On Thu, 01-Dec-2011 at 10:13AM -0800, lglew wrote:
|> Hi R-users,
|>
|> I'm trying to produce decompositions of a multiple time-series, grouped by a
|> factor (called "area"). I'm modifying the code in the STLperArea function of
|> package ndvits, as this function only plots produces stl plots, i
No.
If you want a more informative answer, try following the recommendations in the
posting guide mentioned below.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#.
On Dec 3, 2011, at 10:23 PM, grttt nbbfg wrote:
Sorry for my English, is not my first language..
I have some trouble in terms of using integrate function in R.
fx is a function of m and x where m is supposed to be a unknown
parameter.
R is not an algebraic solver (and R-help is not a homew
On Dec 3, 2011, at 5:42 PM, napps22 wrote:
Dear R users,
I'm trying to carry out monte carlo integration of a posterior density
function which is the product of a normal and a gamma distribution.
The
problem I have is that the density function always returns 0. How
can I
solve this proble
Sorry for my English, is not my first language..
I have some trouble in terms of using integrate function in R.
fx is a function of m and x where m is supposed to be a unknown parameter.
>fx=function(m,x){
+ x*2/(3*m)*(1-x/(3*m))
+ }
The problem is in upper bound, it depends on parameter m.
>in
Hi -
First post, so excuse any errors in protocol:
Wanted to ask if there's an easy way to use 'predict' with objects of class
'censReg', 'maxLik', 'maxim' or 'list'.
Have a left-censored dataset, attempting to use a Tobit model and am working
with the censReg package. I like how easy it is to
Hi,
I apologies for my naive doubts
I have been trying it from a while. I need the area between the curve based
on conditions i wrote (whenever red line is above the blue line then area
between curves to be grey color and whenever blue line is above red line
then area between curves to be red col
Greetings, I have a question that I'd like to get input on. I have a
classic toxicology study where I artificially fertilized and exposed
embryos to a chemical and counted defects. In addition, I kept track of
male-female pairs that I used to artificially fertilize and generate
embryos with. I need
Dear R users,
I'm trying to carry out monte carlo integration of a posterior density
function which is the product of a normal and a gamma distribution. The
problem I have is that the density function always returns 0. How can I
solve this problem?
Here is my code
#generate data
x1 <- runif(100
Hello,
I use the command rnorm, and I feed these results into a lmer command.
Since I am using the rnorm command I expect to get different results for
each iteration, yet for each iteration I am getting the same answer. If
someone understands why I am getting the same answer every time with a
rand
On Dec 3, 2011, at 5:28 PM, avinash barnwal wrote:
Hi Weylandt,
I tried it but i was not successful.
Here is the full code
Any help would be great.
time<-c("02-Jan-05","09-Jan-05","16-Jan-05","23-Jan-05","30-
Jan-05",
A great function for extracting pattern matches is 'm()'
library(caroline)
vect <- m('([xX][0-9])',df$Input)
toupper(vect) #in case you really want all upper case x's
It does the hard work of using 'sub' to remove the non-matching parts
(sub, grep, regexpr, etc aren't very good for this sort
Forgot, you can also do this:
> dcast(x.m, Cluster ~ value, fun = length)
Cluster ind1 ind2 ind3
1 1110
2 2101
3 3110
>
On Sat, Dec 3, 2011 at 7:02 PM, jim holtman wrote:
> try this:
>
>> x <- read.table(text = "Cluster Member1 Member
try this:
> x <- read.table(text = "Cluster Member1Member2
+ 1ind1 ind2
+ 2ind3 ind1
+ 3ind2 ind1", as.is = TRUE, header = TRUE)
> require(reshape2)
> x.m <- melt(x, id = "Cluster")
> x.m
Cluster variable value
1 1 Me
Pivot tables are an Excel concept, not an R concept.
That means you must give an example of your starting pivot table as an R
object (use dump() so we can pick it up from the email and execute it
immediately).
and an example of the R object you want as the result.
Use a trivial but complete exampl
use a < ? > to get help on a function; example:
?read.table
If you do this you will see an option called "header"...
use header=T if your top row contains column names.
Learn how to read these help pages. Also, read thru a few beginner R
manuals and see this website:
http://www.statmethods.
