Dear List,
I am trying to estimate a 3 dimensional density through the DPpackage.
For example
# model
sigma <- matrix(c(0.1,0.05,0.05,0.05,0.1,0.05,0.05,0.05,0.1), ncol=3)
rnormm<- rmvnorm(n=100, mean=c(5,100,150), sigma=sigma)
sigma2 <- matrix(c(10,0.05,0.05,0.05,10,0.05,0.05,0.05,10), ncol=3)
I have a raster (which I called glc), which I read
into R as a raster with "raster.create.from.file" from the raster package
(R-forge). Values in glc range between 1 and 27 (whole numbers only). I'd like
to extract all cells with a value of 1 to create
a new raster with only the cells that have a
I'm trying to write a simple function with a data parameter.
tfun <- function(x, y, data = NULL) {
if(missing(data))
dt <- data.frame(x=x, group=y)
else {
dt <- with(data, data.frame(x=x, group=y))
}
return(dt)
}
If I pass variables
Hi everyone,
I'm trying to plot some data across time. I have a list of articles, ranks,
date/times, authors, etc. Someone suggested using zoo and someone suggested
using ts. I'm pretty new at this and have been trying a simple if() plot()
statement, but it doesn't seem to work. I keep getting an
Hi,
I hope you are fine. I am trying to use scan to open a file "prueba"
extension txt. I am using the scan command as
scan("C:/prueba")
and I get the following error:
Error in file(file, "r") : cannot open the connection
In addition: Warning message:
In file(file, "r") : cannot open file 'prueba
one way is the following:
m <- list(A = 1, B = 1:2, C = 1:3, D = 1:4)
n <- max(sapply(m, length))
t(sapply(m, function (x) c(x, rep(NA, n - length(x)
I hope it helps.
Best,
Dimitris
Daren Tan wrote:
I would like to convert a list to matrix. This can be easily achieved via
do.call. The
another option is to use relist(), e.g.,
x <- list(a=runif(10), b=runif(30), c=runif(25))
x <- as.relistable(x)
ux <- unlist(x)
ux[order(ux)[1:5]] <- -1
relist(ux)
Best,
Dimitris
jim holtman wrote:
Here is one way that might work.
x <- list(a=runif(10), b=runif(30), c=runif(25))
# unlist a
Greg wrote:
I'm trying to write a simple function with a data parameter.
tfun <- function(x, y, data = NULL) {
if(missing(data))
dt <- data.frame(x=x, group=y)
else {
dt <- with(data, data.frame(x=x, group=y))
}
return(dt)
}
If I
I think you need to put quotes around the column header name,
something like: NYT["Title"]
example:
> my.df <- data.frame(header1=1:3, header2=1:3)
> my.df[header2]
Error in `[.data.frame`(my.df, header2) : object "header2" not found
> my.df["header2"]
header2
1 1
2 2
3 3
Hope
Hello all,
I'm trying to calculate the standar desviation and I'm using the function
sd(x,na.rm=TRUE) and I have this error: Error in var(x, na.rm = na.rm) : no
complete element pairs . Why happen this?, What can I do to solve it?. x is
list of three numbers which I have from a table.
Thanks
Hi,
I am writing a function that plots many graphs so therefore I have it turn
recording on. The problem is; that when it is run again, the new graphs get
added to the old ones, doubling the amount recorded. I know that
.SavedPlots<-NULL removes the recorded plots but how can this be done from
insi
On Sat, 21 Feb 2009, Charles C. Berry wrote:
On Sat, 21 Feb 2009, Tal Galili wrote:
Hello dear R mailing list members.
I have recently became curious of the possibility applying model
selection algorithms (even as simple as AIC) to regressions of large
datasets.
Large in the sense of many
Hello all,
I'm trying to calculate the standar desviation with sd(x,na.rm=TRUE) and I
don't know why I have this error Error in var(x, na.rm = na.rm) : no complete
element pairs when I try to calculate it, I have been looking for information
about this error but nothing. Why it happens?. What
Exactly what I wanted. Thanks much.
-thomas
On Sat, 21 Feb 2009, Martin Maechler wrote:
"KK" == Ken Knoblauch
on Fri, 20 Feb 2009 14:11:55 + (UTC) writes:
KK> Thomas Lumley u.washington.edu> writes:
>> I want to construct a symmetric band matrix in the Matrix
>> pa
Hi Thomas,
What you just wrote is very interesting to me - do you have any suggestion
then as to how to implement leaps (or any other package/code) to iterate on
the final lm model produced by biglm ?
Any advice would be very welcomed!
p.s: My purpose is to use a different algorithm than that of
On Sun, 22 Feb 2009, Tal Galili wrote:
Hi Thomas,
What you just wrote is very interesting to me - do you have any suggestion
then as to how to implement leaps (or any other package/code) to iterate on
the final lm model produced by biglm ?
