Hi Anna,
How can I change the barplot so that the left hand axis scales from 0 to
15 and the right hand
axis from 0 to 5?
Try this:
par(mfrow=c(1,1), mai=c(1.0,1.0,1.0,1.0))
Plot1-barplot(rbind(Y1,Y2), beside=T, axes=T, names.arg=c(a,b),
ylim=c(0,15), xlim=c(1,9), space=c(0,1),
On 17/11/2010, at 2:37 PM, Brant Inman wrote:
R-helpers,
I have had difficulty installing the pcvsuite package on R version 2.12.0
(2010-10-15). The pcvsuite package is not available on CRAN, but is located
for download at the following website at the University of Washington:
try to quote the special LaTeX symbols, eg
## add more here if needed
sani - function(x) gsub(_, _, x)
so your example:
\documentclass[a4paper]{article}
echo=false, results=hide=
#mytitlevar - Stuff # case 1, everything is find
mytitlevar - Stuff_first # case 2, f is turned into sub-text
Hi
I got an table that contains this colum and i wonder if there is any nice
way to extract all rows that contains all dose that start with A02, so all
the rows that have A020 and A021 etc etc...
| A010 |
| A010 |
| A010 |
| A020 |
| A020 |
| A020 |
| A020 |
| A020 |
| A021 |
| A021 |
Thx
dear Nick,
I do not know your data, however it seems to me that the pattern of relative
abundance
of salamanders should not exhibit a sudden change, but rather a gradual change.
If this is the case, have a look to the segmented package and references
therin. In
particular have a look to the
On 17-Nov-10 00:02:39, José Fernando Zea Castro wrote:
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R.
I found the next bug which I consider important because in
my job everytime We work with datanames like next. Please
Hi everyone,
I wonder, if anyone can help me with this annoying issue, although it is
related to Tinn-R and Windows Vista...
I am running R 2.11.1 and Tinn-R 2.3.5.2 together. I saved 2 R-hotkeys:
ALT+A for sending the whole content of one file and ALT+S for a selection
from one file from Tinn-R
Hi,
Someone recently told me that R is not working well with other programs in
a clustered environment. I believe we are using gensoft and working in a
redhat environment. What was also mentioned was that there are problems when
running multiple R instance at the same time.
Personally I haven't
Hi,
I try to make a xyplot withou colors. I try:
pdf(file=test.pdf)
trellis.device(color=F)
Depth - equal.count(quakes$depth, number=8, overlap=.1)
xyplot(lat ~ long | Depth, data = quakes)
dev.off()
the graphic is showed without colors in a X device but it dont work
saving directly on pdf
Hi,
A reproducible example would have been nice...
Here is one:
vect - c(rep(| A010 |,3), rep(| A020 |,5), rep(| A021 |,2))
df - data.frame(a=vect, b=rnorm(10), d=rnorm(10))
df[grep(pattern=A02, df$a), ]
HTH,
Ivan
Le 11/17/2010 10:16, Joel a écrit :
Hi
I got an table that contains this
Thx alot works perfect!
//Joel
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On 11/17/2010 03:06 PM, anna.richa...@csiro.au wrote:
Hi,
I hope this is a simple question. I am having trouble changing the scale of a
secondary y-axis on a barplot. When I run the code below the limits set for the
first axis are always applied to the second axis as well. I am using the
Hello everyone.
Consider the following number below:
0.9333 0.9333 0.06667
Is it possible to print them with less digits like 0.93 0.93 0.06 (but without
losing the R's accuracy).
I would like to thank you in advance
Best Regards
Alex
Try this:
print(c(0.9333, 0.9333, 0.06667), 1)
On Wed, Nov 17, 2010 at 8:55 AM, Alaios ala...@yahoo.com wrote:
Hello everyone.
Consider the following number below:
0.9333 0.9333 0.06667
Is it possible to print them with
Hi,
I am using the clusters function in the evd package and was wondering if
anyone knew how to set the row names as date/time objects. My data has 1
column of date and time and another of wave height (H).
My two options are:
1 - When in import the data using read.table make sure that the
Thanks a lot.