Hello R-users,
I've got a huge table with about 20.00 rows and 50 columns. The table now
has headers as Members1, Members2 etc. My data are 8 different individuals.
And I've got a column with clusters. So each individual belongs to different
clusters and can occurs multiple times within a cluster
Hi Weylandt,
I tried it but i was not successful.
Here is the full code
Any help would be great.
time<-c("02-Jan-05","09-Jan-05","16-Jan-05","23-Jan-05","30-Jan-05","06-Feb-05","13-Feb-05","20-Feb-05","27-Feb-05","06-Mar-0
Hi there,
I'm looking to analyze a set of data on local gas prices for a single day.
I'm wondering what kind of questions I should be looking to ask and how to
find and answer to them with R. Examples would be:
Do prices differ between brands?
Does location affect (NE, NW, SE, SW) price?
Does th
? polygon
example(polygon)
Michael
On Sat, Dec 3, 2011 at 4:08 PM, avinash barnwal
wrote:
> Hi all,
>
> I have been trying to shade the specific part of the plot
>
> Part to be shaded
>
> 1. Any color whenever a2>a3
>
> 2. Any other color( Not same as 1) whenever a2
> Suggest me some code for th
Hi all,
I have been trying to shade the specific part of the plot
Part to be shaded
1. Any color whenever a2>a3
2. Any other color( Not same as 1) whenever a2<>__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
I don't know the answer, but would suppose not.
You could test this for yourself using:
system.time()
example:
system.time(rnorm(10,0,1))
Chega wrote
>
> This solved my problem - Thanks a lot for your help! Please allow me one
> more question: Works zoo's rollapply on a plain matrix faster
Because lapply() tries to pass an argument to FUN and there's none
that it can receive.
This would work however:
lapply(rep(100, 6), rnorm, mean = 1, sd = 2)
Michael
On Sat, Dec 3, 2011 at 2:42 PM, B77S wrote:
> Interesting and thank you; I'm confused as to why this doesn't work with:
>
> lapp
And with replicate:
replicate(100, rnorm(100, 1,2), simplify = FALSE)
Michael
On Sat, Dec 3, 2011 at 2:38 PM, andrija djurovic wrote:
> Hi Brad. Maybe something like this:
>
> lapply(rep(1,6), function(x) rnorm(10,0,1))
>
> Andrija
>
> On Sat, Dec 3, 2011 at 8:21 PM, B77S wrote:
>
>> Hi Micha
Interesting and thank you; I'm confused as to why this doesn't work with:
lapply(rep(1,6), FUN=rnorm, n=10, mean=1.0, sd=1)
andrija djurovic wrote
>
> Hi Brad. Maybe something like this:
>
> lapply(rep(1,6), function(x) rnorm(10,0,1))
>
> Andrija
>
> On Sat, Dec 3, 2011 at 8:21 PM, B77S <
Hi Brad. Maybe something like this:
lapply(rep(1,6), function(x) rnorm(10,0,1))
Andrija
On Sat, Dec 3, 2011 at 8:21 PM, B77S wrote:
> Hi Michael,
> How would you do this with lapply to return a list?
> I can't seem to get that to work (I haven't used these much and am trying
> to
> learn).
> T
Hi Michael,
How would you do this with lapply to return a list?
I can't seem to get that to work (I haven't used these much and am trying to
learn).
Thanks
Brad
Michael Weylandt wrote
>
> ? replicate
>
> or a for loop
>
> or do all one hundred simulations at once
>
> x <- matrix(rnorm(100^2,
Hi. One approach is using replicate. See ?replicate:
replicate(3,rnorm(100,1,2))
Andrija
On Sat, Dec 3, 2011 at 7:10 PM, Martin Zonyrah wrote:
> Hi,
> I need help. I am trying to iterate this command x <- rnorm(100, 1.0,
> 2.0) one hundred times in R but I don't seem to have a clue.
> Can anyo
On Dec 3, 2011, at 1:10 PM, Martin Zonyrah wrote:
Hi,
I need help. I am trying to iterate this command x <- rnorm(100,
1.0, 2.0) one hundred times in R but I don't seem to have a clue.
