On Thu, Jul 7, 2011 at 9:45 PM, EdBo n.bow...@gmail.com wrote:
Hi Josh,
1) I got hold of the library(optimx) but surprising I seem to be
failing to use it.
On the previous code, that was working, I replaced the optim function with:
out[i, ] - optimx(llik, par = start.par, method =
Dr Eberhard Lisse e...@lisse.na wrote:
Hi,
I am reading payment data like so
2010-01-01,100.00
2010-01-04,100.00
...
2011-01-01,200.00
2011-01-07,100.00
and plot it aggregated per month like so
library(zoo)
df - read.csv(daily.csv, colClasses=c(d=Date,s=numeric))
z - zoo(df$s,
On 07/08/2011 04:54 AM, sderose wrote:
Hi! I fairly new to R, have only done pretty basic things so far, so this may
be a very basic question. But I did search the forums and didn't see a
solution
I'm trying to get going with kripp.alpha(). I'm loading data from a file,
like this:
Dear Sir,
I am doing some work on a population of patients. About half of them are
admitted into hospital with albumin levels less than 33. The other half have
albumin levels greater than 33, so I stratify them into 2 groups, x and y
respectively.
I suspect that the average length of
Hi
[R] Using t tests
Dear Sir,
I am doing some work on a population of patients. About half of them are
admitted into hospital with albumin levels less than 33. The other half
have
albumin levels greater than 33, so I stratify them into 2 groups, x and
y
respectively.
I
On 07.07.2011 20:30, Dimitris.Kapetanakis wrote:
Dear all,
I am trying to use apply or similar functions in order to improve the
efficiency of my code but I guess I am misunderstanding the function of
these commands. What I want to do is possible to see with the following code
that I use a
HI All,
I have written and succesfully uploaded a new package. The licence it is under
is 'GPL' --no version. My assumption is, since all the code is written in R the
licence R used for R would affect the code (hence my GPL stands for whatever
version of the GPL R is under)
I am happy with
I've rolled up R-2.13.1.tar.gz a short while ago. This is a maintenance release
to consolidate various minor fixes to 2.13.0.
The Licencing change to GPL-2 | GPL-3 may be of interest to some.
You can get it from
http://cran.r-project.org/src/base/R-2/R-2.13.1.tar.gz
or wait for it to be
Well, I'm not sure it is really an rgdal problem as much as
having a stale so and I didn;t want to go wreck the thing to capture
the exact error message, the point was that apache build seems to have
come with incompatible expat so. In the apache/lib dir, I apparently had a much
different libs,
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is under is 'GPL' --no
version. My assumption is, since all the code is written in R the licence R used for R would affect
the code (hence my GPL stands for whatever
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is
under is 'GPL' --no version. My assumption is, since all the code is written
in R the licence R used for R would
Not sure if I understand your problem completely.
If so, you could try to run an LDA on the x with y as classlabels
factor(rep(1:3, each=10)) and see if it can perfectly separate the classes.
Uwe Ligges
On 06.07.2011 17:31, Aparna Sampath wrote:
Hi All
This might be something very
Dear r users,
does anyone have a suggestion of a book on graph theory, which may help
designing the analysis of neuroimaging resting-states data?
Thanks,
Guilherme
[[alternative HTML version deleted]]
Dear Sir,
When i am trying to set CRAN mirror to install packages, its showing the
following messages, i can't install the packages
chooseCRANmirror()
Warning message:
In open.connection(con, r) :
unable to connect to 'cran.r-project.org' on port 80.
Best regards
Mahmud
Hi Jim,
Thanks for your response. I was actually thinking in terms of an executable
file which could return the output back to an excel sheet. I was thinking of
having a small script/executable which would read an excel file get the
values... Run the modelget the prediction and return
Hi, everyone,
I need an efficient way to do vectors cross product in R.
Set vectors,
V1 = ai + bj + ck
V2 = di + ej + fk
then the cross product is
V1 x V2= (bf - ce) i + (cd - af) j + (ae - bd) k
As shown here ( http://en.wikipedia.org/wiki/Cross_product ).
Thanks.
Best,
Bai
Dear All,
I have several objects (imported from excel via using “xlsx”) with the field
names: Month/Year, X, Y1, Y2, Y3.
What is the best library/way to generate a graph which will be consist of
multiple plots (Month/Year) that each contain the X, Y1, Y2, Y3 dataset.
Thanks a lot.
Cheers,
Július 7-től 14-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 7 July till 14 July with no access to my
emails.