Any advice would be very welcomed!
If you look at th
see ?'<<-'
hth.
Will Stone schrieb:
Hi,
I am writing a function that plots many graphs so therefore I have it turn
recording on. The problem is; that when it is run again, the new graphs get
added to the old ones, doubling the amount recorded. I know that
.SavedPlots<-NULL removes the recorded
Thanks Thomas!
I'll give it a go and will send updates as to how I am doing.
Cheers,
Tal
On Sun, Feb 22, 2009 at 1:52 PM, Thomas Lumley wrote:
> On Sun, 22 Feb 2009, Tal Galili wrote:
>
> Hi Thomas,
>>
>> What you just wrote is very interesting to me - do you have any suggestion
>> then as t
Hi Sharai,
I think scan() needs the extension as well. Have you tried
scan("C:/prueba.txt")
(this assumes that prueba.txt resides in your c-root dir)
hth.
Sharai Gomez schrieb:
Hi,
I hope you are fine. I am trying to use scan to open a file "prueba"
extension txt. I am using the scan command a
Hi Thomas,
On second thought - I will need some more help:
What I only now noticed is that leaps is "an exhaustive search for the best
subsets of the variables in x for predicting y in linear regression", while
what I am aiming for (for now) is a method to implement forward selection (on
biglm) - a
hi,
I need to draw a line joining graphs,but abline stops within a graph.What do
i do to cover portions between two graphs
--
Rajesh.J
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
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Dear R users,
I have a very simple problem but I can't find the function in R to deal with
it.
I need to split (or decompose) one line into many lines using one field as a
reference.
I have a table with the following format:
A B Frequency
23 3 2
24 2 5
25 1 3
And need to split eac
I am given a text file of records to be converted into a table format.
I have searched related topics or packages, but can't find any similar
cases. Please help.
Sample record is given below. Take note the last element doesn't have
a semi colon.
###-Start of record--
On Sun, 22 Feb 2009, Tal Galili wrote:
Hi Thomas,
On second thought - I will need some more help:
What I only now noticed is that leaps is "an exhaustive search for the best
subsets of the variables in x for predicting y in linear regression", while
what I am aiming for (for now) is a method to
hi,
I am trying to plot a series of graph with varied titles (slightly).
So I was trying to format it and pass it to "main" in the plot
function, but it doesn't seem to work:
for example: i is variable part
plot( main=format(c("graph ", i)) ...)
what is the correct way of handling this?
T
Hi Oliver,
this a classic "paste" task
...main=paste("graph", i)...
hth.
Oliver schrieb:
hi,
I am trying to plot a series of graph with varied titles (slightly).
So I was trying to format it and pass it to "main" in the plot
function, but it doesn't seem to work:
for example: i is variable
Daren Tan gmail.com> writes:
>
> I am given a text file of records to be converted into a table format.
> I have searched related topics or packages, but can't find any similar
> cases. Please help.
>
> Sample record is given below. Take note the last element doesn't have
> a semi colon.
>
> #
Mafalda Viana-2 wrote:
>
>
> I need to split (or decompose) one line into many lines using one field as
> a
> reference.
>
> I have a table with the following format:
>
> A B Frequency
> 23 3 2
> 24 2 5
> 25 1 3
>
>
olddata[rep(1:nrow(olddata),olddata$B),]
repeats the rows o
Here is one way of reading in your data:
> input <- readLines(textConnection("###-Start of
> record--
+
+ Name: John
+ Height: 170cm
+ Weight: 70kg
+ Age: 30
+
+ Status: Married
+ Children: 2
+
+ Employment: Engineer
+
+ ###-End of record--
Thank you Tony. I tried that, but still got an error. This is what I got.
if(df["Title"]=="A")
+ plot(df["B"])
Error in `[.data.frame`(df, "Title") : undefined columns selected
I Think there is something wrong with my file because when I call [,1] It
gives me the whole data set, but when I cal
Hi, there
could you help me coding this problme?
I am just starting to leard the R. So I really need help
Question is from Ross, Simulation, 4th Edition. ch3
14.
with x1=23, x2=66
Xn=3*Xn-1+5*Xn-2 mod(100) n>=3
we will call the sequence Un=Xn/100 n>=1
fi
BTW, if we are on the subject, here is a cool trick I came a cross the other
day about creating a title with multi-coloured words.:
http://blog.revolution-computing.com/2009/01/multicolor-text-in-r.html
Tal
On Sun, Feb 22, 2009 at 4:13 PM, Eik Vettorazzi <
e.vettora...@uke.uni-hamburg.de> wr
You have not really made it clear what you are trying to do, and I don't see
the zoo vs ts involvement in your question.
