Ted Byers wrote:
Thanks Jim,
Alas, it wasn't this. Here is the output from both of your suggestions:
refdata18 = read.csv(K:\\MerchantData\\RiskModel\\Capture.Week.18.csv,
header = TRUE,na.strings=)
str(refdata18)
'data.frame': 341 obs. of 1 variable:
$ X0: int 0 0 0 0 0 0 0 0
The way you've attempted to get this result seems to align with the way
R should work, but it fails in this case.
The fix is to break things up a little bit:
p - ggplot(mydata, aes(x=Est, y=Tri))
p - p + geom_point(aes(colour=factor(Group),shape=factor(Group)))
p - p +
I have been working with the base time series object (ts) and I had a couple of
questions that hopefully this group can help me with:
1) What is the best why to append an observation to an existing time-series?
Suppose I have a time series:
t - ts(1:12, frequency=5)
This would generate two
Hello,
Is there any package in R for balancing matrix
I want to estimate a matrix with
* a initial matrix (1 everywhere for example)
* Row margin
* Col margin
* distance class vector (each cell of the matrix belong to a
distance class) and I want that the distance
Hi Gareth,
My data is transformed to the clr or alr under Aitchison geometry, so I
am essentially working
in Euclidean space.
Great: glad to hear it.
Has anyone had experience doing stepwise LDA?? I can't for the life of
me find any help
online about where to start.
A better option
Hi,
You can treat it as a database and use ODBC to fetch data from the CSV
file using SQL. See the package RODBC for details about database
connections. (I have dealt with similar problems before with RODBC)
Regards,
Yihui
--
Yihui Xie [EMAIL PROTECTED]
Phone: +86-(0)10-82509086 Fax:
Hello, Yihui
You can treat it as a database and use ODBC to fetch data from the CSV
file using SQL. See the package RODBC for details about database
connections. (I have dealt with similar problems before with RODBC)
Thanks for your tip, I have used RODBC before to read data from MSAccess and
I wouldn't call a 4GB csv text file a 'database'.
Obviously, a csv it's not a database itself, I tried to mean (though it
seems I was not understood) that I had a huge database, exported to csv file
by the people who created it (and I dont have any idea of the original
format of the database).
Adaikalavan Ramasamy wrote:
I agree! The best way to learn (and remember for longer) is to teach
someone else about it.
And there is not reason not to repeat some of the anlysis done on SAS
with R. That way you can verify your outputs or compare the
presentations. If you consistently find
Hi Ted (from Ted),
Just to clarify Marc's comments about dataframes in more basic terms.
If you read in data with read.csv() the result returned by the function
is a dataframe. This is a specialised kind of list, which you can think
of as a list of columns all of the same length. You can think of
2008/9/22 José E. Lozano [EMAIL PROTECTED]:
I wouldn't call a 4GB csv text file a 'database'.
It didn't help, sorry. I perfectly knew what a relational database is (and I
humbly consider myself an advanced user on working with MSAccess+VBA, only
that I've never face this problem with
Hello,
Is there any package in R for balancing matrix
I want to estimate a matrix with
* a initial matrix (1 everywhere for example)
* Row margin
* Col margin
* distance class vector (each cell of the matrix belong to a
distance class) and I want that the distance class
What are you going to do with the data once you have read it in? Are
all the data items numeric? If they are numeric, you would need at
least 8GB to hold one copy and probably a machine with 32GB if you
wanted to do any manipulation on the data.
You can use a 'connection' and 'scan' to read the
2008/9/22 José E. Lozano [EMAIL PROTECTED]:
Exactly, raw data, but a little more complex since all the 50 variables
are in text format, so the width is around 2,500,000.
Thanks, I will check. Right now I am reading line by line the file. It's
time consuming, but since I will do it only
Duncan Murdoch wrote:
This is fixed now on R-forge; eventually it will make it into the next
rgl release on CRAN. You should be able to download a binary of the
development version from R-forge sooner. Make sure you get version
0.81.706 or newer.
