Thanks Gabor,
I have got the idear how to aggregate the daily data into monthly data
using the zoo package. However, as I indicated earlier, in the series I am
analysing, 29th February of leap years are excluded (so as to have 365 days
in each year). How do I coerce(force) the the aggregate
Hi
r-help-boun...@r-project.org napsal dne 20.04.2010 21:21:26:
Hi all,
I'm a newbie with R and with a very basic question.
Can I define the minor unit for ylim? For example, I have a y scale
ranging
from 1 to 14, jumping automatically every 2 units, but I want it to jump
1
unit at a
Thanks Peter,
I'm using version 2.8.0 (2008-10-20). This version should be recent enough to
pick up fundamentals such as this, right? I guess the obvious thing to do is
update and try again...
Cheers,
Steve
Date: Tue, 20 Apr 2010 11:55:20 -0600
Dear list
I have a question, I would like to get the following results but I dont know
how to do it:
I have a dataframe like this:
NC8PROD X 1X2
1423 P1342 10
1564
On 04/21/2010 05:21 AM, poliveira wrote:
Hi all,
I'm a newbie with R and with a very basic question.
Can I define the minor unit for ylim? For example, I have a y scale ranging
from 1 to 14, jumping automatically every 2 units, but I want it to jump 1
unit at a time...is it possible? I tried
chrisli1223 wrote:
(1) I have written a script which requires user input. I am using the
readline() command.However, everytime when I run the script, R does not
wait for the user input and proceed to the next line. Is there something
like par(ask=T) to solve this problem?
Depends on how
Tim Clark mudiver1200 at yahoo.com writes:
Yes, I have read about the problems with stepwise algorithms and am looking
for a valid alternative to
narrowing down models when you have a lot of data and a large number of
variables your interested in.
Tim,
This is not an answer to your
Hi,
You should send your data using dput() so that we can copy/paste into
the console.
That said, aggregate() or doBy::summaryBy() might do what you're looking
for.
HTH,
Ivan
Le 21-Apr-10 10:01, n.via...@libero.it a écrit :
Dear list
I have a question, I would like to get the following
DW == David Winsemius dwinsem...@comcast.net
on Mon, 19 Apr 2010 10:49:51 -0400 writes:
DW The OP wrote me privately to say that the errant documantation was at:
http://lib.stat.cmu.edu/S/Harrell/help/Hmisc/html/ecdf.html
DW That is a rather old bit of information. It dates
Hi,
perhaps the example at
http://igraph.sourceforge.net/screenshots2.html#8
helps.
Best,
Gabor
On Tue, Apr 20, 2010 at 6:42 PM, Narges Zarabi narges.zar...@gmail.com wrote:
Hi,
I am trying to fit a line in the log plot of my networks degree distribution
to show that it is a power-law
Where is the problem with your code? You seem to have understood the
principle correct and all what you want to to is possible with your
code. You only seem to have made some typos (or is there any other
reason why you plot num1 twice?), otherwise I do not really understand
why you can not add
If you only have 28 days in Feb in your series then that is what will
be summed if you follow those examples. If you mean that your series
does include Feb 29th but you want to exclude it from the aggregates
then just remove the Feb 29th points first and work with the reduced
series:
The 'border' argument was added in 2.1.10.
On 2010-04-21 1:53, Steve Murray wrote:
Thanks Peter,
I'm using version 2.8.0 (2008-10-20). This version should be recent enough to
pick up fundamentals such as this, right? I guess the obvious thing to do is
update and try again...
Cheers,
Hi,
I'm new to R and S4 classes. I defined a class with two methods (myMethod1 and
myMethod2). I want to call myMethod1 within myMethod2. Why does the code below
not work? The name 'myMethod1' doesn't appear to have meaning inside myMethod2,
even though the two methods belong to the same
On 21/04/2010 7:00 AM, Albert-Jan Roskam wrote:
Hi,
I'm new to R and S4 classes. I defined a class with two methods (myMethod1 and myMethod2). I want to call myMethod1 within myMethod2. Why does the code below not work? The name 'myMethod1' doesn't appear to have meaning inside myMethod2, even
On 2010-04-21 4:35, Peter Ehlers wrote:
The 'border' argument was added in 2.1.10.
