search for: ranks

...1 0 0 0 1 0 then rank(A)=2 what is this function? thanks I did try help.search("rank"), but all the returned help information seem irrelevant to what I want. I would like to know how people search for help information like this. rank(base) Sample Ranks SignRank(stats) Distribution of the Wilcoxon Signed Rank Statistic Wilcoxon(stats) Distribution of the Wilcoxon Rank Sum Statistic friedman.test(stats) Friedman Rank Sum Test kruskal.test(stats) Kruskal-Wallis Rank Sum Test pair...

I have not been able to find a way to do dense rank in R Here is an example of what I need rank() gives the following 5 rank 1 7 rank 2 7 rank 2 9 *rank 4* but I want 5 rank 1 7 rank 2 7 rank 2 9 *rank 3* * * thanks SS [[alternative HTML version deleted]]

Le 22/01/2018 ? 17:40, Keith O'Hara a ?crit?: > This behavior is noted in the qr documentation, no? > > rank - the rank of x as computed by the decomposition(*): always full rank in the LAPACK case. For a me a "full rank matrix" is a matrix the rank of which is indeed min(nrow(A), ncol(A)) but here the meaning of "always is full rank" is somewhat confusing. Does it

Hi, I have noticed different rank values calculated by qr() depending on LAPACK parameter. When it is FALSE (default) a true rank is estimated and returned. Unfortunately, when LAPACK is set to TRUE, the min(nrow(A), ncol(A)) is returned which is only occasionally a true rank. Would not it be more consistent to replace the rank in the latter case by something based on the following pseudo code ?

...ollow this (the question still relates to the previous posts by me): I have provided full steps to reproduce the symptoms. If it matters, I''m running Ruby 1.8.2, Rails 1.0, and MySQL 5.0.15 on Windows XP Professional. 1. First, the setup MySQL ----- create database testdb; create table ranks (id tinyint unsigned not null auto_increment, description varchar(50) not null unique, primary key(id)); create table customers (id mediumint unsigned not null auto_increment, name varchar(50) not null, phone varchar(25), address varchar(50), rank_id tinyint unsigned not null, primary key(id), for...

Le 23/01/2018 ? 08:47, Martin Maechler a ?crit?: >>>>>> Serguei Sokol <sokol at insa-toulouse.fr> >>>>>> on Mon, 22 Jan 2018 17:57:47 +0100 writes: > > Le 22/01/2018 ? 17:40, Keith O'Hara a ?crit?: > >> This behavior is noted in the qr documentation, no? > >> > >> rank - the rank of x as

Hello, May someone help me with how in R I can rank a vector from highest to lowest. i.e rank 1 (smallest rank) is given to the highest value instead of the usual way that it get's the highest rank. Regards, Jessy [[alternative HTML version deleted]]

Does anyone know why the two rank functions gives different results? I need to use the rank function in a "for" loop, so the sequence to be ranked is given values in the form of part (1). How can I use assignment like in part (1) to get correct ranks as in part (2)? Thank You Part (1) i<-1.94 b<-0.95-i c<-1.73-i d<-2.62-i y<-c(0.68,0.95,b,c,d) y 0.68 0.95 -0.99 -0.21 0.68 rank(y) 3 5 1 2 4 Part(2) rank(c(0.68,0.95,-0.99,-0.21,0.68)) 3.5 5.0 1.0 2.0 3.5

...Thanks, Hal. I'll have a look at CGP to see how this might be done. It's > a little more complicated than just pulling the GEP apart, there needs to > be a loop-invariant-aware reassociation to undo the damage done by the > initial merge. > So, this is in fact, just using the ranks reassociation would normally use, to do splitting :) That is, even outside of geps, you end up with chains of operations where, if you moved the operands around, you would expose loop-invariant calculation. Reassociation builds a rank map ordered by RPO traversal of the CFG, and uses it to place...

Hello, First time posting to this mail list. I'd like to use R in the most efficient way. I'm accomplishing what I want, but feel there is a more R'ish way to do it. Just learning R. *My goal: get ranks of value across rows with row names and column names intact.* I'm guessing one of the [?]apply functions will do what I need, but I couldn't sort out which one (after a lot of searching). Here is a simplified example of what I am doing. Again, I get the correct result, but I assume there...