The R-forge version 0.81.706 works as
Hello,
I am calling the auto.arima method in the forecast package at it returns what
seems to be valid Arima output. But when I feed this output to 'predict' I get:
Error in predict.Arima(catall.fit[[.index]], n.ahead = 12) :
'xreg' and 'newxreg' have different numbers of columns
Is there a
Please read the posting guide an tell us:
- Which version of R
- Which OS?
- Which version of the matlab package (I guess you are using that one?)
- If Windows and a binary version of the matlab package: Does the binary
it fit to your version of R?
Uwe Ligges
Mac wrote:
Dear R users£¬
Dear R users
From what I understand, the joint maximum likelihood procedure for Rasch
(availabe in the package MiscPsycho) in R can only be used on binary data.
I was wondering if the code is currently being adapted for application to
ordinal data? I'm trying to replicate results obtained from
Matthew,
As per the CRAN Ubuntu README
http://cran.r-project.org/bin/linux/ubuntu/
install the Ubuntu r-base-dev package to compile R packages from
sources.
Vincent
Le lun. 22 sept. à 00:08, Matthew Pettis a écrit :
Hi,
I'm trying to get the Hmisc module on my Ubuntu Hardy
On Mon, Sep 22, 2008 at 08:48:12AM -0400, Vincent Goulet wrote:
Matthew,
As per the CRAN Ubuntu README
http://cran.r-project.org/bin/linux/ubuntu/
install the Ubuntu r-base-dev package to compile R packages from
sources.
Well there should be a working r-cran-hmisc package. You
José E. Lozano [EMAIL PROTECTED] writes:
Maybe you've not lurked on R-help for long enough :) Apologies!
Probably.
So, how much design is in this data? If none, and what you've
basically got is a 2000x50 grid of numbers, then maybe a more raw
Exactly, raw data, but a little more
H
I want to use the system() command to execute a command and have to
return the result in a r-variable, so I an using intern=TRUE.
On the other hand, I want to evaluate the return value of the command,
to determine if the command was successful.
According to the help, these to objectives are
Christophe LOOTS Christophe.Loots at ifremer.fr writes:
Thank you so much for your help.
The function dbinom seems to work very well.
However, I'm a bit lost with the dnorm function.
Apparently, I have to compute the mean mu and the standard deviation
sd but what does it mean
Hi Juliet,
On Sun, Sep 21, 2008 at 11:47 PM, Juliet Hannah [EMAIL PROTECTED] wrote:
Here is some sample data:
mydata - read.table(textConnection(Est GroupTri
00 4.639644
10 4.579189
20 4.590714
01 4.443696
11
Hello,
I am attempting to use smoothScatter to plot a heatmap of locations of
events in an x-y axis. When I plot the heatmap without passing xlim and ylim
parameters, it fills the plot area but the perspective is a bit skewed. I
would like to standardize these plots to a uniform window size that
Hello,
I guess the solution is rather simple but whatever I tried, I don't manage to
get the result as I want to have it:
I have several vectors of equal length in a list and I'd like to combine all
first elements to a single string, all second elements to a single string, ...,
all n-th
Hi!
I've got a time series as a zoo object which contains hourly values. My problem
is that these values occur in every real hour with regard to daylight savings
time. I.e. the last sunday in march, i'll have 23values whereas the last sunday
in october contains 25 values instead of 24.
Thus
try this:
t1 - c(1, 2, 3)
t2 - c(3.4, 5.5, 1.1)
tl - list(t1, t2)
do.call(paste, c(tl, sep = \t))
I hope it helps.
Best,
Dimitris
Antje wrote:
Hello,
I guess the solution is rather simple but whatever I tried, I don't
manage to get the result as I want to have it:
I have several
Try this:
paste(tl[[1]], tl[[2]], sep=\t)
On Mon, Sep 22, 2008 at 11:08 AM, Antje [EMAIL PROTECTED] wrote:
Hello,
I guess the solution is rather simple but whatever I tried, I don't manage
to get the result as I want to have it:
I have several vectors of equal length in a list and I'd like
Great! That's exactly what I was looking for.