Worked :)
--- On Wed, 11/17/10, Henrique Dallazuanna www...@gmail.com wrote:
From: Henrique Dallazuanna www...@gmail.com
Subject: Re: [R] reduce print accuracy
To: Alaios ala...@yahoo.com
Cc: Rhelp r-help@r-project.org
Date: Wednesday, November 17, 2010, 10:58 AM
Try this:
Hi!
I was wondering if there are any other functions for numerical integration,
besides 'integrate' from the stats package, but which wouldn't require the
integrand to be vectorized. Oh, and must be capable of integrating over
(-inf,+inf).
Thanks in advance,
Eduardo Horta
[[alternative
Hello everoyne,
If you have ever used matlab you should know the variable editor. You click
over the value of a variable in the workspace and it opens like a excel sheet.
Do you know if there is something like that in R . This will make easier for me
to understand what values are stored in a
Take a look on ?edit
edit(iris)
On Wed, Nov 17, 2010 at 10:11 AM, Alaios ala...@yahoo.com wrote:
Hello everoyne,
If you have ever used matlab you should know the variable editor. You click
over the value of a variable in the workspace and it opens like a excel
sheet. Do you know if there
Dear all at the R-project help list.
I have run into a problem when it comes to getting values for total inertia
and R-squared for my DCA using decorana in the package vegan. I have
tried the goodness function, but the reply indicates that it does not work
with decorana class objects. In
On Wed, Nov 17, 2010 at 1:11 PM, Alaios ala...@yahoo.com wrote:
Hello everoyne,
If you have ever used matlab you should know the variable editor. You click
over the value of a variable in the workspace and it opens like a excel
sheet. Do you know if there is something like that in R . This
Hi all,
I am trying to do this type of graph (code taken from the example of xYplot
help).
dfr - expand.grid(month=1:12, continent=c('Europe','USA'),
sex=c('female','male'))
set.seed(1)
dfr - upData(dfr,
y=month/10 + 1*(sex=='female') + 2*(continent=='Europe') +
Dear all,
I am having a hard time to figure out a suitable test for the match
between two nominal classifications of the same set of data.
I have used hierarchical clustering with multiple methods (ward,
k-means,...) to classify my dat into a set number of classesa, and I
would like to compare
Hello is there in R any operator that give you all the data of a matrix
for example in matlab
x(2,3) returns the 2ndth row and 3rdth column
x(2,:) returns all the columns of the 2nd row.
In R now I would like to print all the
CRagent[[i]][2]
CRagent[[:]][2] doesnot work of course. Other
On Wed, 17 Nov 2010 04:11:17 -0800, Alaios wrote:
Hello everoyne,
If you have ever used matlab you should know the variable editor.
You click over the value of a variable in the workspace and it opens
like a excel sheet. Do you know if there is something like that in R .
This will make
Hi Alex,
Is that what you're looking for:
df - data.frame(a=LETTERS[1:5], b=rnorm(5))
df
a b
1 A -0.2401323
2 B -0.9414998
3 C 0.4289836
4 D 1.9802749
5 E -0.6993612
df[3,2]
[1] 0.4289836
df[3,]
a b
3 C 0.4289836
df[,2]
[1] -0.2401323 -0.9414998 0.4289836 1.9802749
There are really no set ways to determine a changepoint, since a
changepoint depends completely on what you decide. Recursive partitioning
will fit a best changepoint, but it will pretty much always fit one. This
function can be found in the package rpart:
fit - rpart(count ~ year, control =
Thanks a lot :)
I am using Rkward but I didnot about that feature
Regards
Alex
--- On Wed, 11/17/10, Lee Hachadoorian lee.hachadooria...@gmail.com wrote:
From: Lee Hachadoorian lee.hachadooria...@gmail.com
Subject: Re: [R] Variable Editor
To: r-h...@stat.math.ethz.ch
Date: Wednesday, November
I've noticed it, but I haven't looked into it much since I rarely work
on Vista. I have found that opening R before I open Tinn-R tends to
work better than using Tinn-R to open the preferred GUI.