Egad! Did I really type that?
I meant 'in R 2.10.0'.
-Peter Ehlers
On 2010-04-21 1:53, Steve Murray wrote:
Thanks Peter,
I'm using version 2.8.0 (2008-10-20). This version should be recent
enough to
Ok thanks Peter! That should solve it.
Thanks again for looking into this.
Steve
Date: Wed, 21 Apr 2010 05:04:33 -0600
From: ehl...@ucalgary.ca
To: smurray...@hotmail.com
CC: r-help@r-project.org; tgstew...@gmail.com
Subject: Re: [R] Unwanted boxes
The Department of Statistics of the Ludwig-Maximilians-University
Munich offers a position as Professor (W3) for Biostatistics and its
Applications in the Life Sciences. R plays a central role both in
teaching and research at our department, hence we would especially
welcome applications from
Store the returned value of partialPlot() in an object and do your own
barplot. Read the Value section in the help page for partialPlot.
Andy
From: Daudi Jjingo
Hello,
I need to draw a partial dependence bar graph.
My the my predictor vectors are continous and so is the
response
Dear Madame, Dear Sir,
I am able to obtain the coefficients from a 'summary' of 'lm', but NOT from a
'summary' of 'aov'.
The following example shows my steps.
## Initialize
rm(list = ls()) # remove (almost) everything in the working environment
utils::data(npk, package=MASS) # get data
model -
Dear R-Help,
my name is Henrik and I am currently trying to solve a Maximum Likelihood
optimization problem in R. Below you can find the output from R, when I use the
BFGS method:
The problem is that the parameters that I get are very unreasonable, I would
expect the absolute value of each
Dear R-Help,
I also send the following post by e-mail to you, however I try to post it
here aswell. My name is Henrik and I am currently trying to solve a Maximum
Likelihood optimization problem in R. Below you can find the output from R,
when I use the BFGS method:
The problem is that the
Hi,
I have v1 (date of test) and v2 (year of birth).
v1 v2
15.5.2008 88
18.6.200954
I want R to use only year in v1; and v2 see as 1988 and calculate age in v3.
any ideas how to do that?
--
**
Vlatka MatkoviÄ PuljiÄ
095/8618 171
Ben
MiscPsycho 1.6 has been placed on CRAN with the bug fix. Should hit mirrors in
a day or so. If for any reason you need it sooner, I can send the tar.gz file.
Thanks for the bug report.
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, April
On Apr 21, 2010, at 8:37 AM, Andrea Bernasconi DG wrote:
Dear Madame, Dear Sir,
I am able to obtain the coefficients from a 'summary' of 'lm', but
NOT from a 'summary' of 'aov'.
The following example shows my steps.
## Initialize
rm(list = ls()) # remove (almost) everything in the working
An updated version of this package has been placed on CRAN. The following are
new/fixed in 1.6:
1) The classical() function now computes design-consistent standard errors for
a one-stage cluster sample with unbalanced data.
2) The jml() function now uses better starting values to reduce
Hi,
I have dataset n1 and v1 (years).
when i ask
median(year)
[1] NA
but if i put summary of dataset n1:
summary(n1)
R produces median (together with min/max/mean)
why it is so?
--
**
Vlatka Matkovic Puljic
095/8618 171
[[alternative HTML version
Hello,
summary() removes NAs by default. You can get the same results using
median(year, na.rm=TRUE)
see ?median
HTH,
Josh
On Wed, Apr 21, 2010 at 6:40 AM, Vlatka Matkovic Puljic
v.matkovic.pul...@gmail.com wrote:
Hi,
I have dataset n1 and v1 (years).
when i ask
median(year)
[1] NA
Hi,
I have v1 (date of test) and v2 (year of birth).
v1 v2
15.5.2008 88
18.6.200954
I want R to use only year in v1; and v2 see as 1988 and calculate age in v3.
any ideas how to do that?
