Unfortunately, this fine control is only hardly possible:
1. [x,y].ticklabs labels the ticks that are in the plot. If there are
less ticks than ticklabs, the ticklabs are recycled and hence overlap.
2. xlim should do the trick to restrict to a certain range of values. It
does not precisely
hi i have the following dataframe
x y
1 345
6 NA
8 123
32 123
12 NA
6 124
7 NA
and i want to extract the data rows which contains NA data, I tried
subset(dataframe,y==NA)
but fail. if you know the answers, please let me know thanks.
typhoong
I need to use an API which requires me set certain attribute=values pairs in
the header.
The postToHost() method allows only to send data in the body.
Please help me.
--
View this message in context:
http://r.789695.n4.nabble.com/setting-headers-in-POST-tp3235542p3235542.html
Sent from the R
I am using calais api in R for text analysis.
But im facing a some problem when fetching the rdf from the server.
I'm using the getToHost() method for the api call but i get just a null
string.
The same url in browser returns an RDF document.
Hello sir:
I have a question about the axis label of scatterplot3d function.
The data is in the attachment.
If I use the command:
scatterplot3d(x,y,z,type=h)
I want the plot's x-axis lab to be 1,2,3,...,13, y-axis lab to be 1,2,3,...11
But if I use the command:
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping consecutive elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in the
community knows this.
The error resulting in the NA was
Hi
r-help-boun...@r-project.org napsal dne 24.01.2011 23:18:35:
Thanks for the reply Erik, As you mentioned, grouping consecutive
elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in
the
community knows this.
The error resulting in the NA
On 25/01/2011 8:07 p.m., typhoong wrote:
hi i have the following dataframe
x y
1 345
6 NA
8 123
32 123
12 NA
6 124
7 NA
and i want to extract the data rows which contains NA data, I tried
subset(dataframe,y==NA)
but fail. if you know the
Hi
r-help-boun...@r-project.org napsal dne 25.01.2011 08:07:10:
hi i have the following dataframe
x y
1 345
6 NA
8 123
32 123
12 NA
6 124
7 NA
and i want to extract the data rows which contains NA data, I tried
subset(dataframe,y==NA)
See
?is.na
however
Hi!
Try
subset(dataframe, is.na(y))
or
df[is.na(df$y),]
HTH,
Ivan
Le 1/25/2011 08:07, typhoong a écrit :
hi i have the following dataframe
x y
1 345
6 NA
8 123
32 123
12 NA
6 124
7 NA
and i want to extract the data rows which contains NA data,
Hello, I have some simulations of financial data, I have 17 variables
simulated 1000 times to three horizons. I am tring to plot the efficient
frontier which I already obtained using th fPortfolio package. I am using
the following commands:
Data=timeSeries(X[1,,])
lppSpec - portfolioSpec()
On Tue, 25 Jan 2011, Petr Savicky wrote:
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping consecutive elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in the
community knows this.
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
df
x y
1 1 102
2 14 500
3 3 40
4 21 101
5 11 189
# Actually I am having dataframe having multiple columns. I am just giving an
example.
I need to subtract all the rows of df
Hi Petr:
I had to do a little bit of finagling, but this seems to work. I basically
did the following:
(i) coordinated the dodging of the points and boxplots by typ within konc.f;
(ii) summarized the group medians in a separate data frame and added an
additional
column to compensate for the
Hi,
May be try this :
data[which(is.na(data[,2])),]
2011/1/25 typhoong graham...@eurus-energy.com
hi i have the following dataframe
x y
1 345
6 NA
8 123
32 123
12 NA
6 124
7 NA
and i want to extract the data rows which contains NA
On Tue, Jan 25, 2011 at 09:05:03AM +0100, Petr Savicky wrote:
[...]
to foreach loop in Perl. If v is a vector, then
for (n in v)
first creates the vector v and then always performs length(v) iterations.
