I am trying to understand how to fit an ARMAX model with the arima
function from the stats package. I tried the simple data below, where
the time series (vector x) is generated by filtering a step function
(vector u, the exogenous signal) through a lowpass filter with AR
coefficient equal to 0.8.
Hey
Sorry if it was not very clear, I assumed few things. Normalization is
actually subjective, I am basically comparing data from two similar
experiments and would like to normalize for the systematic errors which
might affect the actual result and inference.
What I am looking for is some exampl
Hi
r-help-boun...@r-project.org napsal dne 13.11.2009 18:54:05:
>
> Ok Jim it worked, thank you! it´s funny because it worked with the first
> syntax in some cases...
you can use another approach in this case
P<-max(c(P1,P2))
Regards
Petr
>
>
> anna_l wrote:
> >
> > Hello, I am getting a
Dear All,
How to use SQL code in R?
Thanks!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
What do you mean with "normalization"?
Maybe you are looking for "scale" function on R.
El dom, 15-11-2009 a las 16:29 -0800, Abhishek Pratap escribió:
> Hi All
>
> I am looking for some resource to learn data normalization. I understand I
> am talking very broad here, I need something like a p
Hi, David,
The matrix has 20 columns.
Thank you very much for your help. I think it's right, but it seems I need
some time to figure it out. I am a green hand. There are so many functions
here I never used before. :)
Cindy
On Sun, Nov 15, 2009 at 5:19 PM, David Winsemius wrote:
> Assuming that
On Sun, 15 Nov 2009 14:33 -0500, "Jacob Wegelin"
wrote:
>
> Often I perform the same task on a series of variables in a dataframe,
> by looping through a character vector that holds the names and using
> paste(), eval(), and parse() inside the loop.
>
> For instance:
>
> rm(environmental)
> the
gcc has options like -MM, which can generate the dependence files for
a C/C++ file that I can be used by gnu make. I'm wondering if there is
a tool that can generate dependence file for an R script.
For example, I have an R script test.R
#test.R
load('input.RData')
save.image('output.RDat
Hi Ron,
Yes. Take a look at
?tolower
HTH,
Jorge
On Sun, Nov 15, 2009 at 8:45 PM, RON70 <> wrote:
>
> I have a vector of letters like c("a", "B", "c"). Is there any R function
> to
> force all elements to lower-cap ?
>
> Thanks,
> --
> View this message in context:
> http://old.nabble.com/How-
try this:
> x <- c("a", "B", "c")
> ?tolower
> tolower(x)
[1] "a" "b" "c"
>
On Sun, Nov 15, 2009 at 8:45 PM, RON70 wrote:
>
> I have a vector of letters like c("a", "B", "c"). Is there any R function to
> force all elements to lower-cap ?
>
> Thanks,
> --
> View this message in context:
> http
I have a vector of letters like c("a", "B", "c"). Is there any R function to
force all elements to lower-cap ?
Thanks,
--
View this message in context:
http://old.nabble.com/How-to-make-all-elements-all-elements-lower-cap---tp26365794p26365794.html
Sent from the R help mailing list archive at N
On 15/11/2009 8:15 PM, Peng Yu wrote:
Suppose I have a string variable
string='some_string'
Now I want to have a list, where is the same as the string in
the variable string. I'm wondering if this is possible in R.
list(=1:3)
data.frame(=1:3)
The most straightforward way is
x <- list(1:3)
Assuming that the number of columns is 4, then consider this approach:
> prs <-scan()
1: 2 5 1 6
5: 1 7 8 2
9: 3 7 6 2
13: 9 8 5 7
17:
Read 16 items
prmtx <- matrix(prs, 4,4, byrow=T)
#Now make copus of x.y and y.x
pair.str <- sapply(1:nrow(prmtx), function(z) c(apply(combn(prmtx[z,],
2), 2,f
Suppose I have a string variable
string='some_string'
Now I want to have a list, where is the same as the string in
the variable string. I'm wondering if this is possible in R.
list(=1:3)
data.frame(=1:3)
__
R-help@r-project.org mailing list
https://
I could of course be wrong but have you yet specified the number of
columns for this pairing exercise?
On Nov 15, 2009, at 5:26 PM, cindy Guo wrote:
Hi, All,
I have an n by m matrix with each entry between 1 and 15000. I want
to know
the frequency of each pair in 1:15000 that occur togethe
Hi, Charlie,
Thank you for the reply. Maybe I don't need the frequency of each pair. I
only need the top, say 50, pairs with the highest frequency. Is there anyway
which can avoid calculating for all the pairs?