(I see, I still have to learn a lot...)
Thank you!
Antje
Dimitris Rizopoulos schrieb:
try this:
t1 - c(1, 2, 3)
t2 - c(3.4, 5.5, 1.1)
tl - list(t1, t2)
do.call(paste, c(tl, sep = \t))
I hope it helps.
Best,
Dimitris
Antje wrote:
Hello,
See question #1 in the zoo faq:
library(zoo)
vignette(zoo-faq)
Also in the upcoming zoo 1.6-0, not yet on CRAN but in the development version
at R-Forge found here:
http://r-forge.r-project.org/projects/zoo/
there are a set of make.unique functions and a make.unique= argument in
read.zoo which
Try this to append 100 to the end of the series, say:
tt - ts(1:12, frequency=5) # sample data
ts(c(tt, 100), start = start(tt), frequency = frequency(tt))
On Mon, Sep 22, 2008 at 2:17 AM, [EMAIL PROTECTED] wrote:
I have been working with the base time series object (ts) and I had a couple
What are you going to do with the data once you have read it in? Are
all the data items numeric? If they are numeric, you would need at
least 8GB to hold one copy and probably a machine with 32GB if you
wanted to do any manipulation on the data.
Well, I will use only sets of variables to
So is each line just ACCGTATAT etc etc?
Exacty, A_G, A_A, G_G and the such.
If you have fixed width fields in a file, so that every line is the
same length, then you can use random access methods to get to a
particular value - just multiply the line length by the row number you
Nice hint! I
Hi guys,
Suppose I have 2 data frames ie:
values
one0.32
two0.25
three 0.11
and
values
two0.66
one0.74
three 0.19
nb the first column is the row names in both cases
How can I combine them on the row names column? Ie to make something like
Try:
data.frame(merge(df1, df2, by = row.names), row.names = 1)
On Mon, Sep 22, 2008 at 12:34 PM, jimineep [EMAIL PROTECTED] wrote:
Hi guys,
Suppose I have 2 data frames ie:
values
one0.32
two0.25
three 0.11
and
values
two0.66
one0.74
three 0.19
Why don't you make one pass through your data and encode you
characters as integers (it would appear that you only have 16
combinations). You might also want to consider using the 'raw' object
since these only take up one byte of storage -- will reduce your
storage requirements by 4. Then store
2008/9/22 jim holtman [EMAIL PROTECTED]:
Why don't you make one pass through your data and encode you
characters as integers (it would appear that you only have 16
combinations). You might also want to consider using the 'raw' object
since these only take up one byte of storage -- will reduce
OK, I am now at the point where I can use fitdistr to obtain a fit of one of
the standard distributions to mydata.
It is quite remarkable how different the parameters are for different
samples through from the same system. Clearly the system itself is not
stationary.
Anyway, question 1: I
On 22-Sep-08 11:00:30, José E. Lozano wrote:
So is each line just ACCGTATAT etc etc?
Exacty, A_G, A_A, G_G and the such.
If you have fixed width fields in a file, so that every line is the
same length, then you can use random access methods to get to a
particular value - just multiply the
Hello,
I have another stupid question. I hope you can give me a hint how to solve this:
I have a list and one element is again a list containing matrices, all of the
same dimensions. Now, I'd like to set the dimnames for all matrices:
example code:
m1 - matrix(1:25, nrow=5)
m2 -
Hi there,
I am looking for a way to use wildcards in a subset, this is not
working:
subset(data, colname-1==valuecolname2==value*,
select=colx:coly)
is there a way to use wildcards here?
Thanks for your help,
Daniel
[[alternative HTML version deleted]]
Hi,
If all your matrices have the same size, you should work with an array
and not with a list. Then you can use dimnames to set the names of the
rows, columns, and so on..
Alain
Antje wrote:
Hello,
I have another stupid question. I hope you can give me a hint how to
solve this:
I have
Thank you for the input.