I forgot that ‘break’ may stop the loop. See ?for for further
information. In
Dear Erik,
Thanks for the mapply idea. I never got around to understand all those apply
functions.
I am still curious as to why the other loop didnt work. I even tried the
debug but doesnt help.
Anyway I will leave that for now.
Thanks a lot for your help.
Regards,
Roy
On Mon, Jan 24, 2011 at
Hi:
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
apply(df, 2, function(x) x - x[1])
x y
[1,] 0 0
[2,] 13 398
[3,] 2 -62
[4,] 20 -1
[5,] 10 87
HTH,
Dennis
On Tue, Jan 25, 2011 at 1:20 AM, Vincy Pyne vincy_p...@yahoo.ca wrote:
Dear R helpers
I have a
Hi,
Try this:
df_new - as.data.frame(lapply(df, FUN=function(x) x-x[1]))
I hope it works!
Ivan
Le 1/25/2011 10:20, Vincy Pyne a écrit :
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
df
x y
1 1 102
2 14 500
3 3 40
4
I always forget about sapply():
df_new - sapply(df, FUN=function(x) x-x[1])
Ivan
Le 1/25/2011 10:33, Ivan Calandra a écrit :
Hi,
Try this:
df_new - as.data.frame(lapply(df, FUN=function(x) x-x[1]))
I hope it works!
Ivan
Le 1/25/2011 10:20, Vincy Pyne a écrit :
Dear R helpers
I have a
If you are foolish enough not to be
following R Bloggers via RSS or twitter,
you might miss:
http://www.portfolioprobe.com/2011/01/24/review-of-r-graphs-cookbook-by-hrishi-mittal/
Executive summary:
Extremely useful for new users, informative
to even quite seasoned users.
--
Patrick Burns
Mr Ripley,
May I ask why seq_len() and seq_along() are better than seq()?
Thanks,
Ivan
Le 1/25/2011 09:58, Prof Brian Ripley a écrit :
On Tue, 25 Jan 2011, Petr Savicky wrote:
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote:
Thanks for the reply Erik, As you mentioned, grouping
Thank you
I did not realise I can simply add data from different source data frame
to constructed graph. It even works with stat_smooth() quite
straightforward. One is always learning new tricks.
Best regards
Petr
r-help-boun...@r-project.org napsal dne 25.01.2011 10:24:42:
Hi Petr:
I
On 24.01.2011 23:53, Peter Langfelder wrote:
On Mon, Jan 24, 2011 at 2:47 PM, Yanika Borgakina...@gmail.com wrote:
I am using the function permutations from the package *gregmisc*. However, I
am also making use of the package *e1071*, which also contains a function
called permutations. I
Dear useR,
although I admit that getting the log likelihood is important, you must
concede that obtaining the parameter estimates is not bad either.
Regarding craze, well there are crazier things in the world than this,
just look at the political situation in Italy.
Anyway, the loglik has always
Hello,
I need to integrate the absolute difference between two lines measured
on different points.
# For example :
x - seq(0, 1, 1/100)
f_x - runif(101) + x
y - seq(0, 1, 1/23)
f_y - runif(24) + (1 - y)
plot(x, f_x, type=l)
lines(y, f_y)
Then I would like to compute Integral( | f_x - f_y |
Dear All,
I would like to ask you help with the following:
Assume that I have an area of 36 cells (or sub-areas)
sr-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)
sr
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123456
[2,]789 10 11 12
[3,] 13 14
On 2011-01-25 01:20, Vincy Pyne wrote:
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
df
x y
1 1 102
2 14 500
3 3 40
4 21 101
5 11 189
# Actually I am having dataframe having multiple columns. I am just giving an
Try this:
sweep(as.matrix(df), 2, as.matrix(df[1,]))
On Tue, Jan 25, 2011 at 7:20 AM, Vincy Pyne vincy_p...@yahoo.ca wrote:
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
df
x y
1 1 102
2 14 500
3 3 40
4 21 101
What version of caret and R? We'll also need a reproducible example.