In urgent cases please contact Ms.
Hello all,
I am facing difficulty in deciding how to go about analysis of a situation
explained as follows:
I have two variables Y (response) and X (explanatory)
(There are several other potential candidate explanatory variables, but
right now I am looking at only one variable.)
Y is binary
Dear list,
I am new to R and am using it to develop and test my own neural network
codes.
I need some training datasets that have the prediction results that
should (approximately) appear when the datasets are passed through a
good neural network, in order to test whether my code is working
i have option prices of 1 month option contract for several days in a csv
file. The data is in chronological order of dates. Within each date, price
quotes for each strike appears (about 30 strikes per month- so there are in
30 rows per day for many days and many months). i want only the quote on
Hello,
I wonder if someone can elaborate on why in the first graph I am able to set
labels vertical to the x-axis but not in the second.
I tried to select the window but it didnt really help.
Many Thanks
Paolo
ExtAvgCWV = rnorm(200)
ExtAvgDemand = rnorm(200)
ExtGasDays = seq(from =
A little more specificity required; sample of the data, do you want
multiple plots each with a single variable, or do you want them all on
one plot as lines/areas/points/etc. Do you want then broken down by
year, month, or some other way. It is easy to generate plots once you
know what you want.
Dear Prof. Therneau
I was impressed to discover that the 'survConcordance' now handles Surv()
objects in counting format (example below to clarify what I mean). This is not
documented in the help page for the function. I am very curious to see how a
c-index is estimated in this case, using
Thank you! Exactly what I wanted.
Ramzi
--
View this message in context:
http://r.789695.n4.nabble.com/How-do-I-overlay-two-trellis-plots-of-lme-fitted-lines-produced-by-plot-augPred-tp3652204p3653988.html
Sent from the R help mailing list archive at Nabble.com.
It is a warning: it will fall back to the version used in your
unstated version of R. Nothing you show justifies your claim
i can't install the packages
However, most likely you have a problem connecting to the internet at
all: see the FAQ relevant to your unstated OS or ask your local IT
Hi,
how about this:
mm-cbind(V1,V2)
xy-sapply(1:3,function(x)det(mm[-x,])*(2*(x%%2)-1))
#some checks
all.equal(0,as.vector(xy%*%V1))
all.equal(0,as.vector(xy%*%V2))
Am 08.07.2011 08:27, schrieb Bai:
Hi, everyone,
I need an efficient way to do vectors cross product in R.
Set vectors,
V1
Hey,
from what I see, you try to use the par(las=3) function after the plot
command. You should use it before it, though. Somewhat liek this:
ExtAvgCWV = rnorm(200)
ExtAvgDemand = rnorm(200)
ExtGasDays = seq(from = as.Date(2010-8-4, %Y-%m-%d), along.with =
ExtAvgCWV, by = days)
op - par(las
Hi Berry,
True, it works. Thanks for this and teh general advice. I have been doing
things wrong from day 1 and never realised it!
Cheers,
Paolo
On 8 July 2011 13:43, Berry Boessenkool berryboessenk...@hotmail.comwrote:
Hey Paolo,
you should specify las BEFORE you send the axis command.
Dear all,
I have encountered the following problem where coxme seems to allow
model with only random effect in R 2.11.1 but not in R 2.13.0. Following
is the error message using rat example data. Any comment on this is
appreciated.
In R2.13
library(coxme)
rat1 - coxme(Surv(time,
-Original Message-
[mailto:r-help-boun...@r-project.org] On Behalf Of francisco.ahued
Subject: [R] Return invisible list
x=read.table(data.txt,header=T)
goa=(x[,3])
meplot(goa)
I can see the plot, but I would like to see the values of the
x and y axis.
Something returned
Hey Paolo,
you should specify las BEFORE you send the axis command.