--
**
Vlatka MatkoviÄ PuljiÄ
095/8618 171
Thank you!
2010/4/21 Joshua Wiley jwiley.ps...@gmail.com
Hello,
summary() removes NAs by default. You can get the same results using
median(year, na.rm=TRUE)
see ?median
HTH,
Josh
On Wed, Apr 21, 2010 at 6:40 AM, Vlatka Matkovic Puljic
v.matkovic.pul...@gmail.com wrote:
Hi,
In what format are you dates?
Are they always like this: two digits for day, then one or two digits
for month, and four digits for the year?
Also - are any of your people born after 1999?
Dimitri
2010/4/21 Vlatka Matković Puljić vlatk...@gmail.com:
Hi,
I have v1 (date of test) and v2 (year
Dear list,
I have a question about the selection of Variables in dataframes.
I have a dataframes like this:
CFISCA FIRMS
YEARVAR VALUE
20345nike
2005
Two possible problems:
(a) If you're working with a normal likelihood---and it seems that you
are---the exponent should be squared. As in:
... + (1/(2*stdev*stdev))*sum( ( y-(rev/12)-lag(y)*exp(-lap/12) )^2 )
(b) lag may not be working like you think it should. Consider this silly
example:
And what is the format for year of birth in Excel?
Can't you use concatenate to concatenate a new column (that contains
only 19) and the column that contains a 2-digit year of birth?
And then change all those who are born (in the new column) before 1910
to XX+100 - which will give you 2009 (or
Hm, well.. I can do it in xls but I wanted to see if there is some (simple)
way to do this in R.
2010/4/21 Dimitri Liakhovitski ld7...@gmail.com
And what is the format for year of birth in Excel?
Can't you use concatenate to concatenate a new column (that contains
only 19) and the column that
Give what you've given us, you can try:
x-data.frame(date=c(39804,9527,39917),birth=c(1988,1911,1969)) # your
years have to be in 4-digit format
# formatting date as Date. What you have is, I think, number of days
since 1900 - can anyone correct me?
x$date-as.Date(x$date,origin=1900-01-01)
It's not very clear from what you pasted into your message what your
data frame looks like.
Does it always have 2 lines per record or just one?
Also - in what format is your YEAR?
On Wed, Apr 21, 2010 at 9:54 AM, n.via...@libero.it n.via...@libero.it wrote:
Dear list,
I have a question about
Dear list,
I would like to format the result of the 'cut' function to perform a subsequent
frequency distribution table (fdt) suitable for publications.
Below an reproducible example:
set.seed(1)
x - c(rnorm(1e3, mean=10, sd=1), 50, 100)
start - 0
end - 110
h -10
c1 - cut(x,
Dear list,
I have a question about the selection of Variables in dataframes.
I have a dataframes like this:
CFISCAFIRMSYEAR VARVALUE
20345 nike2005EC01 34
20345 nike2006
n - 63
a - 1:n
x - a-1
y - outer(x,a)
matplot(x,y,type='l')
Warning message:
In matplot(x, y, type = l) :
default 'pch' is smaller than number of columns and hence recycled
Why is it complaining if I specifically ask for type=l, so no pch
involved?
Annoyance or feature?
The fix (if needed)
Hi all!
I would like to generate random numbers between 0 and 1. How can I do this?
I downloaded it single RNG but it generates ones between only 1 and
1...:(
Thank you for the help!
Tamas
[[alternative HTML version deleted]]
__
On Wed, Apr 21, 2010 at 4:37 PM, tamas barjak tamas.bar...@gmail.com wrote:
Hi all!
I would like to generate random numbers between 0 and 1. How can I do this?
I downloaded it single RNG but it generates ones between only 1 and
1...:(
Thank you for the help!