Benjamin
-Original Message-
From: r-help-boun...@r-project.org
Thanks a lot for your helpful answer.
In my case now I have implemented some struct with the following structure
Mystruct.Map
Mystruct.xy
If I do Mystruct[2].$xy I get correctly the xy values of the second item.
I want now to print all the $xy fields of the Mystruct[[]]
I tried
I have a file, each column of which is a separate year, and each row of each
column is mean precipitation for that month. Looks like this (except it goes
back to 1964).
monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008
X2009
11.600 1.010 4.320 2.110
A list contains several matrices. Over all matrices (list elements) I'd like to
access one matrix cell:
m - matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3]))
l - list(m1=m, m2=m*2, m3=m*3)
l[[3]] # works
l[[3]][1:2, ] # works
l[[1:3]][1, 1] # does not work
How can I slice all C-c
Well, without knowing what is the structure of Mystruct.Map and
Mystruct.xy, it's really difficult to know how to help you.
And what is Mystruct by the way!?
Could you send us the output from dput(your_objects) and/or
str(your_objects)?
I think you're really confused about the structure of
To: carus...@gmail.com
From: jda...@usgs.gov
Date: Wed, 17 Nov 2010 08:45:01 -0500
CC: r-help@r-project.org; r-help-boun...@r-project.org
Subject: Re: [R] Population abundance, change point
There are really no set ways to determine a changepoint, since a
changepoint depends completely
try this:
m - matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3]))
l - list(m1=m, m2=m*2, m3=m*3)
lapply(l, [, 1, 1)
# or
sapply(l, [, 1, 1)
I hope it helps.
Best,
Dimitris
On 11/17/2010 3:01 PM, soeren.vo...@eawag.ch wrote:
A list contains several matrices. Over all matrices
Indeed I have looked into various non-standard changepoint analysis
methods. I figured the OP was more interested in traditional methods since
you have to spend less time justifying your methodology. Wavelets are one
potential nontraditional method, as is Significant Zero Crossings (R
package
one approach is the following:
# say 'Data' is your data frame
DataNew - reshape(Data, direction = long, varying = list(2:length(Data)))
DataNew$year - rep(2000:2009, each = 12)
DataNew
I hope it helps.
Best,
Dimitris
On 11/17/2010 3:03 PM, Graves, Gregory wrote:
I have a file, each column
It seems that I am confusing something:
List of 50
$ :List of 2
..$ CRmap: logi [1:100, 1:100] NA NA NA NA NA NA ...
..$ xy : num [1:2] 21 11
$ :List of 2
..$ CRmap: logi [1:100, 1:100] NA NA NA NA NA NA ...
..$ xy : num [1:2] 80 68
--- On Wed, 11/17/10, Ivan Calandra
Please see below.
On Wed, 2010-11-17 at 04:41 -0500, Ted Harding wrote:
On 17-Nov-10 00:02:39, José Fernando Zea Castro wrote:
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R.
I found the next bug which I consider
Hello everyone.
In matlab (again) there is a fucntion find that returns you the indexes where
the condition in find was met. I want the same functionality in R i.e
find(Mydata2) to return all the indexes where the condition is met. Do you
know something like that?
Also when I try to search in
Maybe
lapply(l,function(x){x[1,1]})
or
unlist(lapply(l,function(x){x[1,1]}))
does what you want?
Benno
Am 17.Nov.2010 um 15:01 schrieb soeren.vo...@eawag.ch:
m - matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3]))
l - list(m1=m, m2=m*2, m3=m*3)
l[[3]]
There are several statistics used to compare nominal classifications, or
_partitions_ of a data set. A partition isn't quite the same in this
context because partitioned data are not restricted to a fixed number of
classes. However, the statistics used to compare partitions should also
work for
Sorry I am a bit late to this discussion, but I can't see if you ever
got an answer. Anyhow, on your first question:
On 31/10/10 19:14, Lorenzo Isella wrote:
Dear All,
I have some questions about probit regressions.
