of matrices. They look like this (pardon using fruit for examples, my actual data tables are far too enormous): Matrix1 Apples Oranges Pears A 5 6 7 B 5 3 4 C 8 9 10 D 11 13 14 E 15 3 8 F 1 4 5

I have been trying to copy the row names of one matrix to another matrix but having difficulty. The original matrix contains a row name which I would like to replicate in the new matrix. I use the following approach? The two matrices have identical dimensions. rN <- row.names(origMatrix) row.names(newMatrix) <- rN However the new matrix does not take on the labels. I have also tried,

Dear R-users I need to convert a matrix with three columns in a new array with multiple columns. For example, oldmatrix 1 4 5 1 54 52 1 9 43 2 32 5 2 54 6 2 76 6 3 54 54 3 543 7 3 54 6 newmatrix 5 5 54 52 6 7 43 6 6 if the first column have a new value then add a column to the new matrix and the new[i,j] <- old[,3][i] I write this code, but my initial matrix is very

Dear all, i am stuck with a syntax problem. i have a matrix which has about 500 rows and 6 columns. now i want to kick some data out. i want create a new matrix which is basically the old one except for all entries which have a 4 in the 5 column AND a 1 in the 6th column. i tried the following but couldn´t get a new matrix, just some wierd errors:

Hi all, I need to replace missing values in a matrix by 10 % of the lowest available value in the matrix. I've got a function I've used earlier to replace negative values by the lowest value, in a data frame, but I'm not sure how to modify it... nonNeg = as.data.frame(apply(orig.df, 2, function(col) # Change negative values to a small value, close to zero { min.val =

Hi together I'm pretty new to R, so excuse me if it is a basic question. I have a big dataset (extract of it found in the attachment) of returns from firms. I'd like to compute the Pearson correlation of each firm with the "Market" and the corresponding p-Value. So I thought of making a list of 'cor.test's and then extract the needed values with a for loop. What I did so

Hi when I fit a glm by glm.fit(x,y,family = binomial()) and then try to use the object for prediction of newdata by: predict.glm(object, newdata) I get the error: Error in terms.default(object) : no terms component I know I can use glm() and a formula, but for my case I prefer glm.fit(x,y)... thanks for a hint christoph $platform [1] "i686-pc-linux-gnu" $arch [1]

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

Dear all, given I have data in a data.frame which indicate the number of people in a specific year at a specific age: n <- 10 mydf <- data.frame(yr=sample(1:10, size=n, replace=FALSE), age=sample(1:12, size=n, replace=FALSE), no=sample(1:10, size=n, replace=FALSE)) Now I would like to make a matrix with (in this simple example) 10 columns (for the

Hi all~ I have a bit of stupid question. And I promise, I have struggled through this and it sort of pains me to post this question... I have the following vector which is are the estimated parameter values output from "optim". $Rates [1] 0.006280048 0.330934659 I want to match the order in the vector by the number in the following matrix: $Index.matrix ALPHA 1 1 BETA 2

This post is NOT a question, but an answer. For readers please disregard all earlier posts by myself about this question. I'm posting for two reasons. First to say thanks, especially to Dimitris, for suggesting the use of errorest in the ipred library. Second, so that the solution to this problem is in the archives in case it gets asked again. If one wants to run a k-fold cross-validation

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

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

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

I am grateful to Prof. Ripley for his explanation. Indeed, rounding explains it all. I take the difference between two vectors and call it "z" method.a <- c(6.3, 6.3, 3.5, 5.1, 5.5, 7.7, 6.3, 2.8, 3.4, 5.7, 5.6, 6.2, 6.6, 7.7, 7.4, 5.6, 6.3, 8.4, 5.6, 4.8, 4.3, 4.2, 3.3, 3.8, 5.7, 4.1) method.b <- c(5.2, 6.6, 2.3, 4.4, 4.1, 6.4, 5.4, 2.3, 3.2, 5.2, 4.9,

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",