similar to: remove NA rows and columns

Just received the news from Tom Mornini. Congrats Ezra for the new Engine Yard site and the blog you will be collaborating. Hope to read you there soon. http://www.engineyard.com/ Jonathan -------------- next part -------------- An HTML attachment was scrubbed... URL: http://rubyforge.org/pipermail/backgroundrb-devel/attachments/20060925/3f251fa4/attachment.html

I'm trying to test if a correlation matrix is positive semidefinite. My understanding is that a matrix is positive semidefinite if it is Hermitian and all its eigenvalues are positive. The values in my correlation matrix are real and the layout means that it is symmetric. This seems to satisfy the Hermitian criterion so I figure that my real challenge is to check if the eigenvalues are all

Greetings. I've been experimenting with some algorithms for document classification (specifically, a Naive Bayes classifier and a kNN classifier) and I would now like to calculate some inter-rater reliability scores. I have the data in a PostgreSQL database, such that for each document, each measure (there are 9) has three variables: ap_(measure), nb_(measure), and knn_(measure). ap is me

Hello, I am attempting to write a function that assigns a value to a row or column within the matrix that it runs on. I am, however, having trouble accessing the row or column within this dynamic variable. I have looked through the archives (nothing there) and read the rules for posting to this list. Please help me out! Here is a sample of the problem: testMatrix = matrix(data=2, nrow=10,

Dear list, Please find below an example of odd behaviour of the by function. It occurs both under GNU/Linux R 2.6.2 and Windows R 2.7.0alpha. Respective sessionInfo()'s are given below. I hope I do not overlook anything. testFactor <- factor(sample(LETTERS[1:6], size = 42, replace = TRUE)) testMatrix <- matrix(rnorm(42 * 6), nrow = 42) testDf <- as.data.frame(testMatrix) by(data

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

I'm having trouble making contour lines for this attached, sparse dataset (low data:NA ratio!). Is it the high number of NA's, or funny layout of the densities, or something else that's causing this? This is a subsample of the dataset, and I get the same problem when using the full data. The only reason I get any contours is because I'm forcing specific levels. If no levels are

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Thank you all for the useful suggestion. I did some of my homework. library(data.table) DFM <- read.table(header=TRUE, text='obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004',stringsAsFactors = FALSE) DFM DFM$D =as.numeric(difftime(as.Date(DFM$end,format="%m/%d/%Y"),

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Thank you Jeff and All, Within a given time period (say 700 days, from the start day), I am expecting measurements taken at each time interval;. In this case "0" means measurement taken, "1" not taken (stopped or opted out and " -1" don't consider that time period for that individual. This will be compared with the actual measurements taken (Observed-

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

Hi there, I have a question concerning the behaviour of the colouring with levelplot. (I hope, I manage to explain) If I give the parameters "at" and "col.regions" like this: at <- c(1,2,3,4,5,6) col.regions <- c("blue","blue","blue","yellow","yellow","yellow") Which color would have the value 3.5? I would

Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice fit or not. So I have to apply them to observation data. But how does the method know about my observed

Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =

# read.table is NOT part of the data.table package #library(data.table) DFM <- read.table( text= 'obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004 ',header = TRUE, stringsAsFactors = FALSE) # cleaner way to compute D DFM$start <- as.Date( DFM$start, format="%m/%d/%Y" ) DFM$end

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