?replicate
==
David Winsemius, MD
West Hartford, CT
__
On Thu, Dec 1, 2011 at 10:32 AM, Douglas Esneault
wrote:
> I am new to R but am experienced SAS user and I was hoping to get some help
> on counting the occurrences of a character within a string at a row level.
>
> My dataframe, x, is structured as below:
>
> Col1
> abc/def
> ghi/jkl/mno
>
> I
? replicate
or a for loop
or do all one hundred simulations at once
x <- matrix(rnorm(100^2, 1, 2), 100)
It's going to depend on what you want to do with the numbers.
Michael
On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah wrote:
> Hi,
> I need help. I am trying to iterate this command x <-
Hi,
Thank Michael for your help.
Komine
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.
Hi,
I need help. I am trying to iterate this command x <- rnorm(100, 1.0, 2.0) one
hundred times in R but I don't seem to have a clue.
Can anyone help?
Your help is very much appreciated.
Martin
[[alternative HTML version deleted]]
__
R-help@
I can tell that you are puzzled and confused. Unfortunately, I am not psychic,
so I cannot tell what you did, and therefore cannot tell where you went astray.
The solution is for you to read the posting guide mentioned at the bottom of
every R-help message. Spend a little time to create a small
Great, thanks for answering this basic question.
On Sat, Dec 3, 2011 at 5:12 AM, Duncan Murdoch-2 [via R] <
ml-node+s789695n4152880...@n4.nabble.com> wrote:
> On 11-12-02 10:25 PM, stillerfan wrote:
> > I'm a new user to R and recently installed R 2.14 for Windows 7 (64
> bit). I
> > have downl
On Dec 3, 2011, at 8:41 AM, syrvn wrote:
Hello,
imagine the following data.frame:
ID name
1 *_A
2 *_A
3 *_B
4 *_B
* = can be any pattern
I want to replace every row which ends with _A by 1 and every row
which ends
by _B with a 0
You can use grep(patt, x, value=TRUE) to return vlaues t
Hello,
imagine the following data.frame:
ID name
1 *_A
2 *_A
3 *_B
4 *_B
* = can be any pattern
I want to replace every row which ends with _A by 1 and every row which ends
by _B with a 0
so that the data.frame looks like the following:
ID name
1 1
2 1
3 0
4 0
Which function do I use best
Thanks for your reply.
I tried to use the postForm function of RCurl as below but do not have much
clue as to how to go further with what it does.
library(RCurl)
stn<-"03015795"
myurl<-paste("http://ida.water.usgs.gov/ida/available_records.cfm?sn=",stn,sep="";)
mypage1 = readLines(myurl)
# Get
Dear R users,
I am a beginner in R trying to apply the thin plate spline method to my climate
data. I used the example in R to do so, and the lines seem to run fine ( I am
not getting errors) but I am not getting any output in the form of graph or
anything. I got a warning message saying that
Thanks for your suggestions. I will try them.
The "-" in my original post was actually only there to serve as a separator
so that it is easier for you to see the data structure but apparently it
rather confused you... sorry :)
--
View this message in context:
http://r.789695.n4.nabble.com/Data-a
I just noticed that the question has already been answered on the R-help list
before:
http://r.789695.n4.nabble.com/Sweave-font-problems-with-Signif-codes-lines-td977346.html
Adding the following R line solves the problem:
options(useFancyQuotes = FALSE)
Mark
>
>
> On Fri, Dec 2, 2011 at 4:
Does this do it for you:
> #Creating example data in similar format to data I have
> sub <- rep(1,10)
> trial <- seq(1,10,1)
> size <- rep(3,10)
> shortest <- rep(444,10)
> startlab <- rep(444,10)
> endlab <- rep(444,10)
> startconf <- rep(444,10)
> endconf <- rep(444,10)
> steps <- sample(1:30,10
try this:
> match(x$Name> x <- read.table(text = "Name - Value
+ A - 400
+ A - 300
+ B - 200
+ B - 350
+ C - 500
+ C - 350
+ D - 450
+ D - 600
+ E - 700
+ E - 750
+ F - 630
+ F - 650", header = TRUE, as.is = TRUE)
> map <- data.frame(key = c("A", "B", "C", "D", "E", "F")
+ , valu
You can use gsub() to replace parts of strings.