On Mon, Jan 24, 2011 at 12:44 PM, Neeti nikkiha...@gmail.com wrote:
Hi,
I am trying to construct a svmpoly model using the caret package (please
see code below). Using the same data, without changing any setting, I am
just
g - function(x) abs(f1(x)-f2(x))
now you have one function and you can integrate it.
Rich
Sent from my iPhone
On Jan 25, 2011, at 7:32, Xavier Robin xavier.ro...@unige.ch wrote:
Hello,
I need to integrate the absolute difference between two lines measured
on different points.
# For
Dear Carlos,
please refrain from posting the same question umpteen times. Please
consider that code is hard to read and people might not have the time to
run your simulation etc. etc..
As I told you privately in response to your message on 18/1,
Re: generating correlated effects, I tried this
The xlsReadWrite[Pro] package allows to natively read and write Excel
files (.xls) on the Win 32-bit platform.
Changes:
o fix bug with integer conversion (http://dev.swissr.org/issues/113)
PROBLEM: values outside the integer range (i.e. 12345678901) didn't give an NA
(and a warning
Am 24.01.2011 14:08, schrieb Bart Joosen:
I think this is a problem with quotes.
If you look good, you see:
seiz.df- sqlFetch(chnl, 'source.MAIN')
... 'source.main': table not found on channel
You asked MAIN, but your db can't find main.
If you use seiz.df- sqlFetch(chnl, '\source\.\MAIN\')
Xavier Robin Xavier.Robin at unige.ch writes:
Hello,
I need to integrate the absolute difference between two lines measured
on different points.
# For example :
x - seq(0, 1, 1/100)
f_x - runif(101) + x
y - seq(0, 1, 1/23)
f_y - runif(24) + (1 - y)
plot(x, f_x, type=l)
lines(y,
Le 25.01.2011 15:23, Rmh a écrit :
g - function(x) abs(f1(x)-f2(x))
now you have one function and you can integrate it.
Thank you Rich.
Unfortunately I have no f1 and f2 functions, only a set of observed
points on two lines - and no idea about the underlying distribution to
create a
fayazvf wrote:
I am using calais api in R for text analysis.
But im facing a some problem when fetching the rdf from the server.
I'm using the getToHost() method for the api call but i get just a null
string.
You haven't told us nearly enough for us to be able to reproduce
what you are
Greetings Friends,
I would be grateful if you can help me undestand how to make my R code more
efficiently.
I have read in R intoductory tutorial that a for loop is not used so ofter (and
is not maybe not that efficient) compared to other languages.
So I am trying to build understanding how to
Version:
R = 2.11.1
CARET = 4.68
--
View this message in context:
http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3236251.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
From: ggrothendi...@gmail.com
Date: Mon, 24 Jan 2011 22:43:55 -0500
To: megh700...@gmail.com
CC: r-help@r-project.org
Subject: Re: [R] Downloading data from internet
On Mon, Jan 24, 2011 at 8:48 PM, Megh Dal wrote:
Dear all, I need to download an excel file from net, on which I
ooh.. I have another question.
What if I want to add the value in the vector a to the hello each time it
prints.
Here is your output
a - c(2,3,5,5,5,6,6,7)
mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE)
[[1]]
[1] hello
[[2]]
[1] hello
[[3]]
[1] hello hello hello
[[4]]
[1] hello hello
Dear all,
Does anybody know why the following code returns an error message?
library(MatchIt)
library(Zelig)
data(lalonde)
m.out1-matchit(treat~age+educ+black+hispan+nodegree+married
+re74+re75, method=full, data=lalonde)
z.out1-zelig(re78~age+educ+black+hispan+nodegree+married+re74+re75,
I am trying to do binning on three variables (3d binning). The bin boundaries
are specified by the user separately for each variable. I used the bin2
function in the 'ash' package for 2d binning that involves only two
variables but didn't any package for similar binning with three variables.
