Thanks for your answers,
Best,
Gildas
Brian Diggs a écrit :
On 7/22/2010 5:01 AM, Allan Engelhardt wrote:
There are so many ways Here is one:
aggregate(v ~ u, data=X, function(...) length(unique(...)))
# u v
# 1 T1 2
# 2 T2 1
Hope this helps
Here is one other way, using the
Hello, for my data preparation and administration (data, labels, etc.) I use
OpenOffice.org ODS spreadsheet files with several sheets in one file. However,
I find it inconvenient to export every single sheet to a csv file whenever I
apply changes to the labelling or so, and I haven't found a
Dear All,
I fear that this is a really easy question but I do seem to go around in
circles..
I have 2 points on a plot and would like to calculate the slope of the
line drawn through these 2 points.
that cant be so hard?!
Thank you in advance,
Katrin
--
Katrin Fleischer
Vrije Universiteit
On 07/23/2010 12:56 AM, Sam Albers wrote:
Hello,
I am trying to create a plot often seen in hydrodynamic work than includes a
contour plot representing the water speed with arrows pointing in the
direction of flow. Does anyone have any idea how I might add arrows based on
wf$angle (in the
Hi List,
I have start and end times of events
structure(list(start = c(15:00, 15:00, 15:00, 11:00,
14:00, 14:00, 15:00, 12:00, 12:00, 12:00, 12:00,
12:00, 12:00, 12:00, 12:00, 12:00, 12:00, 12:00,
12:00, 12:00), end = c(16:00, 16:00, 16:00, 12:00,
16:00, 15:00, 16:00, 13:00, 13:00, 13:00,
Hello,
Is it possible to import some (large enough) polynomial expression
from java (or mathematica) to R as a function?
I found very interesting package rJava but I did not see this feature there.
We used Mathematica as calculation engine for our program with
statistical calculations and now
HI list,
I am using auto.arima from forecast package, I wonder whether its
possible to save model orders to a seperate file
Thanks
nuncio
--
Nuncio.M
Research Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
[[alternative HTML
On Thu, Jul 22, 2010 at 7:56 AM, Sam Albers tonightstheni...@gmail.com wrote:
Hello,
I am trying to create a plot often seen in hydrodynamic work than includes a
contour plot representing the water speed with arrows pointing in the
direction of flow. Does anyone have any idea how I might add
On 23-Jul-10 09:46:28, Katrin Fleischer wrote:
Dear All,
I fear that this is a really easy question but I do seem to go
around in circles..
I have 2 points on a plot and would like to calculate the slope
of the line drawn through these 2 points.
that cant be so hard?!
Thank you in
Dear R users,
I need to find the coordinates for the point (midpoint) located half
way between two pairs of coordinates (lon1,lat1 and lon2,lat2)
assuming a straight line between them. What would be the best way? I
tried to find an answer in the help archives but without success. I
would greatly
Nordlund, Dan (DSHS/RDA) wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Peter Dalgaard
Sent: Thursday, July 22, 2010 3:13 PM
To: Pat Schmitz
Cc: r-help@r-project.org
Subject: Re: [R] Question about a perceived irregularity
hi all,
i'm have a small sample (N = 6, n(groups) = 3) and hence the number of
possible combinations is small. but if i resample with replacement i have a
large number of combinations and get a reasonable result.
what are the considerations in this context - resampling with or without
Dear RGL experts,
I haven't been able to add greek letters to my rgl plot3d.
I have tried expression with no success.
Here is the interested bit:
library(rgl)
cb - cube3d()
plot3d(cb,xlab=expression(alpha),ylab=,zlab=,box=FALSE,alpha=0.5)
The expression(alpha) appears as alpha, rather than
I have a vector a = c(1,0,0,0,1,0,4,0,0,0)
Now I want to get a vector of positions in a vector a where the value is non
zero..
How to do it??
