Dear R users,
I am fronting my firts time series problem. I have hourly temperature data
for 3 years (from 01/01/2013 to 5/02/2016). I would like to use those in
order to PREDICT TEMPERATURE OF THE NEXT HOURS according to the
observations.
A subset of the data look like this:
date <- rep(seq(as
Try the auto.arima function in the forecast package..
Regards,
DR SEAN PORTER
Scientist
South African Association for Marine Biological Research
Direct Tel: +27 (31) 328 8169 Fax: +27 (31) 328 8188
E-mail: spor...@ori.org.za Web: www.saambr.org.za
1 King Shaka Avenue, Point, Durban 4001 KwaZu
Some good references:
https://www.otexts.org/fpp
http://link.springer.com/book/10.1007%2F978-0-387-88698-5
http://www.statoek.wiso.uni-goettingen.de/veranstaltungen/zeitreihen/sommer03/ts_r_intro.pdf
Best,
--
GG
This Communication is Ericsson Confidential.
We only send and receive email on
Folks
TPP, and in a European context, TTIP are very dangerous not only to open
source software but to any public service and no satisfactory response has
been forthcoming.
There are ways of circumventing it I guess or opposing it (maybe using
ISPs in China, Russia or North Korea???). The issue r
Does that mean I could poison the Archive by posting IP on this list, or poison
someone else's code if they use some of mine that I post here?
B.
On Feb 4, 2016, at 7:48 PM, Spencer Graves
wrote:
> It's not clear if the TPP would ever directly impact the R project.
> However, it coul
> I'm doing error analysis of predictive models and I need to calculate global
> error, this is, I need to calculate the resultant error from propagation of
> indirect measurements errors.
If the problem is propagation of variance on inputs through a known function,
you could look at uncert() in
Thank everyone, I have found some good, but limited infomation about it. As
Ross Ihaka mention in his presentation: "Houston, We Have a Problem".
The R software is a important part of the Free Software Fundation, they
have been fighting back TPP long time, but last weeks in Chile was not so
good.
Hello,
I have a set of 12 individuals with thousands of variables measured.
I understand that when I'm using the cor() function on my matrix I'm
calculating the correlation between the different variables according to
their values for the different individuals.
What I'm willing to do is to ca
Assuming your data is called BIG and has 12 rows and thousands of columns:
cor(t(BIG))
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.o
Assuming your dataset is in a matrix you want to transpose it. So you
can go t(mesdonnees) and then call cor on that.
On 05/02/2016 14:06, emmanuelle morin wrote:
Hello,
I have a set of 12 individuals with thousands of variables measured.
I understand that when I'm using the cor() function on
R's subscripting operators do not "guess" the value of a missing
argument: a missing k'th subscript means seq_len(dim(x)[k]).
I bet that you use syntax like x[,1] (the entire first column of x)
all the time and that you don't want this syntax to go away.
Some languages use a placeholder like '.' o
Dear there,
Here is a snipped code,
> rm(list = ls())
> x <- 123
> save.image("abc.RData")
> rm(list = ls())
> load("abc.RData")
> sample(10)
[1] 3 7 4 6 10 2 5 9 8 1
> rm(list = ls())
> load("abc.RData")
> sample(10)
[1] 3 7 4 6 10 2 5 9 8 1
you will see that, after loading
Hi Rgonauts,
I am plotting 3 variables from one data set on one plot. Two of them as a
stacked bar and a ratio on a completely different scale, so I need to put one
of the axes on the top of the plot for clarity. Whilst this is not good
visualisation practice, there is a valid reason why, in
On 05/02/2016 11:14 AM, Jinsong Zhao wrote:
Dear there,
Here is a snipped code,
> rm(list = ls())
> x <- 123
> save.image("abc.RData")
> rm(list = ls())
> load("abc.RData")
> sample(10)
[1] 3 7 4 6 10 2 5 9 8 1
> rm(list = ls())
> load("abc.RData")
> sample(10)
On 02/05/2016 05:25 PM, Duncan Murdoch wrote:
On 05/02/2016 11:14 AM, Jinsong Zhao wrote:
Dear there,
Here is a snipped code,
> rm(list = ls())
> x <- 123
> save.image("abc.RData")
> rm(list = ls())
> load("abc.RData")
> sample(10)
[1] 3 7 4 6 10 2 5 9 8 1
> rm(list =
One more question (see below). I cannot use macro-var, mvar, for creating new
name, as shown below. Any advice is highly appreciated!
> mvar<-"pop"
> new.pop<-tab[[mvar]]; new.pop
[1] 698 1214 1003 1167 2549 824 944 1937 935 570 0
> new.tab[[mvar]]<-d$pop;
Error in new.tab[[mvar]] <-
On 05/02/2016 11:49 AM, Dénes Tóth wrote:
On 02/05/2016 05:25 PM, Duncan Murdoch wrote:
> On 05/02/2016 11:14 AM, Jinsong Zhao wrote:
>> Dear there,
>>
>> Here is a snipped code,
>>
>> > rm(list = ls())
>> > x <- 123
>> > save.image("abc.RData")
>> > rm(list = ls())
>> > load("abc.RData
Yes, you can use a name stored in a variable to create a new column in a
data frame (guessing that's what you want). Here's an example:
> df <- data.frame(a=1:5)
> df[['b']] <- 2:6
> df
a b
1 1 2
2 2 3
3 3 4
4 4 5
5 5 6
> mvar <- 'c'
> df[[mvar]] <- 0:4
> df
a b c
1 1 2 0
2 2 3 1
3 3 4 2
4 4 5
If 'tab' is a data.frame then new.tab <- tab[[mvar]] is a column from that
data.frame, not a data.frame with one column. new.tab <- tab[ , mvar,
drop=FALSE ] will give you a data.frame that you can add to with either of
nvar <- "newName"
new.tab[ , nvar] <- newColumn
new.tab[[nvar]] <-
You REALLY NEED to read the "Introduction to R" document discussion of
indexing.
tab is a variable. It is apparently a data.frame. tab[[mvar]] is an expression
that retrieves part of the data in the tab data.frame. The data it returns is a
vector, not a data.frame.
The "[[" operator extracts
On 05/02/2016 17:10, emmanuelle morin wrote:
ok, it will still make sense ?
Whether it makes sense to correlate the people rather than the variables
depends on the underlying science which (a) we do not know, and (b) is
not really an R question.
Le 05/02/2016 15:31, Michael Dewey a écr
ok, it will still make sense ?
Le 05/02/2016 15:31, Michael Dewey a écrit :
Assuming your dataset is in a matrix you want to transpose it. So you
can go t(mesdonnees) and then call cor on that.
On 05/02/2016 14:06, emmanuelle morin wrote:
Hello,
I have a set of 12 individuals with thousands
You are trying to use shortcuts where shortcuts are not appropriate
and having to go a lot longer around than if you did not use the
shortcut, see fortune(312).
You should really reread the help page: help("[[") and section 6.1 of
An Introduction to R.
Basically you should be able to do something
Hello,
I have over 1000 csv data sets I need to read into R, so I want to read
them in using a loop. The data sets are named as
pheno_1000ind_4000m_add_h70_prog_1_2.csv,
pheno_1000ind_4000m_add_h70_prog_1_3.csv, ... so I need 2 loops (for the
last 2 numbers in the names). What I would like to do is
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