Are
--- begin included message ---
Im an honours student at Monash University. I'm trying to analyse some
data for my project, which involved 2 treatments. My subjects were
exposed to both treatments, and i gave them 60 minutes to perform a
certain behaviour. 3 of my subjects performed the behaviour
Try this:
expand.grid(seq(startpos, endpos, by = diff(c(startpos, endpos)) /
nrow(sr)),
seq(startpos, endpos, by = diff(c(startpos, endpos)) / nrow(sr)))
On Tue, Jan 25, 2011 at 1:29 PM, Alaios ala...@yahoo.com wrote:
Greetings Friends,
I would be grateful if you can help me
Apologies for this simple question -
Given the number of comparisons I need to do it has become somewhat
laborious to compute the SSE manually. I first have to extract the
coefficients, build the model and run the model on the data. So far I
haven't found any method in R that will do this for me.
I have a problem integrating the 'standard map' (
http://en.wikipedia.org/wiki/Standard_map
http://en.wikipedia.org/wiki/Standard_map ) with deSolve:
By using the modulo-operator '%%' with 2*pi in the ODEs (standardmap1), the
resulting values of P and Theta, should not be greater than 2pi.
I would like to produce a bar plot with varying-width bars. Here is an example
to illustrate:
ww - c(417,153,0.0216,0.0065,556,256,0.0162,0.0117,
+ 726,379,0.0358,0.0501,786,502,0.0496,0.0837,
+ 892,591,0.0785,0.0795)
yy-t(t(array(ww,c(2,10
Hi
r-help-boun...@r-project.org napsal dne 25.01.2011 10:58:36:
ooh.. I have another question.
What if I want to add the value in the vector a to the hello each time
it
prints.
Here is your output
a - c(2,3,5,5,5,6,6,7)
mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE)
[[1]]
[1]
Dear List,
I am using function distCosine from package geosphere to a list of lat/lon
coordinates, and I want to calculate the great circle distance between a
pair of coordinates in the list and all other pairs --- essentially, the
output should be a matrix. I have been able to achieve this with
Well, I'm not Prof. Ripley, but the answer is: Look at the code.
seq_len, seq.int, and seq_along call Primitives, which are implemented
in C, and therefore MUCH faster than seq(), which is implemented as
pure R code (and is also a generic, so requires method dispatch).
Though for small n (up to a
Good Evening,
I would like to work with multivariate polynomials (x and y variables).
I know that there is a package called multipol but I am not sure that supports
my needs.
I use a function (in reality legendre.polynomials) which creates me the
polynomials I want.
For example the following
Inline below.
-- Bert
On Tue, Jan 25, 2011 at 7:48 AM, Henrique Dallazuanna www...@gmail.com wrote:
Try this:
expand.grid(seq(startpos, endpos, by = diff(c(startpos, endpos)) /
nrow(sr)),
seq(startpos, endpos, by = diff(c(startpos, endpos)) / nrow(sr)))
On Tue, Jan 25, 2011 at
Can anyone illuminate the following for me?
How can I get rid of the blue line in the key in the second plot?
## Create a simple data frame
df=data.frame(x=1:1000, y=2*1:1000+rnorm(1000,sd=1000),
type=sample(letters[1:2],1000, replace=TRUE))
## Very nice! Almost what I want
qplot(x, y,
Greetings all,
I am failing to install the package rggobi on windows 7 with R 2.12.0.
On R 2.11.1, the package was installed fine.
I asked for help on the rggobi google group 4 days ago, and didn't receive
any help, so I was wondering if someone here might have a suggestion.