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-get-vector-of-poitions-in-a-vector-tp227p227.html
Sent from the R help
Hi
a = c(1,0,0,0,1,0,4,0,0,0)
which(a!=0)
[1] 1 5 7
Beware of floating point near zero values.
Petr
r-help-boun...@r-project.org napsal dne 23.07.2010 13:12:44:
I have a vector a = c(1,0,0,0,1,0,4,0,0,0)
Now I want to get a vector of positions in a vector a where the value is
non
Hello,
I noticed that model.tables fails when applied to an aov() fit if called
inside a function. The problem seems to occur when as.formula is used
inside a function on a string containing
formula + Error( x / y )
The reason I tried to use as.formula is to generate dynamic calls to aov().
Hello to the community .
First post :)
I would like to ask you which text editor do you use in Linux and how did
you setup the syntax highlightning?
one more question is it possible to debug any program in R by inserting
breakpoints?
I would like to thank you in advance for your help
Best
check out the coxme and the kinship packages, both have the capability to fit
the Cox proportional hazard model in a multi-level setting, or you could use
glmer in lme4 to fit discrete-time (logistic) models with random intercepts.
CS
-
Corey Sparks, PhD
Assistant Professor
Department of
On 23/07/2010 8:08 AM, alaios wrote:
Hello to the community .
First post :)
I would like to ask you which text editor do you use in Linux and how did
you setup the syntax highlightning?
I don't use Linux, but I think the most popular editors there are vim
and emacs, the latter with ESS.
I would like to thank you for your immediate reply.
Actually I do not like vim and emacs.. I am trying to find an editor with a gui
that will work fine in Linux.
One more question. If I am editing a file using my external text editor is it
possible to execute directly one of the functions that
i use gedit. It has nice text highlighting.
Stephen Sefick
On Fri, Jul 23, 2010 at 7:13 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 23/07/2010 8:08 AM, alaios wrote:
Hello to the community .
First post :)
I would like to ask you which text editor do you use in Linux and how did
On Jul 23, 2010, at 7:13 AM, Mafalda Viana wrote:
Dear R users,
I need to find the coordinates for the point (midpoint) located half
way between two pairs of coordinates (lon1,lat1 and lon2,lat2)
assuming a straight line between them. What would be the best way? I
tried to find an answer in
try this:
char2hr - function(time){
+ mat - do.call(rbind, strsplit(time, :))
+ mode(mat) - 'numeric'
+ mat %*% c(1, 1/60) # convert to hours
+ }
# convert to hours
x.hr - apply(x, 2, char2hr)
# generate a set of sequences to set values
x.seq - apply(x.hr, 1, function(.hr)
It would be nice if you could post what the data looks like that you
want to import. R can import any text file and then you have string
manipulation that you can do to parse it. So the basic answer is
probably yes, but we do need to understand the format of the data to
give a more precise
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 23/07/2010 14:17, Alaios wrote:
I would like to thank you for your immediate reply.
Actually I do not like vim and emacs.. I am trying to find an editor with a
gui
that will work fine in Linux.
Well - you are missing out with not using emacs
I have the following code to write the output from auto.arima function. The
issue is not in finding the model but to divert its out put
fit to a file order_fit.txt. code runs but nothing is written to
order_fit.txt
where am I going wrong
library(forecast)
for (i in 1:2) {
filen =
The error message actually is
Error in inherits(object, formula) : object 'frmlc' not found
and it is the omitted-by-you part after the colon that matters: the
body of model.tables does not have 'frmlc' in its scope.
Here is one approach to using dynamic formulae that works in a lot of
(1:length(a))[a != 0]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of vikrant
Sent: 23 July 2010 12:13
To: r-help@r-project.org
Subject: [R] How to get vector of poitions in a vector ?
