similar to: Retrieving an evaluated gradient value (UNCLASSIFIED)

Classification: UNCLASSIFIED Caveats: NONE Hello, R users, Has anybody written a function for random test based on the length of longest run of same events. I really appreciate your help. Kyong Park Classification: UNCLASSIFIED Caveats: NONE [[alternative HTML version deleted]]

Classification: UNCLASSIFIED Caveats: NONE Dear R users, I'm interested in finding a random effect of the Block in the data shown below, but 'lme' does not work without the random effect. I'm not sure how to group the data without continuous value which is shown in the error message at the bottom line. If I use 'aov' with Error(Block), is there a test method comparing

Classification: UNCLASSIFIED Caveats: NONE Thanks for your quick response. The program you mentioned below available from R is based on number of runs (up or down) not based on a longest length of runs of same events. To be more specific, for example, from a series, HHTHTTTTHHH, the number of runs are 5, and the longest length of runs of the same events is 4. I'll check for the website you

All, I'm re-running some analysis that has been augmented with additional data. When I use the exact same code for the augmented data, the behavior of the aggregate function is very strange, viz., one of the resulting variables is now coded as a factor while it was coded as numeric for the original data. Unfortunately, I cannot provide a reproducible code example since it only seems to occur

I am trying to define a general R function that has a function as the output that depends on the user's input arguments (this may make more sense by looking at the toy example below). My real use for this type of code is to allow a user to choose from many parameterizations of the same general model. My "issue" is that when I compile a package with this type of code in it I get a

Hi All: I have been looking for an elegant way to do the following, but haven't found it, I have never had a good understanding of any of the "apply" functions. A simplified idea is I have an array, say: junk(5, 10, 3) where (5, 10, 3) give the dimension sizes, and I want to reverse the second dimension, so I could do: junk1 <- junk[, rev(seq_len(10), ] but what I am

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

How about this: f <- function(a,wh){ ## a is the array; wh is the index to be reversed l<- lapply(dim(a),seq_len) l[[wh]]<- rev(l[[wh]]) do.call(`[`,c(list(a),l)) } ## test z <- array(1:120,dim=2:5) ## I omit the printouts f(z,2) f(z,3) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

Hello, R users, I am a very beginner of R and tried read.csv to import an excel file after saving an excel file as csv. But it added alternating rows of fictitious NA values after row number 16. When I applied read.delim, there were trailing several commas at the end of each row after row number 16 instead of NA values. Appreciate your help. Kyong [[alternative HTML version deleted]]

Hi R users, I’d like to get an expected value for a minimum value from order statistics of sample size, say, 5 for standard normal distribution, and the formula would be 5* Integral (from -inf to Inf) x*f(x)* (1-F(x))^4,where f(x) and F(x) are a standard normal density and a cumulative distribution function respectively. After searching R posts, I applied the following double integral

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter

Dear R users, I'm using R 2.1.0 with Windows 2000. In the middle of a R session, an error message pops up saying Rgui.exe generated errors and shuts down the R session. How can I correct this problem? Appreciate your help. Kyong [[alternative HTML version deleted]]

Classification: UNCLASSIFIED Caveats: NONE A similar question has been posted in the past but never answered. My question is this: for probit analysis, how do you program a 95% confidence interval for the LD50 (or LC50, ec50, etc.), including a heterogeneity factor as written about in "Probit Analysis" by Finney(1971)? The heterogeneity factor comes into play through the chi-squared

Classification: UNCLASSIFIED Caveats: NONE R users, This is a summary for the responses I got from Sandra Dorai-Raj and Simon Blomberg followed by my question. Sandra suggested using lm instead of lme for a model without a random effect, and Simon suggested anova.lme instead of anova.lm for comparing two models with and without a random effect. For the anova test, both anova.lme and anova

Hello (and Happy New Year), When I create a factor with labels in the order I want, write the data as a text file, and then retrieve them, the factor levels are no longer in the proper order. Here is what I do (I tried many variations): # educ is a numeric vector with 1,001 observations. # There is one NA # Use educ to create a factor feducord <- factor(educ, labels = c('Elem',