Which command in the spatstat package am I looking for? The documentation is
unclear to me.
milton ruser wrote:
Dear J.J.Harden
I think that on spatial stat you will find several ways of simulate
spatial
pattern that (point or line) that may be what you are
Hi,
Bioconductor.org is the home of the geneplotter package. You get a
quicker response if you ask there.
/Henrik
On Mon, Sep 22, 2008 at 7:06 AM, Jason Pare [EMAIL PROTECTED] wrote:
Hello,
I am attempting to use smoothScatter to plot a heatmap of locations of
events in an x-y axis. When I
I am in a situation where I have to fit a distrution, such as cauchy or
normal, to an empirical dataset. Well and good, that is easy.
But I wanted to assess just how good the fit is, using ks.test.
I am concerned about the following note in the docs (about the example
provided): Note that the
Try this:
read.table(pipe(/Rtools/bin/gawk -f cut.awk bigdata.dat))
where cut.awk contains the single line (assuming you
want fields 101 through 110 and none other):
{ for(i = 101; i = 110; i++) printf(%s , $i); printf \n }
or just use cut. I tried the gawk command above on Windows
Vista with
Hi, all,
I was wondering if there is a way to change the offset of axis labels
from the axis. In other words, I need the axis labels closer to the
acis than the default. Thanks for the help.
Best wishes,
Art Roberts
University of Washington
Seattle, WA
I am was going to look at the as.yearmon function in the zoo package
and write a as.day function to aggregate a time series of 96
observations per day into the mean for each day, but I don't know how
to look at the code so that I can convert it into something I can use.
On top of that I believe
chron values are represented as day + fraction of a day so:
try this:
aggregate(z, floor, mean)
On Mon, Sep 22, 2008 at 12:56 PM, stephen sefick [EMAIL PROTECTED] wrote:
I am was going to look at the as.yearmon function in the zoo package
and write a as.day function to aggregate a time series
perfect thanks
On Mon, Sep 22, 2008 at 1:07 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
chron values are represented as day + fraction of a day so:
try this:
aggregate(z, floor, mean)
On Mon, Sep 22, 2008 at 12:56 PM, stephen sefick [EMAIL PROTECTED] wrote:
I am was going to look at
Sorry, I misunderstood what I was doing and misspoke. I don't think there's
a bug. I had called COMMAND w/in read.delim.
Thanks for all of your help and sorry for the misinformation.
Sincerely,
Mike
-
Department of Ecology Evolutionary
If one of the goals is the normality test, then there may be better
alternatives to the Kolmogorov-Smirnov test.
See an explanation on:
http://graphpad.com/FAQ/viewfaq.cfm?faq=959
The R implementation:
?shapiro.test
A casual search also turned this up:
Thank You All,
I think all of this may have been due to shared library conflict
headaches. At one point, I inadvertently upgraded my Perl install to
5.10, and I think that messed up a lot of my libraries. I have now
started with a clean Ubuntu install, and am going to see if I can work
my way
Look at ?par and scroll down to the section on 'mgp'. Or you can suppress the
axis when you make the plot, then use the axis function to include it with more
control (see ?axis).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL
Hi, all,
Could anyone give me advise on who the execute external programs with
R? It would be greatly appreciated.
Art Roberts
University of Washington.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
On 9/22/2008 2:50 PM, Arthur Roberts wrote:
Hi, all,
Could anyone give me advise on who the execute external programs with
R? It would be greatly appreciated.
The system() or shell() functions can do this; Windows also has
shell.exec().
Duncan Murdoch
On an (Intel Leopard) Mac I try to build a package (mxFinance) which
depends on another package (mxGraphics). The dependendy is 1) a
'Depends:' in DESCRIPTION and 2) an import in NAMESPACE.
- The build fails if the dependent package (mxGraphics) is not
installed in the R.framework
Do I need to
Good afternoon
Has anyone tried using Dr. Elith's BRT script? I cannot seem to run
gbm.step from the installed gbm package. Is it something external to gbm?
