?glht says
with 'print', 'summary', 'confint', 'coef' and 'vcov' methods
being available.
try:
example(glht)
summary(glht(amod, linfct = mcp(tension = Tukey)))
confint(glht(amod, linfct = mcp(tension = Tukey)))
On 3/19/07, Nair, Murlidharan T [EMAIL PROTECTED] wrote:
I used the multcomp
try:
mtext(substitute(R^2 * : * GoodnessOfFits[i],list(i=graphNumber)))
HTH.
On 3/16/07, Bob Farmer [EMAIL PROTECTED] wrote:
Hi all:
I would like to create a line of plot margin text (using mtext() ) that
features both a superscript and a subset of an object. However, I
cannot seem to do
you can use nls:
see
?nls
and its example.
HTH.
On 3/16/07, Hufkens Koen [EMAIL PROTECTED] wrote:
Hi list,
I was wondering how I should go about fitting a sigmoid curve to a dataset.
More specifically how I estimate parameters a and b in the following
equation:
1 / 1+exp(-(x-a)*b)
Internally, labs in persp are drawn as when you use text function.
So you cannot change the sizes by cex.lab, but you can change by cex:
persp(x, y, z, cex=1.5)
gives larger labs in persp 3d plot.
Of course there may be some side effect because it may change the size
something other than labs.
Probably, you have to do it by hand.
Exactly, I do not know the reason, but I can imagine.
Once you define factor, the empty factor is not meaningless.
The simple way to do it is refactorize:
f-factor(1:3)
f
[1] 1 2 3
Levels: 1 2 3
factor(f[f!=2])
[1] 1 3
Levels: 1 3
On 2/9/07, Roger Leigh
try:
mx[,new.col.names]
HTH.
On 2/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi R users,
I would like to know how to sort a matrix according a different order of
colnames (or rownames) ,e.g.,
mx = matrix(rnorm(1:20),5,4)
colnames(mx) = letters[1:4]
rownames(mx) = letters[1:5]
mx
summary.lm tells you F, df, and p value.
try:
summary(regression9)
On 2/6/07, Jason R. Finley [EMAIL PROTECTED] wrote:
Hello,
I have spent a good deal of time searching for an answer to this but
have come up empty-handed; I apologize if I missed something that is
common knowledge.
I am
:
On Tue, 6 Feb 2007, talepanda wrote:
For size, maybe:
# Dimensions(in characters) of the internal pager.
#pgrows = 25
#pgcolumns = 80
pgrows = 48
pgcolumns = 128
in Rconsole, but location cannot be handled.
For the very good reason that you can have multiple pagers and I at least
do
I think several ways can do that.
my code is:
data.frame(t(apply(dat,1,function(x)as.numeric(sub(\\*,,x[sort(1:2,grep(\\*,x)==2)])
HTH
On 2/7/07, Dale Steele [EMAIL PROTECTED] wrote:
Given two columns of type character in a dataframe of the form:
col1col2
31* 66
0 0*
I'm sorry, I forgot my definition of dat.
dat in code of previous post is:
dat-data.frame(col1=c(31*,0,102*,71*,31,66,47),col2=c(66,0*,66,80,2*,31*,38*))
dat
col1 col2
1 31* 66
20 0*
3 102* 66
4 71* 80
5 31 2*
6 66 31*
7 47 38*
On 2/7/07, talepanda [EMAIL PROTECTED
If I correctly understand,
hist(rep(1, 100), col=lightblue, border=black,breaks=0:10*0.1)
see:
breaks arg in ?hist
On 2/7/07, Dong H. Oh [EMAIL PROTECTED] wrote:
Dear expeRts,
I'd like to picture histogram of ones.
For example,
hist(rep(1, 100), col=lightblue, border=black)
A bin is
d-data.frame(id=1:4,x=c(0.1,0.5,0.2,0),y=c(3,1,9,5))
d
id x y
1 1 0.1 3
2 2 0.5 1
3 3 0.2 9
4 4 0.0 5
d[order(d$y),]
id x y
2 2 0.5 1
1 1 0.1 3
4 4 0.0 5
3 3 0.2 9
On 2/6/07, XinMeng [EMAIL PROTECTED] wrote:
Hello sir:
How can I sort a dataframe by sorting one of its column?
For size, maybe:
# Dimensions(in characters) of the internal pager.
#pgrows = 25
#pgcolumns = 80
pgrows = 48
pgcolumns = 128
in Rconsole, but location cannot be handled.
On 2/6/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi,
Using the Rconsole file I can specify the size and location of
Dear R-er:
I want to use invisible function in some packages.
I know that triple colon operater is available.
However, how can I use all invisible functions in some packages.
One solution is , of course, to use ::: in every use, but is there
any workaround?
It is similar to using namespace
Currently, both color and plotting symbol does not change with the
grouping variable g1, that is, when
g1-rep(5:8,each=64)
,plotxy still generates same plot.
subscripts argument is useful for your purpose.
see:
panel section and details in
?xyplot
try:
plot-xyplot(y~x|f,
try:
plot(1:10,pch=1:10,cex=1:10)
pch specifies the types and cex specifies the sizes of symbols.
