Hi Mary,
One can use arrows too...
Here is the code :
x<-seq(75,225,0.1)
plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
*arrows(149,0,149,dnorm(149,140,15),length=0)
*par(new=T)
plot(x,dnorm(x,mean=150, sd=15), type='l', col='orange',axes=F)
Regards
Radha
On Sun, Jul 26, 2009 at 5:09
Mary A. Marion schrieb:
> Hello,
>
> I am plotting two distributions and want to draw a vertical line at
> the critical point 149.
> How can I stop it from going further up than the norm(140,15) curve?
>
> x<-seq(75,225,0.1)
> plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
> abline(v = 149,
lanc...@fns.uniba.sk schrieb:
> Good day,
>
> I'm trying to get more time series in one plot. As there are bigger
> differences in values of variables I need logaritmic y axis.
>
> The code I use is the following:
>
> nvz_3_data <- read.csv('/home/tomas/R_outputs/nvz_3.csv')
> date <- (nvz_3_data$d
Dear experts:
I am a newbie to R. Recently, I try to make prediction models with R and the
Design library.
I have read Prof. Harrell's excellent book. But I did not quite understand.
I have two problems about the validation and calibration of prediction
models:
1. Can someone explain the results o
Another thing you can do to save time is to use the --no-vignettes switch
when you build the package.
On Sat, Jul 25, 2009 at 11:05 PM, Shige Song wrote:
> Dear All,
>
> I have been using Sweave (mainly via the Sweave.sh script) and really like
> it. I am working a paper (using Sweave, of course)
Dear All,
I have been using Sweave (mainly via the Sweave.sh script) and really like
it. I am working a paper (using Sweave, of course) which includes several
time-consuming computations, and it gets tedious to re-compile the whoel
thing every time I made changes. Then I discover the "cacheSweave"
It is true that R does not offer support for custom likelihood functions
in nonlinear mixed models. However you can switch to R and use
AD Model Builder's random effects module http://admb-project.org
This is freely available software and it is more flexible than
proc nlmixed. I'm sure there are pe
I should have mentioned that I am using the lmer library for my analyses,
just in case other methods provide results differently.
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-he
Hi, using polynomial contrasts for the ordered factors in an experiment
leads to much nicer covariance structure than using treatment contrasts. It
is easy to assess the mean effect for each of the experimental groups.
However, standard errors are provided only for the components of the
orthogonal
Thanks guys! I appreciate the pointers.
Cheers,
Mark
On Sat, Jul 25, 2009 at 5:57 PM, Gabor
Grothendieck wrote:
> See ?layout
>
> opar <- par(mar = c(2, 2, 2, 2))
> m <- matrix(c(1, 1, 1, 2, 2, 2,
> 1, 1, 1, 2, 2, 2,
> 3, 4, 5, 9, 10, 11,
> 6, 7, 8, 12, 13, 14), 4, byrow = TRUE)
See ?layout
opar <- par(mar = c(2, 2, 2, 2))
m <- matrix(c(1, 1, 1, 2, 2, 2,
1, 1, 1, 2, 2, 2,
3, 4, 5, 9, 10, 11,
6, 7, 8, 12, 13, 14), 4, byrow = TRUE)
layout(m)
for(i in 1:14) plot(i)
On Sat, Jul 25, 2009 at 8:34 PM, Mark Knecht wrote:
> Hi,
> I made the attached pictur
Use 'layout' to define the regions: may need to change the 'mar'
layout(rbind(c(1,1,2,2), c(3,4,7,8), c(5,6,9,10)), height=c(2,1,1))
layout.show(10)
plot(0)
plot(1)
for (i in 1:4) plot(i)
for (i in 11:14) plot(i)
On Sat, Jul 25, 2009 at 8:34 PM, Mark Knecht wrote:
> Hi,
> I made the atta
There is an example of this in the examples section of ?plot.zoo
after the comment: shade a portion of a plot and make axis fancier
On Sat, Jul 25, 2009 at 8:36 PM, stvienna wiener wrote:
> Hi all,
>
> I am plotting a financial time series, but I need a more detailed X-Axis.
>
> Example:
> x <- zo
Hi all,
I am plotting a financial time series, but I need a more detailed X-Axis.