When I run the script itself
- gbm.step(data=model.data,
gbm.x = colx:coly,
gbm.y = colz,
family = bernoulli,
Hi,
I try to change the panel name in a xyplot without success.
Look this example from xyplot manual:
xyplot(Murder ~ Population | state.region,data=states)
The panel title are:
Northeast, South, North Central, West, that are factor from state.region.
I need do change some names and, for
Try this:
xyplot(Murder ~ Population | state.region,
data = states,
strip = strip.custom(factor.levels = c(expression(italic(A)),
B, C, D)))
On Mon, Sep 22, 2008 at 4:33 PM, Ronaldo Reis Junior [EMAIL PROTECTED] wrote:
Hi,
I try to change the panel name in a xyplot without
Dear R users,
the help for findInterval(x,vec) suggests a logarithmic dependence on N
(=length(vec)), which would imply a binary search type algorithm.
However, when I test this hypothesis, in the following manner:
set.seed(-3645);
l - vector();
N.seq - c(5000, 50, 100, 1000,
Dear Hans,
Thanks for your reply.
I will read that book.
Cheers!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Hi Steve,
If you read your netCDF files into R you end up with sp-classes which
can be displayed using spplot. But you do not seem to use rgdal.
If you can make a data.frame with the x, y and z coordinates this can
quite easily be transformed into an sp-class:
library(sp)
dat =
Hello All,
In general when we use Rprof for performance evaluation on
Multicore systems the output provides the time on the basis of the user
time and the sampling time is equal to the the user time as reported by
system.time. This does not seem right behavior when R is linked to
On Mon, 22 Sep 2008, Martin Morgan wrote:
José E. Lozano [EMAIL PROTECTED] writes:
Maybe you've not lurked on R-help for long enough :) Apologies!
Probably.
So, how much design is in this data? If none, and what you've
basically got is a 2000x50 grid of numbers, then maybe a more raw
Hi All,
After rebuilding my Ubuntu image, I followed the instruction in this
thread, and everything worked out fine -- thank you again.
So, I'll just add: if you use R and perl, and don't have to download
perl5.10, then don't do it, at least not yet. Or, if you do, then you
will have a lot of
On 9/22/2008 1:51 PM, Markus Loecher wrote:
Dear R users,
the help for findInterval(x,vec) suggests a logarithmic dependence on N
(=length(vec)), which would imply a binary search type algorithm.
However, when I test this hypothesis, in the following manner:
R is open source. Why test things
Hi,
I'm analysing a dataset in which the same 5 subjects (male.pair) were subjected
to two treatments (treatment) and were measured for 12 successive days within
each treatment (layingday). Overall 5*2*12=120 observations.
I want to test the effect of treatment, time (layingday) and their
on 09/22/2008 11:26 AM Bert Chan wrote:
Warranty on Accuracy, Precision, Legality, ... of R in Research
(These questions may well have been raised.)
What is the implied warranty of using R for research publications,
consulting, etc.?
Alternately, how does one obtain such a warranty?
Dear R-users,
After running a logistic regression, I need to calculate OR by exponentiating
the coefficient, and then I need the 95% CL for the OR as well. For the
following example (taken from P. Dalaagard's book), what would be the most
straightforward method of getting what I need? Could
Dear Luciano,
See ?logistic.display in the epicalc package. If glm1 is your model,
something like
logistic.display(glm1)
should do the job.
HTH,
Jorge
On Mon, Sep 22, 2008 at 5:28 PM, Luciano La Sala
[EMAIL PROTECTED]wrote:
Dear R-users,
After running a logistic regression, I need to
Hi All,
i have searched the web for a simple solution but have been unable to find
one. Can anyone recommend a neat way of deleting multiple variable?
I see, i need to use dataframe$VAR-NULL to get rid of one variable,
In my situation i need to delete all vars between two points.
I've used the
Mike,
how about
M_UC - M_UC[,-(myvars[1]:myvars[2])]
?