I have a vector a = c(1,0,0,0,1,0,4,0,0,0)
On 23/07/2010 7:14 AM, Duncan Murdoch wrote:
Nordlund, Dan (DSHS/RDA) wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Peter Dalgaard
Sent: Thursday, July 22, 2010 3:13 PM
To: Pat Schmitz
Cc: r-help@r-project.org
Hi,
I use heatmap.2 and heat.colors. Is it possible to specify the colors in
that way that
all values below, for instance, 1.5 should be coloured red, values between
1.5 and 1.7 green
and above 1.7 black?
Many thanks
--
View this message in context:
The arithmetic mean was my first approach and to nearby points it
doesn't make much difference. However, when the distance between the 2
points gets bigger this is no longer accurate enough. So yes, I was
thinking on spherical geometry, midpoint considering the great circle
distance or similar.
On Jul 23, 2010, at 8:58 AM, Mafalda Viana wrote:
The arithmetic mean was my first approach and to nearby points it
doesn't make much difference. However, when the distance between the 2
points gets bigger this is no longer accurate enough. So yes, I was
thinking on spherical geometry,
I like loops for this kind of thing so here is one:
df- structure(list(start = c(15:00, 15:00, 15:00, 11:00,
14:00, 14:00, 15:00, 12:00, 12:00, 12:00, 12:00,
12:00, 12:00, 12:00, 12:00, 12:00, 12:00, 12:00,
12:00, 12:00), end = c(16:00, 16:00, 16:00, 12:00,
16:00, 15:00, 16:00, 13:00, 13:00,
Thank you all for your kind reply and help.
I changed the legend function as Mr. Peter Ehlers suggested and
the plots look good.
Thank you again.
Hannah
2010/7/22 Peter Ehlers ehl...@ucalgary.ca
On 2010-07-21 22:06, li li wrote:
Hi all,
I am have some difficulty
Hi all,
I want to add 6 plots in the format of 2 columns and 3 rows as one
figure in latex. The plots are in .eps file.
I know how to add 2 plots side by side, but could not figure out how to do
multiple rows.
I know this may not be the right place to ask such a question. But I do
not know
I have been trying emacs with ess, Kate, plain gedit..nothing really
satisfied me..
but then I found a nice plugin for gedit: Rgedit !
it supports split screen with R terminal window (you can also personalize
colour if, like me, do like to
stick with old habits :) ), moreover gedit has a brackets
On Jul 23, 2010, at 9:43 AM, li li wrote:
Hi all,
I want to add 6 plots in the format of 2 columns and 3 rows as one
figure in latex. The plots are in .eps file.
I know how to add 2 plots side by side, but could not figure out how
to do
multiple rows.
I know this may not be the right
I would like oto thank everyone for the replies.
I instaled rkward.. It looks like an editor.. Is it possible to execute also
your code from the text editor.
If there is a function call myfunc inside my file test.R
Is it possible to exectute the function from the rwkard?
Best Regards
Alex
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 23/07/2010 15:56, Alaios wrote:
I would like oto thank everyone for the replies.
I instaled rkward.. It looks like an editor.. Is it possible to execute
Yes
also your code from the text editor.
Look into the Run menu
If there is a function
Dear David and R users,
Thank you very much for the help. There is a function in the suggested
package called midPoint which does what I need.
df- data.frame(lon1=c(-4.568,-4.3980), lat1=c(59.235,56.369),
lon2=c(-5.123,-4.698), lat2=c(60.258,59.197) )
library(geosphere)
p1 -
Dear List
I have the POSIX timestamps of two events, A and B respectively. I want
an expression that tests if A occurs within 4 weeks of B.
Something like:
if (ts_A is between ts_B + 4(weeks) and ts_B - 4) ...
Can probably done by regex but the regex docu for R says nothing about
if
[cc'd to package maintainer]
This feels like a fairly straightforward question, so I'm surprised
not to have found an answer so far (RSiteSearch, package help,
vignette ...)
It seems fairly hard (without some fairly serious hacking) to
hard-code the axis limits in a hexbin plot from the
help(difftime) is probably what you want.