Here are the
Hello R user,
I have following data frame:
df=data.frame(id=c(1:10),strata=rep(c(1,2),c(5,5)),y=c(
10,12,10,NA,15,70,NA,NA,55,100),x=c(3,4,5,7,4,10,12,8,3,15))
and I would like to replace NA's with:
instead of first NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[1]*
*7
try this:
qplot(x, y, data=df, colour=factor(type), size=I(1)) + geom_smooth()
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
http://www.fws.gov/redbluff/rbdd_jsmp.aspx
- Original Message
From: Gene Leynes
after using options(error=utils::recover) option, following is the output. if
i am correct this means that in ksvm there is some problem, but really could
not understand. could anyone please tell me what is wrong...
any help will be great
thank you so much..
Enter a frame number, or 0 to
With large numbers of points you might want to consider hexagonal binning
instead of scatter plots. I don't know of any tools that both do the binning
and take groups into account, but you could think it through and work something
out.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Hi
I would really appreciate some help with my code for coxme...
My data set
I'm interested in survival of animals after an experiment with 4
treatments, which was performed on males and females. I also have two
random factors:
Response variable: survival (death)
Factor 1:
Now I understand what the difference between a primitive and a
non-primitive!
Thanks for the clarification!
Ivan
Le 1/25/2011 18:03, Bert Gunter a écrit :
Well, I'm not Prof. Ripley, but the answer is: Look at the code.
seq_len, seq.int, and seq_along call Primitives, which are implemented
in
On 1/25/2011 9:44 AM, Felipe Carrillo wrote:
try this:
qplot(x, y, data=df, colour=factor(type), size=I(1)) + geom_smooth()
Felipe very nicely answered the how of your question. I thought I'd
followup with the why.
Using qplot, it assumes that you are giving a set of aesthetic mappings.
Thank you both, very much.
Using the identity function I() is a very nice trick, but it still feels
like a trick.
Using ggplot makes the most sense to me.
ggplot(df, aes(x=x, y=y, colour=factor(type))) +
geom_point(size=1) +
geom_smooth()
Thank you very much for taking the time to
It is not clear what you are doing or why you are doing it. If you tell us
your ultimate goal we may be able to help you find a way that does not require
all the computing that you are doing.
How do you get your coefficients? Are you using lm? Have you looked at the
resid function?
--
On 2011-01-25 08:08, Brian J Mingus wrote:
Apologies for this simple question -
Given the number of comparisons I need to do it has become somewhat
laborious to compute the SSE manually. I first have to extract the
coefficients, build the model and run the model on the data. So far I
haven't
Dear list,
When I use the predict.glm method on a glm fitted object, I get the
following warning message:
In addition: Warning message:
In predict.lm(object, x) :
prediction from a rank-deficient fit may be misleading
As the documentation says this happens if the fit is rank-deficient, some
Hi R-users,
I'm trying to find an elegant way to count the number of rows in a dataframe
with a unique combination of 2 values in the dataframe. My data is
specifically one column with a year, one with a month, and one with a day.
I'm trying to count the number of days in each year/month
If you want count:
xtabs( ~ x + y, X)
or sum:
xtabs(z ~ x + y, X)
On Tue, Jan 25, 2011 at 5:25 PM, Ryan Utz utz.r...@gmail.com wrote:
Hi R-users,
I'm trying to find an elegant way to count the number of rows in a
dataframe
with a unique combination of 2 values in the dataframe. My
Dear colleagues, I have a dataset that looks as below.
I would like to make a new dataset that excludes the cases which are joint
conjunctions of particular state names and years, so Connecticut and 2010,
Maryland and 2010 and Vermont and 2010.
I'm trying the following subset code:
newdata-
I know a lot of people asked similar questions like this. I have tried using
read.xls ()
Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for ReadXls
read.table or read.csv (Wrong table format)
odbcConnectExcel have
Hi!! Im doing my graduated work in Onion Curves Growth with Nonlinear Models,
I'm amateur in R so i have doubt how i put or program next models,
http://r.789695.n4.nabble.com/file/n3236748/96629508.png
Also, i cant derivate for Gauss Model, and Richard Model dont have funtion,
If someone
Hello ;
Do you know what is the default value of starting value in glm ? glm(...,
start=c(),... )
I know that it is NULL by default but it need a value to start iteration .
what is this value?