Tamas
Hi tamas,
I am
Or:
sample(1:1,100) # to generate 100 random numbers between 1 and 1
On Wed, Apr 21, 2010 at 10:42 AM, Gustaf Rydevik
gustaf.ryde...@gmail.com wrote:
On Wed, Apr 21, 2010 at 4:37 PM, tamas barjak tamas.bar...@gmail.com wrote:
Hi all!
I would like to generate random numbers between 0
On 20.04.2010 12:02, danda wrote:
Dear Pete,
Thanks, it works now!
I did as you suggested:
--internet flag to the target line (right click, properties) e.g.
C:\Program Files\R\R-2.8.1\bin\Rgui.exe --internet2
Strangely enough, I can now easly download packages but I still get these
Take a look at the 'ti' stuff in the tis package on CRAN. If I
understand you correctly, you want something like this:
weekNumber - function(aDate){
aTi - ti(aDate, tif = wfriday)
may1ymd - 1*year(aTi) + 501
baseWeek - ti(may1ymd, tif = wfriday)
return(aTi - baseWeek + 1)
}
On Apr 21, 2010, at 9:40 AM, Michael Hosack wrote:
Thank you David, this approach is a start in the right direction but
it does
not yield the needed results. I need new week numbers to only begin
on Saturdays. The only exception will be for the first date (May 01)
which will start week 1 on
I have additional Q:
v1 is gender (M=1 and F=2)
v2 is age
I want R to calculate median only for M (1),
but my comand is not good :)
while(v1=1){median(v2,na.rm=TRUE)}
Error: unexpected '=' in while(Q2=
2010/4/21 Vlatka Matkovic Puljic v.matkovic.pul...@gmail.com
Thank you!
2010/4/21 Joshua
Hi,
Try this median(v2[v1==1])
Mohamed
Regards
Vlatka Matkovic Puljic a écrit :
I have additional Q:
v1 is gender (M=1 and F=2)
v2 is age
I want R to calculate median only for M (1),
but my comand is not good :)
while(v1=1){median(v2,na.rm=TRUE)}
Error: unexpected '=' in while(Q2=
Look at ?by
for example
by(data=v2, INDICES=v1, FUN=median, na.rm=TRUE)
This will calculate the median of v2 (age) for each level of the
indices v1 (in your case M and F).
If you are only interested the median for a single level, Mohamed's
solution is simpler.
Josh
On Wed, Apr 21, 2010 at
Perfect! Thanx a lot!
:)
2010/4/21 Joshua Wiley jwiley.ps...@gmail.com
Look at ?by
for example
by(data=v2, INDICES=v1, FUN=median, na.rm=TRUE)
This will calculate the median of v2 (age) for each level of the
indices v1 (in your case M and F).
If you are only interested the median for
One option might be to turn the sequence into a character string, and then
use something like grep(). Kind of a kludge, but possibly easy.
--
View this message in context:
http://n4.nabble.com/Count-matches-of-a-sequence-in-a-vector-tp2019018p2019161.html
Sent from the R help mailing list
Hi,
I am working with the package nlme, and I tried creating a new correlation
class (which, according to the help pages, is possible if you write a few
new method functions). Anyways, I think I am 99% of the way there, but I
have a recurring problem with R crashing on seemingly innocuous
On Tue, Apr 20, 2010 at 7:59 PM, Michael Hosack mhosa...@hotmail.com wrote:
R experts,
How could I extract the week number from a date vector (in Date class)
such that week numbering (week 1...2...) begins (May 01) and ends
(October 31) on the same specific dates each year? Week numbering
If I understand what you're looking for, I think this
may work:
seq=c(2,3,4)
v=c(4,2,5,8,9,2,3,5,6,1,7,2,3,4,5)
lseq = length(seq)
lv = length(v)
matches = sapply(1:(lv-lseq+1),function(i)all(v[i:(i+lseq-1)] == seq))
sum(matches)
[1] 1
- Phil Spector
On Tue, 20 Apr 2010, Noah Silverman wrote:
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the conditional probability within group. Neither the lp or the
risk options seem
Here is one way of doing it:
x - read.table(textConnection(CFISCAFIRMS
YEAR VARVALUE
+ 20345 nike2005EC01 34
+ 20345 nike2006 EC01 45
+ 56779
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not given by
s.npk.aov$coef[block,Pr(F)] ?
