> The actual data have more than 5 thousand data points, Just want to plot a few
> #selected randomly.
Based on this, it seems you're looking to pull a random sample of 5 from your
dataset.
Does this link help?
http://stat.ethz.ch/R-manual/R-patched/library/base/html/sample.html
~Nicole Ford
As to your xyplot question, I found a multiline solution. (Below)
I am pretty certain there is a multipolt/ graph solution in the effects package
(I haven't used that functionality in some years).
http://r.789695.n4.nabble.com/XYplot-in-Lattice-Package-td878511.html
~Nicole Ford
Ph.D. Studen
Hi all,
I have gone through some basics of R in eclipse environment. What I need to
work is more than basic level so I could not find out the solution by
myself. could anyone help me ?
My case,
I have climate data (precipitation) in nc format with single layer for each
time interval in a day. Th
I have several linear models on the same data:
m1 <- lm(y ~ poly(x,1))
m2 <- lm(y ~ poly(x,2))
m3 <- lm(y ~ poly(x,3))
What I don't understand is why
anova(m1, m2, m3, test="F")
- yields the same RSS and SS values, but a different p-value from anova(m1,
m2, test="F")
- when it also yields the S
On 02/23/2013 08:55 PM, Robert Zimbardo wrote:
I have several linear models on the same data:
m1 <- lm(y ~ poly(x,1))
m2 <- lm(y ~ poly(x,2))
m3 <- lm(y ~ poly(x,3))
What I don't understand is why
anova(m1, m2, m3, test="F")
- yields the same RSS and SS values, but a different p-value from an
hello Rui,
Many thanks. That really helped me.you are totally right about[[4]]
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-merge-two-functions-into-one-tp4659365p4659446.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi Peter
Try the 2 examples below - it may give you some ideas
useOuterStrips(strip =
strip.custom(factor.levels = paste("Year", c(2010:2012)),
par.strip.text = list(cex = 0.75)),
strip.left =
strip.custom(factor.levels = paste("Lat
Hi all,
I have gone through some basics of R in eclipse environment. What I need to
work is more than basic level so I could not find out the solution by
myself. could anyone help me ?
My case,
I have climate data (precipitation) in nc format with single layer for each
time interval in a day. Th
Part of the space between the boxplots is the margin area, you can remove
that. You can also adjust the width of the box within the plot:
layout(matrix(c(1,2,3), 3, 1, byrow = TRUE), heights=c(0.3,0.3,0.6))
op <- par(mar=c(0,4,0,2)+0.1)
boxplot(rnorm(100), horizontal=TRUE, axes=FALSE, width=1)
bo
I have a longitudinal competing risk data of the form:
IDCOMPL SEX HEREDITY
1 0 1 2
1 0 1 2
1 3 1 2
2 0 0 1
2 1 0 1
2 2 0 1
2 2 0 1
3 0 0 1
3 0 0 1
3 0
Dear R Help Members,
I need to draw a pie chart for someone and there are 3 small slices in it
identifying different species, all 0.7%, whoâs labels overlap so you cannot
read them. Is there a way to have only 1 label the three slices, so as to merge
the three labels into one without merging
On 23.02.2013 17:06, Melanie Zoelck wrote:
Dear R Help Members,
I need to draw a pie chart for someone and there are 3 small slices in it
identifying different species, all 0.7%, who’s labels overlap so you cannot
read them. Is there a way to have only 1 label the three slices, so as to me
Hi,
Try this:
dat1<- read.table(text="
ID COMPL SEX HEREDITY
1 0 1 2
1 0 1 2
1 3 1 2
2 0 0 1
2 1 0 1
2 2 0 1
2 2 0 1
3 0 0 1
3 0 0 1
3 0 0 1
3 0 0 1
3
On 22.02.2013 11:41, Bob Green wrote:
Hello,
In SPSS the cluster analysis output includes an agglomerations schedule,
which details the stages when cases are joined.
Is it possible to obtain such output when performing cluster analysis in
R? If so, I'd appreciate advice regarding how to obta
On 15.02.2013 13:05, Amir Kasaeian wrote:
Dear all,
Good day!
