search for: 0.33333

I have the joint distribution of three discrete random variables z1, z2 and z3 which is captured by "z" and "prob" as described below. For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. Also, the probability adds up to 1. > head(z) z1 z2 z3 [1,] -0.46667 -1.0000 -1.0000 [2,] -0.33333 -0.9333 -0.9333 [3,] -0.20000 -0.8667 -0.8667

Most likely, previous computations have ended up giving slightly different values of say 0.13333. A pragmatic way out is to round to, say, 5 digits before applying unique. In this particular case, it seems like all numbers are multiples of 1/30, so another idea could be to multiply by 30, round, and divide by 30. -pd > On 28 Jul 2017, at 17:17 , li li <hannah.hlx at gmail.com> wrote:

I would like to create a tree model with at most one split point per variable using tree, rpart or other routine. Its OK if a variable enters at more than one node but if it does then all splits for that variable should be at the same point. The idea is that I want to be able to summarize the data as binary factors with the chosen split points. I don't want to have three level or more

Hello all and thanks in advance for any advice. I am very new to R and have searched my question but have not come up with anything quite like what I would like to do. My problem is: I have a data set for individuals (rows) and values for behaviours (columns). I would like to know the proportion of shared behaviours for all possible pairs of individuals. The sum of shared behaviours divided by

Hi, Today I was flabbergasted to see something that looks like a rounding error in the very basic seq function in R. > a = seq(0.1,0.9,by=0.1) > a [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 > a[1] == 0.1 [1] TRUE > a[2] == 0.2 [1] TRUE > a[3] == 0.3 [1] FALSE It turns out that the alternative > a = (1:9)/10 works just fine. Are there any good guides out there on how to deal

Hi all- I am trying to use the time function for ts class objects and do not understand the return value. I want to use it to set up a time trend in arima fits. It does not seem to return a correct linear sequence that matches the underlying time series. I am running: R version 2.8.1 (2008-12-22). For example: R> ## create a time series R> x <- rnorm(24) R> (xts <-

Hi, I want to apply ^ operator to a vector but it is applied to some of the elements correctly and to some others, it generates NaN. Why is it not able to calculate -6.108576e-05^(1/3) even though it exists? tmp [1] -6.108576e-05 4.208762e-05 3.547092e-05 7.171101e-04 -1.600269e-03 > tmp^(1/3) [1] NaN 0.03478442 0.03285672 0.08950802 NaN > -6.108576e-05^(1/3) [1]

I have a log table which has a column recording the number of hours worked on a task. So there''ll be multiple lines for a particular task. I want to retrieve the total number of hours worked on a task, so that I can include the information when one views the task details. I can write a sql query to do this easily enough, like "SELECT SUM(hours) AS total FROM tasklog WHERE

A non-text attachment was scrubbed... Name: not available Type: text Size: 3126 bytes Desc: not available Url : https://stat.ethz.ch/pipermail/r-help/attachments/19991130/5cb00c0f/attachment.pl

Hi all, First post for me here, but I have been reading on the forum for almost two years now. Thanks to everyone who contributed btw! I have a dataset of 4000 observations of count of a mammal and I am trying to predict abundance from a inflated-zero model as there is quite a bit of zeros in the response variable. I have tried multiple options, but I might do something wrong as every