On the other, the procedure to extract coefficients from a summary of lm or aov
should be the same.
Andrea
On 21 Apr, 2010, at 3:20 PM, David Winsemius wrote:
On
Dear vegan-helpers,
I calculated an NMDS with metaMDS and then displayed the results with ordiplot.
The NMDS consist of 4 axes. I want to plot two diagrams: 1st vs. 2nd and 3rd
vs. 4th axis. I used the ordiplot-command choices = c(1,2) and c(3,4),
respectively. 1st vs. 2nd does not make any
Hello all,
I would like to compare the similarity of two cluster solutions using a
validation criteria (such as Hubert's gamma coefficient, the Dunn index the
corrected rand index and so on)
I see (from here:http://www.statmethods.net/advstats/cluster.html) that
the function cluster.stats() in
Try this:
set.seed(1)
x - c(rnorm(1e3, mean=10, sd=1), 50, 100)
start - 0
end - 110
h -10
c1 - cut(x, br=seq(start, end, h), right=TRUE)
levels(c1)
[1] (0,10](10,20] (20,30] (30,40] (40,50] (50,60]
(60,70] (70,80]
[9] (80,90] (90,100] (100,110]
# I get:
# [1]
gsubfn is like gsub except instead of a replacement string it uses a
replacement function whose input is the string matched by the regular
expression and whose output replaces the match. The replacement
function can optionally be specified as a formula as we do here. If
there is no left hand
The R version is 2.9.2.
nlme version is 3.1-93
The OS is Windows XP.
This is for my work computer but I get the same behavior at home using
Vista.
-- Forwarded message --
From: Sarah Goslee sarah.gos...@gmail.com
Date: Wed, Apr 21, 2010 at 11:50 AM
Subject: Re: [R] R crashing oddly
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by s.npk.aov$coef[block,Pr(F)] ?
??? That's not a coefficient. It's a p-value.
On the other, the procedure to extract coefficients from a
On 21.04.2010 18:19, Michael Steven Rooney wrote:
The R version is 2.9.2.
Please try with R-2.11.0 RC (release candidate) to be released tomorrow
as well as a recent version of nlme.
If its still fails, please send code (and data, if you do not generate
them in the code) that reproduces
I provided a minimized version of my dataframe at the bottom of this message
containing the results of David's code in variable ('wkoffset') and Jeff
Hallman's code in ('WEEK'). Jeff's code produced the correct results (thank you
Jeff) though I have been unable to understand it. David, as
On 21.04.2010 18:15, Tal Galili wrote:
Hello all,
I would like to compare the similarity of two cluster solutions using a
validation criteria (such as Hubert's gamma coefficient, the Dunn index the
corrected rand index and so on)
I see (from
On Apr 21, 2010, at 12:36 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by s.npk.aov$coef[block,Pr(F)] ?
??? That's not a coefficient. It's a p-value.
On 21.04.2010 16:36, Mario Valle wrote:
n - 63
a - 1:n
x - a-1
y - outer(x,a)
matplot(x,y,type='l')
Warning message:
In matplot(x, y, type = l) :
default 'pch' is smaller than number of columns and hence recycled
Why is it complaining if I specifically ask for type=l, so no pch
involved?
Thanks for the fast reply Uwe.
My hope in posting this was to find if anyone had already done work (in R)
in this direction. So far I wasn't able to find any such relevant code, so
I turned to the mailing list.