I have a question in my codes, would you please help me how to rectify it?
these are my coded but at the last line I received the error!
mac_30 = read.dta("MAC results4.dta")
mac_30
map_30 = read.dta("MAP results4.dta")
map_30
mac_
On 14.02.2013 20:29, James Jong wrote:
What is the function call interface for predict in the package party for
cforest? I am looking at the documentation (the vignette) and ?cforest and
from the examples I see that one can call the function predict on a cforest
classifier. The method predict s
On 13.02.2013 05:38, Marino David wrote:
Dear all mailing listers,
I failed in install source packages in Windows system. I have installed
Rtools in order to install the source package based on the discussion
available on the following link:
http://r.789695.n4.nabble.com/Installing-a-Package-t
On Feb 23, 2013, at 6:28 AM, Tasnuva Tabassum wrote:
> I have a longitudinal competing risk data of the form:
>
> IDCOMPL SEX HEREDITY
> 1 0 1 2
> 1 0 1 2
> 1 3 1 2
> 2 0 0 1
> 2 1 0 1
> 2 2 0 1
> 2
On Feb 23, 2013, at 10:06 , Rolf Turner wrote:
>>
>> What am I missing?
>
> A basic understanding of the theory of linear models. This really has little
> to do with R. Go and read a good intro to linear modelling.
>
> Insofar as your question has anything to do with R:
>
> When you do
>
>
Hi,
You can also use:
do.call(rbind,lapply(split(dat1,dat1$ID),function(x) head(x[x$COMPL!=0,],1)))
# ID COMPL SEX HEREDITY
#1 1 3 1 2
#2 2 1 0 1
#3 3 2 0 1
- Original Message -
From: Tasnuva Tabassum
To: r-help@r-project.org
Cc:
Sent: Sat
which method in statistics is completely free from model misspecification?
Thanks and regards,
_
Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
__
Hi Jannetta,
as far as I can see, your implementation was almost ;-) correct, except
that:
init[2] <- init[1] + d
should be:
init[2] <- init[2] + d
The root function with:
return(init[1]-30)
is correct, because this triggers an event when init[1]-30 crosses the
zero line, i.e. when ini
On 23.02.2013 19:33, Arun Kumar Saha wrote:
which method in statistics is completely free from model misspecification?
The data.
Uwe Ligges
Thanks and regards,
_
Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Vis
Hello,
You can use ?aggregate and ?head to do what you want. Try the following.
dat <- read.table(text="
IDCOMPL SEX HEREDITY
10 1 2
10 1 2
13 1 2
20 0 1
21 0 1
22 0 1
22 0 1
30 0
The server removed any attached files
It is better to put everything in the actual email.
https://github.com/hadley/devtools/wiki/Reproducibility
John Kane
Kingston ON Canada
> -Original Message-
> From: berega...@gmail.com
> Sent: Sat, 23 Feb 2013 02:40:29 +0300
> To: r-help@r-projec
Hello,
Why do you think your data is gaussian? For what it's worth,
qqnorm(small) # doesn't look
qqline(small) # gaussian
Hope this helps,
Rui Barradas
Em 22-02-2013 23:27, Samantha Warnes escreveu:
Hello,I'm still working with this data set, and trying to fit it with a
nonlinear model. H
Try this:
dat$sequence <- as.vector(unlist(lapply( aggregate(dat$ID, by=list(x),
FUN=length)$x, FUN=function(x){seq(1, x
dat0 <- dat[dat$sequence==1, ]
HTH, X
On Sat, Feb 23, 2013 at 1:07 PM, Rui Barradas wrote:
> Hello,
>
> You can use ?aggregate and ?head to do what you want. Try the fol
To account for COMP,
dat$sequence <- as.vector(unlist(lapply( aggregate(dat$ID, by=list(dat$ID),
FUN=length)$x, FUN=function(x){seq(1, x
dat0 <- dat[dat$sequence==1 & dat$COMPL!= 0, ]
HTH, X
On Sat, Feb 23, 2013 at 1:15 PM, Xiaogang Su wrote:
> Try this:
> dat$sequence <- as.vector(unlist
On Feb 23, 2013, at 11:09 AM, Rui Barradas wrote:
> Hello,
>
> Why do you think your data is gaussian? For what it's worth,
>
> qqnorm(small) # doesn't look
> qqline(small) # gaussian
It's a bit hard to say with such a small sample, isn't it? Here's a poor man's
functional data analysis:
For an application, I have formulas defined in terms of a left Kronecker
product of matrices,
A,B, meaning
A \otimes_L B = {A * B[i,j]} -- matrix on the left multiplies each
element on the right.
