Hello,
By running the code of the OP, I've come to the conclusion that
print.forecast calls print.data.frame:
forecast:::print.forecast
function (x, ...)
{
print(as.data.frame(x))
}
But I'm unable to get what the OP wants by call as.data.frame(...).
Here is the code with some made up da
Dear Jim,
Thank you for your valuable replies. So variable mf5 has the forecasts for
a fitted nnetar model.
Below is the class and the str(mf5) output. Maybe with this information I
am sharing with you, hopefully you can give me some more guidance.
Again, thank you so much!
> class(mf5)
[1] "fo
Hi Paul,
When manipulating any R object, the first thing to ascertain is what it is:
class(TSmodelForecast)
should give you useful information.
str(TSmodelForecast)
should give you more. Because of the wealth of defined data structures
in R, it is difficult to manipulate them without this infor
Dear All:
it seems I have solved my reordering issue, and to thank all those who
answered, I briefly list the mistake I made.
I stared out from some data.frame holding values x and a grouping
factor. Actually, my factor was a number (but not numeric) which was,
however, not ordered. So it w
But note that m:n does not always produce an integer sequence. It does when
the output can be accurately represented as an integer sequence, it
does not care
if the inputs were integer or numeric.
> class( (2^31-10):(2^31-2) )
[1] "integer"
> class( (2^31-10):(2^31) ) # biggest integer is as
Hello,
Not an answer to the OP's statement, but
> class(1L:4L)
[1] "integer"
> class(1:4)
[1] "integer"
When using m:n there's no need for mL or nL.
Rui Barradas
Em 17-03-2017 16:58, Jeff Newmiller escreveu:
Reprex confirming Bert:
A <- data.frame( y = 1L:4L )
B <- data.frame(
Evan:
Yes, I stand partially corrected. You have the concept correct, but R
implements it differently than SAS.
I think what you want for your approach is diff():
evens <- (seq_len(nrow(mydata)) %% 2) == 0
newdat <-data.frame(exp=mydata[evens,1 ],reslt= diff(mydata[,2])[evens[-1]])
... which s
Evan:
You misunderstand the concept of a lagged variable.
Ulrik:
Well, yes, that is certainly a general solution that works. However,
given the *specific* structure described by the OP, an even more
direct (maybe more efficient?) way to do it just uses (logical)
subscripting:
odds <- (seq_len(
Hi Evan
you can easily do this by applying diff() to each exp group.
Either using dplyr:
library(dplyr)
mydata %>%
group_by(exp) %>%
summarise(difference = diff(rslt))
Or with base R
aggregate(mydata, by = list(group = mydata$exp), FUN = diff)
HTH
Ulrik
On Fri, 17 Mar 2017 at 17:34 Evan C
Reprex confirming Bert:
A <- data.frame( y = 1L:4L )
B <- data.frame( x = 1L:4L )
A$x <- B$x
plot(B$x)
#' 
Care to demonstrate for us, Karl?
https://cran.r-project.org/web/packages/reprex/README.html
On Fri, 17 Mar 2017, Bert Gunter wrote:
You are
Suppose I have a dataframe that looks like the following:
n=2
mydata <- data.frame(exp = rep(1:5,each=n), rslt =
c(12,15,7,8,24,28,33,15,22,11))
mydata
exp rslt
11 12
21 15
327
428
53 24
63 28
74 33
84 15
95 22
10 5 11
The variabl
Dear Karl,
This is hard to investigate without a reproducible example.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgiu
You are wrong. No reordering occurs.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Fri, Mar 17, 2017 at 8:05 AM, Karl Schilling
wrote:
> Dea
Dear all:
I have two data.frames A and B of the same number of rows (about
40,000). I realized that when I copy column x from data.frame A to B,
the order of this column gets changed. This seems to affect only values
in rownumbers > ~ 35/36,000. It also happens in any of the following
three
If I understand you correcty, I suggest you consult a local
statistician. Individual data points *cannot* have "confidence
intervals." So you clearly need some statistical guidance, which is
not what this list is about.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that peo
Just to do a better job in what is perhaps more "R-ish" w.r.t least squares
calculations, here is perhaps a better example
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2, 10, 20, labels = c("Ctl","Trt"))
weight
Hi John
Thierry already pointed me to propper direction. The key for such „zooming“ is
to use coord_cartesian(xlim ...) which is stated, as I finally found, in xlim
help page.
Thanks
Petr
From: John Kane [mailto:jrkrid...@yahoo.ca]
Sent: Friday, March 17, 2017 2:44 PM
To: PIKAL Petr ; r-help@
Hi Petr,It "seems" to be linked to the =/- 14 days that you are using for the
limits.
The code below should restore one more bar. After that I don't know.
p2 <- p+geom_segment(alpha=.4, size=3)+xlim(c(today()-21, today()+21))+
geom_vline(xintercept=as.numeric(today()), colour="pink", size=5)
Dear Thierry
Thanks, works great. My bad, after your advice I went through help page for
xlim and found, that this behaviour is quite clearly stated there. I focused on
geom_segment where I did not (obviously) find any clue.
Best regards
Petr
> -Original Message-
> From: Thierry Onkeli
Kevin,
The short answer is no.
The function table() takes in the vectors provided as arguments, counts the
number of occurrences of each category by adding the integer 1L to a bin (one
for each category or factor level), and at the end it returns the counts in
each bin. Since it does not retur
Is this what you want?
http://opensourceconnections.com/blog/2016/09/17/expanding-data-frequency-table-r-stata/
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Kevin E.
> Thorpe
> Sent: Friday, March 17, 2017 1:23 PM
> To: r-help
> Subj
Dear Petr,
Don't use xlim() but rather coord_cartesian(xlim = ...). See
https://rpubs.com/INBOstats/zoom_in
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kl
Dear all
I try to limit geom segment plot by xlim and encounter weird behaviour.
Here is my data
> dput(temp)
temp <- structure(list(ukol = c("externí kalcinace", "povrchová úprava UVS30",
"Testy mletí UVS30", "změna koncentrace olejových disperzí",
"Mletí povrchově upravené UVS30", "mikronizace
I am wondering if there is a way to undo the results of table().
For example if you had a table that looked like the result of table(x, y) or
table(x, y, z) is there a simple/elegant way to reverse the process to get the
"original" x, y and z vectors?
Thanks,
Kevin
--
Kevin E. Thorpe
Head of
Dear Ross,
On 17 March 2017 at 06:25, wrote:
> I am having a problem running the "impute" function from the bnlearn
> package.
I assume you referred to the online documentation instead of the
manuals that come with the package? I included impute() in bnlearn
4.1, so you will have to upgrade to
A slightly more ³R-ish² way of doing
S <- solve(t(X)%*%X)
Is to instead use
S <- solve(crossprod(X))
And the idea idea of inverting the SSCP matrix only and not actually
solving the linear system is not so great, which is why it is better to do
as Rolf is suggesting and get all things you need
dear Bert,
the code as I read it draws 30 median lines, one for each cluster.
what I am looking for is a function that will plot 190 individual
confidence intervals, one for each measurement. would the code cover
even that instance?
regards
luigi
On Thu, Mar 16, 2017 at 2:41 PM, Bert Gunter wrote
Dear All,
Thank you for helping me out.
First, I am sorry for adding a HTML element. I was not aware of it. I simply
copy and pasted the output of mat_data from R studio.
Second, no it is not homework. It was part of a side project at work.
Third, thank you Jeff. Your comment was really produc
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