If not, please post a standalone example
Allan
On 23/07/10 15:18, Jim Hargreaves wrote:
Dear List
I have the POSIX timestamps of two events, A and B respectively. I
want an expression that tests if A occurs within 4 weeks of B.
Something like:
if
Hi,
I am trying to recode the output from a matrix(here is a small snippet of it):
HGlt10RawPerc2008[1:20]
[1] -5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
Hi,
I am trying to recode the output from a matrix(here is a small snippet of it):
HGlt10RawPerc2008[1:20]
[1] -5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
Hi Tom,
This seems to work like I'd expect:
HGlt10RawPerc2008 - read.table(textConnection(
-5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
))
recode(HGlt10RawPerc2008 ,-100:0.0 = 10; 0:1.0 = 8; 1.001:3.0 = 6;
3.001:4.0 = 4; 4.001:5.0 = 2; else = 0)
I wonder if your data might not
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was hoping someone could share some insight.
I don't know the
Hi folks,
I've got a POSIXct datum as follows:
Sys.time()
[1] 2010-07-23 11:29:59 BST
I want to convert this to the nearest half-hour, i.e., to 2010-07-23 11:30:00
BST
(If the time were 11:59:ss, I want to convert to 12:00:00).
How to achieve this?
Thanks,
Murali
On 23/07/2010 11:16 AM, Brigid Mooney wrote:
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was hoping
On Jul 23, 2010, at 11:16 AM, Brigid Mooney wrote:
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was
In addition, there are 'theoretical' reasons for excluding intercept
from the model that must be considered. The reasons related to the
regressor(s) and depend on the phenomenon being modelled. For example,
whereas the intercept can be excluded in a bivariate model on the
expenditure of an
Hi R-List,
I have a question regarding R-language formats, I think. I am producing a
series of graphs (using plot, barplot, barchart, and bwplot, using either
text or mtext to place values on the graphs) and tables for a Francophone
country. In fact, I have already done so. However, while they
On Jul 23, 2010, at 11:20 AM, murali.me...@avivainvestors.com murali.me...@avivainvestors.com
wrote:
Hi folks,
I've got a POSIXct datum as follows:
Sys.time()
[1] 2010-07-23 11:29:59 BST
I want to convert this to the nearest half-hour, i.e., to
2010-07-23 11:30:00 BST
(If the time
On Jul 23, 2010, at 11:32 AM, Jennifer Sabatier wrote:
Hi R-List,
I have a question regarding R-language formats, I think. I am
producing a
series of graphs (using plot, barplot, barchart, and bwplot, using
either
text or mtext to place values on the graphs) and tables for a
On Jul 23, 2010, at 11:35 AM, David Winsemius wrote:
On Jul 23, 2010, at 11:20 AM, murali.me...@avivainvestors.com murali.me...@avivainvestors.com
wrote:
Hi folks,
I've got a POSIXct datum as follows:
Sys.time()
[1] 2010-07-23 11:29:59 BST
I want to convert this to the nearest
Worked like a charm! Thanks a lot!
Jen
On Fri, Jul 23, 2010 at 11:48 AM, Alain Guillet
alain.guil...@uclouvain.bewrote:
Hi,
I am sure there is a better solution but a possibility is that you use
sub(\\.,,,as.character(newdata)) instead of newdata in the text calls.
HTH,
Alain
If you have a zoo series this should work. If it doesn't then please
tell me because I think it works.
snap2min - function(zoo, min=00:15:00){
min15 - times(min)
a - aggregate(zoo, trunc(time(zoo), min15), function(x) mean(x, na.rm=TRUE))
}
hth
Stephen Sefick
On Fri, Jul 23, 2010 at 11:00 AM,
Wow, that's even better!
Thanks to you both. I know both options will come in handy!
Jen
On Fri, Jul 23, 2010 at 11:58 AM, David Winsemius dwinsem...@comcast.netwrote:
On Jul 23, 2010, at 11:32 AM, Jennifer Sabatier wrote:
Hi R-List,
I have a question regarding R-language formats, I
nbsp;Hi there,
Sorry to bother those who are not interested in this problem.