The standard kronecker() function is the right Kronecker product,
A \otimes_R B = {A[i,j] * B} -
On 23-02-2013, at 20:46, Michael Friendly wrote:
> For an application, I have formulas defined in terms of a left Kronecker
> product of matrices,
> A,B, meaning
> A \otimes_L B = {A * B[i,j]} -- matrix on the left multiplies each element
> on the right.
>
> The standard kronecker() function
HI,
Tried your approach:
dat1$sequence <- as.vector(unlist(lapply( aggregate(dat1$ID,
by=list(dat1$ID),FUN=length)$x, FUN=function(x){seq(1, x)})))
dat0 <- dat1[dat1$sequence==1 & dat1$COMPL!= 0, ] #your second solution
dat0
#[1] ID COMPL SEX HEREDITY sequence
#<0 rows> (or 0-l
My bad. I didn't try it out with the real data. Here you go. HTH, X
dat <- read.table(text="
IDCOMPL SEX HEREDITY
10 1 2
10 1 2
13 1 2
20 0 1
21 0 1
22 0 1
22 0 1
30 0 1
30
Hello Uwes,
Thanks. Re-reading the hclust pages I found that using the hclust
'USArrests' data that the command > plot (hc1) will generate the
order in which cases joined. however, I still can't see how to obtain
the respective height at which each case joined each cluster or the
height whe
FWIW (It may not be much):
1. Data **are never** Gaussian. Failure to reject the hypothesis that
the data are sampled from a Gaussian does not mean that the data can
be assumed to be drawn from a Gaussian. That depends on the
statistical methodology and the application context.
2. Given a large e
Hello, All:
What facilities exist for plots and tables of data in the
standard MediaWiki software beyond the obvious standards?
The standard MediaWiki tables and graphics capabilities seem to
me to be rather clumsy, and I wonder if I'm missing some available
extensions?
If you look at this question objectively, Spencer, would you not agree that
this question is entirely off topic here?
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#.
Hi, Jeff:
On 2/23/2013 3:12 PM, Jeff Newmiller wrote:
If you look at this question objectively, Spencer, would you not agree that
this question is entirely off topic here?
I can understand why you might think that. However, I believe
this lists entertains questions dealing with inte
You didn't show what the tabular summary should look like.
However, look at the height and merge components of
an hclust object:
> hc3 <- hclust(dist(USArrests[1:8, c(1,2,4)]))
> data.frame(hc3[2:1])
height merge.1 merge.2
1 9.297849 -1 -8
2 13.609188 -2 -5
3 23.77919
Willam,
Many thanks. I'll check this against my data tomorrow when I'm back
at work. This looks just what I wanted.
Regards
Bob
At 09:27 AM 24/02/2013, William Dunlap wrote:
You didn't show what the tabular summary should look like.
However, look at the height and merge components of
an h
Ugh. So simple! Thanks, Berend.
On 2/23/2013 2:58 PM, Berend Hasselman wrote:
On 23-02-2013, at 20:46, Michael Friendly wrote:
For an application, I have formulas defined in terms of a left Kronecker
product of matrices,
A,B, meaning
A \otimes_L B = {A * B[i,j]} -- matrix on the left multi
Fortune candidate?
I hear the landlord is hell, but the company good.
Bryan
> I've
> already got an apartment reserved for me in one of Pat Burns's "R
> Inferno" levels, and I don't want to descend even further.
>
> Best,
> Bert
__
R-help@r-project.o
The universe of efficient ways to generate graphics for Wikipedia would be on
Wikipedia talk pages somewhere. The universe of ways to make SVG graphics with
R could be on topic here. They might intersect, and if other graphics formats
that R can generate for Wikipedia are identified through appr
Can anyone help with this ?
On 14/02/2013 14:07, Robert Long wrote:
I am working with some survival data with missing values.
I am using the mice package to do multiple imputation.