Example:
x <- zoo(rnorm(1:6000), as.Date("1992-11-11")+c(1:6000))
plot(x)
The X-Axis is labeled "1995", "2000" and "2005".
I would need either "1995", "1997", etc. or maybe yearly
I used google first, then look a
summary(lm1) will show you R2 and adjusted R2 too.
Ronggui
2009/7/26 Steve Lianoglou :
> Hi Sarah,
>
> On Jul 25, 2009, at 8:25 AM, Buckmaster, Sarah wrote:
>
>> Hi everyone,
>>
>> I have a question about calculating r-squared in R. I have tried searching
>> the archives and couldn't find what I
Try this:
> ix <- c(1, 3, 4, 2)
> mapply("[", dimnames(mydatastructure), ix)
[1] "S1" "T3" "U4" "V2"
On Sat, Jul 25, 2009 at 5:12 PM, Sauvik De wrote:
> Hi:
> How can I extract the dimension-names of a pre-defined element in a
> multidimensional array in R ?
>
> A toy example is provided below:
Thanks for the answer Tal!
But I can't get it to work correctly! :(
Please bear with me this is the first time I am using R! and I am in a rush
to correct a paper
in fact on the plane I am plotting a table
> fullpointed=read.table("fullpoints_backup.txt",h=F)
> plot(range(-2.5,0.95),range(0.00,1.0
hey
thanks for the answer but I couldn't achieve it? would you explain a bit
more?
I have like 300 points to label!
thanks
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PLE
Good day,
I'm trying to get more time series in one plot. As there are bigger
differences in values of variables I need logaritmic y axis.
The code I use is the following:
nvz_3_data <- read.csv('/home/tomas/R_outputs/nvz_3.csv')
date <- (nvz_3_data$date)
NO3 <- (nvz_3_data$NO3)
NH4 <- (nvz_3_da
Thanks so much. You saved me a lot of programming time. FYI, here is the
code that produces what I need.
library(survival)
### Data
time<-rexp(100)
grp<-rep(0:1,times=50)
cens<-rbinom(100,size=1,p=.9)
### Score Test From iter.max=0
fit0 <- coxph( Surv(time,cens)~grp , iter.max = 0 )
vcov(fit0)
On Sat, 25 Jul 2009, Sean wrote:
Does anyone know how get the score and information under the null from
coxph? I know that I can get the chi-square value of the score from coxph,
but I need the two components separately. I have a function that computes
the two components when I do not have ties
Hi:
How can I extract the dimension-names of a pre-defined element in a
multidimensional array in R ?
A toy example is provided below:
I have a 4-dimensional array with each dimension having certain length. In
the below example, "mydatastructure" explains the structure of my data.
mydatastructure
Hi Mary,
On Jul 25, 2009, at 7:39 PM, Mary A. Marion wrote:
Hello,
I am plotting two distributions and want to draw a vertical line at
the critical point 149.
How can I stop it from going further up than the norm(140,15) curve?
x<-seq(75,225,0.1)
plot(x,dnorm(x,mean=140, sd=15), type='l',
Try with ?segments,
x<-seq(75,225,0.1)
plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
#abline(v = 149, col = "black")
segments(149, 0, 149, dnorm(149,140,15))
curve(dnorm(x,mean=150, sd=15),from=75, to=225, col='orange', add=TRUE)
HTH,
baptiste
2009/7/26 Mary A. Marion
> Hello,
>
>
Hello,
I am plotting two distributions and want to draw a vertical line at the
critical point 149.
How can I stop it from going further up than the norm(140,15) curve?
x<-seq(75,225,0.1)
plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
abline(v = 149, col = "black")
curve(dnorm(x,mean=15
Does anyone know how get the score and information under the null from
coxph? I know that I can get the chi-square value of the score from coxph,
but I need the two components separately. I have a function that computes
the two components when I do not have ties but I would like to leverage the
op
Victor Landeiro wrote:
>
> Hi all,
> I wrote a piece of code that generates simulated variables. after variable
> generation I use them in several analyzes.
> However, when I use a for to repeat the procedure 1000 times I get an erro
> message in one of the "for" steps, precisely at this time:
Waverley,
if you want to modify components of the ROCR plot, you need to direct
the parameters to the component functions by prefixing them with the
name of that component function. In your case, you should add
"boxplot.outline=FALSE" as follows:
plot(perf, avg= "vertical", spread.estimate="boxpl
Hi all,
I wrote a piece of code that generates simulated variables. after variable
generation I use them in several analyzes.