I'm dealing with a large data set, more than 6 GB file, and doing regression
test with those data. I was wondering are there any efficient ways to read
those data? Instead of just using read.table()? BTW, I'm using a
Rkward is great.
If you are a programmer then have a look at the StatET plugin for eclipse
(Java). Not so simple to set up.
On Friday 23 July 2010 02:17:36 pm Alaios wrote:
I would like to thank you for your immediate reply.
Actually I do not like vim and emacs.. I am trying to find an
On Jul 23, 2010, at 12:04 PM, stephen sefick wrote:
If you have a zoo series this should work. If it doesn't then please
tell me because I think it works.
snap2min - function(zoo, min=00:15:00){
min15 - times(min)
a - aggregate(zoo, trunc(time(zoo), min15), function(x) mean(x,
na.rm=TRUE))
By entering trunc.POSIXt at the R commandline, you can see the standard
truncate implementation. Riffing on this,
roundhalfhour - function( x ) {
x - as.POSIXlt( x + as.difftime( 15, units=mins ) )
x$sec - 0
x$min - 30*(x$min %/% 30)
as.POSIXct(x)
}
The as.double approach ought to work
This truncates to quarter-hour rather than rounding to half-hour.
stephen sefick ssef...@gmail.com wrote:
If you have a zoo series this should work. If it doesn't then please
tell me because I think it works.
snap2min - function(zoo, min=00:15:00){
min15 - times(min)
a - aggregate(zoo,
The arithmetic that David describes should work fine (POSIXct is
internally in UTC) unless you are in the Chatham Islands (which has a
UTC+12:45 time zone [1]) or Nepal (UTC+05:45 [2]) or some such place
with a funny idea about when the 1/2 hour mark is.
The formatting of the output may be
David, Stephen,
You're right - it's the time zone conventions that threw me as well. I tried
those round() operations earlier, but inevitably ended up being either an hour
behind. Even when I specified my time zone, it didn't make any difference. So
there's something else that I'm missing. I'll
On 23/07/2010 12:10 PM, babyfoxlo...@sina.com wrote:
nbsp;Hi there,
Sorry to bother those who are not interested in this problem.
I'm dealing with a large data set, more than 6 GB file, and doing regression
test with those data. I was wondering are there any efficient ways to read
those
Hello!
I have a data set similar to the data set monthly in the example below:
monthly-data.frame(month=c(20090301,20090401,20090501,20100301,20100401,20090301,20090401,20090501,20100301,20100401),monthly.value=c(100,200,300,101,201,10,20,30,11,21),market=c(Market
A,Market A, Market A,Market A,
read.table is not very inefficient IF you specify the colClasses=
parameter. scan (with the what= parameter) is probably a little more
efficient. In either case, save the data using save() once you have it
in the right structure and it will be much more efficient to read it
next time. (In
Hi,
Strange, there seems to be different behavior of old style classes and S4
classes.
This worked just like you expected (but it's not the same thing, no S4
classes)
f2=function(x,...) UseMethod(fxy)
fxy.default=function(x,...){
print(default)
}
fxy.X=function(x,...) {
Hi,
does anyone know a tutorial for Tinn-R? During all my search I only found
R-Tutorials...
The problem now is: I would like to make Tinn-R an autosave. But since I had
several questions before concerning Tinn-R (for example, how to have Tinn-R
and R in one window or how do the new versions
On 23/07/10 17:36, Duncan Murdoch wrote:
On 23/07/2010 12:10 PM, babyfoxlo...@sina.com wrote:
[...]