I have found code in this thread which handles pooling of the MI results:
https://stat.ethz.ch/pipermail/r-help/2
On 2/23/2013 4:35 PM, Jeff Newmiller wrote:
The universe of efficient ways to generate graphics for Wikipedia would be on
Wikipedia talk pages somewhere. The universe of ways to make SVG graphics with
R could be on topic here. They might intersect, and if other graphics formats
that R can gene
I've not done any of this, but you might look at (a variation of) one
of the following tool chains:
knitr markdown --pandoc--> mediawiki
brew --pander--> mediawiki
pander is at http://rapporter.github.com/pander/
Best,
Ista
On Sat, Feb 23, 2013 at 4:59 PM, Spencer Graves
wrote:
> Hello, All:
>
On 02/24/2013 03:06 AM, Melanie Zoelck wrote:
Dear R Help Members,
I need to draw a pie chart for someone and there are 3 small slices in it
identifying different species, all 0.7%, who’s labels overlap so you cannot
read them. Is there a way to have only 1 label the three slices, so as to m
I haven't seen anyone solve this. I think it would be reasonable to do a
time point by time point averaging (over multiple imputations) of the
underlying survival curve, although there is some question about whether to
"freeze" the centering constant (sum of beta times covariate mean). What
will
Hi All
I'm just wondering if there is a quick way of filling a way with the
following.
I want to declare array I with a specific length and then alternatively
fill it with 10 zeros and 10 specified values:
v<- 14
I <- c(0,length(t))
But in stead of just filling I with 0 I want 10 zeros and the
On 02/24/2013 02:56 PM, Jannetta Steyn wrote:
Hi All
I'm just wondering if there is a quick way of filling a way with the
following.
I want to declare array I with a specific length and then alternatively
fill it with 10 zeros and 10 specified values:
v<- 14
I<- c(0,length(t))
But in stead o
Hi Jannetta,
Try this:
rep(c(0, v), each = 10)
See ?rep for details
Cheers,
Josh
On Sat, Feb 23, 2013 at 7:56 PM, Jannetta Steyn wrote:
> Hi All
>
> I'm just wondering if there is a quick way of filling a way with the
> following.
>
> I want to declare array I with a specific length and th
Hi
Thank you very much, but I forgot to tell that I also want to include the
patients for which no complication occurred. That is, for my data I want to
include patient no. 4, for which the COMPL value will be 0.
In that case, what R function should I write?
On Sat, Feb 23, 2013 at 12:23 PM, Xia
Hi,
Try this:
#dat1
do.call(rbind,lapply(split(dat1,dat1$ID),function(x) if(any(x$COMPL!=0))
head(x[x$COMPL!=0,],1) else head(x,1)))
# ID COMPL SEX HEREDITY
#1 1 3 1 2
#2 2 1 0 1
#3 3 2 0 1
#4 4 0 1 2
A.K.
_
Hi,
i have a very huge number of data with the size 2375ko, i want to import
them for R to xlsx
but the size of excel is limited
How can i resolve this problem?
And please how can i define the frame
Note that i have a table
rows times c(1.100) columns WLc(200...1000) and inside S c(15,...
Hi,
d3<-structure(list(m1 = c(2, 3, 2), n1 = c(2, 2, 3), cterm1_P0L = c(0.9025,
0.857375, 0.9025), cterm1_P1L = c(0.64, 0.512, 0.64), cterm1_P0H = c(0.9025,
0.9025, 0.857375), cterm1_P1H = c(0.64, 0.64, 0.512)), .Names = c("m1",
"n1", "cterm1_P0L", "cterm1_P1L", "cterm1_P0H", "cterm1_P1H"), row.nam
Hi,
You can also use ?rowMins() or ?rowMaxs() from library(matrixStats)
library(plyr)
res2<- join(res1,d3,by=c("m1","n1"),type="inner")
p0L<-0.05
p0H<-0.05
p1L<-0.20
p1H<-0.20
res2<- within(res2,{p1<- x/m; p2<- y/n;term2_p0<-dbinom(x1,m1, p0L, log=FALSE)*
dbinom(y1,n1,p0H, log=FALSE)*db
Hello,
I tried to manipulate the order of boxplots using the "reorder" below and
worked.
However, please kindly advise how to specify the colors of each boxplot
(bird group),
like red, green, and blue, from the left to the right.
Thanks again.
code
bwplot(GE_distance~OF, data=dataN,
index.c
There is a reason why merge is not designed to work directly with arrays: the
matching of rows can easily yield a result with a different number of rows than
the initial data had. This plays havoc with the nature of arrays. I would
recommend that if a merge is really what you want then unrolling
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