Andrew
On Mon, Sep 22, 2008 at 11:04:34PM +0100, Michael Pearmain wrote:
Hi All,
i have searched the web for a simple solution but have been unable to find
one. Can anyone recommend a neat way of deleting multiple variable?
I see, i
Hello,
I'm estimating an ordered logit model on a probability weighted survey
sample. polr permits case weights with the weights option, but I cannot
figure out from existing documentation what it actually does with these
weights. I'm concerned about this because I get somewhat different
I am an R beginner and trying to run a SUR model in R framework.
subset(esasp500, Obs =449 Obs=197, select = -Date) -ev13sub
c(Obs=397) c(Obs=399) -d13
c(Obs=400) c(Obs=449) -f13
SP500*f13 -SP500f13
BBC~SP500+d13+SP500f13 -sur132
BOW~SP500+d13+SP500f13 -sur133
Hello,
I am using forecast() in the forecast package to predict future values of an
ARIMA model fit to a time series. I have read most of the documentation for
the forecast package, but I can't figure out how to obtain the forecast
variance for the predicted values. I tried using the argument
On Mon, 22 Sep 2008, Gregory Wawro wrote:
Hello,
I'm estimating an ordered logit model on a probability weighted survey
sample.
You could use svyolr() in the survey package.
polr permits case weights with the weights option, but I cannot
figure out from existing documentation what it
On Mon, Sep 22, 2008 at 4:07 PM, Marc Schwartz
[EMAIL PROTECTED] wrote:
on 09/22/2008 11:26 AM Bert Chan wrote:
Warranty on Accuracy, Precision, Legality, ... of R in Research
(These questions may well have been raised.)
What is the implied warranty of using R for research publications,
Sorry, I am resending in plain text.
Hello,
I am using forecast() in the forecast package to predict future values
of an ARIMA model fit to a time series. I have read most of the
documentation for the forecast package, but I can't figure out how to
obtain the forecast variance for the predicted
Dear all,
If I have a data frame x-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2)):
x1 x2 x3
1 4 8
7 6 2
I want to sort the whole data and get this:
x1 1
x3 2
x2 4
x2 6
x1 7
x3 8
If I do sort(X), R reports:
Error in order(list(x1 = c(1, 7), x2 = c(4, 6), x3 = c(8, 2)),
If I have the string below. does someone know a regular expression to
just get the BLC.NYSE. I bought the O'Reilley
book and read it when I can and I study the solutions on the list but
I'm still not self sufficient with these things. Thanks.
One possibility is:
x - data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2))
names - t(matrix(rep(names(x),times=nrow(x)),nrow=ncol(x)))
m - as.matrix(x)
ind - order(m)
df - data.frame(name=names[ind],value=m[ind])
df
name value
1 x1 1
2 x3 2
3 x2 4
4 x2 6
5 x1 7
6 x3
I tried downloading R-2.7.2
(http://cran.cnr.berkeley.edu/bin/windows/base/R-2.7.2-win32.exe, both from
Berkeley and cran) and both times I got a warning from Computer Associates
eTrust Antivirus (version 7.1.710) that the Win32/Adclicker.JO trojan was
detected:
The Win32/Adclicker.JO was
On Mon, Sep 22, 2008 at 6:54 PM, zhihuali [EMAIL PROTECTED] wrote:
Dear all,
If I have a data frame x-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2)):
x1 x2 x3
1 4 8
7 6 2
I want to sort the whole data and get this:
x1 1
x3 2
x2 4
x2 6
x1 7
x3 8
If I do
This is exactly what I wanted!
Thank you so much!