Thanks,
Cindy
On Sun, Nov 15, 2009 at 4:18 PM, cls59 wrote:
>
>
>
> cindy Guo wrote
On Sun, Nov 15, 2009 at 6:05 PM, David Winsemius wrote:
>
> On Nov 15, 2009, at 6:35 PM, Peng Yu wrote:
>
>> My question was from replacing a pattern by '\\'. How to replace '/'
>> in string by '\'?
>>
>> string='abc/efg'
>> gsub('/','\\',string)
>
> No, that was most definitely _not_ your posed
Hi All
I am looking for some resource to learn data normalization. I understand I
am talking very broad here, I need something like a primer to give me a jump
start. If you happen to know any good resource please do let me know.
Cheers,
-Abhi
[[alternative HTML version deleted]]
___
Hope this help:
> m <- matrix(c(2,1,3,9,5,7,7,8,1,8,6,5,6,2,2,7),4,4)
> p <- c(2, 6)
> apply(m == p[1], 1, any) & apply(m == p[2], 1, any)
[1] TRUE FALSE TRUE FALSE
If you want the number of rows which contain the pair, sum() could be used:
> sum(apply(m == p[1], 1, any) & apply(m == p[2], 1,
cindy Guo wrote:
>
> Hi, All,
>
> I have an n by m matrix with each entry between 1 and 15000. I want to
> know
> the frequency of each pair in 1:15000 that occur together in rows. So for
> example, if the matrix is
> 2 5 1 6
> 1 7 8 2
> 3 7 6 2
> 9 8 5 7
> Pair (2,6) (un-ordered) occurs toget
On Nov 15, 2009, at 6:35 PM, Peng Yu wrote:
My question was from replacing a pattern by '\\'. How to replace '/'
in string by '\'?
string='abc/efg'
gsub('/','\\',string)
No, that was most definitely _not_ your posed question. If you want
now to change your question and supply a reproducib
Regular expression needs double the '\' again, so try this:
gsub('/','',string)
On Mon, Nov 16, 2009 at 7:35 AM, Peng Yu wrote:
> My question was from replacing a pattern by '\\'. How to replace '/'
> in string by '\'?
>
> string='abc/efg'
> gsub('/','\\',string)
> ___
My question was from replacing a pattern by '\\'. How to replace '/'
in string by '\'?
string='abc/efg'
gsub('/','\\',string)
On Sun, Nov 15, 2009 at 5:07 PM, David Winsemius wrote:
> ?cat
>
>> cat("\\")
> \
>
>
> On Nov 15, 2009, at 5:30 PM, Peng Yu wrote:
>
>> I can not get the string '\'. Cou
Hi, All,
I have an n by m matrix with each entry between 1 and 15000. I want to know
the frequency of each pair in 1:15000 that occur together in rows. So for
example, if the matrix is
2 5 1 6
1 7 8 2
3 7 6 2
9 8 5 7
Pair (2,6) (un-ordered) occurs together in rows 1 and 3. I want to return
the val
I have some beginner's questions regarding local, in the docs, it says that
"local evaluates an expression in a local environment".
Q1: why is B different from A? In B, is a<-a+1 getting evaluated
before eval proceeds?
#A
a=0
eval(quote(a<-a+1),new.env())
a # 0
#B
a=0
eval(a<-a+1,new.env())
a #
?cat
> cat("\\")
\
On Nov 15, 2009, at 5:30 PM, Peng Yu wrote:
I can not get the string '\'. Could somebody let me know how to get
it?
print('\')
+
+
print('\\')
[1] "\\"
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/list
There are probably better ways but cant you subset each parameter? So create
new variables for parameter 1, 2, ... and look at the summary data for those
which will include a min and max for all variables.
Joe King
206-913-2912
j...@joepking.com
"Never throughout history has a man who lived a life
I can not get the string '\'. Could somebody let me know how to get it?
> print('\')
+
+
> print('\\')
[1] "\\"
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/p
On Nov 15, 2009, at 4:11 PM, martin08 wrote:
Hi all,
I am new to R. Can someone please give me some hints in how to do the
following things:
1- Get ONE permutation of a set. I have looked at the gregmisc
package's
permutations() method, but I just want to get one permutation at a
time.
Hi all,
I am new to R. Can someone please give me some hints in how to do the
following things:
1- Get ONE permutation of a set. I have looked at the gregmisc package's
permutations() method, but I just want to get one permutation at a time.
2- Simulate a Markov chain in R. For instance, I wan
The term "line" in R refers to a sequence of text ending in and
marker, which I doubt is what you meant. Please stop referring
to the items or elements of a list as "lines" if you hope to
communicate with R-users.
