similar to: GLS 95% CI

Dear all, I have just constructed the GLS model in R . How do I print the 95% confidence intervals for the vector of predicted values "R code/s for GLS CI"? See the model below: Please advice Regards Peter South Africa +2712 422 7357 +2782 456 4669 Generalized least squares fit by maximum likelihood Model: log(y) ~ log(x) + d1 + d3 Data: NULL

Dear "R Gurus", I am having two dummy csv data sets A and B containing 19 and 15 cases/observations respectively. From the two data set 13 cases are intersection. From one of the two (any) data set, How do I then retrieve the unmerged data ? let's take A for example, six cases must appear in our results. See the R codes below. Please assist. Looking forward to hearing

Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields: ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the

A better thanks would be to post the solution that met your needs... particularly since in spite of our efforts it appears we may not have understood your problem. A small sample data set generated with dput along with your code would be helpful in making sure others learn along with you. -- Sent from my phone. Please excuse my brevity. On July 11, 2017 12:56:45 AM PDT, Mangalani Peter

Thank you very much for the support. I have just used the reshape library and my problem was solved. Kind regards, Peter -----Original Message----- From: Jeff Newmiller [mailto:jdnewmil at dcn.davis.ca.us] Sent: 10 July 2017 03:52 PM To: r-help at r-project.org; Mangalani Peter Makananisa; r-help at r-project.org Cc: Vito Ricci (vito_ricci at yahoo.com) Subject: Re: [R] dplyr help I am

See FAQ #4 on the sqldf github home page. On Fri, Aug 11, 2017 at 9:21 AM, Mangalani Peter Makananisa <pmakananisa at sars.gov.za> wrote: > Dear all, > > I recently read the book " R data preperation and manipulation using sqldf package" by Djoni Darmawikarta > However, I have a problem with manipulation of dates using this package, I do not get the expected results.

I am pretty sure that this is not a question about dplyr... it is a question about tidyr. Look at the help file ?tidyr::spread. If I understand your question (I may not, because you gave no example of input/output data), the answer is no, the column names come from the column named by the key parameter and the values that fill those columns come from the column named by the value parameter.

Dear all, I'm using garchFit from fSeries package and I am not getting the desired results (error message : could not find function "garchFit" ). Would you please advise as to how I can build an ARIMA(p, d, q) - GARCH(p,q) model using R see the attached data and R-output. Thanking you in advance Kind regards Mangalani Peter Makananisa Statistical Analyst South

Dear R-Gurus, I am trying to read in data with 24,349,113 rows to R-3.3.3 (64 bit) and have used the library "data.table" and It managed to read 23,347,070 rows and the remainder was 2,043 rows only. Could you please advise me as to which library/R-commands is suitable to read the full data in to R? Kind regards, Mangalani Peter Makananisa (5786) South African Revenue Service

Dear all, I recently read the book " R data preperation and manipulation using sqldf package" by Djoni Darmawikarta However, I have a problem with manipulation of dates using this package, I do not get the expected results. Do I need to install some packages to be able to subset the data by dates in sqldf? I am not getting Djoni Darmawikarta email address. Please see the practice

> On Sep 11, 2017, at 1:05 AM, Mangalani Peter Makananisa <pmakananisa at sars.gov.za> wrote: > > Hi all, > > I am trying to create a new variable called Fiscal Year (FY) using case expression in sqldf and I am getting a null FY , see the code below . > > Please advise me as to how I can do this mutation. > > library(zoo) > library(lubridate) >

Hi all, I am trying to create a new variable called Fiscal Year (FY) using case expression in sqldf and I am getting a null FY , see the code below . Please advise me as to how I can do this mutation. library(zoo) library(lubridate) library(stringr) library(RH2) library(sqldf) cr$ReportDate = as.Date(cr$ReportDate, format ='%Y-%m-%d') > cr2 = sqldf(" select

2018-03-3 in your code should be 2018-03-31. The line then'201415' needs to be fixed. When posting please provide minimal self-contained examples. There was no input provided and library statements not relevant to the posted code were included. Fixing the invalid date and bad line, getting rid of those library statements that are unnecessary and providing some test input, it works

Dear all, Please help me with the R code which compute SCHEFFE TEST Thanking you in advance Kind regards Mangalani Peter Makananisa Statistical Analyst South African Revenue Service (SARS) Segmentation and Research : Data Modelling Tel: +2712 422 7357 Cell: +2782 456 4669 Fax:+2712 422 6579 Please Note: This email and its contents are subject to our email legal notice

HI all, Is it possible to use one column spread on multiple columns values. example spread( Key_col, value1:value7) spreading Key_col to variable value1, value2, ....... Value7 Please advise, Kind regards Mangalani Peter Makananisa (5786) South African Revenue Service (SARS) - HO +2782 456 4669 / +2712 422 7357 Please Note: This email and its contents are subject to our email legal notice

Dear R gurus, I am currently using the chron package for date manipulation in R and I am able to control the date which take the format - 1998/07/03, and - 2006-09-15 (date without time) Library(chron) As.Date(2006-09-15, format = '%Y-%m-%d') However I have a problem with the date containing only the year and month (i.e. 2006-09, 2007/08, 200609).

Hi, I am trying to plot the original data with the line of the model using the predict function. I want to add SE to the graph, but not sure how to get them out as the predict function for gls does not appear to allow for SE=TRUE argument. Here is my code so far: f1<-formula(MaxNASC40_50~hu3+flcmax+TidalFlag) vf1Exp<-varExp(form=~hu3) B1D<-gls(f1,correlation=corGaus(form=Lat~Lon,

Dear R-help, I don't understand why Gls gives me an error when trying to fit a model with AR(2) errors, while gls (from nlme) does not. For example: library(nlme) library(rms) set.seed(1) d <- data.frame(x = rnorm(50), y = rnorm(50)) gls(y ~ x, data=d, correlation = corARMA(p=2)) #This works Gls(y ~ x, data=d, correlation = corARMA(p=2)) # Gives error # Error in

Hi, I have fit a model using the rms package with the Gls() function. Is there a way to get the model estimates, std errors, and p-values (i.e. what you get with print(fit)) into latex format? I have tried: f <- Gls(...) latex(f, file='') ... but I get the following error Error in replace.substring.wild(s, old, new, test = test, front = front, : does not handle > 1 * in

Hi everyone: Is there a function in R to calculate the confidence intervals for the predictions of a GLS(Generalized Least Square) model? The function "predict" gives confidence intervals for the predictions of other types of models (lm, glm, etc) but not gls. Any input will be much appreciated, Best, Camilo Camilo Mora, Ph.D. Department of Biology Dalhouisie University