Thanks;
[[alternative HTML version deleted]]
__
Hi,
I am creating a correlation based graph for the data 27 X 3040.
I am not sure if I am allocating enough space for the data.
I am using the following .R file (which is an example for geneData)!
http://r.789695.n4.nabble.com/file/n3236801/sampleClimate.R
sampleClimate.R/nabble_a
and the data
Hi Ryan,
One option would be
X$a - paste(X$x, X$y, sep=.)
table(X$a)
Best,
Ista
On Tue, Jan 25, 2011 at 2:25 PM, Ryan Utz utz.r...@gmail.com wrote:
Hi R-users,
I'm trying to find an elegant way to count the number of rows in a dataframe
with a unique combination of 2 values in the
On Tue, Jan 25, 2011 at 2:06 PM, cameron raymond...@invesco.com wrote:
I know a lot of people asked similar questions like this. I have tried using
read.xls ()
Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for
Dear List,
I think I'm going crazy here...can anyone explain why do I get the same
predictions in train and test data sets below when the second has a missing
input?
y - rnorm(1000)
x1 - rnorm(1000)
x2 - rnorm(1000)
train - data.frame(y,x1,x2)
test - data.frame(x1)
myfit - glm(y ~ x1 + x2,
Hi Simon,
You almost had it! Just need to move the negation outside the rest of
the logic, and remove the quotes from year. Not actually tested (no
data), but I think
newdata- subset(bpa, !(State==Connecticut year2010))
should do it.
Best,
Ista
On Tue, Jan 25, 2011 at 1:34 PM, Simon Kiss
See the note in the help page for ?predict.glm
Best,
Ista
On Tue, Jan 25, 2011 at 2:59 PM, Axel Urbiz axel.ur...@gmail.com wrote:
Dear List,
I think I'm going crazy here...can anyone explain why do I get the same
predictions in train and test data sets below when the second has a missing
Hi,
I am trying to manipulate a gls regression model output to adjust for use of
two-stage least squares. Basically, I want to estimate a model, then feed in
a new set of residuals, then re-calculate all of the model output (i.e. the
standard errors of the estimators, etc.). I have found some
Good afternoon,
I am working on a plot that requires custom legends to be placed in some
panels of the plot; other panels do not contain legends. The problem
that I run into is positioning of the legend in individual panels. In
particular, the 'x' and 'y' elements of the key-list are ignored by
Your question has some similarities this paper: Alison Smith, Brian
Cullis, and Arthur Gilmour. The analysis of crop variety evaluation
data in Australia. Aust. N. Z. J. Stat., 43:129--145, 2001.
In that paper, the authors fit a mixed model with several random
effects. The variances are then
Hello
I would like to ask you if it is possible In R Cran to change the default way
of addressing a matrix.
for example
matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row numbering) # not having R at
this pc
will create something like the following
1 2
3 4
the way R address this matrix is from
From: Leitch, Matthew C.
Sent: Monday, January 24, 2011 6:53 PM
To: 'i...@network-theory.co.uk'
Subject: question about the pt() calculation
Hello
Thank you for your time. I am a graduate student at the University of Texas
Medical Branch, and I was wondering if you could help me with a R
On Tue, Jan 25, 2011 at 06:00:36AM -0800, vioravis wrote:
I am trying to do binning on three variables (3d binning). The bin boundaries
are specified by the user separately for each variable. I used the bin2
function in the 'ash' package for 2d binning that involves only two
variables but
Suppose I have a function, like list, that takes a variable number of
arguments, and I have those arguments in some vector x. How can execute the
function with the *contents* of x as its arguments? I.e., I don't want
list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to
From: Leitch, Matthew C.