> lista[1:2]
[[1]]
[1] 2 4 5 5 6
[[2]]
[1] 3 5 4 2
> lista[ which(sapply(
It's excellent!
Now, if I have a vector k=c( TRUE, TRUE, FALSE) how I may get lines from
list?
which (list ?? k) ?
David Winsemius wrote:
>
>
> On Nov 15, 2009, at 10:47 AM, Grzes wrote:
>
>>
>> But it's not what I wont
>>
>> I need get a number of line my list
>> 5 is in: list[[1]][1] and
Hi,
I am trying to find a model of best fit, with 8 parameters. As such, I have
created a table with 2^8=256 rows with every variable either in or out of
the model(denoted by 1 or 0), and for each row, I have computed the Adjusted
R^2, AIC, CP and Press. I know I can use the leaps package to find
this works perfectly...
new_data5 <- new_data4[nchar(new_data4$date_abandoned) != 8, ]
...and i can now think of a few different ways to manipulate my data with
what ive learned from these tricks, thanks alot David!
David Winsemius wrote:
>
>
> On Nov 15, 2009, at 11:00 AM, frenchcr wrote
Dear Antonio,
Yes, I am currently working on these extensions. It will take some time before
those functions are sufficiently tested and documented and can become part of
the metafor package. My goal is to do so within the next year, but I cannot
give any specific date for when this will be fin
Often I perform the same task on a series of variables in a dataframe,
by looping through a character vector that holds the names and using
paste(), eval(), and parse() inside the loop.
For instance:
rm(environmental)
thesevars<-names(environmental)
environmental$ToyReal <-rnorm(nrow(environmen
Hi Everyone,
I am trying to understand the unstack() function but after struggling for
two days, I have given up. More specifically, I am trying the exercises at
the end of Chapter 1 of Data Analysis and Graphics Using R by Maindonald
and Braun, 2nd ed. Exercise 18 (p. 41) asks to unstack the Rab
On Sat, Oct 3, 2009 at 9:18 AM, chi ball wrote:
>
>
>
> Hi, I'm not able to find a rpm of "gregmisc" library (>2.0.0) for Linux
> Mandriva 2008 Spring.
> Any suggestion?
> Thanks
>
If you don't find an up to date RPM, either you have to learn how to
build an RPM or just install the package yours
Dimitris,
Thanks, I shall give this a try as an alternative.
Graham
2009/11/15 Dimitris Rizopoulos :
> try the following:
>
> LL <- c(12.5,17,12,11.5,9.5,15.5,16,14)
> n <- length(LL)
> N <- 1000
> threshold <- 10
>
> smpls <- sample(LL, N*n, replace = TRUE)
> dim(smpls) <- c(n, N)
> cnt <- sum(
On Sun, Nov 15, 2009 at 7:19 AM, Green, Gerwyn (greeng6)
wrote:
> Dear all
>
> this is a question of model specification in lme which I'd for which I'd
> greatly appreciate some guidance.
>
> Suppose I have data in long format
>
> gene treatment rep Y
> 1 1 1 4.32
> 1 1 2
David,
Thanks, its me getting mixed up I actually meant less than or equal to
10. That apart, I guess the code is OK, I just expected, especially
as I increased N that I might have got some means less than 10, but
having gone back to it , I see I need a million iterations before
getting two means
try the following:
LL <- c(12.5,17,12,11.5,9.5,15.5,16,14)
n <- length(LL)
N <- 1000
threshold <- 10
smpls <- sample(LL, N*n, replace = TRUE)
dim(smpls) <- c(n, N)
cnt <- sum(colMeans(smpls) > threshold)
cnt
I hope it helps.
Best,
Dimitris
Graham Smith wrote:
I am trying to modify some cod
Thanks Duncan and David
for opening my eyes :-). It took quite a while but I think I learned a
lot about lm today. I used your advice to produce "added variable
plots" as mentioned here [1], [2]. I would bet someone did it in R
already (may leverage.plot in car) but it was worth doing it myself.
On Nov 15, 2009, at 12:12 PM, Graham Smith wrote:
I am trying to modify some code from Good 2005.
I am trying to resample the mean of 8 values and then count how many
times the resampled mean is greater than 10. But my count of means
above 10 is coming out as zero, which I know isn't correct.
On 15/11/2009 11:28 AM, Duncan Murdoch wrote:
On 15/11/2009 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit <- lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term g
I am trying to modify some code from Good 2005.
I am trying to resample the mean of 8 values and then count how many
times the resampled mean is greater than 10. But my count of means
above 10 is coming out as zero, which I know isn't correct.