On 1/28/07, vinod gullu [EMAIL PROTECTED] wrote:
I want to plot variation of more than one variable in
single plot with different point types and points
sizes . Can someone help me to do that,
First question: Yes
Second question: z in func.int2 is z in myfunc (z=y^3).
You can easily test:
f0-function(){
z-1
f1-function()print(f1)
f2-function(z){f1();print(z)}
f2(z)
}
z-0
f0()
R language definition may be helpful
It occurs why start or stop could not be converted into integer of
length 0 by using as.interger().
More presicely,
if( !isInteger(sa) || !isInteger(so) || k == 0 || l == 0 ) # c code
where sa is start, so is stop, k is length(start), l is length(stop)
For example:
substr(orz,character(0),0)
see:
?exists
HTH.
On 1/24/07, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote:
Hi,
Is there any way to check whether an R object exists or not? Say
example: a data frame.
Thanks,
Shubha
[[alternative HTML version deleted]]
__
non-elegant solution:
matrix(c(V1,rep(NA,-length(V1)%%5)),nrow=5)
HTH.
On 1/24/07, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote:
How to suppress the recycling of items in a matrix..instead NA can be
filled.
-Original Message-
From: Chuck Cleland [mailto:[EMAIL PROTECTED]
For Date class, *original* (that is, contents in memory) is 12853,
and 2005-03-11 is one expression of the original.
So you have to convert from the original to the charecter expression as follows.
s[1]-format(date)
s
[1] 2005-03-11 FALSE FALSE
s[1]-as.character(date)
s
[1] 2005-03-11
rownames()- is what you want.
dat-data.frame(V1=sample(10),V2=sample(10))
dat
V1 V2
1 2 5
2 3 8
3 8 4
4 9 6
5 6 2
6 5 7
7 10 3
8 4 9
9 1 10
10 7 1
dat-dat[order(dat$V2),]
dat
V1 V2
10 7 1
5 6 2
7 10 3
3 8 4
1 2 5
4 9 6
6 5 7
2 3 8
8 4
for the column names. For a data frame, 'rownames'
and 'colnames' are calls to 'row.names' and 'names' respectively,
but the latter are preferred.
On Wed, 24 Jan 2007, talepanda wrote:
rownames()- is what you want.
dat-data.frame(V1=sample(10),V2=sample(10))
dat
V1 V2
1
?Date seems to say that weekdays() is appropriate for that:
weekdays(as.Date(2006-12-01))
see:
?weekdays
On 1/25/07, John McHenry [EMAIL PROTECTED] wrote:
Hi WizaRds,
What is the standard way to get the day of the week from a date such
as as.Date(2006-12-01)? It looks like fCalendar has
see:
?commandArgs
or more detail for R startup mechanisms:
?Startup
On 1/22/07, Deepak Chandra [EMAIL PROTECTED] wrote:
Hi All,
A simple and naive question from a newbie. How can one access command-line
arguments in R i.e. equivalent of argv in C?
Have spent a lot of time on finding it.
play below after your code and look at tk window:
tkentryconfigure(editMenu,0,state=disable)
tkentryconfigure(editMenu,0,state=active)
tkentryconfigure(topMenu,1,state=disable)
tkentryconfigure(topMenu,1,state=active)
HTH
On 1/22/07, Jarno Tuimala [EMAIL PROTECTED] wrote:
Hi!
I've
Usually (that is, not limited in R language), when error occurs in
try, stacks are rollbacked, so the variables defined in try no longer
exists after calling try.
One non-elegant solution is:
fit-NULL
try ( (fit = lm(y~x, data = data_fitting)), silent =T)
if(!is.null(fit)){
coeffs =
TeamInfo
TEAMNAME LEVEL WORKTIME BONUS
1 batch sunan B 135 9,818
2 batch Chenqi E 121 6,050
3 batch jiangxu F 97 4,189
4 online zhouxi F 63 2,720
5 online chenhe H 36 1,064
## try:
factor(TeamInfo$TEAM)
[1] batch batch batch
Probably, you have to manually create, but it is easy:
# for beside=F
mp - barplot(VADeaths)
text(mp,y-t(apply(VADeaths,2,cumsum)),y) # labels are values itself
# or
mp - barplot(VADeaths)
text(mp,t(apply(VADeaths,2,cumsum)),VADeaths) # labels are cumsum of values
# for besides=T
mp -
In R language, one solution is:
a-c(3,4,6,2,3)
which(a==max(a))
On 1/18/07, Feng Qiu [EMAIL PROTECTED] wrote:
Hi all:
A short question:
For example, a=[3,4,6,2,3], obviously the 3rd entry of the array
has the maxium value, what I want is index of the maxium value: 3. is
It's not R problem but specification of windows file system.
see MSDN:
http://msdn2.microsoft.com/en-us/library/aa365247.aspx
On 1/15/07, Brandt, T. (Tobias) [EMAIL PROTECTED] wrote:
Hi
I cannot seem to create any files that have the name CON before the file
extension, i.e. all of the
col.region changes both colors in plot and colorkey.