I saw a nice introduction at
http://bit.ly/bU9xL5
and I mainly have two
I am trying to download and open an on-line netcdf file.
I'm using Windows XP and R 2.11.1
Here's my script
library(ncdf)
link -
http://ibis.grdl.noaa.gov/SAT/SeaLevelRise/slr/slr_sla_gbl_free_all_66.nc;
dest - C:/temp/slr_sla_gbl_free_all_66.nc
download.file(url=link,destfile=dest)
Dear Gregory,
Is there an easier, cleaner way to do this? Thanks.
There are of course several ways...
(assuming yearmonth to be a data.frame)
--- 1 ---
year - colnames (yearmonth) [-1]
year - gsub (^[^[:digit:]]*([[:digit:]]*[^[:digit:]]*$), \\1, year)
year - as.numeric (year)
month -
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data - data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 =
I understand now!
But I don't know of any easy way to do it.
You would expect that something like these would work:
Mystruct[[1:2]][[xy]] ## but error: subscript out of bounds, which
makes sense because [[ accept only a single subscript
Mystruct[1:2][[xy]] ## but NULL, which makes sense
Thanks Mat,
I have in the meantime identified the Rand index, but not the others. I
will also have a look at profdpm, that did not pop-up in my searches.
Indeed, the interpretation is going to be critical... Could you please
elaborate on what you mean by the bootstrap process?
Thanks a lot
On Wed, Nov 17, 2010 at 2:54 PM, D Kelly O'Day ko...@processtrends.com wrote:
I am trying to download and open an on-line netcdf file.
I'm using Windows XP and R 2.11.1
Here's my script
library(ncdf)
link -
http://ibis.grdl.noaa.gov/SAT/SeaLevelRise/slr/slr_sla_gbl_free_all_66.nc;
Dickison, Daniel ddickison at carnegielearning.com writes:
The documentation for agrep says it uses the Levenshtein edit distance,
but it seems to get this wrong in certain cases when there is a
combination of deletions and substitutions. For example:
agrep(abcd, abcxyz, max.distance=1)
Hi again!
The Introduction to R or other documentation might help you to get
started.
And you should REALLY read the posting guide and provide a reproducible
example.
In this case, some sample data corresponding to the type of objects you
have, plus the result you expect would be enough (I
Elegant or not ... This approach worked as intended, the objective being to
create individual 'years' each of which was 24 months long (not 12) such that
each 'year' reflected the antecedent rainfall condition. THANKS.
I needed to do this to help us identify years that were similar to one
Hi Alex,
This ought to do what you are after. For searching, I found the most
helpful thing to be thinking about what I was really after---this
helps with generating several key words to look for, which is more
likely to turn up results. I also included some functions/packages to
help with R
Dear people at the R-help-list.
I have run into a problem in the package vegan, while using the function
vegan,
and would greatly appreciate any advice how to solve the problem.
I have put in my data R, using the decorana function in the package vegan
and received nice coordinates for the DCA
Em 16/11/2010 04:55, Tal Galili escreveu:
Hello Cesar,
Thank you for the reply.
Another question I have is if it is possible to detect the library path of
an old R install, from the terminal of the new R install.
Cheers,
Tal,
An at large reply would be no. The old install can be in a
Following up again. I found on the forums for the Google Apps API this
thread that seems to be about a similar issue:
http://www.google.com/support/forum/p/apps-apis/thread?tid=1c22cb44eb5cbba6hl=ensearch_impression_id=ab161b010ecf8803%3A12c5a65ce83search_source=related_question
It's using Java
On Wed, 17 Nov 2010, Barry Rowlingson wrote:
On Wed, Nov 17, 2010 at 2:54 PM, D Kelly O'Day ko...@processtrends.com wrote:
I am trying to download and open an on-line netcdf file.
I'm using Windows XP and R 2.11.1
Here's my script
library(ncdf)
link -
hi all,
after fitting a multiple linear regression
model - lm(y ~ a + b+ c+d)
i wanted to plot diagnostics
plot(model)
but get the error message
Error in object$coefficients : $ operator is invalid for atomic vectors.
which does not make a lot of sense, since there is no $ - i am working with
an
On 17.11.2010 10:11, Bernd Jagla wrote:
Hi,
Someone recently told me that R is not working well with other programs in
a clustered environment.