I would appreciate it if someone could look at the c
I have updated R 2.9.1 to 2.10.0. and JGR GUI 1.7. I am running Windows XP.
I can't seem to get the JGR "Print" or "Help" functions to work. The system
locks and requires me to "stop the process".
In the past I have preferred the opreation and "feel" of JGR GUI. I realize
that this help forum
On Nov 15, 2009, at 11:15 AM, Karsten Weinert wrote:
Hello David,
of course I read those help pages and searched the r-help archive.
But did you look at the as.formula page?
I agree that the way should be clear, but I am stuck and looking for
help. Here is reproducible code
And here is
On Nov 15, 2009, at 11:00 AM, frenchcr wrote:
Yes they are not in date format, theyre just characters.
the earliest date is 1601 i originally had one of 0101 00 00
(101 years
BC)...this was a software problem.
table(nchar(new_data4$date_abandoned))
2 8
315732263
The
On Nov 15, 2009, at 10:47 AM, Grzes wrote:
But it's not what I wont
I need get a number of line my list
5 is in: list[[1]][1] and list[[2]][1] so
I would like to get a vector k = 1,2
I am sorry. I do not understand what you want the second solution
offered gave you numbers (and they were
Yes they are not in date format, theyre just characters.
the earliest date is 1601 i originally had one of 0101 00 00 (101 years
BC)...this was a software problem.
> table(nchar(new_data4$date_abandoned))
2 8
315732263
The 315732 are empty fields i thought.
The 263 are da
But it's not what I wont
I need get a number of line my list
5 is in: list[[1]][1] and list[[2]][1] so
I would like to get a vector k = 1,2
David Winsemius wrote:
>
>
> On Nov 15, 2009, at 10:01 AM, Grzes wrote:
>
>>
>> I heve got a list:
>>
>> lista=list()
>> a=c(2,4,5,5,6)
>> b=c(3,5,4,2)
So here's the funny thing: I've now ran my function 5 times, and in each one
of them R crashes after I got around 20.000 distances. It could be google,
but then as soon as I quit and launch R again, I manage to get another
20.000 distances. So maybe it does have something to do with the memory
usag
On 15/11/2009 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit <- lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as string, like
removeTerm <- function(li
Hello David,
of course I read those help pages and searched the r-help archive.
I agree that the way should be clear, but I am stuck and looking for
help. Here is reproducible code
removeTerm1 <- function(linModel, termName) { update(linModel, .~. - termName) }
removeTerm2 <- function(linModel,
Thanks to all
R is fantastic but ... not easy to know all possible terms ;-)
Knut
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
On Thu, Nov 12, 2009 at 11:29:50AM -0500, Mark Kimpel wrote:
> I am using code that previously worked to remove stopwords using package "tm".
Thanks for reporting. This is a bug in the removeWords() function in
tm version 0.5-1 available from CRAN:
> require(tm)
> myDocument <- c("the rain in Spa
On Nov 15, 2009, at 10:18 AM, wrote:
This is a very simple question but I couldn't form a site search
quesry that would return a reasonable result set.
Say I have a vector:
x <- c(0,2,3,4,5,-1,-2)
I want to replace all of the values in 'x' with the log of x.
Naturally this runs into pro
G'day Kevin,
On Sun, 15 Nov 2009 7:18:18 -0800
wrote:
> This is a very simple question but I couldn't form a site search
> quesry that would return a reasonable result set.
>
> Say I have a vector:
>
> x <- c(0,2,3,4,5,-1,-2)
>
> I want to replace all of the values in 'x' with the log of x.
>
On Nov 15, 2009, at 10:01 AM, Grzes wrote:
I heve got a list:
lista=list()
a=c(2,4,5,5,6)
b=c(3,5,4,2)
c=c(1,1,1,8)
lista[[1]]=a
lista[[2]]=b
lista[[3]]=c
lista
[[1]]
[1] 2 4 5 5 6
[[2]]
[1] 3 5 4 2
[[3]]
[1] 1 1 1 8
I would like to know where is number 5 (which line)?
For example I ha
Google "R graph grallery"
Google "R ggplot2"
Google "R lattice"
and good luck
milton
On Sun, Nov 15, 2009 at 7:48 AM, Sunita22 wrote:
>
> Hello
>
> My data contains following columns:
>
> 1st column: Posts (GM, Secretary, AM, Office Boy)
> 2nd Column: Dept (Finance, HR, ...)
> 3rd column: Tasks
This is a very simple question but I couldn't form a site search quesry that
would return a reasonable result set.
