Hi all (again),
I have a data frame "pop":
xloc yloc yield
1 101295
2 111081
3 121120
4 121110
And I want to get the sum of yield for the cell (pop$xloc, pop$yloc) in
a matrix as follows:
xloc
10 11 12
10 0 81 0
yloc
That worked.. thanks Peter
On 2/22/2011 5:40 PM, Peter Ehlers wrote:
popm <- as.matrix(pop)
recm <- matrix(rep(rec, n), nr=n, byrow=TRUE)
newpop <- data.frame(rbind(popm, recm))
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/
Hi i am doing an environmental research
The equation is as follow:
gam(y1 ~ x1 + s(x2) + s(x3) + s(x4), family = gaussian, fit = true)
I would like to obtain the beta coefficient and 95CI of x4 (or s(x4)), what
should I do?
Thanks,
Lung
--
View this message in context:
http://r.789695.n4.nab
How is the curve is represented? That's more important that its
organ-of-origin. If you have values of y=f(x) at discrete time points, then
y-(x+2) will change sign sometimes... the intersection point is at some time x'
in between. Am I missing something subtle here?
You could interpolate the
plyr is very useful to aggregate data... I strongly recommend it.
On 2/22/2011 5:59 PM, Jeff Newmiller wrote:
Use ?plyr::ddply
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. D
On Wed, Feb 23, 2011 at 8:01 AM, Peter Ehlers wrote:
> On 2011-02-22 11:52, Cory Champagne wrote:
>>
>> Hello all,
>> my first post to this list. I do a lot of experiments using a paired
>> sampling design and I would get a lot of mileage out of figures like
>> this, if I can make it work! Any a
On 2011-02-22 11:52, Cory Champagne wrote:
Hello all,
my first post to this list. I do a lot of experiments using a paired
sampling design and I would get a lot of mileage out of figures like
this, if I can make it work! Any advice would be appreciated.
my email is: cory.champ...@gmail.com.
Tha
On 2011-02-22 12:51, Vlatka Matkovic Puljic wrote:
Well, it should be difference by ID and TIME for q1:
something like:
for ID 1187
in TIME 1 q1=3
and TIME 2 (for same ID) q1=3
so diff would be 3-3=0
TIME ID q1
1 1187 3
1 1187 3
And I don't know how to make
Hi,
You could add arrows with geom_segment; however if you want even
spacing along the path it might get tricky,
library(ggplot2)
d <- data.frame(x=seq(0, 10, length=100),
y=sin(seq(0, 10, length=100)))
N <- 10
dN <- 2
ind <- seq(1,nrow(d),by=N)
ind <- ind[-c(1,length(ind))]
On 2011-02-22 14:48, Nicolas Gutierrez wrote:
Hi All,
I have a data frame "pop":
> id xloc yloc size
> 1 1 101295
> 2 211 1081
And I want to add the vector "rec" to the data frame "n" times (without
using a loop):
> rec=c(3, 5, 5, 10)
> n=2
Th
On Feb 22, 2011, at 7:10 PM, Ista Zahn wrote:
Hi,
This is R, so there are bound to be severay ways to do it. This would
be my first choice:
library(ggplot2)
ggplot(d, aes(x=V2, y=V3, color=V1)) + geom_step()
Best,
Ista
On Tue, Feb 22, 2011 at 3:08 PM, Techni X
wrote:
Hi all,
I have a qu
I'm doing a path plot with ggplot2, the result is looking very nice, but I want
to give some indication of which direction the lines are going. I thought of
using colour gradients, but it doesn't look right. What would be ideal is a
line type that indicated direction, something like ">>>". I
Thanks. How about this?
DT$B = factor(DT$B)
firststep = DT[,cbind(expand.grid(B,B),v=1/length(B),C=C[1]),by=A][Var1!
=Var2]
setkey(firststep,Var1,Var2,C)
firststep = firststep[,transform(.SD,cv=cumsum(v)),by=list(Var1,Var2)]
setkey(firststep,Var1,Var2,C)
DT[, {x=data.table(expand.grid(B,B),C[1]-1
John,
What version of odfWeave and OO are you using?