Sarah
On Saturday, December 3, 2011, arunkumar wrote:
> Hi
>
> I have column name as given below
>
> If the variable is in log(X1 + 1) pattern it should be removed and i need
> only X1
>
> Input
> log(x1 + 1)
> x2
> log(X3 +1)
>
> Expected Outp
I doubt your data frame looks like that, with all the -, but regardless you
can use ifelse() to construct your column.
Sarah
On Saturday, December 3, 2011, syrvn wrote:
> Hello!
>
> I have a data.frame which looks like:
>
> Name - Value
> A - 400
> A - 300
> B - 200
> B - 350
> C - 500
> C - 350
It is not possible in ecodist, and I'd be wary of doing it in other
packages. Instead, I would consider the model matrix approach described in
Legendre and Fortin 1989.
Sarah
On Saturday, December 3, 2011, Nevil Amos wrote:
> I would like to perform partial mantel tests on only within group
val
David Epstein umich.edu> writes:
>
> Hello,
>
> I want to create side-by-side maps of similar attribute data in two
> different cities using a single legend.
>
> To simply display side-by-side census block group boundary
> (non-thematic) maps for Minneapolis & Cleveland I do the following:
>
Hello!
I have a data.frame which looks like:
Name - Value
A - 400
A - 300
B - 200
B - 350
C - 500
C - 350
D - 450
D - 600
E - 700
E - 750
F - 630
F - 650
I want to add another column where all A,B should get an index 1, all C,D an
index of 2 and all E,F an index of 3 so that the data.frame loo
On 12/03/2011 06:04 AM, Hadley Wickham wrote:
Hi all,
I was wondering if any one had scripts that they could share for
capturing the current version of R packages used for a project. I'm
interested in creating a project local library so that you're safe if
someone (e.g. the ggplot2 author) updat
On 11-12-02 10:25 PM, stillerfan wrote:
I'm a new user to R and recently installed R 2.14 for Windows 7 (64 bit). I
have downloaded and ran Rtools 2.14 but do not see any new packages in the
install packages section of R. I installed Rtools in a subdirectory to the
/library/ directory. Does a
On Sat, Dec 3, 2011 at 2:38 AM, Michael wrote:
> Hi all,
>
> Could you please help me?
>
> I am having the following weird problem when debugging R programs
> using "browser()":
>
> In my function, I've inserted a "browser()" in front of Step 1. My
> function has 3 steps and at the end of each ste
AFAICS what you mean is 'how can I fill in an HTML form using R'.
Answer: use package RCurl.
Do study the posting guide: none of the 'at a minimum' information was
given here.
On 03/12/2011 04:47, HC wrote:
Hi all,
I am trying to download some tab separated data from the internet. The data
I would like to perform partial mantel tests on only within group values, with
"between group" values assigned to NA.
This is possible in package ncf partial.mantel.test, however this sues a
different permutation to that used in ecodist.ecodist will not accept data with
NA values, returning a "
I have a multiple text files, separated by a single space, that I need to
combine into a single data.frame. Below is the track I'm on using
list.files to capture the names of the files and then lapply with
read.table. But I run into problems making a usable dataframe out of the
data.
#Creating
Hi
I have column name as given below
If the variable is in log(X1 + 1) pattern it should be removed and i need
only X1
Input
log(x1 + 1)
x2
log(X3 +1)
Expected Output X1 X2 X3
Please help me
--
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I'm a new user to R and recently installed R 2.14 for Windows 7 (64 bit). I
have downloaded and ran Rtools 2.14 but do not see any new packages in the
install packages section of R. I installed Rtools in a subdirectory to the
/library/ directory. Does anyone have any idea what I could be doing
Hi all,
I am trying to download some tab separated data from the internet. The data
is not available directly at the URL that could be known apriori. There is
an intermediate form where start and end dates have to be given to get to
the required page.
For example, I want to download data for a st
>
> Please note that your "df" and "M" are undoubtedly different
> objects by now:
>
> Right. Not my most coherent day.
thanks
W
> > M <- matrix(runif(5*20), nrow=20)
> > colnames(M) <- c('a', 'b', 'c', 'd', 'e')
> > l1 <- lm(e~., data=as.data.frame(M))
> > l1
>
> Call:
> lm(formula = e ~ .,
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