Also, your test data and code snippet you give are not quite consistent.
Thus, my advice is really a long-shot guess.
Assume your data looks like:
Time Date Rank Topic Titl
Hi Michelle,
The r-sig-geo list would be the ideal place for this question since
the authors of 'raster' would be likely to see it.
I haven't used the raster package before, but this seems to work:
#install.packages("raster", repos="http://R-Forge.R-project.org";)
r <- raster()
r <- setValues(r
R-listers
I am still finding my way with R - and feel that I am making a complete dogs
dinner of something that should be pretty simple.
What I'd like to do is to create a simple function that i can use to
calculate compound growth rates (CAGRs) over a data frame. I'd like the
function to be fle
Great. That works.
thanks
Oliver
On Feb 22, 9:13 am, Eik Vettorazzi
wrote:
> Hi Oliver,
> this a classic "paste" task
>
> ...main=paste("graph", i)...
>
> hth.
>
> Oliver schrieb:
>
>
>
> > hi,
>
> > I am trying to plot a series of graph with varied titles (slightly).
> > So I was trying to fo
On Sun, 2009-02-22 at 20:52 +1100, Jim Lemon wrote:
> Greg wrote:
> > I'm trying to write a simple function with a data parameter.
> >
> > tfun <- function(x, y, data = NULL) {
> > if(missing(data))
> > dt <- data.frame(x=x, group=y)
> > else {
> > dt <- with(data, d
On 23/02/2009, at 6:57 AM, Ssophia wrote:
could you help me coding this problme?
I am just starting to leard the R. So I really need help
Question is from Ross, Simulation, 4th Edition. ch3
14.
with x1=23, x2=66
Xn=3*Xn-1+5*Xn-2 mod(100) n>=3
we will call the sequence Un
Hi all,
I have the next matrix:
ab c
1223
2342
3054
4523
5022
6721
7120
8919
I want to filter the rows with the values of b higher than 1 in a way tha
Hello all,
I've been able to separate the columns now so that when I call NYT[,9] it
gives me just that column. I'm still trying to plot however. When I used the
previous if()plot() statement it plots something, but its not what I want. I
want to plot the rank vs time, but its just plotting the sa
Sorry, forgot to attach the error it is giving me.
> if(NYT["Title"]=="A")
+ plot(NYT["B"],ylim=c(0,50),xlim=c(0,50),ylab="B",xlab="A")
Warning message:
In if (NYT["Title"] == "A") plot(NYT["B"], :
the condition has length > 1 and only the first element will be used
Thank you
Tony Breyal w
Dear Juan Pablo,
If "mymat" is your matrix, you could use something like this:
# Option 1
mymat[mymat$b>1,]
# Option 2
mymat[with(mymat,b>1),]
HTH,
Jorge
On Sun, Feb 22, 2009 at 1:10 PM, Juan Pablo Fededa wrote:
> Hi all,
>
> I have the next matrix:
>
>
> ab c
>
> 12
Did you read the Introduction to R and the use of indexing to 'filter'?
> x <- read.table(textConnection("ab c
+
+ 1223
+ 2342
+ 3054
+ 4523
+ 5022
+ 6721
+ 7120
+ 8
A not so elegant way of obtaining your result (compared to merge()) would be:
> t1$ val3<-rep(t2$val3, table(t1$loc))
> t1$ val4<-rep(t2$val4, table(t1$loc))
Eugen.
--- On Tue, 2/17/09, Monica Pisica wrote:
> From: Monica Pisica
> Subject: Re: [R] joining "one-to-many"
> To: ggrothendi...
What is NYT? Is it a dataframe? Is "Title" a column? What exactly
are you wanting to plot? Is it just the values in column "B" where
"Title" is equal to "A"? The posting guide does suggest that you
provide a reproducible set of code/data. You can probably do what you
want if you can explain w
Inspired by the exchange between Rolf Turner and Wacek Kusnierczyk, I
thought I'd clear up for myself the exact relationship among the
various sequence concepts in R, including not only generic vectors
(lists) and atomic vectors, but also pairlists, factor sequences,
date/time sequences, and diffti
hi,
you can use par(xpd=TRUE) and draw the joining line in every sub graph
- don't know if there is another way with simple graphs
eg:
par(mfcol=c(1,2))
plot(1,1,xlim=c(0,5),main="1st")
op<-par(xpd=TRUE)
abline(h=1)
plot(1,1,xlim=c(0,5),main="2nd")
abline(h=1)
hth.
rajesh j schrieb:
hi,
I
rajesh j gmail.com> writes:
> I need to draw a line joining graphs,but abline stops within a graph.What do
> i do to cover portions between two graphs
try par(xpd=NA)
?segments may be useful too
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R-help@r-project.org mailing list
https://stat.e
Dear all -
I would like to adjust the line type of specific contours in
contourplot from the lattice package, but it seems like lty does not
take a list in the call.