However, when I use a for to repeat the procedure 1000 times I get an erro
message in one of the "for" steps, precisely at this time:
gls.temp<- gls(y2 ~ x2,correlation=corE
Be carefull with this, but:
rm(list=ls())
q("yes")
:-) milton
On Sat, Jul 25, 2009 at 7:19 AM, Paul Emberson wrote:
> Hi John,
>
> You can use
>
> rm(object)
>
> to delete a specific object from the current environment and then save
> your workspace again without those objects
>
> ls()
>
> to s
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Waverley
> Sent: Saturday, July 25, 2009 10:38 AM
> To: r-help; r-help@r-project.org
> Subject: [R] ROCR package question
>
> Thanks for the quick reply. That is very clear for
Thanks for the quick reply. That is very clear for my question 1, 2.
How about question 3? When I plot, is there way not to show the
whisker plot outliers for evaluating the multiple runs? I have tried
to put the option from boxplot command outline=FALSE, however, it did
not work.
Can you help?
Christopher Bare wrote:
Hi,
I'm trying to put together an R package. My library has dependencies
on three other libraries: RSQLite, gaggle (a component of
Bioconductor), and another package that is available as a download
from the author's website.
Is it possible to specify each of those depe
Natural splines (nsin the splines package rcs in the Hmisc or Design package)
are linear outside of the range of the knots. You could use those and just
specify the knots to be within the portion that you want to allow to be
non-linear.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statis
Hi Sarah,
On Jul 25, 2009, at 8:25 AM, Buckmaster, Sarah wrote:
Hi everyone,
I have a question about calculating r-squared in R. I have tried
searching the archives and couldn't find what I was looking for -
but apologies if there is somewhere I can find this...
I carried out a droughtin
Sure,
Here is an example:
# get some random data to play with
x <- runif(100)
y <- runif(100)
labels.to.plot <- sample(c("A","B"), 100, replace = T)
# set up the window, play them one by one to see what they do
plot.window(ylim = c(0,1), xlim = c(0,1))
plot.new()
axis(1)
axis(2)
box()
# plot th
Waverley, see help('performance-class') for a description of the slots.
Your AUCs will be in p...@y.values, which itself is a list (one list
element per run).
Thus, you can use functions like unlist or s/lapply to access them, e.g.
mean(unlist(p...@y.values))
Kind regards,
Tobias
On Sat, Jul
I want to estimate a VAR-model with 4 endogenous variables.
One of the included times series is stationary, two are stationary in first
differences and the last is stationary in second differences. By building
the differences I obtained NAs for the first one or two observations. So my
first idea w
Thanks for the reply. I am not sure I am following:
1. for the sample code. I tried p...@auc but get auc object not found
2. I am SPECIFICALLY interested in the averaged auc value of the
multiple runs. How to get that out? I typed perf and it comes out as
a list.
3. as for the plot using whisk
jlfmssm wrote:
>
> Sorry, I didn't get it.
> I made a simple example to explain what I want
>
>> x=c(1,2,4,4)
>> y=c("A",1,2,"A")
>> test<-as.data.frame(cbind(x,y))
>> test
> x y
> 1 1 A
> 2 2 1
> 3 4 2
> 4 4 A
>> test[test$x==4,]$y<-'B'
> Warning message:
> In `[<-.factor`(`*tmp*`, iseq, val
Hi,
You can add extra levels using levels().
e.g.
> test <- data.frame(x=c(1,2,4,4), y=c("A", 1, 2, "A"))
> levels(test$y) <- c(levels(test$y), 'B')
> levels(test$y)
[1] "1" "2" "A" "B"
> test[test$x==4,]$y<-'B'
> test
x y
1 1 A
2 2 1
3 4 B
4 4 B
Regards,
Paul
jlfmssm wrote:
Sorry, I didn
Hi Jack,
Maybe this helps.
# make some data
set.seed(123)
condition <- factor(rep(c("a","b"), each = 5))
score <- rnorm(10);
lg <- data.frame(condition, score)
# Carry out commands
a <- subset(lg,condition=="a")["score"]
b <- subset(lg,condition=="b")["score"]
t.test(a,b,paired=TRUE)
#Error
Sorry, I didn't get it.