Thanks,
Max
On Feb 22, 2011, at 3:17 PM, "Prof. John C Nash" wrote:
> Using R2.12.1 on Ubuntu 10.04.1 I've tried to run the following code chunk in
> odfWeave
>
> <>=
> x<-seq(1:100)/10
> y<-sin(cos(x/pi))
> imageDefs <- getImageDefs()
>
If you want to get honest estimates of accuracy, you should repeat the feature
selection within the resampling (not the test set). You will get different
lists each time, but that's the point. Right now you are not capturing that
uncertainty which is why the oob and test set results differ so mu
Hi,
This is R, so there are bound to be severay ways to do it. This would
be my first choice:
library(ggplot2)
ggplot(d, aes(x=V2, y=V3, color=V1)) + geom_step()
Best,
Ista
On Tue, Feb 22, 2011 at 3:08 PM, Techni X wrote:
> Hi all,
>
> I have a question, that might be a „rookie“ question – but
The documentation of aggregate tells you that your way will not work. Why don't
you aggregate/sum the columns separately? I would doubt that what you want to
try to achieve in one go is already implemented somewhere
Jannis
--- Hongwei Dong schrieb am Di, 22.2.2011:
> Von: Hongwei Dong
>
Use ?plyr::ddply
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
Hi All,
I have a data frame "pop":
>id xloc yloc size
> 1 1 101295
> 2 211 1081
And I want to add the vector "rec" to the data frame "n" times (without
using a loop):
> rec=c(3, 5, 5, 10)
> n=2
The result I want:
>id xloc yloc size
> 1 1 10
try data.table:
x a b
1 1 1 21
2 2 2 22
3 3 3 23
4 4 4 24
5 5 5 25
6 1 6 26
7 2 7 27
8 3 8 28
9 4 9 29
10 5 10 30
11 1 11 31
12 2 12 32
13 3 13 33
14 4 14 34
15 5 15 35
> require(data.table)
Loading required package: data.table
Quick start guide : vignette("datatable-intro")
Gary/Hongwei writes:
> I'm wondering how I can aggregate data in R with different functions for
> different columns. For example:
>
> x<-rep(1:5,3)
> y<-cbind(x,a=1:15,b=21:35)
> y<-data.frame(y)
>
> I want to aggregate "a" and "b" in y by "x". With "a", I want to use
> function "mean"; with "b",
Hi Gary,
I'm not sure if you can do that with aggregate. You might have to
resort to something like
tmp <- cbind(aggregate(y$a, by=list(y$x), mean), aggregate(y$b,
by=list(y$x), sum)$x)
names(tmp) <- c("x", "mean.a", "sum.b")
tmp
Or, using the plyr package:
library(plyr)
ddply(y, .(x), summari
Gabor, that worked great! I was unaware of the sqldf package, but it is great
for data manipulation. Thanks!
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
___
Hi all,
I have a question, that might be a „rookie“ question – but I’m trying now for
days and cannot get my head around. The general question is:
How can I plot a stepped line chart with multiple lines from a subset of a
dataframe?
An example:
d <- matrix(rep(0,24), ncol=3, nrow=8)
d <- as.da
Thanks, Max.
Yes, I did some feature selections in the training set. Basically, I
selected the top 1000 SNPs based on OOB error and grow the forest using
training set, then using the test set to validate the forest grown.
But if I do the same thing in test set, the top SNPs would be different th
>> The first column gives the answer that I am looking for but I just
>> thought their maybe a better way without using the long variable
>
> Sounds like you want the frequency-weighted summary statistics.
> For the mean, look at ?weighted.mean.
> For more comprehensive stats, check ?wtd.mean in th
Hi,
I am trying to make a palaeoenvironmental transfer function using the R
package rioja that predicts the water-table (measured as depth to the water
table) of an area given the testate amoebae that are found there. I've
carried out weighted averaging of the data and am trying to produce a graph
I'm not sure I'm using appropriate tools for what I'm trying to do; a
reason for my lack of certainty is that I'm encountering difficulty with
what I thought would be fairly simple: labeling the X axis of my graphs
with human-readable timestamps.
I'm using R 2.12.1, running in a FreeBSD 8.2-PREREL
I've tried grouping the real and imaginary parts as a single complex number,
e.g.
> z1 <- complex(real=r1, imaginary=i1)
Although complex variables are not allowed in lm, I can run regsubsets just
fine:
> z <- cbind(z1, z2, z3, ..., z10)
> bestsubs <- regsubsets(z, y, nbest=1, nvmax=5, intercep
On 19 February 2011 07:24, Duncan Temple Lang
wrote:
Do you mean that the R variable page gives information about the request
error
and contains the 500 error code? Not sure what you mean by "screen"
here.
