Or, you may use this approach:
> attach(achtergrond)
> spits <- ifelse(uurenminuut >= 5.30 & uurenminuut < 9.30, "morning",
+ ifelse(uurenminuut >=16.30 & uurenminuut < 19.0, "evening",
+ "between"))
> table(spits)
spits
between evening morning
1636 142 579
>
I personally like the app
Hello Mark,
Thank you for your timely reply. All I want to know is if the xreg argument
must contain the same amount of historical observations as the dependent
variable and if newxreg has to have a forecast of the exogenous variable
equal to the number of periods to be forecasted.
Best regards
This question seems to be the problem specific to Ubuntu. What if you
post the message to ?? I hope you get
answers from that mailing list.
Chel Hee Lee
On 12/02/2014 11:10 AM, VG wrote:
Hi everyone,
I was having trouble with R i installed some time ago on my local ubuntu
machine. So i rem
Hello everyone,
I am just trying to understand how the Arimax function works, so my
questions are:
1. If I have a univariate time series of sales and sales are dependent
upon, say inflation rates, then my xreg would be inflation rates right?
2. Now is I have historial data on sales (from january
Hello Chel and David,
Thank you very much for providing new insights into this issue. Here is one
more question. Why does the mutate () give incorrect results here?
# The following gives INCORRECT results - mutated()ed object
na.date.cases = ifelse(!is.na(oiddate),1,0)
# The following gives
In your function 'nbars()', I see the line:
tX = rbind(tX, as.data.frame(cbind(GId = " ",Grp = names(sG[n]),
S = fm, T = fm)))
It seems that you wish to have a data frame that has numeric variables
'S' and 'T'. The reason why you have character variables of 'S' and 'T'
from
> frame1
ID GROUP PROP_AREA
1 1 A 0.33
2 2 A 0.33
3 3 A 0.33
4 4 B 0.50
5 5 B 0.50
6 6 C 1.00
7 7 D 1.00
> frame2
GROUP VALUE1 VALUE2
1 A 10 5
2 B 20 10
3 C 30 15
4 D 40 20
>
The output in the object 'new1' are apparently same the output in the
object 'new2'. Are you trying to compare the entries of two outputs
'new1' and 'new2'? If so, the function 'all()' would be useful:
> all(new1 == new2, na.rm=TRUE)
[1] TRUE
If you are interested in the comparison of two ob
On Dec 3, 2014, at 2:10 PM, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
> Hello,
>
> Two alternative approaches - mutate() vs. sapply() - were used to get the
> desired results (i.e., creating a new column of the most recent date from 4
> dates ) with help from Arun and Mark on this forum. I now fi
solved; sorry for the spam.
library(devtools)
install_github("RcppCore/RcppArmadillo")
On Wed, Dec 3, 2014 at 4:51 PM, stephen sefick wrote:
> I would appreciate any help that you may be able to give. Please let me
> know if any more information is required.
>
> I get a the following error when
You should at least look at the facilities in the 'circular' package. Also the
Envirometrics Task View:
http://cran.r-project.org/web/views/Environmetrics.html which mention another
package that up until today I had not heard of:
http://cran.r-project.org/web/packages/CircStats/index.html
--
I would appreciate any help that you may be able to give. Please let me
know if any more information is required.
I get a the following error when I try to install RcppArmadillo in a
session started with R --vanilla using the
install.packages("RcppArmadillo") command.
make: *** [RcppArmadillo.o]
Hello,
Two alternative approaches - mutate() vs. sapply() - were used to get the
desired results (i.e., creating a new column of the most recent date from 4
dates ) with help from Arun and Mark on this forum. I now find that the two
data objects (created using two different approaches) are no
Posting in HTML format doesn't work nearly as well as you think it does... Your
email is pretty mixed up. Please use plain text format and use dput to make
your data usable in R.
I expect the best answer to your problem is going to be to use the merge
function instead of your for loops.. but th
I apologize if this is a relatively easy problem, but have been stuck on
this issue for a few days. I am attempting to combine values from 2
separate dataframes. Each dataframe contains a shared identifier (GROUP).
Dataframe 1 (3272 rows x 3 columns) further divides this shared grouping
factor into
I'd also suggest plotting a wind rose for each month (try openair) to
understand the statistical test results.
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360
This question is more about statistics than R. I suggest that you post it
to Cross Validated instead, http://stats.stackexchange.com/.
Jean
On Wed, Dec 3, 2014 at 5:40 AM, Dries David wrote:
> Hey
>
> In my data set i have two variables: month (march or april) and wind
> direction (N,NE,E,SE,
Nice, David!!
Worked like a charm!!
Thank you very much.