I made a simple example to explain what I want
> x=c(1,2,4,4)
> y=c("A",1,2,"A")
> test<-as.data.frame(cbind(x,y))
> test
x y
1 1 A
2 2 1
3 4 2
4 4 A
> test[test$x==4,]$y<-'B'
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = c("B", "B")) :
invalid factor level,
Hi Khaled.
I believe using "text" on an empty plot would do the trick...
Tal
On Sat, Jul 25, 2009 at 3:13 PM, Khaled OUANES wrote:
> I created a 2 D plan:
>
> > plot(range(-2.5,0.95),range(0.00,1.00),type="n",axes=TRUE)
>
> I made a projection of points with their coordonates (X,Y) in that
Duncan Murdoch wrote:
On 24/07/2009 6:35 PM, Frank E Harrell Jr wrote:
Dear Group:
I want to create a function having a ... argument and to have the
default arguments evaluated, as thus:
g <- function(a, b, ...) a+b
formals(g) <- alist(a=,b=2+3,...=)
g
function (a, b = 2 + 3, ...)
a + b
But
Try this. The last function removes everything but the
last part, i.e. the part after the -. Then we order the
columns, except the first column, by their last part
and finally split the result by the last part of the ID.
This gives a list of data frames, two in this case:
> last <- function(x) s
Does this do what you want:
> x
ID S1-a S1-b S2-a S2-b
1 001-A1234
2 001-B5678
3 002-A9 10 11 12
4 002-B 13 14 15 16
> # split the column names
> t.n <- strsplit(names(x)[-1], '-')
> t.n <- sort(sapply(t.n, function(a) paste(a[2],a[1])))
> # c
Hi,
The file did not make it through the mailing list. Maybe you are looking
for ?read.dcf
Can you describe the way your application interacts with R.
Romain
On 07/25/2009 10:35 AM, bed.si...@oracle.com wrote:
Hi,
I am using the R-2.9.1 with Window XP.
Queries:
1. I am r
I created a 2 D plan:
> plot(range(-2.5,0.95),range(0.00,1.00),type="n",axes=TRUE)
I made a projection of points with their coordonates (X,Y) in that plan
> fullpoints=read.csv2("fullpoints.csv",h=T)
> plot(fullpoints)
The points are listed in that .csv file it is organized this way:
0,4887
i have a trend that it is linear at first and then it has a particular trend.
There is a mathematic model, but it describes only the linear part. I have
tried to use a spline regressione, but i have obtained a particular fit
trend that it is unreal. Can i create a spline regression that it is of a
Have you thought about fourier filtering?
here is some code that may help.
library(StreamMetabolism)
library(mFilter)
data(DOTemp)
#this makes it the same everytime
set.seed(100)
#making a time series with the frequency = 96
#reading in one unit in other words this is 15min data
#so there are 9
Hi,
I am using the R-2.9.1 with Window XP.
Queries:
1. I am running the java application which needs to load property file in
R.
So can you please tell me how I can load my property file in R session
so that my application can find that property file?
A
Hi everyone,
I have a question about calculating r-squared in R. I have tried searching the
archives and couldn't find what I was looking for - but apologies if there is
somewhere I can find this...
I carried out a droughting experiment to test plant competition under limited
water. I had:
-
Hi John,
You can use
rm(object)
to delete a specific object from the current environment and then save
your workspace again without those objects
ls()
to see a list of objects could also be useful
Regards,
Paul
John Seppänen wrote:
> Hi all! I have accidentially saved few objects when I hav
Hi all! I have accidentially saved few objects when I have closed workspace
and clicked from "save workspace image" "Yes". Now I would like to delete
the .RData files so that workspace wouldnt restore the unwanted objects
everytime I open the workspace.
I know i could delete the corresponding .RDa
Oscar Bayona wrote:
Hell,
I want to plot two column vectors in a plot, line type, but they
have different scale, how can I plot both series on the same graph but one
on the left axis and the other scale reflecting on the right axis?