You probably won't get much faster than read.table with all of the
colClasses specified. It will be a lot slower if you leave that at
the default NA setting, because then R needs to figure
On Fri, Jul 23, 2010 at 12:35 PM, murali.me...@avivainvestors.com wrote:
David, Stephen,
You're right - it's the time zone conventions that threw me as well. I tried
those round() operations earlier, but inevitably ended up being either an
hour behind. Even when I specified my time zone, it
?read.zoo
You didn't specify the index column correctly.
On Jul 23, 2010, at 12:36 PM, Dimitri Liakhovitski wrote:
Hello!
I have a data set similar to the data set monthly in the example
below:
monthly-
data
.frame
(month
=
c
Google on tinn-r tutorial. However, I believe that you will find that the
answer is no, there isn't anything useful out there. Moreover, TINN-R's help
files are minimal and terrible, even though the package itself is quite
useful (I find). Please Do NOT post further TINN-R questions to this list
Strange, I did attach. Attaching again. Maybe the file just doesn't go through?
I have:
names(OrigData):
[1] Brand Month Value
I read ?read.zoo
According to that index should be the column number.
I thought it should be split = 1 in my case - because I am splitting by Brand.
But neither split =
Dear list:
I'm using bwplot to compare concentrations by location and treatment as in:
# using built in data
bwplot( conc ~ Type : Treatment, data = CO2 )
I would like the order of the plots to be: 3,4,1,2. I can't seem to figure
this out with index.cond or permc.cond.
Any help is
But, but, but Did you read my message about the need to correctly
specify index columns?
The problem is that read.zoo is reading your first column as an index
and it's actually the second column that should be used for that
purpose.
--
David.
On Jul 23, 2010, at 1:01 PM, Dimitri
Hello,
2010/7/23 jim holtman jholt...@gmail.com:
It would be nice if you could post what the data looks like that you
want to import. R can import any text file and then you have string
manipulation that you can do to parse it. So the basic answer is
probably yes, but we do need to
On Fri, Jul 23, 2010 at 10:02 AM, Eck, Bradley J
brad@mail.utexas.edu wrote:
Dear list:
I'm using bwplot to compare concentrations by location and treatment as in:
# using built in data
bwplot( conc ~ Type : Treatment, data = CO2 )
I would like the order of the plots to be: 3,4,1,2.
Very sorry - I mistunderstood and confused split with index.column -
totally my fault.
Ok, now I've run this line:
z - read.zoo(OrigData, index.column = 2, split = Brand)
And I am getting:
Error in merge.zoo(` Plus` = c(NA, 98L, 95L, 97L, NA, 98L, 97L, 98L, NA, :
series cannot be merged with
On Jul 23, 2010, at 1:39 PM, Dimitri Liakhovitski wrote:
Very sorry - I mistunderstood and confused split with index.column -
totally my fault.
Ok, now I've run this line:
z - read.zoo(OrigData, index.column = 2, split = Brand)
And I am getting:
Error in merge.zoo(` Plus` = c(NA, 98L, 95L,
I am expecting to see the week names as row labels of z and the
corresponding values (like in the monthly example). I am pretty sure
- in order to get it one needs to install the latest version of zoo.
I've done it just a couple of days ago.
I am getting the error - and nothing is produced. Can it
On Fri, Jul 23, 2010 at 1:39 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Very sorry - I mistunderstood and confused split with index.column -
totally my fault.
Ok, now I've run this line:
z - read.zoo(OrigData, index.column = 2, split = Brand)
And I am getting:
Error in
Hi,
I am trying to fit the following model:
sr.reg.s4.nore - survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Where age_sym4 is the age that a subject develops clinical thought
problems; sym4 is whether they develop clinical thoughts problems (0 or
1); and lifedxm is
On 2010-07-23 11:36, Deepayan Sarkar wrote:
On Fri, Jul 23, 2010 at 10:02 AM, Eck, Bradley J
brad@mail.utexas.edu wrote:
Dear list:
I'm using bwplot to compare concentrations by location and treatment as in:
# using built in data
bwplot( conc ~ Type : Treatment, data = CO2 )
I would
On Jul 23, 2010, at 1:50 PM, Dimitri Liakhovitski wrote:
I am expecting to see the week names as row labels of z and the
corresponding values (like in the monthly example). I am pretty sure
- in order to get it one needs to install the latest version of zoo.