Why not, was is the problem supposed to be?
I believe we are using gensoft and working in a
redhat environment. What was also mentioned was
On Nov 17, 2010, at 7:33 AM, Martin Tomko wrote:
Dear all,
I am having a hard time to figure out a suitable test for the match between
two nominal classifications of the same set of data.
I have used hierarchical clustering with multiple methods (ward, k-means,...)
to classify my dat into
On 17/11/2010 10:28 AM, Manderscheid Katharina wrote:
hi all,
after fitting a multiple linear regression
model- lm(y ~ a + b+ c+d)
i wanted to plot diagnostics
plot(model)
but get the error message
Error in object$coefficients : $ operator is invalid for atomic vectors.
which does not make a lot
rowsum(value, paste(factor1, factor2, factor3))
That is dangerous in general, and always inefficient. Imagine factor1
is c(a, a b) and factor2 is (b c, c). Use interaction with
drop = T.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
On Nov 17, 2010, at 4:49 AM, Ronaldo Reis Junior wrote:
Hi,
I try to make a xyplot withou colors. I try:
pdf(file=test.pdf)
trellis.device(color=F)
Depth - equal.count(quakes$depth, number=8, overlap=.1)
xyplot(lat ~ long | Depth, data = quakes)
dev.off()
the graphic is showed without
Follows is the exact solution to this:
v - NULL
#note that decreasing is FALSE so preceding year preceeds
for(i in 2:46) {
kk - melt(yearmonth[, c(1, i, i+1)], id.vars=month, variable_name=year)
v[[i-1]] - kk[order(kk$year, decreasing=FALSE),]
}
x - do.call(cbind, v)
Barry Prof Ripley:
Thanks!!
mode=wb solve the problem.
BTW, it's he.
D Kelly O'Day
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-downloading-and-opening-netcdf-file-tp3046897p3047002.html
Sent from the R help mailing list archive at Nabble.com.
Dear R helpers,
I have two time series, and I want to perform the cross-spectral analysis
for these time series.
I would like to know whether there is any R function to generate
cross-spectrum such as co-periodogram and quadrature periodogram.
Can anyone give me an advice for cross spectral
Hello
How can I get multipanel conditioning graphics with rgl as I do with lattice
|
For example I have three variables x, y, z, w. Where x,y,z are continuous
and w is categorical or discrete.
I want to use plot3d(x,y,z) for each value of w in a panel
something like plot3d(z~x*y|w)
cheers
Hello
How can I get multipanel conditioning graphics with rgl as I do with lattice
|
For example I have three variables x, y, z, w. Where x,y,z are continuous
and w is categorical or discrete.
I want to use plot3d(x,y,z) for each value of w in a panel
something like plot3d(z~x*y|w)
cheers
Another useful measure to compare partitions is the adjusted Rand
index which is implemented in the library(e1071) within the
classAgreement function.
If you have your data partitions to be compared in a matricial form
(where each column is a different partition), the syntax is
MyStruct - list(list(CRmap = rnorm(50), xy = 1:50), list(CRmap =
rnorm(50), xy = 51:100))
I wonder if it would be possible to extend indexing to accept lists.
For example:
MyStruct[[c(1, 2)]] extracts level1[1] and level2[2], so if the vector
was a list, perhaps
MyStruct[[list(c(1, 2), 2)]] ##
On Nov 17, 2010, at 6:44 AM, Eduardo de Oliveira Horta wrote:
Hi!
I was wondering if there are any other functions for numerical
integration,
besides 'integrate' from the stats package, but which wouldn't
require the
integrand to be vectorized. Oh, and must be capable of integrating
Thank you Matta for the great suggestion,
I will try the additional tests. I have just been experimenting with the
e1071 package and the adjustedRand. It works perfectly, The only
outstadning question is interpretation - is there any rule of thumbs for
the level of agreement that needs to be
Hi, I have a plot and I would like to overlay a PNG image over it. I'm
using the rasterImage function to do this, but the problem I'm facing
is working out the coordinates of the upper right corner of the final
image in user coordinates.