It would also be good to set greater margins, this needs to be set before the
entire plot, thus in following order:
par(las = 3, mar=c(6,2,2,2))
plot(ExtAvgCWV, ann=FALSE, xaxt=n, yaxt=n )
axis(side =1 , at = tickplaces,
I'm using the mcexact function from the exactLoglinTest package on data
comparing performance of rapid and laboratory tests for detection of H1N1
flu. My setup is as follows:
ridt.res - c(A-B-, A+B-, A-B+)
pcr.res - c(Negative, AH3, B)
xtab - expand.grid(ridt = ridt.res, pcr = pcr.res)
Dear all,
I have a file and I am using a code-
df=read.table(Case2.pileup,fill=T,sep=\t,colClasses = character)
to create a data frame that is df.
now in 4th column I have ASCII values which I want to convert in decimal
values.the data frame is like this-
V1 V2 V3
RSiteSearch(cross product)
library(pracma)
?cross
Speed is usually desired in the context of many similar computations, and is
normally achieved in R by vectorizing computation, so storing the large
number of 3d vectors together in a structure like a Nx3 matrix so the code
can be vectorized
On 7/8/2011 4:26 AM, Federico Calboli wrote:
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is under is 'GPL' --no
version. My assumption is, since all the code is
Thank you for your suggestion Gabor. I hope they can fix this in later
versions.
Victor
--
View this message in context:
http://r.789695.n4.nabble.com/gdata-read-xls-values-format-problem-tp3652494p3654334.html
Sent from the R help mailing list archive at Nabble.com.
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I can only
release my code under GPL-v2 because the code is written in R and probably
qualifies as a derivative work.
Did you include someone else's GPL-vx code (possibly
On 07/08/2011 07:58 AM, Spencer Graves wrote:
On 7/8/2011 4:26 AM, Federico Calboli wrote:
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it
is under is 'GPL' --no
On Fri, Jul 8, 2011 at 4:07 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I can only
release my code under GPL-v2 because the code is written in R and probably
qualifies as a
On 8 Jul 2011, at 16:12, Barry Rowlingson wrote:
On Fri, Jul 8, 2011 at 4:07 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I can only
release my code under GPL-v2 because the
On Jul 8, 2011, at 9:58 AM, Spencer Graves wrote:
On 7/8/2011 4:26 AM, Federico Calboli wrote:
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is
under is 'GPL'
On 7/8/2011 8:07 AM, Federico Calboli wrote:
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I
can only release my code under GPL-v2 because the code is written
in R and probably qualifies as a derivative work.
Did you include someone
On 08/07/2011 10:56 AM, Spencer Graves wrote:
On 7/8/2011 4:26 AM, Federico Calboli wrote:
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is under is 'GPL' --no
On Fri, Jul 8, 2011 at 4:16 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
The vast majority of CRAN libraries seem to be released under some sort of
GPL version. I never seen a license though.
You're right. The GNU people say should:
http://www.gnu.org/licenses/gpl-howto.html
---
On Fri, Jul 8, 2011 at 11:16 AM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
On 8 Jul 2011, at 16:12, Barry Rowlingson wrote:
On Fri, Jul 8, 2011 at 4:07 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though
Hello all,
I'm trying to get a specific source file to run at a certain time each day with
WindowsScheduler
http://windows.microsoft.com/en-US/windows7/schedule-a-task
I've tried a number of methods, none of which work:
My best guess was:
1. Associate the script.R file with R in FileTypes.
2.
I just started using R last week.
I have a dataset with 3 columns - Plate, Well and Raw
I need to make a simple plot(Well~Raw) but only when Plate = 101
Thanks for your help.
--
View this message in context:
http://r.789695.n4.nabble.com/Simple-conditional-plot-tp3654300p3654300.html
Sent
Hi There,
I'm facing one problem to construct a vector using the for command:
I have one matrix named 'dados' (same as /data/ from portuguese), for
example:
dados[140:150,]
[,1] [,2] [,3]
[1,] 212.7298 0.14 0.11
[2,] 213.3778 0.14 0.11
[3,] 214.0257 0.15 0.11
[4,] 214.6737 0.15
You can do:
as.numeric(charToRaw(paste(df[,4], collapse=)))
If you somehow want each row to be its own sequence of integers, you
could do something like:
lapply(df[,4], function(c) as.numeric(charToRaw(c)));
~Matthew Maycock
-Original Message-
From: r-help-boun...@r-project.org
On Thu, Jul 07, 2011 at 02:18:17PM -0700, m.marcinmichal wrote:
Hi,
Currently I testing the packets that contain built-in features for
classification. Actually I looked packages such as: e1071, Klar, Caret,
CORElearn. However, from what I noticed when building a naive Bayesian
classifier,
On Thu, Jul 7, 2011 at 6:07 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Thu, Jul 7, 2011 at 4:23 PM, Victor11 yao.z...@invesco.com wrote:
Dear All,
When I use read.xls() in gdata package to read xls files, I noticed an issue
and couldn't find any solutions after I serched all
Looks like FAQ 7.31 to me.