Here is my call to contourplot:
contourplot(preds~size+trt|Size.Name,
data=pred.dat,layout=c(2,4),
at=c(0.02
I think this was posted to the wrong list, so my followup is going to
R-devel.
On 22/02/2009 3:42 PM, Stavros Macrakis wrote:
Inspired by the exchange between Rolf Turner and Wacek Kusnierczyk, I
thought I'd clear up for myself the exact relationship among the
various sequence concepts in R, in
Hello, i need to find in time serie, k=1,2,3...(how if possible) most domain
frequencies and than smooth original serie by Fourier row y=mean +
a1*Cos(2*pi/freq*t)+b1*Sin(2*pi/freq*t)+a2..
numbers a1,b1 ... i will have from simple regresion, i need only to find k
msot domain frequensies a th
The only way I can figure out to do this is to use two calls to
panel.contourplot:
library(lattice)
x <- seq(-2, 2, length = 20)
y <- seq(-2, 2, length = 20)
grid <- expand.grid(x=x, y=y)
grid$z <- dnorm(grid$x) * dnorm(grid$y)
contourplot(z ~ x * y, grid,
panel = function(at, lty, col
Sundar -
Thanks for the help! Here is my modified code. Just what I wanted.
However, it is odd that the at= call in contourplot won't take a
single value - I have had to trick it into plotting the "0" contour
along with the 0.5 to make this work.
contourplot(preds~size+trt|Size.Name,
I have used xlsReadWrite to read data from an Excel spreadsheet.
I had a problem with converting times of the day so that I could create
POSIXct date-time objects. I was wondering if there was a better
solution.
Excel stores times of the day as fractions of a day so I wrote a function
to co
The chron package stores times as a fraction of the day. See R News 4/1.
On Sun, Feb 22, 2009 at 8:26 PM, David Scott wrote:
>
> I have used xlsReadWrite to read data from an Excel spreadsheet.
>
> I had a problem with converting times of the day so that I could create
> POSIXct date-time objects
Hi all,
I was just radomly playing with R and got the following error when
trying to filter a data frame using a string:
> Angel <- c(7,8,6,9,10)
> Buffy <- c(8,9,4,9,10)
> Firefly <- c(9,9,10,10,10)
> DrHorrible <- c(10,9,9,10,10)
> my.df <- data.frame(Angel, Buffy, Firefly, DrHorrible)
> my.df[
Hi Tony,
I "GUESS" my.df[-"DrHorrible"] does not tell R which column "NUMBER" would like
to remove.
As we know, we could use my.df[-4] and it exactly tells R which column must
remvoer from your data.
> my.df[-4]
Angel Buffy Firefly
1 7 8 9
2 8 9 9
3 6 4
I've had a very long file written out by R with write.table, with
fields of time values, converted from POSIXlt as.numeric. Among 2.5
million values, very few had 6 trailing zeroes, and those were output
in scientific notation as in the subject. Is this the default
behavior for long integ
Hi all,
I'm on Mac OS X 10.4...
So I've got a small python script I need to run from R. However, to get
the python script working, I need to have the shell that R calls get
settings from my .bash_profile file, which apparently it doesn't
currently (the shell R calls uses an older version of
I don't think you had 'integers' but integer-valued doubles: try
as.integer to get an integer variable. E.g.
x <- 1.1*(10^(1:8))
write.table(data.frame(x, as.integer(x)), "")
"x" "as.integer.x."
"1" 11 11
"2" 110 110
"3" 1100 1100
"4" 11000 11000
"5" 11 11
"6" 110 110
"7" 1.1e+
On Sun, 22 Feb 2009, Nick Matzke wrote:
Hi all,
I'm on Mac OS X 10.4...
For which there is a separate mailing list, R-sig-Mac .
So I've got a small python script I need to run from R. However, to get the
python script working, I need to have the shell that R calls get settings
from my .ba
Peterko gmail.com> writes:
> Hello, i need to find in time serie, k=1,2,3...(how if possible) most domain
> frequencies and than smooth original serie by Fourier row y=mean +
> a1*Cos(2*pi/freq*t)+b1*Sin(2*pi/freq*t)+a2..
>
> numbers a1,b1 ... i will have from simple regresion, i need only
Hi
r-help-boun...@r-project.org napsal dne 23.02.2009 03:44:41:
>
> Hi Tony,
>
> I "GUESS" my.df[-"DrHorrible"] does not tell R which column "NUMBER"
would
> like to remove.
> As we know, we could use my.df[-4] and it exactly tells R which column
must
> remvoer from your data.
>
> > my.df
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