That is I can place the image so the lower left is located
On Nov 17, 2010, at 9:26 AM, Alaios wrote:
It seems that I am confusing something:
List of 50
$ :List of 2
..$ CRmap: logi [1:100, 1:100] NA NA NA NA NA NA ...
..$ xy : num [1:2] 21 11
$ :List of 2
..$ CRmap: logi [1:100, 1:100] NA NA NA NA NA NA ...
..$ xy : num [1:2] 80 68
When a polytomous variable (nominal scale) is used as an explanatory
variable in GLM (generalized linear models), the result object or
summary of glm function does not show the effect of this variable. In
this case, anova function can be used. My question is on the appropriate
procedure(s) for
Hello when my code executes I receive the message that were some warnings. I
want to catch warning messages at run time so to print some local variables and
try to understand why this warning happens.
I searched on internet and I tried withCallingHandlers(
which seems to work but as I used
On Nov 17, 2010, at 10:37 AM, Graves, Gregory wrote:
Follows is the exact solution to this:
v - NULL
#note that decreasing is FALSE so preceding year preceeds
for(i in 2:46) {
kk - melt(yearmonth[, c(1, i, i+1)], id.vars=month,
variable_name=year)
v[[i-1]] - kk[order(kk$year,
Bill -
An excellent point, especially if you're concerned
about efficiency:
Y = list(sqrt,sin,function(u)u/2)
Ybar0 = function(u)mean(sapply(Y,function(fun)fun(u)))
Ybar1 = function(u) rowMeans(sapply(Y, function(fun) fun(u)))
system.time(one - Vectorize(Ybar0)(seq(0,1,length=1)))
skan juanpide at gmail.com writes:
How can I get multipanel conditioning graphics with rgl as I do with lattice
|
For example I have three variables x, y, z, w. Where x,y,z are continuous
and w is categorical or discrete.
I want to use plot3d(x,y,z) for each value of w in a panel
Actually, after thinking about this, what I want to do is to fit a
curve (Weibull and LogLogistic) to the first 9 data points in the
series. The numbers represent percent of development, and I don't
trust the 1.000 point, so I want to fit these curves to get 2
estimates of the tail, both a thick
Dear All,
I made a function which gives 3 plots in one window(I used
par(mfrow=c(1,3)) in the function).
Using that function 3 times, I want to produce 9 plots in one window.
I tried par(mfrow=c(3,1)) or par(mfrow=c(3,3)) but it didn't work.
For example,
pf - function(p) {
par(mfrow=c(1,3))
Hi:
Try this:
pf - function(p) {
plot(c(p:(p+10)),c(1:11))
plot(c(p:(p+10)),c(2:12))
plot(c(p:(p+10)),c(3:13))
}
par(mfrow = c(3, 3))
for(i in 1:3) pf(i)
par(mfrow = c(1, 1))
HTH,
Dennis
On Wed, Nov 17, 2010 at 8:56 AM, Soyeon Kim yunni0...@gmail.com wrote:
Dear All,
I made a
Dear R-users,
I am trying to make a plot in R where x and y are plotted in a regular way, but
the x axis corresponds to another set of values.
For example I have x,y and T (all 29 values)
x- c( -1.31846232, -1.04744756, -0.87034853, -0.72883370, -0.60618971,
-0.49501845, -0.39128988,
I think this is what you want for the 'axis' command:
axis(1, at = x, labels = T, las=2)
On Wed, Nov 17, 2010 at 12:19 PM, Pelt van, Saskia (KNMI)
saskia.van.p...@knmi.nl wrote:
Dear R-users,
I am trying to make a plot in R where x and y are plotted in a regular way,
but the x axis
Hi,
I need to produce an ordinary scatter plot and it is vital that the aspect
ratio equals 1.
I set the axis as:
plot(x, y, type=n, asp=1, ,ylim=c(-80,70),xlim=c(0,100)).
