similar to: Need to aggregate large dataset by week...

I cannot understand the following error printed out when I try to get the extrema of my time series. I would appreciate some suggestion as I really cannot interpret the error. I might not be using a proper set of parameters in calling such functions. I am learning by doing ... > aa.peak <- wavCWTPeaks (aa.tree) Error in `row.names<-.data.frame`(`*tmp*`, value = c("1",

Dear list, I am trying to create a three-way table with percent occurrence instead of raw frequencies. However, I cannot get the results I expected: I have the following table: > ftable(table( mannerDF$agem, mannerDF$target, mannerDF$manner )) <snip> 50 bak 0 0 0 0 1 0 pak 0 0 0 0 3 0 sak

Hi, I have the precision values of a system on two different data sets. The snippets of these results are as shown: sample1: (total 194 samples) 0.6000000238 0.8000000119 0.6000000238 0.2000000030 0.6000000238 ... ... sample2: (total 188 samples) 0.80000001 0.20000000 0.80000001 0.00000000 0.80000001 0.40000001 ... ... I want to check if these results are statistically significant? Intuitively,

Hi All, I have a function that is used with data frames having multiple id's per row and it aggregates the data down to 1 id per row. It also randomly selects one of the within-id values of a variable (mod), which often differ within-id. Assume this data frame (below) is much larger and I want to repeat this function, say 100 times, and then derive the mean values of r over those 100

Hello, I have a dataframe (tab separated file) which looks like the example below - two values separated by a comma, and tab separation between each of these. [,1] [,2] [,3] [ ,4] [1,] 0,1 1,3 40,10 0,0 [2,] 20,5 4,2 10,40 10,0 [3,] 0,11 1,2 120,10 0,0 I would like to calculate the percentage of the smallest number separated by the comma by: 1) summing the values e.g. for

maybe it's in the data? So here it comes. > sv.growth Grouped Data: length ~ meas | box_id meas spec comp water box_id sprouts leaves length long.sprout 1 1 Sv control moist 1 8.800000 37.80 211.2000 60.6 2 1 Sv xfull moist 2 7.000000 8.00 174.8000 62.8 3 1 Sv control moist 3 9.000000

Dear All, I would like to ask a couple of questions on a LME model. I tested 4 selection lines at 4 food concentrations against a standard competitor stock. I had 3 replicate cages per selection line. In each cage I have 10 vials. I counted the number of wild type flies and competitor stock emerging in each vial. My main question is: is there any difference between selection lines? I did fit

Dear all, I have a matrix with positive and negative values. >From this I would like to produce 2 matrices: 1st - retaining positives and putting NA in other positions 2nd - retaining negatives and putting NA in other positions and then apply rowMeans for both. I am trying to use the function ifelse in the exemplified form: ifelse(A>0,A,NA) but by putting A as a 2nd parameter it

Hi, I am using package deSolve to run some ordinary differential equations (ODE) as part of a mathematical modeling project. I have solved for the following equilibrium states: Seq1<-a*(1-Neq1)/(f*Veq1+m+d) Ceq1<-(f*Seq1*Veq1+g*Ieq1+r*(1-Neq1)-b1*Veq1*Ieq1)/(b2+m+d+g) Ieq1<-(-b2*Ceq1)-r*(1-Neq1)/(b1*Veq1-g-u) Veq1<-o*(Ceq1+Ieq1)/e I want to take the differential of both sides of

Hi, I appologise if this is a rudimentary question and long winded but I just wanted to let ye know where I'm comming from. I'm new to R and I'm trying to use the 'randomForest' package to classify and predict. The Error message that is troubling me is: > pr<-predict(predictors,rf1, ext=ext) Error in x[...] <- m : NAs are not allowed in subscripted assignments In

I am caught in a mental trap. Why isn't the between groups variance estimated (0.0038) to be around the value with which I generated the data (0.0002)? Thanks Toby set.seed(76589437887) fph = 0.4 Sigh = sqrt(0.0002) Sigi = sqrt(0.04) ci = 1 fpi = matrix(,7200,3) for (i in 1:90) { fph = rnorm(1, fph, Sigh) for (k in 1:80) { fpi[ci,1:3] = matrix(c(i, k, rnorm(1, fph, Sigi)),1) ci

R gurus: I'm trying to get another round of rconifers out and I need some advice/help crushing differences in the examples test. I'm trying to make sure the max sdi values are being respected. I've added a tests/rconifers-Ex.Rout.save (from windows i386-pc-mingw32) and when I ran R CMD check (both R-2.13.0), I got the following results: * using log directory

Fellow R Users: I'm not extremely familiar with lda or R programming, but a recent editorial review of a manuscript submission has prompted a crash cousre. I am on this forum hoping I could solicit some much needed advice for deriving a classification equation. I have used three basic measurements in lda to predict two groups: male and female. I have a working model, low Wilk's lambda,