Try all.equal() instead of ==.
Sarah
On Fri, Jul 8, 2011 at 11:06 AM, VictorDelgado
victor.m...@fjp.mg.gov.br wrote:
Hi There,
I'm facing one problem to construct a vector using the for command:
I have one matrix named 'dados' (same as /data/ from portuguese),
Try
plot(Well~Raw, subset= Plate==101)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of rstudent
Sent: Friday, July 08, 2011 8:56 AM
To: r-help@r-project.org
Subject: [R] Simple conditional plot
I just started using R last week.
Are your users willing to install an Excel plug-in? If so, look at the RExcel
project, it does what you describe and the common user thinks they are just
using Excel with a plug-in without realizing that they have installed and are
using a useful tool in the background.
-Original
On Jul 8, 2011, at 10:55 AM, rstudent wrote:
I just started using R last week.
I have a dataset with 3 columns - Plate, Well and Raw
I need to make a simple plot(Well~Raw) but only when Plate = 101
?subset
# Some plotting paradigms allow you to use a subset = argument but
any program
In case anyone else is interested... I just found another solution, which has
some pros and cons compared to Dennis' solution...
plot(comparePred(fit.lme, update(fit.lme, y ~ log(age
# Pros - Automatically adds a legend
# Cons - Only works for comparing two models
# Dennis' solution extends
How are you measuring length of stay? A chi-square test suggests that you have
it categorized, a t-test assumes it is continuous (and relatively symmetric
with the amount depending on sample size).
Do you have any censoring? (patients dying or transferring before discharge) if
so you should
Just to add onto Greg's comments, you may want to review this thread over on
MedStats, since this topic was just discussed extensively this week, initially
as a query about using LOS as a covariate:
http://groups.google.com/group/medstats/browse_thread/thread/f875fdeeaf48dc38?hl=en
It is
Hello Dan,
I reckon that you need to path a batch-file to the scheduler, i.e. something
along the lines
R CMD BATCH script.R
shall be included in, say, 'RBatchjob.bat' and this file shall then be called
by the task scheduler.
Best,
Bernhard
-Ursprüngliche Nachricht-
Von:
In Windows Vista, here is what I do:
Create a .bat file that calls R to run the script. Then set the task
scheduler to run the batch file. It's a round about way but it works
for me.
# begin example of *.bat file contents
cd C:\R\R-2.13.0\bin
R CMD BATCH C:\MyBats\MyBat.R
Hello,
I have tried to generate numbers randomly which follow normal, Student-t and
skewed Student-t distributions. However, when I check those series for
heteroscedastisity test (ARCH) results are showing that there is no
heteroscedastisity.
As we all know, returns (financial returns) usually
Tried this and received this error:
Error in hist.default(x = integer(0), plot = FALSE) :
invalid number of 'breaks'
--
View this message in context:
http://r.789695.n4.nabble.com/Condional-Density-Plot-from-different-data-tp3080615p3654634.html
Sent from the R help mailing list archive at
ty S. Goslee,
It's helpfull to test the condition:
all.equal(s[4],0.15)
[1] TRUE
instead the previous FALSE answer obtained with
s[4]==0.15
[1] FALSE
but I still need get it to vector r:
Victor Delgado wrote:
for (w in 1:length(s)){
r[w] - dados[,3][dados[,2]==s[w]][1]
}
I just
On 08/07/2011 12:15 PM, UnitRoot wrote:
Hello,
I have tried to generate numbers randomly which follow normal, Student-t and
skewed Student-t distributions. However, when I check those series for
heteroscedastisity test (ARCH) results are showing that there is no
heteroscedastisity.
As we all
Command I am using for the plot:
plot(Raw[][Plate==101]~well[][Plate==101], xlab=Well, ylab=Raw,
main=Plate 101)
I only want to label points on the graph where well equals B13, G13, K14 and
N14 with the name of the well.
Thank you for your help.
--
View this message in context:
Another possibility without using subset():
plot(Well[][Plate==101]~Raw[][Plate==101])
This also works to '=' '=' and other conditions.