Regarding new implementations - thanks for offering! - I have already came
around
You cannot specify a dfunction for z, but need to compute the values in
the matrix yourself as in:
persp(data_for_time, data_for_s,
outer(data_for_time, data_for_s, plot_R_i_3d))
Uwe Ligges
On 20.04.2010 17:35, Jin wrote:
Hello Dear,
I have a function, like z=f(x,y), and try a
On Apr 21, 2010, at 12:50 PM, Michael Hosack wrote:
I provided a minimized version of my dataframe at the bottom of this
message containing the results of David's code in variable
('wkoffset') and Jeff Hallman's code in ('WEEK'). Jeff's code
produced the correct results (thank you
Dear Tal,
I took the definition of the Hubert gamma- and Dunn-index from the Gordon
book. They are actually not about comparing two clusters, at least not in
that reference, and they require dissimilarities.
The adjusted Rand index and Meila's VI, as implemented in
cluster.stats, compare
I've got a problem with the sparseby command (reshape
library), and I have reached the peak of my R knowledge (it isn't really that
high).
I have a small data frame of 23 rows and 15 columns, here
is a subset, the first four columns are factors and the rest are numeric (only
one, line54 is
Hi list!
I have prepared a custom function (below) in order to calculate separability
indices (Divergence, Bhattacharyya, Jeffries-Matusita, Transformed divergene)
between two samples of (spectral land cover) classes.
I need help to cross-compare results to verify that it works as expected
On 19.04.2010 13:51, Viechtbauer Wolfgang (STAT) wrote:
Dear All,
Suppose I want to write a method for the generic function confint():
args(confint)
function (object, parm, level = 0.95, ...)
So, it looks like the second
really???
and third argument have been predefined in the
You forgot to rebuild or reinstall ?
Uwe Ligges
On 18.04.2010 00:25, BXC (Bendix Carstensen) wrote:
I am updating the Epi package.
I added functions named pc.points and pc.matpoints.
Erroneously I wrote pc.plot and pc.matplot in the NAMESPACE file and of course
got an error from Rcmd
optimize(function(x) -scores(x), interval=c(0.5, 0))
$minimum
[1] 0.1830174
$objective
[1] -11.67820
So the maximum is at x=0.183 with a score of 11.7, numerically.
Uwe Ligges
On 17.04.2010 16:30, Akito Y. Kawahara wrote:
Hi, I am new to R, and have a quick question regarding an R script
I have a series of tables, one for each environment indicating a date (row)
and a sample at each hour of the day (0 to 23)
Test1 Table:
Date,Hour1,Hour2,...Hour23
1/1/10,123,123,...,123
I would like to model this as a time series but how can I translate the
table into a list such that I can get:
##I am trying to test for fixed factor main effects in an unbalanced mixed
effects model but when I fit the reduced model for mic factor effects, the
extra degrees of freedom are being allocated to a nested term rather than the
residuals. The model has inc, mic and spp are independent
I've generated a levelplot showing the density distribution of a species
derived from survey transects, with lon, lat co-ordinates.
I'd like to overlay this on a map of the study region specified by:
map('worldHires', xlim = range(mlon), ylim = range(mlat)), where mlon, mlat
specifies the study
Phil's algorithm is a good one, unless you're worried about optimizing for
speed. It makes N * M comparisons, where N is the length of the first
vector and M is the length of the second. Explicitly iterating through the
longer vector, you could reduce the number of comparisons to M. As is
Dear List members,
I am using R to generate MCMC time series plots.
This is the code I use; however, everytime an error message will come out
saying could not find function plot.mcmc.
I have coda package and Lattice package installed.
Does anyone know what may get wrong here? Thanks so much for
On 21.04.2010 20:31, Xiao D wrote:
Dear List members,
I am using R to generate MCMC time series plots.
This is the code I use; however, everytime an error message will come out
saying could not find function plot.mcmc.
I have coda package and Lattice package installed.
Does anyone know what
I'm guessing that you are using the words table and
list to mean data frame. If that's the case, something
like this might get you started:
dfnew = reshape(Test1,varying=list(paste('Hour',1:23,sep='')),
timevar='Hour',idvar='Date',direction='long')
dfnew =
Slight addition below;
On 2010-04-21 10:51, David Winsemius wrote:
On Apr 21, 2010, at 12:36 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by
David, there is still a problem. The data in variable 'wkoffset' should
be equivalent to the data in 'WEEK'. May 07 should begin wkoffset=2 and
the preceeding days should all be assigned wkoffset=1. I did make sure to
adjust the year within day.of.week(). Could you please explain what you
Gustaf,
That is correct. Schedule3 does contain all of the Saturdays between April 30
and Nov. 01 for a given year.
