similar to: Confidence intervals for relative risk

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 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

Hi All, I have a data set wit three size classes (pico, nano and micro) and 12 different sites (Seamounts). I want to produce a table with the mean and standard deviation values for each site. Seamount Pico Nano Micro Total_Ch 1 Off_Mount 1 0.0691 0.24200 0.00100 0.31210 2 Off_Mount 1 0.0938 0.00521 0.02060 0.11961 3 Off_Mount 1 0.1130 0.20000 0.06620 0.37920 4 Off_Mount 1

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

I have a file: x y z 0.025 0.025 1.65775 0.025 0.050 1.62602 0.025 0.075 1.63683 0.025 0.100 1.91847 0.025 0.125 2.00913 0.025 0.150 1.82222 0.025 0.175 1.70901 0.025 0.200 1.39759 0.025 0.225 1.39089 0.025 0.250 1.04762 If I read the file like this: data<-read.table("file.dat") How do I access the vectors x,y,z that are inside the dataframe data? I studied Venables and