--
View this message in context:
http://r.789695.n4.nabble.com/Simple-conditional-plot-tp3654300p3654666.html
Sent from the R help mailing list archive at
Hi,
I have a set of nodes and a dissimilarity matrix for them, as well as a csv
file in which the diss matrix has been converted to [node_1, node_2,
dissimilarity] format. I would like to visualize this as a graph in
Euclidean space (that is, similar nodes clumped together in clusters),
rather
Dear All,
I am working on a time series of hourly river flow measurements from 2000 -
2003 and have been trying to compute the extremum index from the original
series as well as a new series from sampling with replacement . The extremum
Index produced from the original data series looks
On Jul 8, 2011, at 1:31 PM, rstudent wrote:
Command I am using for the plot:
plot(Raw[][Plate==101]~well[][Plate==101], xlab=Well, ylab=Raw,
main=Plate 101)
I only want to label points on the graph where well equals B13, G13,
K14 and
N14 with the name of the well.
?%in%
?text
Thank
On Jul 8, 2011, at 1:44 PM, FMH wrote:
Dear All,
I am working on a time series of hourly river flow measurements from
2000 - 2003 and have been trying to compute the extremum index from
the original series as well as a new series from sampling with
replacement . The extremum Index
I can use the text() command to label all points but how do you specify only
to label those four specific points on the graph?
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-label-specific-points-on-a-scatterplot-tp3654697p3654777.html
Sent from the R help mailing list
On Jul 8, 2011, at 2:02 PM, rstudent wrote:
I can use the text() command to label all points but how do you
specify only
to label those four specific points on the graph?
No context, no example, evidence that you read the rest of my message,
no further response.
--
View this message
Thank you Gerrit for the quick reply! And yes, i'm Matti.
I can get the coeffs now, though i'm not sure whether i'm doing
something wrong or whether poly is just not the right method for what
i'm trying to find. I will look into this more closely and give it
another try.
Is poly best for
I want to call in files from subdirectories without changing my working
directory within my script. Is that possible? And if so what would be the
simplest way to do that?
for example, if I do setwd(C:/Users/Hello)
but I have files in C:/Users/Hello/Data1 and C:/Users/Hello/Data2, can I
still
On 08/07/2011 2:47 PM, ssathnur wrote:
I want to call in files from subdirectories without changing my working
directory within my script. Is that possible? And if so what would be the
simplest way to do that?
for example, if I do setwd(C:/Users/Hello)
but I have files in C:/Users/Hello/Data1
I noticed that there is a newer version of Tom Short's cheat sheet than
the version currently posted on CRAN.
Personally I like the newer version, but maybe keeping the old version is
deliberate. Anyway, I was wondering if there's someone that I can notify
that can update the content.
New
Answering
http://r.789695.n4.nabble.com/How-to-generate-heteroscedastic-random-numbers-td3654534.html
I think best way to generated Heteroscedastic data would be, first
fix or assume some DGP of your choice, you may assume any arbitrary
model parameter(s) provided forms of those parameters are
Hi there, given a numeric vector, I can select numbers within a
specific range. However presently, I have something related but
different problem. Suppose I have a numeric vector. Now take an
arbitrary number. Goal to to chose a specific subset with a given
length, from that given vector, so that
On Jul 8, 2011, at 3:16 PM, Nipesh Bajaj wrote:
Hi there, given a numeric vector, I can select numbers within a
specific range. However presently, I have something related but
different problem. Suppose I have a numeric vector. Now take an
arbitrary number. Goal to to chose a specific subset
I have only been using R for a very short time and I'm trying to learn.
What information do you need to help me?
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-label-specific-points-on-a-scatterplot-tp3654697p3654903.html
Sent from the R help mailing list archive at
rstudent,
one solution could be given by spliting your graph in two parts:
part_1 - # take the desired four points subset #
part_2 - # all rest points not to label
You will also need one vector with your names. I supose:
list - c(B13, G13, K14, N14)
So
plot(part_1$x, part_1$y)
text(part_1$x,
On Jul 8, 2011, at 2:53 PM, rstudent wrote:
I have only been using R for a very short time and I'm trying to
learn.
What information do you need to help me?
As it says at the bottom of every message to Rhelp:
read the posting guide http://www.R-project.org/posting-guide.html
--
View
Hi:
Do you want the N points in the vector closest to a fixed point? If
so, then try this:
f - function(vec, center, nselect) {
d - (vec - center)^2
vec[order(d)][1:nselect]
}
v - rnorm(1000)
f(v, 0, 10)# select the ten points in v closest to zero
Your goal is subject to multiple
Dear sir,
I am struggling with a problem.Please help me.