Yes that's exactly what I mean
Client-server interactions are hard to debug
Hello all,
my first post to this list. I do a lot of experiments using a paired
sampling design and I would get a lot of mileage out of figures like
this, if I can make it work! Any advice would be appreciated.
my email is: cory.champ...@gmail.com.
Thanks!
#define dummy variables and a data
Hi,
Sorry for the very short explanation about the problem of intersection.
I have a wave function monitored from the heart beat in a particular interval
of
times. Apart fom that, there is a line with positive slope (e.g: y = x+2) which
lies across the wave and intersect on a number of point
Hi,
Sorry for the very short explanation about the problem of intersection.
I have a wave pattern monitored from the heart beat in a particular interval of
times. Apart fom that, there is a line with positive slope (e.g: y = x+2) which
lies across the wave and intersect on a number of points.
My surmise would be that you have not analyzed the situation correctly, and you
are making a false assumption about your code. Since you can't show the code,
it's pretty hard to figure out what that is. I think you're going to have to
produce a simple example that you can share that has the sa
How is the curve defined? If the curve is y=f(x) and the line is y=mx+b, you
look for the roots of f(x)-mx-b.
?polyroot
?uniroot
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of FMH
Sent: Tuesday, February 22, 2011 6:28 AM
To: r-he
Hi, R users,
I'm wondering how I can aggregate data in R with different functions for
different columns. For example:
x<-rep(1:5,3)
y<-cbind(x,a=1:15,b=21:35)
y<-data.frame(y)
I want to aggregate "a" and "b" in y by "x". With "a", I want to use
function "mean"; with "b", I want to use function "
Hi,
Sorry for the very short explanation about the problem of intersection.
I have a wave function monitored from the heart beat in a particular interval
of
times. Apart fom that, there is a line with positive slope (e.g: y = x+2) which
lies across the wave and intersect on a number of points
In addition's to Max's suggestion about caret, look at ROCR which visualizes
ROC charts for any binary classifier. I have an example of e1071::SVN and ROCR
here
https://heuristically.wordpress.com/2009/12/23/compare-performance-machine-learning-classifiers-r/
-Original Message-
From:
Hi Max and Andrew,
Thanks so much for your reply. Indeed I found your link last night using
steps shown below.
My first question is if the following two steps are right and AUC is 51% as
shown below.
My seond question is that currently I am using cost parameter=1 (the default
in R-SVM; http://ww
Dear All,
I would like any information about lacunarity analyis in R. Does anyone know
if the sliding-box lacunarity analysis of Allain and Cloitre (1991) or Plotnick
et al. (1996) has been implemented by anyone in R? Or any R package useful for
that? I have seen anything in R mail list fil
Well, it should be difference by ID and TIME for q1:
something like:
for ID 1187
in TIME 1 q1=3
and TIME 2 (for same ID) q1=3
so diff would be 3-3=0
TIME ID q1
1 1187 3
1 1187 3
And I don't know how to make R to find pairs and calculate diff?
2011/2/21 Dennis M
By luck I was able to solve my own problem! (Big "phw.")
As a reference, I simple change my value of z in image.plot to be X[,n:1],
where X = t(lower). Everything else remains the same.
Cheers,
Manussawee
On Tue, Feb 22, 2011 at 10:03 AM, Manussawee Sukunta <
msuku...@illinoisalumni.org> w
Using R2.12.1 on Ubuntu 10.04.1 I've tried to run the following code chunk in
odfWeave
<>=
x<-seq(1:100)/10
y<-sin(cos(x/pi))
imageDefs <- getImageDefs()
imageDefs$dispWidth <- 4.5
imageDefs$dispHeight<- 4.5
setImageDefs(imageDefs)
X11(type="cairo")
plot(x,y)
title(main="sin(cos(x/pi))")
savePlot
The objects functions for kernel methods are unrelated to the area
under the ROC curve. However, you can try to choose the cost and
kernel parameters to maximize the ROC AUC.
See the caret package, specifically the train function.
