similar to: error message: only first element in each line of matrix used

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cberry at tajo.ucsd.edu> >Cc: "simona.racioppi at

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, and I do not get the right answer. I tried

Hi all (really probably just Deepayan): In the plot below I want to add text on either side of each violin plot that indicates the number of observations that are either positive or negative. I'm trying to do this with ltext() and I've also monkeyed about with panel.text(). The code below is generally what I want but my calls to ltext() are wrong and I'm not sure how to fix them.

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

Dear community I am using ggraph to plot a network analysis. See part 2 in the working example. Besides different colors for different groups of nodes: --> geom_node_point(aes(size = V(network)$hub_score*200, color= as.factor(V(network)$community))) I additionally want to consider different colors for different edge groups The grouping is defined in the edge_list$relationship: negative

hello, I am using following command classify_polarity(documents,algorithm="bayes",verbose=TRUE) output is: [1] "DOCUMENT 1" [1] "WORD: excited CAT: positive POL: strongsubj SCORE: 8.44419229853175" [1] "WORD: happy CAT: positive POL: strongsubj SCORE: 8.44419229853175" [1] "WORD: optimistic CAT: positive POL: weaksubj SCORE: 7.7510451179718" [1]

How can I generate LLVM IR for both logical NEG (!)? For example, if I have Int32Ty a, For the bitwise NEG(~): c = ~a ; I can use the following API from LLVM: BinaryOperator *neg = BinaryOperator::CreateNeg(nbits, "bitwiseNEG", insertBefore); How, if I want to generate logical NEG: c = !a; what should I do for this? Thanks -------------- next part -------------- An HTML

Dear Sibylle, your example is not working! E.g. no data for "aes_collapsed". Best, Kimmo ke, 2024-03-20 kello 19:28 +0100, SIBYLLE ST?CKLI via R-help kirjoitti: > Dear community > > I am using ggraph to plot a network analysis. See part 2 in the working > example. > Besides different colors for different groups of nodes: > --> geom_node_point(aes(size =

Dear all, ## I have 4 tables like this: satu <- array(c(5,15,20,68,29,54,84,119), dim=c(2,4), dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", "Blond"))) dua <- array(c(50,105,30,8,29,25,84,9), dim=c(2,4), dimnames=list(c("Negative", "Positive"),

Dear useRs, I got stuck trying to generate a palette of topographic colors that would satisfy these two requirements: - the pallete must be 'anchored' at 0 (just like on a map), with light blue/lawn green corresponding to data values close to 0 (dark blue to light blue for negative values, green-yellow-brown for positive values) - the brown must get darker for higher positive

Hi, I have a problem sorting and selecting entries within a data-frame and I don't know if it is possible to solve it with R ... (probably yes, but I have no idea how). Following Data; row1 row2 a 12 pos NA a 3 neg NA a 5 neg NA a 11 pos NA I want to extract the values in row 2 in the lines with an "a" in row1. But I want to have two vectors: vector x with all

Thanks, Bruce. So, what is the easiest way to check if there is any bit set to 1 in a <N x i1> vector type? I used bitcast instruction to cast it into "iN" first and them compare iN to 0. Do you have a better way to do it? Thanks again. On Fri, Apr 10, 2015 at 5:22 PM, Bruce Hoult <bruce at hoult.org> wrote: > LLVM doesn't have a "logical neg" (or

Dear community Find enclosed the full working example. Many thanks Sibylle Test_cat.csv Names Subcategory_type sources.cyto source Factor A.A material "A" A 1 B.B material "B" B 1 C.C regulation "C" C 1 D.D regulation "D" D 1 E.E habitat "E" E 1 F.F cultural "F" F 1 Test_adjac.csv

Hi, this seems to work (assuming that your problem was the setting of colours...): --- snip --- network %>% ggraph(., layout = "auto") + # This produces an error... # geom_edge_arc(curvature=0.3, aes(width=(E(network)$weight/10), color=c("darkblue", "red")[as.factor(edge_list$relationship)], alpha=0.5)) + # ... this works :-)

Hi, I want to fit a GEE with a negative binomial distribution. I have uesd already a poisson glm and then neg binommial to deal with alot of dispersion. In my neg binomial residuals i have some patterns so i have implemented a GEE, but only with a poisson family as i couldnt with neg binomial. However the residual patterns in fact look worse here. When i try and put neg binomial family it

Hi R community! I've fitted three mixed-effects regression models to a thousand bootstrap samples (case-resampling regression) using the lme4 package in a custom-built for-loop. The only output I saved were the inferential statistics for my fixed and random effects. I did not save any output related to the performance to the machine learning algorithm used to fit the models (REML=FALSE).

Yes, but my point is that there would be some overhead to do cast the <N x i1> vectortype to an integerNty. Is there any good way to check not all of these N bits in the vectortype are 0s? On Fri, Apr 10, 2015 at 5:37 PM, Bruce Hoult <bruce at hoult.org> wrote: > Sure, if you actually just want an i1 saying whether or not at least one > bit is set to 1, then comparing against 0

My small brain is having trouble getting to grips with lme() I wonder if anyone can help me correctly set the random = argument to lme() for this kind of setup with (I think) 9 variance/covariance components ... Study.1 Study.2 ... Study.10 Treatment.A: subject: 1 2 3 4 5 6 etc. 28 29 30 Treatment.B: subject: 31

Hi userRs! I am trying to fit some GLM-poisson and neg.binomial. The neg. Binomial model is to account for over-dispersion. When I fit the poisson model i get: (Dispersion parameter for poisson family taken to be 1) However, if I estimate the dispersion coefficient by means of: sum(residuals(fit,type="pearson")^2)/fit$df.res I obtained 2.4. This is theory means over-dispersion since

Hello I have a data set that looks like this; > b2 dato chr status PRRSvac PRRSsanVac PRRSsanDk PRRSdk 33 2007-12-03 090432 R?d SPF 34 2007-02-09 090432 R?d SPF+sanDK 35 2002-12-17 090432 R?d SPF+DK 36 2002-11-27 090432 R?d SPF+sanDK 37 2002-07-23