Say I have a vector:
x <- c(0,2,3,4,5,-1,-2)
I want to replace all of the values in 'x' with the log of x. Naturally this
runs into problems since some of the values are negative
I heve got a list:
lista=list()
a=c(2,4,5,5,6)
b=c(3,5,4,2)
c=c(1,1,1,8)
lista[[1]]=a
lista[[2]]=b
lista[[3]]=c
> lista
[[1]]
[1] 2 4 5 5 6
[[2]]
[1] 3 5 4 2
[[3]]
[1] 1 1 1 8
I would like to know where is number 5 (which line)?
For example I have got a loop:
k= vector(mode
You need to review:
?update
?update.formula
?as.formula
The way should be clear at that point. If not, then include a
reproducible data example to work with.
--
David
On Nov 15, 2009, at 9:23 AM, Karsten Weinert wrote:
Hello,
at the Rgui command line I can easily remove a term from a fitt
On Nov 14, 2009, at 7:02 PM, Greg Riddick wrote:
Hello all,
I'm trying to run lda() from the MASS library but the Help example
generates
the following error:
#Code from example in lda Help file
# Resulting Error
Error in if (targetlist[i] == stringname) { : argument is of length
zero
Hello,
at the Rgui command line I can easily remove a term from a fitted lm
object, like
fit <- lm(y~x1+x2+x3, data=myData)
update(fit, .~.-x1)
However, I would like to do this in a function with term given as string, like
removeTerm <- function(linModel, termName) { ??? }
removeTerm(fit, "x1")
On Nov 14, 2009, at 8:43 PM, frenchcr wrote:
sorry David,
im really new to R (my first week) and appreciate your help. Also I
dont
always know what info to give people on the forum (although im
starting to
catch the drift).
heres what i get...
summary(new_data4$date_abandoned)
Min.
David Winsemius comcast.net> writes:
>
> It's really not that difficult to get the variance covariance matrix.
> What is not so clear is why you think differential weighting of a set
> that has a perfect fit should give meaningfully different results than
> a fit that has no weights.
Aga
Dear all
this is a question of model specification in lme which I'd for which I'd
greatly appreciate some guidance.
Suppose I have data in long format
gene treatment rep Y
11 1 4.32
11 2 4.67
11 3 5.09
.. ..
.. ..
.
Hello
My data contains following columns:
1st column: Posts (GM, Secretary, AM, Office Boy)
2nd Column: Dept (Finance, HR, ...)
3rd column: Tasks (Open the door, Fix an appointment, Fill the register,
etc.) depending on the post
4th column: Average Time required to do the task
On Sun, Nov 15, 2009 at 11:57 AM, Dimitri Szerman wrote:
>
> Thanks. The reason I didn't want to do something like that is because, in
> the event of a crash, I'll loose everything that was done. That's why I
> though of appending the results often.
Oops yes, I missed the 'append=TRUE' flag. Th
Peter Dalgaard biostat.ku.dk> writes:
>
> The point is that R (as well as almost all other mainstream statistical
> software) assumes that a "weight" means that the variance of the
> corresponding observation is the general variance divided by the weight
> factor. The general variance is still
How are you executing your script? Are you doing a cut/paste into the
command window? On WIndows using Tinn-R, this procedure will invoke
'browser' and it will then continue to read the rest of the script
since it is coming from the standard input. The way around it is to
put the script in a fil
2009/11/15 Barry Rowlingson
> On Sun, Nov 15, 2009 at 11:10 AM, Dimitri Szerman
> wrote:
> > Hello,
> >
> > This is what I am trying to do: I wrote a little function that takes
> > addresses (coordinates) as input, and returns the road distance between
> > every two points using Google Maps. Cat
On Sun, Nov 15, 2009 at 11:10 AM, Dimitri Szerman wrote:
> Hello,
>
> This is what I am trying to do: I wrote a little function that takes
> addresses (coordinates) as input, and returns the road distance between
> every two points using Google Maps. Catch is, there are 2000 addresses, so I
> have
Hello,
This is what I am trying to do: I wrote a little function that takes
addresses (coordinates) as input, and returns the road distance between
every two points using Google Maps. Catch is, there are 2000 addresses, so I
have to get around 2x10^6 addresses. On my first go, this is what I did:
sorry David,
im really new to R (my first week) and appreciate your help. Also I dont
always know what info to give people on the forum (although im starting to
catch the drift).
heres what i get...
summary(new_data4$date_abandoned)
Min.1st Qu.Median Mean 3rd Qu. Max.
I need to use browser() to stop a while loop to input some value for the
loop. But the browser() just will not stop until the last line of the code.
Does anyone know the possible reason? I use ggobi in the loop, and open a
few ggobi windows before the browser(), will that be the reason?
Thanks A
78 matches
Mail list logo