Z
Date: Mon, 22 Sep 2008 19:21:43 -0500
From: [EMAIL PROTECTED]
Subject: RE: [R] sort a data matrix by all the values and keep the names
To: [EMAIL PROTECTED]
Hi: there might be a quicker way but you can use stack and order. stack
On Sun, Sep 21, 2008 at 5:25 PM, Tom Bonen [EMAIL PROTECTED] wrote:
hi,
is there any way to suppress the legend in ggplot(data, aes(y=Y,
x=X,fill=Z)) ? i'd like the values to be displayed in different colors
as specified by fill= and this works just fine. but i do not want to
have the legend
Hi Mark,
stock-/opt/limsrv/mark/research/equity/projects/testDL/stock_data/fhdb/US/BLC.NYSE
gsub(.*/([^/]+)$, \\1,stock)
[1] BLC.NYSE
--- On Tue, 23/9/08, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
From: [EMAIL PROTECTED] [EMAIL PROTECTED]
Subject: [R] perl expression question
To:
Hello,
I have been using:
model - lm(y~x+I(x^2))
I am namely interested in the values of the residuals. If I use the 'names'
command I get the following:
names(model)
[1] coefficients residuals effects rank
[5] fitted.values assignqrdf.residual
[9] xlevels
Is something missing in the melt()?
x-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2))
require(reshape)
Loading required package: reshape
dfm - melt(x, id = c())
Error in if (!missing(id.var) !(id.var %in% varnames)) { :
missing value where TRUE/FALSE needed
dfm[order(dfm$value), ]
Error: object
Hi,
I would like to know how to plot the implicit function. For example,
f(x,y)=0. I'd like to plot x-y figure.
Thanks,
Ying
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PLEASE do read the posting guide
Hi Mark,
do you mean the regex to get the portion of the address after the
final slash? Something like
gsub(.*/([^/]*$), \\1, stock, fixed=FALSE)
Cheers
Andrew
On Mon, Sep 22, 2008 at 07:29:25PM -0500, [EMAIL PROTECTED] wrote:
If I have the string below. does someone know a regular
Try this:
sub(.*/, , stock)
[1] BLC.NYSE
On Mon, Sep 22, 2008 at 8:29 PM, [EMAIL PROTECTED] wrote:
If I have the string below. does someone know a regular expression to just
get the BLC.NYSE. I bought the O'Reilley
book and read it when I can and I study the solutions on the list but I'm
By the way, although a regular expression solutions was asked for
if one expands that to any solution then R does have a function
specifically for this case:
basename(stock)
[1] BLC.NYSE
On Mon, Sep 22, 2008 at 9:23 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Try this:
sub(.*/, , stock)
If this is a path name, then 'basename' will work for you:
stock-/opt/limsrv/mark/research/equity/projects/testDL/stock_data/fhdb/US/BLC.NYSE
basename(stock)
[1] BLC.NYSE
On Mon, Sep 22, 2008 at 8:29 PM, [EMAIL PROTECTED] wrote:
If I have the string below. does someone know a regular
Hmm, maybe it only works in my development version (to be released v. v. soon)
Hadley
On Mon, Sep 22, 2008 at 8:02 PM, Steven McKinney [EMAIL PROTECTED] wrote:
Is something missing in the melt()?
x-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2))
require(reshape)
Loading required package: reshape
Consider:
x - array(1:12,dim=12)
x[13]
[1] NA]
m - array(1:12,dim=c(3,4))
m[3,5]
Error: subscript out of bounds
Can anyone tell me it there is a Good Reason for the difference in
behaviour
between 1 dimensional and higher dimensional
Dave DeBarr wrote:
I tried downloading R-2.7.2
(http://cran.cnr.berkeley.edu/bin/windows/base/R-2.7.2-win32.exe, both from
Berkeley and cran) and both times I got a warning from Computer Associates
eTrust Antivirus (version 7.1.710) that the Win32/Adclicker.JO trojan was
detected:
The
Hello,
Below are the first two rows from my dataset and the header. This dataset
has 5749 rows and I want to select only certain rows to be used based on
existing grouping values. I am trying to group the data based on the values
under 'ex_bin'. (e.g a group for 250, 251, 252, 500, 501, 502) I
could this be an intentional attack to compromise a very popular
download, and infect thousands of people.what could be the
motivations...i hope its not some corporate thug here
What exactly does the Win32/Adclicker.JO trojan do ???
Ajay
www.decisionstats.com
www.iwannacrib.com
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