Max
On Mon, Feb 21, 2011 at 5:34 PM, Angel Russo wrote:
> *Hi,
>
On 2011-02-22 08:10, Daniel Harris wrote:
Thanks for the replies
The file is below
Height Frequency
62 3
63 20
64 24
65 40
66 85
67 122
68 139
69 179
70 139
71 107
72 55
73 47
74 22
75 12
76 5
77 1
I use the follo
I found an easy way to do it using trans3d() and the transformation matrix.
Maybe in the future somebody would ask about it:
# For plotting the bivariate gaussian density
library(mvtnorm)
x<-seq(-5,5,by=0.1)
y<-seq(-3,7,by=0.1)
# Joint density
f<-function(x,y)
dmvnorm(cbind(x,y),mean=c(0,2),sigm
And how to read a file with a BOM is actually discussed in detail in
the 'R Data Import/Export' manual of 2.12.2 RC.
What should be ASCII files with BOMs seem to be cropping up rather
frequently these days: the recent culprits are Mac applications with
origins on Windows (SPSS was one, some ve
Another way is with ifelse:
z<-data.frame(x,y)
z$y2 <- ifelse(z$x==5,z$y-1,z$y)
HTH,
Ivan
Le 2/22/2011 18:27, Erik Iverson a écrit :
Is this what you mean?
z[which(z[,"x"] == 5) - 1, "y"]
?which is probably what you're looking for...
Hongwei Dong wrote:
Hi, R users,
I'm wondering if I can
Hi Gary,
Another possibility besides Erik's (although I suspect his is what you
are really after):
## easier way to data
z <- cbind(x = 1:10, y = 11:20)
z[z[,"x"] == 5, "y"] - 1
## To see what is "going on", break it into pieces
## logical; does column 'x' of 'z' equal 5?
z[, "x"] == 5
## all va
Hi Gary,
Try
transform(z, y = ifelse(x == 5, y-1, y))
HTH,
Jorge
On Tue, Feb 22, 2011 at 12:18 PM, Hongwei Dong <> wrote:
> Hi, R users,
>
> I'm wondering if I can identify an element in a column by an element in
> another column. For example:
>
> x<-1:10
> y<-11:20
> z<-cbind(x,y)
> z
> x
try this:
x <- 1:10
y <- 11:20
z <- cbind(x, y)
ind <- x == 5
z[ind, "y"] <- z[ind, "y"] - 1
z
I hope it helps.
Best,
Dimitris
On 2/22/2011 6:18 PM, Hongwei Dong wrote:
Hi, R users,
I'm wondering if I can identify an element in a column by an element in
another column. For example:
x<-1:
Is this what you mean?
z[which(z[,"x"] == 5) - 1, "y"]
?which is probably what you're looking for...
Hongwei Dong wrote:
Hi, R users,
I'm wondering if I can identify an element in a column by an element in
another column. For example:
x<-1:10
y<-11:20
z<-cbind(x,y)
z
x y
[1,] 1 11
[
Hi, R users,
I'm wondering if I can identify an element in a column by an element in
another column. For example:
x<-1:10
y<-11:20
z<-cbind(x,y)
z
x y
[1,] 1 11
[2,] 2 12
[3,] 3 13
[4,] 4 14
[5,] 5 15
[6,] 6 16
[7,] 7 17
[8,] 8 18
[9,] 9 19
[10,] 10 20
What I want to do i
searching using
library(sos)
h <- ???"Parzen fractional estimator"
h
leads to
library(fracdiff) # on CRAN
?fdSperio
On Wed, Feb 16, 2011 at 5:19 PM, Fologo Dubois wrote:
> Does R have a function for Parzen fractional degree of differencing
> estimator? I am referring to the non-parametric kern
You could try
install.packages("sos", dep=T)
library(sos)
h <- finfFn("formal concept analysis")
h
but it does not seem to find anything relevant.
On Wed, Feb 16, 2011 at 6:21 PM, Mark Heckmann wrote:
> I am looking for an R package for formal concept analysis
> (http://en.wikipedia.org/wiki/F
Dear R-users,
I am in the process of creating new custom functions and am quite puzzled by
some discrepancies in execution time when I run some R scripts that call
those new functions. So here is the situation:
- let's assume I have created two custom functions, called myg and myf;
- myg is mostly
You do not need the regions with the double densities near the center
(1/2,1/2), you can use just the parallel 45 degree borders in that
region also.
> x-x
> |/*\\\|
> |---/***\\|
> |--/*\|
> |-/**/
Thanks.
I made it!
Best wishes,
S.