Now I have a dataframe with these columns-
V7 V8 V9 V10
0 1 G82
0 1 CGT
Hi,
I really feel I've looked everywhere, although I know this can't be a hard
problem. I'd like to be able to call the graph below as a function, but I
can't get the function to recognize variables beyond 'dframe'. I've read
through many papers on writing functions in R, but I can't get this to
Dear jholtman,
Thanks for the reply sorry for the been unclear before.
My desire graph is to have multiply plots showing Y1,Y2,Y3 with the same X,
were each plot is month-year (e.g., 5-2001, 6-2001, etc). It ill be great if
each Y can have a different line and point style.
The Lattice
---BeginMessage---
Hi Vikas. Apply the logic in the example below in your dataframe
dta
V1 V1
1 85 32
2 80 33
3 77 11
4 75 56
5 96 43
6 99 12
7 94 32
8 97 44
dta[,1]
[1] TRUE TRUE FALSE FALSE TRUE TRUE TRUE TRUE
which(dta[,1])
[1] 1 2 5 6 7 8
ref - which(dta[,1])
dta[ref,]
I am trying to create a set of wavelets in frequency space--namely Cauchy
wavelets for an intensity analysis (von Tscharner, 2000). The wavelets are
defined by the following formula:
[(f/cf)^(cf*scale)]*[e^((-f/cf)+1)^(cf*scale)]
where *f *is frequency of length *n*, *cf* is center frequency
At 07:12 09/07/2011, you wrote:
Dear jholtman,
Thanks for the reply sorry for the been unclear before.
My desire graph is to have multiply plots showing Y1,Y2,Y3 with the same X,
were each plot is month-year (e.g., 5-2001, 6-2001, etc). It ill be great if
each Y can have a different line and
Dear all,
I am really sorry for not giving the input file because in my mail,I did not
explain my problem in a best way.
I have a file that is summary.txt(I have attached it) .we can read this file
using-
dfa=read.table(summar.txt,fill=T,colClasses = character,header=T)
In V10 column I have
Dear all,
I am really sorry for not giving the input file because in my mail,I did not
explain my problem in a best way.
I have a file that is summary.txt(I have attached it) .we can read this file
using-
dfa=read.table(summar.txt,fill=T,colClasses = character,header=T)
In V10 column I
Hello,
Sorry if it's a simple question, this is my first script complex, but can
not find solution in the list
I have many TXT files with daily data of meteorological stations and a
table that connects it with each of these stations. (More than one table per
station and date).
With the Script
On 7/7/2011 9:14 PM, michael.laviole...@dhhs.state.nh.us wrote:
Are there any packages with functions that can fit quasi-symmetry and
quasi-indepedence models to square contingency tables?
M. Laviolette
Yes.
See the gnm package for convenience functions that work with
glm() and gnm().
See
I get something entirely different when I execute that input command
with the attached file:
This is what I see as the first 14 lines for a displayed value for dfa:
dfa
V7 V8 V9 V10
10 1 G`
20 1 Ta
30 1 Ca
40 1 Aa
50 1 G_
60 1
Happy weekend helpeRs!
As usual, I'm stumped by R...
My plan was to take an integer number, convert it to binary and wind
up with a data.frame where each column is either 1 or 0 so I can see
which bits are changing:
bb-function(i) ifelse(i, paste(bb(i %/% 2), i %% 2, sep=), )
dear R wizards---more ignorance on my part, exacerbated by too few
examples in the function documentations.
d - data.frame( id=rep(1:3,3), x=rnorm(9), y=rnorm(9))
Question 1: how do I work with the output of by? for example,
b - by( d, d$id, function(x) coef(lm( y ~ x, data=x ) ))
b
d$id:
Yes sir.you are right.after this I use this code to convert ASCII values in
column V10 to decimal numbers-
dfa$V10=lapply(dfa[,4], function(c) as.numeric(charToRaw(c)))
now u will get output something like this-
V7 V8 V9 V10
Dear expeRts,
How can I vertically adjust an axis tick label so that it is nicely aligned with
the other labels?
library(lattice)
xyplot(0~0, xlim=c(0,3), scales=list(x=list(at=c(1,1.1),
labels=c(expression(hat(theta)[italic(n)]),expression(theta)
## aim: move the leftmost expression up so
On Jul 8, 2011, at 6:46 PM, Bansal, Vikas wrote:
Yes sir.you are right.after this I use this code to convert ASCII
values in column V10 to decimal numbers-
dfa$V10=lapply(dfa[,4], function(c) as.numeric(charToRaw(c)))
now u will get output something like this-
V7 V8
V9
1 - 100 of 110 matches
Mail list logo
