similar to: Delivery Failure: Ziqxzb (PR#6609)

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Hi. This is the qmail-send program at provida.org.br. I'm afraid I wasn't able to deliver your message to the following addresses. This is a permanent error; I've given up. Sorry it didn't work out. <musical@provida.org.br>: vdeliver: ERRO: Usu?rio n?o existe: 'musical'. --- Below this line is a copy of the message. Return-Path: <r-bugs@r-project.org>

I am attempting to graph 12 months of temperatures, delineate the months with a vline and place the names of the months at the top of the graph. So far I have gotten everything to work except the names, despite getting a similar graph to work yesterday the day before yesterday with Baptise A's help. Can anyone suggest what I am doing wrong. Data set is below code. Thanks. Code

Hadley, I don't know if I am doing something wrong or if it is ggplot please see the two graphs at the bottom of the page (code). melt.nut <- (structure(list(RiverMile = c(119L, 119L, 119L, 119L, 119L, 119L, 119L, 119L, 119L, 148L, 148L, 148L, 148L, 148L, 148L, 148L, 179L, 179L, 179L, 179L, 179L, 179L, 179L, 185L, 185L, 185L, 185L, 185L, 185L, 185L, 190L, 190L, 190L, 190L, 190L, 190L,

Dear all, My data is by minutes and I can see it has seasonal trend by daily and weekly. How do I decompose the minute data into daily and weekly some data: > dput(tail(dt_train,100))structure(c(11L, 11L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 10L,

Hallo you can extract POSIX object tv <- as.POSIXct(index(dt_train)) and use cut together with aggregate cut(tv, "hour") aggregate(dt_train, list(cut(tv, "hour")), mean) 2014-10-06 21:00:00 9.807692 2014-10-06 22:00:00 8.666667 Cheers. Petr ?t 3. 10. 2024 v 17:25 odes?latel roslinazairimah zakaria < roslinaump at gmail.com> napsal: > Dear all, > > My

Putting names on a ggplot Can anyone suggest what I am doing wrong here. I am plotting daily temperatures at Ottawa Ontario for 2008 broken down by months, I seperate them by lines and want to put the names of the months at the top of the chart ( with in the graphing area) Everything is working as I want until I try to add the names of the months. I did something similar a few days ago and I

I have a dataframe df with columns x, y, and height. I want to create a heatmap-like plot that creates a grid of x by y, and then color codes the grid depending on the value of height. Is there a ggplot2 object to do this? I'm able to easily do this in Excel with pivot tables and conditional formatting so I'm including an image that is close to the output I want. I want to be able to

Hi: I need some help with the legend. I got 14 samples(Muestreo) and I am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance. fish_ByMuestreo <- structure(list(data = structure(list(SampleDate = structure(c(3L, 3L, 3L, 3L,

Dear r-users, I would like to construct 3-day moving average for block maxima series. I tried this: bmthree <- lapply(split(dt, dt$Year), function(x) max(sapply(1:(nrow(x)-2), function(i) with(x, mean(Amount[i:(i+2)],na.rm=TRUE))))) bmthree and got the following output. $`1971` [1] 70.81667 $`1972` [1] 68.94553 $`1973` [1] 102.7236 $`1974` [1] 73.6625 $`1975` [1]

I had a problem annotating a graph last year ( see http://n4.nabble.com/Putting-names-on-a-ggplot-td907158.html#a907158 for the discussion) Stefan (smu) provided a solution using annotate(). However I apparently did not update the graph file and,now, when I go back to the thread and try to use Stefan's solution it does not seem to work although I am sure that it did then. The problem

Hi all: I am trying to use the ddply function to estimate the mean of 'Total','Fry','Smolt' and 'Fry.Eq' columns without success. I have the dput of my dataset below. I wonder if someone can give me a hand with this function. # dput(winter) winter <-structure(list(IDDate = structure(c(37L, 48L, 59L, 62L, 63L, 64L, 65L, 66L, 67L, 38L, 39L, 40L, 41L, 42L, 43L,

I am aware of the inherent risks of having plots with more than two axes, but I am trying to produce the graphs that I have been tasked with. That being said I am having a hard time figuring out how to have two axes onto a boxplot. below is the sample code. I would like BC on the plot produced with this code to be on a second axis with all of the others being on the first axis. This will be

Dear R Group I have the following data for which I am trying to create subject wise lattice plot for a given attribute and product . though the lattice plot is generated, for some reasons that i dont understand in each plot the subject panels take a random order, I would rather want all the plots to display the subject order in the same order as how i have ordered this particular factor level.

> On Dec 6, 2017, at 5:07 AM, Paul Bernal <paulbernal07 at gmail.com> wrote: > > Dear friends, > > I have a weekly time series which starts on Jan 4th, 2003 and ends on > december 31st, 2016. > > I set up my ts object as follows: > > MyTseries <- ts(mydataset, start=2003, end=2016, frequency=52) > > MyModel <- auto.arima(MyTseries, d=1, D=1)

Thank you very much David. As a matter of fact, I solved it by doing the following: MyTimeSeriesObj <- ts(MyData, freq=365.25/7, start=decimal_date(mdy("01-04-2003"))) After doing that adjustment, my forecasts dates started from 2017 on. Cheers, Paul 2017-12-06 12:03 GMT-05:00 David Winsemius <dwinsemius at comcast.net>: > > > On Dec 6, 2017, at 5:07 AM, Paul

d <- structure(list(Site = structure(c(8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L,

Hello list, I am not sure if the terminology that I am using here is widely used, however, I provide an example in the hopes that my problem will become clear. My basic problem is that I am unsure of how to 'constrain' my model estimates to reproduce the aggregate (by factor levels) observed dependent variable for a negative.binomial model. I realize this sounds rather vague, so I provide

Hello everyone, I have a dataset with 5 colums (4 colums with thresholds of weather stations and one with month - data of 5 years). Now I would like to calculate the average for each month. I tried this unsuccessfully: lf.med <- sapply(LF[,1:4],mean,LF[,5]) Error in mean.default(X[[1L]], ...) : 'trim' must be numeric and have length 1 With lf.med <- by(LF[,1:4],LF[,5],mean)

hello, i found it most convenient to use package contrast for planned comparisons on mixed models. for instance i have a model with 2 fixed factors, one with 4 levels (stage) and one with 2 levels (gap) and a nested random factor (site) and i tested gap within level A of factor stage: library(contrast) library(nlme) m1<-lme(rich ~ stage*gap, random=~1|site,data=richness) contrast(m1, a =