Dear R Users,
I have struggled with the following problem for days, which I thought was
simple, although it would likely be basic to most of you.
I am working with time series data.
In my script, my intention is to create first differences of the variables in
the file so that I end up estimat
Dear R Users,
Sorry for what seems like I am re-posting. When I was typing my initial
posting, I intended to copy and paste the commands from my script, but ended up
forgetting. I am now pasting the commands in this email.NB: Below is a copy of
'all' the relevant commands in my script
###
Dear Sarah,
Thanks very much for the assistance. I will go through what you have suggested
to understand.
Lexi
- Original Message -
From: Sarah Goslee
To: "Lekgatlhamang, lexi Setlhare"
Cc: "r-help@r-project.org"
Sent: Friday, June 15, 2012 3:58 PM
Subjec
On Friday, June 15, 2012, Lekgatlhamang, lexi Setlhare wrote:
Dear R Users,
>Sorry for what seems like I am re-posting. When I was typing my initial
>posting, I intended to copy and paste the commands from my script, but ended
>up forgetting. I am now pasting the commands in this email
e's a more R-ish way:
# Differences of Cred
DCred <- list()
for(i in 1:5){
print(DCred[[i]]<- diff(DCred, lag=i, difference=1))
}
DCred[[1]]
Sarah
On Friday, June 15, 2012, Lekgatlhamang, lexi Setlhare wrote:
Dear R Users,
>Sorry for what seems like I am re-posting. When I was t
stand much
better. I will turn to testing the second code.
Cheers.
Lexi
- Original Message -
From: Sarah Goslee
To: "Lekgatlhamang, lexi Setlhare"
Cc: "r-help@r-project.org"
Sent: Friday, June 15, 2012 3:58 PM
Subject: Re: [R] Help with Sequential Differencing
I m
Dear R Users,
I ask the following question in order to learn more on the use of 'assign' and
'paste' functions and for loop; otherwise what I am asking could be solved by
binding the various first differences of the series using the 'ts.union'
operator.
The problem is:
I have several varia
Thanks a lot Peter. It is a learning process for me.
From: Petr PIKAL
Cc: R-Help
Sent: Thursday, June 28, 2012 6:20 PM
Subject: Re: [R] Adjusting length of series
Hi
I use R for quite a long time and as I remember I did not use such assign
paste i loop
In the equation with error message, only the last term is specified by
referencing its name (ie., Dcr$Dbobc3[1:29]. Also note that the error occurs
whether I append '[1:29]' to Dcr$Dbobc or not.
How do I resolve this?
Thanks. Lexi
- Original Message -
From: "Lekgatlh
Hi
I have a follow up question, relating to subsetting to list items. After using
the list and min(sapply()) method to adjust the length of the variables, I
specify a dynamic regression equation using the variables in the list. My list
looks like this:
Dcr<-
list(Dcre1=DCred1,Dcre2=DCred2,Dcre3
Thanks a lot David and A.K. for the detailed and helpful suggestions.
Lexi
Cc: R help
Sent: Sunday, July 1, 2012 3:39 AM
Subject: Re: [R] Adjusting length of series
Hello,
Try this:
Dcr<-lapply(1:5,function(x) rnorm(10,15))
names(Dcr)<- c("Dcre1","D
Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I
encountered some challenges.
Â
As for David's suggestions, I could only implement it after some
modifications. Using an abridged version of my data, I dpud my dataset and
then show my steps below.
Â
> dp
Noted David, and thanks very much.
Â
Lexi
From: David Winsemius
Sent: Monday, July 2, 2012 4:26 PM
Subject: Re: [R] Adjusting length of series
On Jul 2, 2012, at 5:13 AM, Lekgatlhamang, lexi Setlhare wrote:
> Hi David and AK,
> I have been try
Thanks very much A.K. I have to admit that my problem was not clearly stated,
with the structure of my data provided. Now all is well.
Â
Cheers
Lexi
Cc: R help
Sent: Monday, July 2, 2012 4:40 PM
Subject: Re: [R] Adjusting length of series
Hello,
The
Dear all,
I have a challenge with a supposedly simple graph (a scatter). I wanted to use
R to create a plot/graph with a positive y-axis intercept but it does not seem
to give me what I expect to see. As an example, I used the following values
> X<- c(0, 4, 8, 11)
> Y<- c(8, 12, 15, 22)
then
intercept
Your question is not entirely clear to me, but I think you might want:
plot(X, Y, xlim=c(0, 25), ylim=c(0, 25))
Sarah
On Thu, Jul 19, 2012 at 5:31 PM, Lekgatlhamang, lexi Setlhare
> Dear all,
>
> I have a challenge with a supposedly simple graph (a scatter). I wanted to
>
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