I've done it just a couple of days
David/Gabor,
you helped me a lot.
Gabor - I've run table(OrigData$Month) - and it looked weird.
I went back to my file and changed the format of the date (Month) in
Excel. I have no idea what it does - but after I saved it again, it
worked.
Thanks a lot!
I'll open another thread where I'll ask
On Fri, Jul 23, 2010 at 2:07 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
David/Gabor,
you helped me a lot.
Gabor - I've run table(OrigData$Month) - and it looked weird.
I went back to my file and changed the format of the date (Month) in
Excel. I have no idea what it does
http://nixtricks.wordpress.com/2009/11/09/latex-multiple-figures-under-the-same-caption-using-subfigure/
It will create two rows of subfigures with two subfigures on each row
On Fri, Jul 23, 2010 at 6:43 AM, li li hannah@gmail.com wrote:
Hi all,
I want to add 6 plots in the format of 2
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Hi,
I am trying to fit the following model:
sr.reg.s4.nore - survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Next time include a reproducible example. i.e. something we can run.
Now, Google Hauck Donner Effect to
I found the user guide available at Help Main User Guide Html to
be quite useful for getting started. I don't think this was available
when first I looked at Tinn-R a couple years ago, at which point I too
felt the help files were very poor. But under 2.3.5.2 the file is
there, and Tinn-R is
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
anova(sr.reg.s4.nore)
Df Deviance Resid. Df-2*LL P(|Chi|)
NULL NA NA45
to make the desired within R's plotting device rather than in latex try:
par(mfrow = c(3, 2))
and you'll probably want to adjust other mar, parameters as well (e.g.
mar, cex.lab, etc).
Or, take advantage of the flexibility offered by the graphics package
by studying ?layout
hth,
Kingsford
Hi All,
I would like to estimate the sparsity function using Hall Sheater
bandwidth.
Is there any command for this?
Thanks,
NGS
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R-help@r-project.org mailing list
Hannah: I am not sure if this is what you
need but you can use an array to do that.
Copy and paste the below code to your latex code.
\newpage
\begin{landscape}
\begin{figure}[h]
\begin{center}$
\begin{array}{cc}
\includegraphics[width=2in]{yourgraphicname}
The functions {summary,bandwidth}.rq() in the package quantreg looks
promising, but it is not really my area...
http://lmgtfy.com/?q=Hall-Sheater+sparsity+%22r-project%22
Hope this helps a little.
Allan
On 23/07/10 21:02, Nathalie Gimenes wrote:
Hi All,
I would like to estimate the
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
anova(sr.reg.s4.nore)
Df Deviance Resid. Df
Here's one way - you can reorder the levels of a factor variable by specifying
the levels the way you want them.
require(lattice)
Loading required package: lattice
bwplot( conc ~ Type : Treatment, data = CO2 )
str(CO2$Type)
Factor w/ 2 levels Quebec,Mississippi: 1 1 1 1 1 1 1 1 1 1 ...
Here is an example dataset:
ZoneCover.df- data.frame(Value=c(1,2))
row.names(ZoneCover.df) - c(Floodplain1.Tree, Floodplain1.Shrub)
I want to Export the Row.Names to a column in the dataframe:
ZoneCover.df$ID - names(ZoneCover.df)
which yields this:
ZoneCover.df
Value
On Jul 23, 2010, at 6:18 PM, chipmaney wrote:
Here is an example dataset:
ZoneCover.df- data.frame(Value=c(1,2))
row.names(ZoneCover.df) - c(Floodplain1.Tree, Floodplain1.Shrub)
I want to Export the Row.Names to a column in the dataframe:
ZoneCover.df$ID - names(ZoneCover.df)
which
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