What is the easiest way to calculate a percent rank “by” an index key? Foe example, I have a dataset with 3 fields: Year, State, Income , I wish to calculate the rank, by year, by state. I also wish to calculate the “percent rank”, where I define percent rank as rank/n. (n is the number of numeric data points within each date-state grouping.) This is what I am

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,

...-value is: p-value = 0.003892, which is different from the p-value computed in the Wilcox test, I am using the following steps to compute the z-value:   diff = c(endprice0 - endprice1) diffNew = diff[diff !=0] diffNew.rank = rank(abs(diffNew)) diffNew.rank.sign <-  diffNew.rank  *  sign(diffNew) ranks.pos <- sum(diffNew.rank.sign[diffNew.rank.sign >0]) = 1819 ranks.neg <- -sum(diffNew.rank.sign[diffNew.rank.sign<0]) = 1751   v = ranks.neg n = 100 z= (v - n *(n+1)/4)/sqrt(n*(n+1)*(2*n+1)/24) = -2.661263 Which p-value should I take for the Wilcox test then?   Hix   the data sets use...

Hey Everyone, Im fresh new in R, and Im supposed to write a code to give me a correlation between two rankings. So I have two ranking lists, which contain file names, e.g.: Ranking list 1: file1.java file3.java file2.java Ranking list 2: fiile2.java file4.java file1.java I need to see how much are these two ranking lists are alike, get a correlation between them. I dont even know where to

Hi, I have a data.frame which is ordered by score, and has a factor column: Browse[1]> wc[c("report","score")] report score 9 ADEA 0.96 8 ADEA 0.90 11 Asylum_FED9 0.86 3 ADEA 0.75 14 Asylum_FED9 0.60 5 ADEA 0.56 13 Asylum_FED9 0.51 16 Asylum_FED9 0.51 2 ADEA 0.42 7 ADEA 0.31 17

...1 22 23 24 25 26 375.0 222.0 68.5 68.5 402.5 222.0 33.5 33.5 465.5 222.0 500.0 402.5 222.0 402.5 222.0 222.0 378.5 222.0 222.0 222.0 222.0 222.0 402.5 222.0 33.5 222.0 As you can see, first 4 values of 'fmodel' are 180.3763, so after running rank(fmodel) I expected the ranks of first 4 to be the same, but they are not. What am I doing wrong? Please let me know. Thanks. [[alternative HTML version deleted]]

>>>>> Serguei Sokol <sokol at insa-toulouse.fr> >>>>> on Mon, 22 Jan 2018 17:57:47 +0100 writes: > Le 22/01/2018 ? 17:40, Keith O'Hara a ?crit?: >> This behavior is noted in the qr documentation, no? >> >> rank - the rank of x as computed by the decomposition(*): always full rank in the LAPACK case. > For a

Dear R users,   I have a dataframe with lot of duplicate cases and I want to delete duplicate ones which have low rank and keep that case which has highest rank. e.g   > df1   cno      rank 1  1342    0.23 2  1342    0.14 3  1342    0.56 4  2568    0.15 5  2568    0.89   so I want to keep 3rd and 5th  cases with highest rank (0.56 & 0.89) and delete rest of the duplicate cases. Could

Could someone give me an idea on how to do rolling ranking, i.e. rank in the moving window of last 100 numbers in a long vector? I tried naive solution like roll.rank<-function(v, len){ r<-numeric(length(v)-len+1) for(i in len:length(v)) r[i-len+1]<-rank(v[(i-len+1):i])[len] r } However, it turns out pretty slow even on my rather able Linux box. For

Hi! Maybe this is a silly question, but I need the column rank (http://en.wikipedia.org/wiki/Rank_matrix) of a matrix and R function 'rank()' only gives me the ordering of the elements of my matrix. How can I compute the column rank of a matrix? Is there not an R equivalent to Matlab's 'rank()'? I've been browsing for a time now and I can't find anything, so any