try:
x - seq(pi/4, 5 * pi, length = 100)
y - seq(pi/4, 5 * pi, length = 100)
levelplot(z~x*y) #default
levelplot(z~x*y,col.regions=rainbow(24)) #custom color
On 1/11/07, Bram Kuijper [EMAIL PROTECTED] wrote:
Hi all,
I try to make a
because given data is a part of your data, I cannot examine,
however, try:
##out.block-identify(tb_ncs$y,tb_ncs$Slide)
out.block-identify(tb_ncs$Slide,tb_ncs$y)
On 1/11/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear R-users,
Following is part of my data, where slide has 36
levels and
the observation number
everytime I clicked on any point.What I want is
instead of obervation numbers it would be block and/or
slide numbers.
Any other idea how I can make it works ?
Thanks
--- talepanda [EMAIL PROTECTED] skrev:
because given data is a part of your data, I cannot
examine
try:
readLines(n=1)-str
On 1/13/07, Tong Wang [EMAIL PROTECTED] wrote:
Hi all,
Sorry about the simple question, but I have searched the web with
prompt , input etc. and never got the answer .
thanks a lot
tong
__
NIR seems to be not dataset. It is a member of dataset yarn.
Try:
library(pls)
data(yarn)
str(yarn)
On 1/10/07, Carmen Meier [EMAIL PROTECTED] wrote:
I did just the download of the pls package, but the NIR dataset is not
available
require(pls)
[1] TRUE
data(NIR)
Warning message:
data
See changelog of pls 2.0-0
- The 'NIR' and 'sensory' data sets have been renamed to 'yarn' and 'oliveoil'.
you can see it in package source from CRAN
http://cran.r-project.org/src/contrib/Descriptions/pls.html
HTH
On 1/10/07, Carmen Meier [EMAIL PROTECTED] wrote:
talepanda schrieb:
NIR
Maybe I misunderstand what you want to do, one solution is:
l1
$a
[1] 1 2
$b
[1] 1 2 3
$c
[1] 1 2 3 4
l2
V1 V2
1 d nd
2 c nc
3 b nb
4 a na
names(l1)-sapply(names(l1),function(n)l2[l2$V1==n,2])
l1
$na
[1] 1 2
$nb
[1] 1 2 3
$nc
[1] 1 2 3 4
On 1/10/07, Christoph Heibl [EMAIL
try apply() :
par(new=F);
apply(s,2,function(x){plot(x[[1]],x[[2]],type=o);par(new=T)})
On 1/8/07, Antje [EMAIL PROTECTED] wrote:
Hi all,
I've got the following problem. I have a vector containing file names. I
want to read these files as csv and calculate the density-function for
each file
It can be explained.
class(A)
[1] data.frame
length(A)
[1] 5
class(A==0)
[1] matrix
length(A==0)
[1] 10
class(-A*log(A))
[1] data.frame
length(-A*log(A))
[1] 5
as you can see, the result of A==0 is matrix with length=10, while the
result of -A*log(A) is still data.frame with length=5.
You have to create *formula* object from string and pass it to lda().
See my previous post.
On 12/29/06, Feng Qiu [EMAIL PROTECTED] wrote:
Hi Gabor:
Thank you! But it didn't work. Since lda() takes the variable
name as the input parameter. So what I was trying to do is make the
Generally, you can create formula from string:
lda(formula(paste(names(iris)[5],~.)),iris)
On 12/29/06, Feng Qiu [EMAIL PROTECTED] wrote:
Hi Gabor:
Thank you! But it didn't work. Since lda() takes the variable
name as the input parameter. So what I was trying to do is make the
Try this:
lda(formula(paste(names(iris)[5],~.)),iris)
You have to create *formula* object from string and pass it to lda().
On 12/29/06, Feng Qiu [EMAIL PROTECTED] wrote:
Hi Gabor:
Thank you! But it didn't work. Since lda() takes the variable
name as the input parameter. So what
Generally, you can create formula from string:
lda(formula(paste(names(iris)[5],~.)),iris)
On 12/28/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try:
lda(iris[-5], iris[,5])
On 12/26/06, Feng Qiu [EMAIL PROTECTED] wrote:
Hi everyone:
I'm trying to compose a string dynamicly
Try:
as.date(Date1)-as.date(Date2)
On 12/28/06, Brian Edward [EMAIL PROTECTED] wrote:
Hello all,
Can somebody point me to references or provide some code on dealing with
this date issue. Basically, I have two vectors of values that represent
dates. I want to convert these values into a
one solution is:
img1-matrix(1:5)
img2-matrix(2:5)
col-1:5 # col-c(green,yellow,...)
image(img1,col=col[sort(unique(img1))])
image(img2,col=col[sort(unique(img2))])
On 12/26/06, Milton Cezar Ribeiro [EMAIL PROTECTED] wrote:
Dear All,
How can I define a color sequence for each image value? I
46 matches
Mail list logo