Am 22.02.2011 17:41, schrieb Erik Iverson:
Sandra Stankowski wrote:
is.na function does'nt seem to work, but maybe I'm just dealing with
it in a wrong way.
here's an example
> m <- c(2, 3, 5, 6, 3, 7, -99, -99, 6)
> n <- c(1,1,1,1,1,2,2,2,2)
so my mat
is.na function does'nt seem to work, but maybe I'm just dealing with it
in a wrong way.
here's an example
> m <- c(2, 3, 5, 6, 3, 7, -99, -99, 6)
> n <- c(1,1,1,1,1,2,2,2,2)
so my matrix contains certain missing values
> m[m==-99] <- NA
> o <- data.frame(m, n)
> o
m n
1 2 1
2 3 1
3 5 1
Thanks for the replies
The file is below
Height Frequency
62 3
63 20
64 24
65 40
66 85
67 122
68 139
69 179
70 139
71 107
72 55
73 47
74 22
75 12
76 5
77 1
I use the following
data <- read.table("file",header=T)
at
On 2011-02-21 02:42, Rosario Garcia Gil wrote:
Hello
I have a data set with outlier and it is not normally distributed. I would
instead like to use a more robust distribution like t-distribution.
My question is if the coefficients of the regression are different from zero,
but assuming a t-di
Dear all,
I need to perform a two-way unbalanced ANOVA on my data set. The unbalancing
is due to missing values (NA) for some subjects. My data set concerns the
evaluation of 4 different prototypes tested by subjects three times and it
is organized as follows (column by column):
number of observ
On Mon, Feb 14, 2011 at 12:22 PM, mathijsdevaan wrote:
>
> Hi,
>
> I have a large dataset with info on individuals (B) that have been involved
> in projects (A) during multiple years (C). The dataset contains three
> columns: A, B, C. Example:
>
> A B C
> 1 1 a 1999
> 2 1 b 1999
> 3 1 c
did you try
RSiteSearch("t-regression")
?
That seems to give some usefull hits.
On Mon, Feb 21, 2011 at 7:42 AM, Rosario Garcia Gil
wrote:
> Hello
>
> I have a data set with outlier and it is not normally distributed. I would
> instead like to use a more robust distribution like t-distribution
Sandra Stankowski wrote:
is.na function does'nt seem to work, but maybe I'm just dealing with it
in a wrong way.
here's an example
> m <- c(2, 3, 5, 6, 3, 7, -99, -99, 6)
> n <- c(1,1,1,1,1,2,2,2,2)
so my matrix contains certain missing values
Thank you for the example.
You're construct
Thanks for all the people that replied my message. The text file indeed has
"\uFEFF". I have fixed the text file by using 'gvim -b'.
On Tue, Feb 22, 2011 at 10:03 AM, Jeff Newmiller
wrote:
> What you describe could be a bug (in which case providing your OS and R
> version info per the posting gui
Did you try
RSiteSearch("QuadTree")
?
It does seem to give some hits.
On Sun, Feb 20, 2011 at 7:54 PM, Jaimin Dave wrote:
> Could any one tell me how to implement QuadTree in R?
> Or are there any packages avaialble to implement it in R.
>
> [[alternative HTML version deleted]]
>
> __
To R community:
I've been pulling my hair for the past few days over this issue now, and I
may be at the point where I just can't spot my error. So I thought to reach
out to the community and hope someone could kindly help me.
I'm trying to create a correlation matrix and plot it. However, I'm
On Tue, Feb 22, 2011 at 7:43 AM, John Edwards wrote:
> Hi,
>
> I have the following input file.
> $ cat main.txt
> CEL_A CELL_B
> 1 4
> 2 5
> 2 6
>
> Then I run read.table in R.
>
>> f=read.table('main.txt', header=T, check.names=F, sep='\t')
>> head(f)
> \ufeffCEL_A CELL_B
> 1 1 4
> 2
What you describe could be a bug (in which case providing your OS and R version
info per the posting guidelines would be a minimum requirement to get it fixed)
or a control character that is actually in your file (which you might need a
binary editor to see).
---
On 22/02/2011 10:43 AM, John Edwards wrote:
Hi,
I have the following input file.
$ cat main.txt
CEL_A CELL_B
1 4
2 5
2 6
Then I run read.table in R.
