similar to: Assigning NA to a rows of a dataframe/datamatrix

I am doing multiple imputation with Hmisc, and can't figure out how to replace the NA values with the imputed values. Here's a general ourline of the process: > set.seed(23) > library("mice") > library("Hmisc") > library("Design") > d <- read.table("DailyDataRaw_01.txt",header=T) > length(d);length(d[,1]) [1] 43 [1] 2666

On 3/11/2018 3:35 PM, Christofer Bogaso wrote: > But for my reporting purpose, I need to generate a bell curve like > plot based on empirical PDF, that also contains original points. > > Any idea would be helpful. Thanks, > Christofer, something like the following may get you what you want: ## get the kernel density estimate dens <- density(Dat) ## estimate the density at

Hello, I have an existing legacy app written in .NET that will be rewritten with Rails. One of the key components of the App is to generate and read datamatrix barcodes from a PDF. I''ve looked online and message posts but haven''t see anything worthwhile for reading / generting datamatrix barcodes. Any recommendations for this? Todd -- You received this message because you are

I have huge data file, and I would like randomize just one column at a time , is there any easy way? Thanks a lot. -- View this message in context: http://www.nabble.com/Randomizing-one-column-in-the-dataMatrix-tf4957535.html#a14197423 Sent from the R help mailing list archive at Nabble.com.

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Dear List: I have a data manipulation problem that I was unable to solve in R. I did it in SQL, and it may be that the solution in R is to do it in SQL, but I wondered if people could imagine a vector-based solution. Imagine a list A[i] of observers who observe some set of events B[j]. Each observer i may observe one or more events, and each event j may have been observed by one or more

Used for celt_pitch_xcorr on aarch64, and celt_fir and celt_iir on both armv7 and aarch64. --- celt/arm/arm_celt_map.c | 17 +++++++++++++ celt/arm/celt_neon_intr.c | 61 ++++++++++++++++++++++++++++++++++++++++++++++- celt/arm/pitch_arm.h | 31 +++++++++++++++++++++++- 3 files changed, 107 insertions(+), 2 deletions(-) diff --git a/celt/arm/arm_celt_map.c b/celt/arm/arm_celt_map.c index

Hi, I'm trying to generate a GLMM for Y following a negative binomial distribution. The first step I'm taking as a beginner in the use of R2Jags is to generate a GLM for Y following a Poisson distribution (I heard it's good in order to get used to the coding language), so here I am and here I face the first error message for which I can't find any mistake of spelling. Could you

Hi all, I'm extracting the name of the term in a regression model that dropterm specifies as the least significant one, and I'm assigning this name to an object. However, when I use update(), it ignores this object. Is there a way I can make it not ignore it? A reproducible example is below: > lm(x1~1+y1*y2+y3+y4,data=anscombe)->my.lm >

I have a vector of 2,1,0 I want to change to 0,1,2 respectively (the data is allele dosages) I have tried multiple nested if/else statements and looked at the ?if help and cannot work out what is wrong, other people have posted code which is identical and they state works. Any help would be greatly appreciated. > A[1:20] [1] 1 1 0 0 1 0 1 0 1 0 0 0 1 1 0 1 1 1 0 0 > B <-

Hi, I am trying to fit a negative binomial model with a number of parasite tapeworms as response variable to geographical coordinates (actually preparing a trend surface before kriging). When I try an anova, I get warnings: > glm4.nb<-glm.nb(wb~X4+Y4+I(X4^2)+I(Y4^2)) > anova(glm4.nb) Analysis of Deviance Table Model: Negative Binomial(0.0463), link: log Response: wb Terms added

But for my reporting purpose, I need to generate a bell curve like plot based on empirical PDF, that also contains original points. Any idea would be helpful. Thanks, On Mon, Mar 12, 2018 at 3:49 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > You need to re-read ?density and perhaps think again -- or do some study -- > about how a (kernel) density estimate works. The points at

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

Hi First post and a relative R newbie.... I am using the vioplot library to produce some violin plots. I have an input CSV with columns off irregular length that contain NAs. I want to strip the NAs out and produce a multiple violin plot automatically labelled using the headers. At the moment I do this Code:? ds1 = read.csv("http://www.lecturematerials.co.uk/data/spelling.csv")

I wish to rearrange the matrix, df, such that all there are not repeated x values. Particularly, for each value of x that is reated, the corresponded y value should fall under the appropriate column. For example, the x value 3 appears 4 times under the different columns of y, i.e. y1,y2,y3,y4. The output should be such that for the lone value of 3 selected for x, the corresponding row entries

Hi, I'm a newbie to the world of HMMs and HMMs in R. I've had a look at the hmm package and the RHmm package but I couldn't see anything straightforward on how a labelled sequential dataset with observed values and underlying states might be used to construct and train a HMM based on that data and no pre-computed values for the transition, emission or initial state distributions. Does

Hi R-experts, suppose I have a list with containing data frame elements: [[1]] (Intercept) y1 y2 y3 y4 -6.64 0.761 0.383 0.775 0.163 [[2]] (Intercept) y2 y3 -3.858 0.854 0.834 Now I want to put them into ONE dataframe like this: (Intercept) y1

RNG in R and Splus 3.4 Prof. Ripley asked the details of the example. We were doing parametric bootstrap, so it is similar to simulation. Anyway here is the details. We start with a sample of 19 positive numbers. We know the sample is from truncated exp(0.3)...only the truncation point, theta, is unknown. In other words, the sample can be generated from something like x1 <- rexp(100,

RNG in R and Splus 3.4 Prof. Ripley asked the details of the example. We were doing parametric bootstrap, so it is similar to simulation. Anyway here is the details. We start with a sample of 19 positive numbers. We know the sample is from truncated exp(0.3)...only the truncation point, theta, is unknown. In other words, the sample can be generated from something like x1 <- rexp(100,

Hi, I would like to create a plot of y1,y2,y3,y4 against x for several subjects such that y1 and y2 are plotted against x in one panel and y3 and y4 against x in another panel. Thus if there are 3 subjects I should end up with 6 panels. Is there a simple way of doing so (i.e. without calling xyplot() several times, and then padding the results together)?? Regards S?ren