Sent: Monday, January 24, 2011 6:53 PM
To: 'i...@network-theory.co.uk'
Subject: question about the pt() calculation
Hello
Thank you for your time. I am a graduate student at the University of Texas
Medical Branch, and I was wondering if you could help me with a R
Suppose I have a function, like list, that takes a variable number of
arguments, and I have those arguments in some vector x. How can execute the
function with the *contents* of x as its arguments? I.e., I don't want
list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to
?do.call
Please provide a reproducible example if the help file is
not sufficient.
- Phil Spector
Statistical Computing Facility
Department of Statistics
You should try different tuning parameters; the defaults are not likely to work
for many datasets. I don't use the polynomial kernel too much but scale
parameter values that are really of could cause this. Unlike the rbf, I don't
know of any good techniques for estimating this.
Max
On Jan 25,
Matthew -
Others will probably tell you about the folly of performing
1733 t-tests on groups with 4 observations each, but an alternative
to your approach would be to use R to solve your problem. (I'm
using var.equal=TRUE because that's what you're calculating, but
you might consider using
Y'all,
I need to get the look of a standard fixed effect ANOVA table:
anova(aov(meas~op*part,data=fs))
Analysis of Variance Table
Response: meas
Df Sum Sq Mean Sq F value Pr(F)
op 22.62 1.308 1.3193 0.2750
part 19 1185.43 62.391 62.9151 2e-16 ***
op:part 38
On Tue, Jan 25, 2011 at 1:14 PM, Akram Khaleghei Ghosheh balagh
a.khaleg...@gmail.com wrote:
Hello ;
Do you know what is the default value of starting value in glm ? glm(...,
start=c(),... )
I know that it is NULL by default but it need a value to start iteration .
what is this value?
Hello, All,
How can I maintain the decimal places when using write.table()?
Jim
e.g.
df:
EFFECT2 PVALUE
1 0.0230.88080
2 -0.260 0.08641
3 -0.114 0.45200
write.table(df,file='df.txt',quote=F,sep='\t',row.names=F)
df.txt:
EFFECT2PVALUE
0.023 0.8808
-0.26 0.08641
-0.114
Dear list,
I´m wondering if there is something analogous to the TukeyHSD function
that could be used for parametric terms in a GAM. I´m using the mgcv
package to fit models that have some continuous predictors (modeled as
smooth terms) and a single categorical predictor. I would like to do
post
On Tue, 25 Jan 2011 16:16:37 -0800,
Jim Moon moo...@ohsu.edu wrote:
Hello, All, How can I maintain the decimal places when using
write.table()?
Have a look at ?format.data.frame
--
Seb
__
R-help@r-project.org mailing list
Dear all,
I always encounter a crash when running hazard.ratio.plot from rms package with
my predictor as a factor.
It works fine when the predictor is a continous score.
Anyone encounters this too? Is this a bug or something?
Thanks,
Lilian
On 2011-01-25 16:16, Jim Moon wrote:
Hello, All,
How can I maintain the decimal places when using write.table()?
Jim
e.g.
df:
EFFECT2 PVALUE
1 0.0230.88080
2 -0.260 0.08641
3 -0.114 0.45200
write.table(df,file='df.txt',quote=F,sep='\t',row.names=F)
write.table(format(df,
Thank you for the response, Peter.
The approach:
write.table(format(df,
drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F)
surprisingly still results in some loss of trailing 0's.
df:
EFFECT2 PVALUE
1 0.0230.88080
2 -0.260 0.08641
3 -0.114 0.45200
df.txt:
EFFECT2
On 2011-01-25 17:22, Jim Moon wrote:
Thank you for the response, Peter.
The approach:
write.table(format(df,
drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F)
surprisingly still results in some loss of trailing 0's.
What version of R?
I'm using R version 2.12.1 Patched
Hi All,
I have a for loop inside the function and I cannot get UUU to give me an
updated grid.dens object when I run the function (it does update when I
run just the for loop). Here's a simplified version of my function:
UUU=function(pop, grid.dens) {
for (i in 1:10){
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