> f=read.table('main.txt', header=T, check.names=F, sep='\t')
> head(f)
\ufeffCEL_A CELL_B
11 4
22 5
32 6
> f$CE
Hi Sandra,
What about ?is.na function ?
Hope this help
Regards,
ML
Le 22/02/11 16:11, Sandra Stankowski a écrit :
> NROW(data[jan,16] != NaN)
--
Mohamed Lajnef,IE INSERM U955 eq 15#
Pôle de Psychiatrie#
Hôpital CHENEVIER
?is.nan
Peter Ehlers
On 2011-02-22 07:11, Sandra Stankowski wrote:
Hey there,
I tried to count the number of rows, where my data isn't NaN in a
certain column.
this was my guess:
(given is a data frame with 2069 rows and 17 cols)
NROW(data[jan,16] != NaN)
("jan" is defined this way: jan<-
On 2011-02-22 05:53, Francesco Nutini wrote:
Sorry, I mean "heading".
Thanks for the tip.
I'm not quite certain what you mean by "heading",
but it sounds to me as though you might want
xyplot( ... your stuff ...,
par.settings = list(
strip.background = list(col = "l
Sandra,
Please provide a small, reproducible example of this issue.
You probably want to use ?is.nan and not the inequality
operator.
Similar example, contrast:
x <- NA
is.na(x)
x == NA
Sandra Stankowski wrote:
Hey there,
I tried to count the number of rows, where my data isn't NaN in a
cer
Hi,
I have the following input file.
$ cat main.txt
CEL_A CELL_B
1 4
2 5
2 6
Then I run read.table in R.
> f=read.table('main.txt', header=T, check.names=F, sep='\t')
> head(f)
\ufeffCEL_A CELL_B
11 4
22 5
32 6
> f$CEL_A
NULL
I'm not sure where the special character
> Date: Tue, 22 Feb 2011 14:53:07 +0100
> From: r.m.k...@gmail.com
> To: marchy...@hotmail.com
> CC: graham.willi...@togaware.com; r-help@r-project.org
> Subject: Re: [R] problem installing R in Ubuntu 10.04 -HELP
>
> -BEGIN PGP SIGNED MESSAGE-
On Mon, Feb 21, 2011 at 10:06 PM, dadrivr wrote:
>
> Thanks Dennis,
>
> The code works for perfectly for the data in the example. For some reason,
> however, I get the following error message when I use a different data set:
>
>> preds <- expand.grid(age = c(30,36,42), Subject = unique(mydata$id)
Hey there,
I tried to count the number of rows, where my data isn't NaN in a
certain column.
this was my guess:
(given is a data frame with 2069 rows and 17 cols)
NROW(data[jan,16] != NaN)
("jan" is defined this way: jan <- which(data[,2]==1, arr.ind= TRUE))
but I only get the number of c
library(zoo)
library(tseries)
library(quantmod) #for access to FRED
require(quantmod)
require(TTR)
secA <- getSymbols("DEXUSEU",src="FRED")
secB <- getSymbols("DEXUSUK",src="FRED")
secA <- zoo(DEXUSEU[,1])
secB <- zoo(DEXUSUK[,1])
t.zoo <- merge(secA, secB, all=FALSE)
t <- as.data.frame(t.zoo
I am having problems with forward prediction using the output of the
Basic Structural Model from StructTS. The following snippet
illustrates the problem:
t_end <- 139
nahead <- 20
data(AirPassengers)
ap <- log10(AirPassengers)-2
fit <- StructTS(ts(ap[1:t_end], freq=12), type="BSM")
p <- stats:::p
Yes John
My editing mistake. Thanks!!
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Sent from the R help mailing list archive at Nabble.com.
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--- On Tue, 2/22/11, D Kelly O'Day wrote:
> From: D Kelly O'Day
> Subject: Re: [R] Calculate a mean for several months for several years
> To: r-help@r-project.org
> Received: Tuesday, February 22, 2011, 12:21 AM
>
> Pete
>
> The original question "I would like to calculate the
> average of
Hi all,
When I detect the spatial point pattern, I want to use the Cramer-von Mises
statistic to assess the curve-wise significance of deviations from null
hypotheses. Who can tell me which function in R package "Spatstat" can do
this work?