Mike
R experts,
How could I extract the week number from a date vector (in Date class)
such that week numbering (week 1...2...) begins (May 01) and ends
(October 31)
On Apr 21, 2010, at 2:27 PM, Simon Goodman wrote:
I've generated a levelplot showing the density distribution of a
species
derived from survey transects, with lon, lat co-ordinates.
I'd like to overlay this on a map of the study region specified by:
map('worldHires', xlim = range(mlon),
Hi
I have mone quick question I am not quite familiar with, for generic
function plot, why some methods are marked by '*', I think plot(as.mcmc())
may dispatch the right method, I try to use get to view the function, but it
seems that get() only works for the one with no *, e.g. get('plot.ecdf')
On Apr 21, 2010, at 1:15 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:50 PM, Michael Hosack wrote:
I provided a minimized version of my dataframe at the bottom of
this message containing the results of David's code in variable
('wkoffset') and Jeff Hallman's code in ('WEEK').
On Apr 21, 2010, at 3:21 PM, Tengfei Yin wrote:
Hi
I have mone quick question I am not quite familiar with, for generic
function plot, why some methods are marked by '*', I think
plot(as.mcmc())
may dispatch the right method, I try to use get to view the
function, but it
seems that get()
Hi David,
Thank you so much, that's just what I want!!
Best
Tengfei
On Wed, Apr 21, 2010 at 2:24 PM, David Winsemius dwinsem...@comcast.netwrote:
On Apr 21, 2010, at 3:21 PM, Tengfei Yin wrote:
Hi
I have mone quick question I am not quite familiar with, for generic
function plot, why
Wonderful Christian, thank you for the (*very*) helpful reply!
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
Try using contour() instead of levelplot. See the examples
in help('contour') for how to add contour lines to an
existing plot.
-Peter Ehlers
On 2010-04-21 13:08, David Winsemius wrote:
On Apr 21, 2010, at 2:27 PM, Simon Goodman wrote:
I've generated a levelplot showing the density
Assuming the date as id is the first column followed by 23 values, try
the read.reps function found here:
http://www.mail-archive.com/r-help@r-project.org/msg92123.html
like this:
DF - read.csv(myfile, as.is = TRUE)
read.reps(DF, 23)
On Wed, Apr 21, 2010 at 2:24 PM, Idgarad idga...@gmail.com
Thomas Lumley wrote:
On Tue, 20 Apr 2010, Noah Silverman wrote:
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the conditional probability within group. Neither the lp or
I am trying to test for fixed factor main effects in an unbalanced
mixed effects model but when I fit the reduced model for mic factor
effects, the extra degrees of freedom are being allocated to a nested
term rather than the residuals. The model has inc, mic and spp are
independent
Dear R-Help,
my name is Henrik and I am currently trying to solve a Maximum Likelihood
optimization problem in R. Below you can find the output from R, when I use
the BFGS method:
The problem is that the parameters that I get are very unreasonable, I would
expect the absolute value of each
Hey there,
I need to count the matches of a sequence seq=c(2,3,4) in a long vector
v=c(4,2,5,8,9,2,3,5,6,1,7,2,3,4,5,).
With sum(v %in% seq) I only get the sum of sum(v %in% 2), sum(v %in% 3) and
sum(v %in% 4), but that's not what I need :(
Who can help me?
Thanks a lot!
--
View this
This sort of calculation can't be vectorized; you'll have to iterate through
the sequence, e.g. with a for loop. I don't know if a routine has already
been written.
--
View this message in context:
http://n4.nabble.com/Count-matches-of-a-sequence-in-a-vector-tp2019018p2019108.html
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