Em Tue Dec 02 2014 at 19:22:48, David L Carlson
escreveu:
> Let's try a different approach. You don't need a loop for this. First we
> need a reproducible example:
>
> > set.seed(42)
> > dadosmax <- data.frame(above=runif(150) + .5)
>
>
I am trying to create groups of barplots from data that have different number
of records in the groups, in such a way that all of the plots will have the
same numbers and sizes of bars represented even when some of the groups will
have some bars of zero height. The goal then would be to display
John M. Chambers Statistical Software Award - 2015
Statistical Computing Section
American Statistical Association
The Statistical Computing Section of the American Statistical Association
announces the competition fo
Hey
In my data set i have two variables: month (march or april) and wind direction
(N,NE,E,SE,S,SW,W,NW). I have to know if there is a difference in wind
direction between these months. What kind of test statistic should i use?
Kind regards
Dries David
Hello,
I have a dataset of asthma patients for which white blood cells gene
expression was measured with one-color Affymetrix microarrays (N~500,
asthma is a factor with 4 levels: control, moderate, severe, severe &
smokers).
I also have an extensive clinical dataset related, but with many m
> Hi,
>
> Is there an easy way to extract the AICc from a model within a glmulti
> object? I see the AIC, but not AICc. For example:
>
> data(mtcars)
> cardata = mtcars
> library(glmulti)
> # create models
> global = glm(mpg ~ ., data=mtcars)
> models = glmulti(global, level=1, crit="aicc", confs
Hi,
You didn't provide enough details (like the error message), but you
could start by calling Rscript with --vanilla option so that it
doesn't read Rprofile, nor environment files and see what happens.
#! Fábio
On Mon, Dec 1, 2014 at 12:09 AM, PO SU wrote:
>
> Dear expeRts,
> These days
Thank you!
> Date: Wed, 3 Dec 2014 08:02:17 -0500
> From: murdoch.dun...@gmail.com
> To: pmassico...@hotmail.com; r-help@r-project.org
> Subject: Re: [R] Substitute initial guesses of parameters in a function
>
> On 03/12/2014 7:37 AM, philippe massicotte wrote:
> > Hi everyone, I have a formula
Ok. I am trying to figure out if it's a bug in my code or in nlme, will let
you know and send you a reproducible example to you, Martin, if the latter, or
apologize profusely for blaming nlme :) if the former.
Thanks again for everyone's input.
From: Ma
If you want this resolved, you are going to have to provide the full
function in a reproducible example. Nearly a half-century with this type
of problem suggests a probability of nearly 1 that nlogL will be poorly
set up.
JN
On 14-12-03 06:00 AM, r-help-requ...@r-project.org wrote:
Message:
On 03/12/2014 7:37 AM, philippe massicotte wrote:
Hi everyone, I have a formula like this:
f <- as.formula(y ~ p0a * exp(-0.5 * ((x - p1a)/p2a)^2))
I would like to "dynamically" provide starting values for p0a, p1a, p2a. Is
there a way to do it?
Just give a named vector of starting values.
Hi everyone, I have a formula like this:
f <- as.formula(y ~ p0a * exp(-0.5 * ((x - p1a)/p2a)^2))
I would like to "dynamically" provide starting values for p0a, p1a, p2a. Is
there a way to do it?
#Params estimates
p <- c(12, 10, 1)
# This is where I have difficulties
mystart <- substitute(...)
Dear colleagues,
This is to inform you that Version 1.4.0 of the R package 'apcluster'
has been released on CRAN earlier this week. This is a major release
that - apart from other important improvements - fulfills a long-term
user request: the genuine support of sparse similarity matrices. For
Comments in line
On 02/12/2014 23:27, Silong Liao wrote:
Dear ALL,
I have a dataset contains 2 variables: mate (mating groups) and ratio (ratio of
number of mothers and fathers). And mate is an identifer which consists three
components: year, flock (flk), and tag.
I am using command "barchar
Katherine,
for a deeper understanding of differing values it makes sense to provide the
list at least with an online description of the corresponding functions used in
Minitab and SPSS…
Best
Simon
On 03 Dec 2014, at 10:45, Katherine Gobin via R-help
wrote:
> Dear R forum
> I sincerely apolo
Dear R forum
I sincerely apologize as my earlier mail with the captioned subject, since all
the values got mixed up and the email is not readable. I am trying to write it
again.
My problem is I have a set of data and I am trying to fit some distributions to
it. As a part of this exercise, I nee
Here is a reproducible bug in nlme (reported in 2008) that still crashes R
today:
https://stat.ethz.ch/pipermail/r-sig-mixed-models/2008q3/001425.html
Seems to be related to memory corruption (as diagnosed by Martin and William
Dunlap at the time):
https://stat.ethz.ch/pipermail/r-sig-mixed-mo
Dear R Forum
I have a set of data say as given below and as an exercise of trying to fit
statistical distribution to this data, I am estimating parameters.
amounts =
c(38572.5599129508,11426.6705314315,21974.1571641187,118530.32782443,3735.43055996748,66309.5211176106,72039.2934132668,21934.884
> Bert Gunter
> on Tue, 2 Dec 2014 14:03:44 -0800 writes:
> Yes, Bill almost always has helpful ideas.
> Just a comment: If indeed the process is gobbling up too much memory,
> that might indicate a problem with your function or implementation. I
> defer to real expert
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