Hi Oscar,
See if twoord.plot in the plotrix package does w
mfreidin wrote:
I have a matrix containing means and CIs (lower and upper in two columns, so
three columns for every data point) for several points. I have to build a
graph of these means accompained by the CIs (as wiskers). No problems with
making the graph of means, but I don't know how to intr
On Sat, Jul 25, 2009 at 5:07 AM, Dieter
Menne wrote:
>
>
>
> losemind wrote:
>>
>> If I use a moving average, it will smooth the choppy time series, but
>> it will lead to lagging...
>>
>
> The lagging can only be removed if you look into the future, otherwise you
> run into causality problems.
> S
How about like this:
for (i in seq_along(a)) {
result <- as.list(a[1:i])
cat("iterator", i, ":\n")
print(result)
}
On Sat, Jul 25, 2009 at 6:48 AM, Alberto Lora M wrote:
> Hi Everybody
>
> I have the following problem
>
> suppose that we
>
> a<-c("uno","dos","tres")
>
> I am working with a
Waverley,
use @ (instead of $) to extract the slots from the performance object
(it's S4 class system).
HTH,
Tobias
On Sat, Jul 25, 2009 at 8:20 AM, Waverley wrote:
> I use ROCR to plot multiple runs' performance. Using the sample code
> as example:
>
> # plot ROC curves for several cross-val
You're right. Using read.csv, the first column is a factor, not string
(or should I use str?). The following is a 2x2 version of the data frame
after read.csv
IDS1-a S1-b S2-a S2-b
1 001-A1 2 3 4
2 001-B5 6 7 8
3 002-A9 101112
4
losemind wrote:
>
> If I use a moving average, it will smooth the choppy time series, but
> it will lead to lagging...
>
The lagging can only be removed if you look into the future, otherwise you
run into causality problems.
So the easiest way is to center your output data on half the window
osbacan wrote:
>
> I want to plot two column vectors in a plot, line type, but they
> have different scale, how can I plot both series on the same graph but one
> on the left axis and the other scale reflecting on the right axis?
>
See
http://markmail.org/thread/tdlsk3rpao7t6k7z
but make s
waverley palo wrote:
>
> I use ROCR to plot multiple runs' performance. Using the sample code
> as example:
>
> I can follow the code and plot without any problem. However, I don't
> know how to extract the averaged ROC area under curve value.
>
>
See the documentation of performance. Amo
Hello Daniel,
Thank you for the extended answer. You gave me one more reason to delve into
mixed models in the future.
I will be honest and say that I believe I will explore Roberts approach more
due to the simplicity of it. But nevertheless, I am very grateful for your
reply.
With best regards,
Are you using 'read.csv'? At least include an 'str' of the object you
are wanting to convert so that we know the structure of it, since we
can not guess at what it is.
On Sat, Jul 25, 2009 at 4:32 AM, Junqian Gordon Xu wrote:
> Actually when I read the spreadsheet from cvs file, "S1-[abcd]" are t
Actually when I read the spreadsheet from cvs file, "S1-[abcd]" are the
header and "T1-[abcd]" are the strings in first column of the data frame.
Gordon
On 07/25/2009 03:13 AM, jim holtman wrote:
It it not entirely clear what the format of your data is. If you have
a dataframe that you would l
Here is one way:
> a<-c("uno","dos","tres")
> x <- list()
> a<-c("uno","dos","tres")
> x <- list()
> for (i in seq_along(a)){
+ # add to the list
+ x[[i]] <- a[i]
+ str(x)
+ }
List of 1
$ : chr "uno"
List of 2
$ : chr "uno"
$ : chr "dos"
List of 3
$ : chr "uno"
$ : chr "dos"
$ : chr "t
If what you are showing is a 'matrix' (you did not include a
reproducible script, nor did you show an 'str' of the data), then it
is a character matrix. I took your data, read it in as a dataframe
and then converted it to 'matrix' (x.mat). You can see below that
this is a character matrix. If I
It it not entirely clear what the format of your data is. If you have
a dataframe that you would like to separate into several different one
based on the value in a column, then something like this will work:
df.list <- split(yourDF, yourDF$column)
This will create a list of dataframes, split acc
Hell,
I want to plot two column vectors in a plot, line type, but they
have different scale, how can I plot both series on the same graph but one
on the left axis and the other scale reflecting on the right axis?
Thanks in advance.
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