The problem is that I get some 'additional' blank plot area (basically, the
lower bound of xlim becomes quite
Don't fixate on avoiding loops. Bury them in a function
so you don't have to see them and then you just want something
that does the right thing quickly enough. (I'm assuming
this is not a homework/puzzle type problem where you are not
allowed to use loops). E.g., the following does the job
Thanks.
I wanted to avoid loops (even inside functions) because I thought they were
much slower than vector operations, and I really need an efficient code (not
an exercise!).
However, after a few tentatives I'm coming to the conclusion that loops are
actually MORE efficient (at least in the
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of William Dunlap
Sent: Wednesday, November 17, 2010 9:42 AM
To: Eduardo de Oliveira Horta
Cc: r-help@r-project.org
Hello,
is there anyone, who works with modelling of time series of count
data. I try to model traffic accidents with inar(1)-models, but I have
a lot of problems with the R-Code. I would be very grateful, if
someone helped me.
Nazli
__
Hi All,
I am trying to figure out how to get the position of the knots in a pspline
used in a cox model.
my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being
continuous
How do I find out where the knot of the spline are? I would like to know to
figure out how many
On Windows R 2.12.0, the following seemed to work for me:
par(pty=s) ## before plotting
plot(1:10,xlim=c(0,20),ylim=c(0,30))
I do not find that setting the asp parameter does anything useful;
indeed, the Help documentation seems to be backwards (asp=x/y not
y/x). I probably misunderstood
Hi all. It will be great if some one will help me to solve my home task. So,
the deal : i have .pcap file, i convert it to csv using tcpdump (tcpdump -tt
-n -r x.pcap x.csv)
CSV file looks like that :
12890084,761659 IP 10.10.20.20.47808 10.10.20.255.47808: UDP, length 12
12890084,761659 IP
Dear R-user
I used lme to fit a linear mixed model inlcuding weights=varPower().
Additionally I wanted to use glht to calculate Tukey-Kramer multiple
comparision.
error:
glht(modelF, linfct=mcp(Species=Tukey))
Error in glht.matrix(model = list(modelStruct = list(reStruct =
list(SubPlot
On Wed, Nov 17, 2010 at 9:42 AM, William Dunlap wdun...@tibco.com wrote:
Don't fixate on avoiding loops. Bury them in a function
so you don't have to see them and then you just want something
that does the right thing quickly enough. (I'm assuming
this is not a homework/puzzle type problem
Eduardo:
On Wed, Nov 17, 2010 at 9:53 AM, Eduardo de Oliveira Horta
eduardo.oliveiraho...@gmail.com wrote:
Thanks.
I wanted to avoid loops (even inside functions) because I thought they were
much slower than vector operations, and I really need an efficient code (not
an exercise!).
This is
I am new comer in R.There r few IDE like Tinn R,VIM etc.I mean How to
use them? Do I need to install R and then install them to use or they
can work alone? Also does one install packages on R or IDEs? Can I
call/use the package from IDEs?
regards
Parth
--
Socrates, proclaimed: I came to know one
Thanks to everyone.
Paolo
On 16 November 2010 20:14, William Dunlap wdun...@tibco.com wrote:
Have you tried filter()?
filter(a, rep(1,7)/7)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
Dear R users,
Here is the coxme output I obtain on my survival dataset having 3 strains
and 2 infection status (i: infected, ni: non infected)
coxme(Surv(lay) ~ infection*strain, data=datalay, random= ~1 |block)
Cox mixed-effects model fit by maximum likelihood
Data: datalay
n= 1194
Each IDE is going to have its own setup that will also usually depend on
what type of computer you are running. EMACS, for instance, requires a
pre-existing R install with the 'ess' package (packages are usually
installed inside R). Geany, on the other hand, requires an R install and,
in some
Hi List,
I'm hoping to get opinions for enhancing the efficiency of the following
code designed to take a vector of probabilities (outcomes) and calculate a
union of the probability space. As part of the union calculation, combn()
must be used, which returns a matrix, and the parallelized version
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