similar to: Find the 50 highest values in a matrix

Dear R-List, I have a dataframe area<-c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10)) type<-c(rep(1:10,5)) a<-rnorm(50) b<-rnorm(50) c<-rnorm(50) d<-rnorm(50) df<-cbind(area,type,a,b,c,d) df area type a b c d [1,] 1 1 0.45608192 0.240378547 2.05208079 -1.18827462 [2,] 1 2 -0.12119506 -0.028078577

Hi, I am new to xslt and I need some help removing the highest value from the 'total time' and then calculating the 'average time' for jmeter reports Thanks, Hema -- View this message in context: http://old.nabble.com/how-ot-remove-the-highest-value-from-total-time-jmeter-xslt-tp26208608p26208608.html Sent from the R help mailing list archive at Nabble.com.

Hi I am trying to evaluate the quality of the CELT codec by using the 0.8.0 testcelt tool to encode and decode the input. I want to test different bitrates and selected the below parameters for 64, 96, 128, 196, 256kB: ./celt-0.8.0/libcelt/testcelt.exe 44100 2 256 46 $1.sw $1-64kb.sw ./celt-0.8.0/libcelt/testcelt.exe 44100 2 192 46 $1.sw $1-96kb.sw ./celt-0.8.0/libcelt/testcelt.exe 44100 2 128

idvalues <- data.frame (ID = c(1, 1, 2, 2, 3, 4, 4, 4, 5, 5), Value = c(0.69, 0.31, 0.01, 0.99, 1.00, NA, 0,1, 0.5, 0.5)) aggregate(Value~ID, data=idvalues, max) ID Value 1 1 0.69 2 2 0.99 3 3 1.00 4 4 1.00 5 5 0.50 -- Best, GG [[alternative HTML version deleted]]

Since the aggregate S3 method for class formula already has got na.action = na.omit, ## S3 method for class 'formula' aggregate(formula, data, FUN, ..., subset, na.action = na.omit) I think that to deal with NA's, it is enough: aggregate(Value~ID, dta, max) Moreover, passing na.rm = FALSE/TRUE is "don't care": aggregate(Value~ID, dta, max, na.rm=FALSE)

http://bugzilla.mindrot.org/show_bug.cgi?id=492 djm at mindrot.org changed: What |Removed |Added ---------------------------------------------------------------------------- Status|NEW |RESOLVED Resolution| |DUPLICATE ------- Additional Comments From djm at mindrot.org 2003-02-12 11:58

No it is one Sun solaris 8 box trying to back up several aix boxes.But apparantly it is not possible to disable the encryption so I was hoping that I could use another -e like ftp or something like that thru rsync. But I cannot seem to get to work. Regards Boris -----Original Message----- From: Jim Kleckner [mailto:jek-rsync@kleckner.net] Sent: 15. januar 2003 19:47 To: Boris Gegenheimer Cc:

Hi, Try to install limma 2.8.1 with: > biocRep <- "http://bioconductor.org/packages/1.9/bioc" > install.packages("limma", repos=biocRep) then try to update it with > repos <- c(biocRep, "http://cran.fhcrc.org") > update.packages(repos=repos) --> it will not get updated (even if there is a 2.9.8 version on CRAN). This is (1)

Dear smart ones, Image a vector 'test' that looks like so: 0 0.5 1.5 2.1 3.1 4 I'm looking for an efficient function that would do something like moddown <- function(test,integer){ for (element %in% test){ if (element%%integer != 0) { "return next lowest integer for which '%%' returns 0" } else { "return element" } } } and

Suppose i want to compute a 95% highest density for a beta distribution beta(a,b) the two end points x1 and x2 shoudl satisfy the following two equations: pbeta(x1,a,b)-pbeta(x2,a,b)=95% dbeta(x1,a,b)=dbeta(x2,a,b) Is there any fast way to compute x1 and x2 in R? [[alternative HTML version deleted]]

I am using the package hdrcde to get the highest density region. I have the data from an unknown distribution. And I used the subroutine hdr from the package to get the highest density region. But I always got a error message. I do not know why. Who can help?Thanks! The codes look like this: j is the data set, > hdr(j) Error in bw.SJ(x) : no solution in the specified range of bandwidths

Hi I have a sorted array ( in ascending order) and I want to find the subscript of a number in the array which is the the next highest number to a given number. For example,if I have 67 as a given number and if I have a vector x=c(23,36,45,62,79,103,109), then how do I get the subscript 5 from x (to get 79 which is the next highest to 67) without using a for loop? Thx -- 'Raghu'

I have a bunch of files named "[identifier1].[identifier2].[yyyymmdd].[hhmmss].txt" and im having trouble updating my read script for the timestamp. I've been using file.exists and a for loop but its highly inefficient and cumbersome. Would there be someway to create an object containing the string of each .txt file in some directory and to sort them by the date and time stamps so

Hi I have a problem for which I would like to know a solution. I have a gene expression data and I would like to choose only lets say top 200 genes that had the highest expression variance across patients. How do i do this in R? I tried x=apply(leukemiadata,1,var) x1=x[order(-1*x)] but the problem here is x and x1 are numeric data , If I choose the first 200 after sorting in descending, so I

Hi, I'm trying to normalize data by fitting a line through the highest density of points (in a 2D plot). In other words, if you visualize the data as a density plot, the fit I'm trying to achieve is the line that goes through the "crest" of the mountain. This is similar yet different to what LOESS does. I've been using loess before, but it does not exactly that as it takes

Please consider the following #Data to use Props<-c("p754921","p754921" ,"p754921","p75506" ,"p75506" ,"p75506","p75506","p75508","p75508","p75508","p75508","p75508") TAZ<-c(38,37,37,171,171,282,171,46,46,169,169,169) Area<-c(109828.04, 128134.71, 46469.57,

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http://bugzilla.mindrot.org/show_bug.cgi?id=492 Summary: Spurious error message from loginrec when attempting to login in with the highest uid for the first time. Product: Portable OpenSSH Version: 3.5p1 Platform: ix86 OS/Version: BSDI Status: NEW Severity: trivial Priority: P2

Hello I am trying to set up a backup server running Solaris 8 with rsync 2.5.5 and ipfilter the latest version. The problem i have is i have about 16 different interfaces that are secured via ipfilter , and i tried running rsync via rsh but ipfilter would not set up a keepstate with rsh which meant i had to open up and that is not acceptable. So what i tried then was via ssh and that worked fine

Assuming your dataframe is in a variable x: > require(dplyr) > x %>% group_by(ID) %>% summarise(maxVal = max(Value,na.rm=TRUE)) On Fri, 2016-04-22 at 13:51 +0000, Saba Sehrish via R-help wrote: > Hi > > > I have two columns in data frame. First column is based on "ID" assigned to each group of my data (similar ID depicts one group). From second column, I