Thanks a lot
Jeff
[[alternative HTML version
> Date: Tue, 22 Feb 2011 11:12:26 +0100
> From: r.m.k...@gmail.com
> To: graham.willi...@togaware.com
> CC: r-help@r-project.org
> Subject: Re: [R] problem installing R in Ubuntu 10.04 -HELP
>
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> On 0
I don't know the term "intestation", but from context it appears you
might need:
?trellis.par.set
--
David.
On Feb 22, 2011, at 7:23 AM, Francesco Nutini wrote:
Hi Dennis and [R]users!
as I said last week, I need more info about xyplot.
Is it possible to change the color of the intestatio
Dear mailing list,
I am using the cforest() method from the party package to train a
randomForest with ten input parameters which sometimes contain "NA"s.
The predicted variable is a binary decision. Building the tree works
fine without warnings or error messages, but when using the predict()
stat
Sorry, I mean "heading".
Thanks for the tip.
Francesco
> CC: djmu...@gmail.com; r-help@r-project.org
> From: dwinsem...@comcast.net
> To: ui...@hotmail.it
> Subject: Re: [R] [r] align xyplot
> Date: Tue, 22 Feb 2011 08:33:07 -0500
>
> I don't know the term "intestation", but from context it app
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 02/22/2011 02:33 PM, Mike Marchywka wrote:
>
>
>
>
>
>
>
>
>> Date: Tue, 22 Feb 2011 11:12:26 +0100
>> From: r.m.k...@gmail.com
>> To: graham.willi...@togaware.com
>> CC: r-help@r-project.org
>> Subjec
What you are describing is a path model (e.g., http://bit.ly/f4siTs). You can
fit such a model with the sem package
(http://cran.r-project.org/web/packages/sem/index.html).
Best,
--
Wolfgang Viechtbauer
Department of Psychiatry and Neuropsychology
School for Mental Health and Neuroscience
Maas
Daniel, how is the data stored? The answer to your question may be as
simple as
> df <- read.csv("filename.csv")
> summary(df)
See ?read.csv for info on reading various file formats.
HTH, Bryan
Prof. Bryan Hanson
Dept of Chemistry & Biochemistry
DePauw University
602 S. Coll
What about using read.table() and applying summary() on the result?
Uwe Ligges
On 22.02.2011 13:20, Daniel Harris wrote:
Hello
Is it possible to get summary statistics (inc mean, sd etc) from a
text file that has the following info stored in it?
Height Frequency
123
I think it should be the negative of the first Dxy but this is all why the
posting guide says to create the simplest self-defined example that shows
the problem. That way I could run it and get to the bottom of this. See
the help file for cph which has examples of simulating test data. Try to
h
From: ui...@hotmail.it
To: djmu...@gmail.com
CC: r-help@r-project.org
Subject: RE: [R] [r] align xyplot
Date: Tue, 22 Feb 2011 12:23:55 +
Hi Dennis and [R]users!
as I said last week, I need more info about xyplot.
Is it possible to change the color of the intestation of xyplot? By
The output for the new example should be:
project v
1 0
2 0.5
3 1.5
4 0.5
The output you calculated was correct for the v per year, but the v per
group would be incorrect. I think the problem lies in the fact that
expand.grid(B,B) doesn't take into account that combinations of B can only
be
On 22.02.2011 12:27, FMH wrote:
Dear All,
I'm looking an appropriate way in R to compute/estimate points of intersection
between a line and a curve and will really appreciate for any suggestion or
ideas?
Sounds like second level school homework. If not, please explain your
problem in more
Dear Jari,
For what it's worth, I could see the need in my own work for species
accumulation curves with variable sampling effort.
Scott Chamberlain
On Tuesday, February 22, 2011 at 2:26 AM, Jari Oksanen wrote:
Vanessa Francisco gmail.com> writes:
>
> >
> > Hello! I'm a PhD student working
Hello
Is it possible to get summary statistics (inc mean, sd etc) from a
text file that has the following info stored in it?
Height Frequency
123 5
124 8
125 3
126 9
127 7
etc etc
Hi Dennis and [R]users!
as I said last week, I need more info about xyplot.
Is it possible to change the color of the intestation of xyplot? By default is
pale-pink, but light-gray is better for a paper.
Thanks,
Francesco
From: nutini.france...@gmail.com
To: djmu...@gmail.com
CC: r-help@r-pr
Dear All,
I'm looking an appropriate way in R to compute/estimate points of intersection
between a line and a curve and will really appreciate for any suggestion or
ideas?
Thank you,
Fir
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