David Scott-6 wrote:
>
> I have been trying to read some data from an Excel workbook without
> success.
> ...
> > faults <- sqlFetch(channel, sqtable = 'Data',
> +colnames = FALSE, as.is = TRUE)
> > faults
> [1] "HY001 -1040 [Microsoft][ODBC Excel Driver] Too many fields de
But using the approproate tool, Sys.glob, whould be much simpler.
Note that 'pattern' in list.files is
- a regexp, and '.' is a special character in a regexp: Phil's
solution also needs to escape it or use fixed = TRUE
- it is documented to match file *names*, not file paths.
One of the autho
I have been trying to read some data from an Excel workbook without
success. The workbook is in .xls format and has multiple sheets, one
with the sheet name Data, which is the sheet I wish to read from. One
complication is that the header row of this sheet is comprised of
dropdown boxes.
I tr
Yes, there are in Europe. And there are summer classes in the US, as well.
And no, this list is not so much about helping beginners to learn R. For
that, there is a myriad of online sources. Rather, this list is for people
who have exhausted their ability to (elegantly) solve a problem.
Also, it s
On Monday, July 18, 2011 05:56:14 peter dalgaard wrote:
> but even this is dubious, since there is no year 0 AD. In Gregorian and
> Julian calendars, 1 BC continues directly into 1 AD.
>
Although this seems to be a widely recognized "problem," I would argue it is
an entirely specious one. It m
On Mon, Jul 18, 2011 at 10:48 AM, a.me...@yahoo.co.uk
wrote:
> Ok thank you Josh.
>
> Basically I have a matrix A with 7 rows and 18 columns.
If i < j (where i is the number of rows in your matrix and j is the
number of columns), then the determinant of the covariance (or
correlation) matrix |Sig
First, it would have helped if you had posted the actual results for us to
see how far they are off (and, more specifically, by which factor).
Second, given your epiphany, you will find that that's exactly what David
(and others before him) said or suggested. It is not about standardizing a
nomina
On 7/19/2011 4:04 PM, Bert Gunter wrote:
On Tue, Jul 19, 2011 at 3:45 PM, David Winsemius wrote:
On Jul 19, 2011, at 6:29 PM, J. wrote:
Thanks for the answer.
#
However, I am still curious about which result I should use? The result
from
R or the one from SPSS?
It
Thanks very much to everyone who replied. Peter got me on my way with
the use diag() hint, and I came with a less pretty version of Dan's
first option almost at the same time as I got that email. Seems I
can't avoid one for loop, but one is better than two.
Just as a note, with this code you hav
Thank you Rolf,
>> Using the analysis of co-variance example from MASS (fourth edition, p
>> 142), what is the correct notation for the formula "Gas, ~ Insul/Temp
> There shouldn't be a comma after ``Gas'' in that formula.
>> - 1"? Obviously, if we fit it as two separate models (as in the
>> exa
Dear all,
I've been trying on and off for the past few months to get SSOAP to work with
chemspider. First I tried the WSDL file:
cs<-processWSDL("http://www.chemspider.com/MassSpecAPI.asmx?WSDL";)
Error in parse(text = paste(txt, collapse = "\n")) :
:1:29: unexpected input
1: function(x, ...,
On 20/07/11 07:24, Carson Farmer wrote:
Dear list, I am currently writing up some of my R models in a more
formal sense for a paper, and I am having trouble with the notation.
Although this isn't really an 'R' question, it should help me to
understand a bit better what I am actually doing when fi
Hi:
Does this work for you?
mydiags <- function(mat) diag(mat[seq_len(ncol(mat)), ])
# Example:
set.seed(103)
u <- matrix(rpois(200, 10), ncol = 10)
# > dim(u)
# [1] 20 10
mydiags(u)
# [1] 7 12 6 13 12 6 5 6 14 6
u[1:10, ] # as a double check
HTH,
Dennis
On Tue, Jul 19, 2011 at 2:15
On Tue, Jul 19, 2011 at 5:20 PM, Dimitri Liakhovitski
wrote:
> Thanks a lot, Sarah.
> I assume, if the values against which I am comparing are REALLY zero
> ("0") - then even the first one (mean(testvec[testvec != 0])) should
> work, right?
> Dimitri
Well, yes. But what's "really" zero?
> ((.2 +
Use the cut() function to produce the interval categories and color
names in the data frame and then pass that variable to the col =
argument in the appropriate plot function. Something like
mydata$mycolors <- cut(mydata$value, c(-Inf, 1, 2, 3, Inf), label =
c('blue', 'green', 'yellow'))
Ask an a
> From: dwinsem...@comcast.net
> To: seoulseoulse...@gmail.com
> Date: Tue, 19 Jul 2011 18:45:47 -0400
> CC: r-help@r-project.org
> Subject: Re: [R] Different result of multiple regression in R and SPSS
>
>
> On Jul 19, 2011, at 6:29 PM, J. wrote:
>
> >
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Peter Lomas
> Sent: Tuesday, July 19, 2011 2:16 PM
> To: r-help@r-project.org
> Subject: [R] Taking all "complete" diagonals of a matrix
>
> Hi R-Help!
>
> I am trying to find a
On Tue, Jul 19, 2011 at 3:45 PM, David Winsemius wrote:
>
> On Jul 19, 2011, at 6:29 PM, J. wrote:
>
>> Thanks for the answer.
>>
#
>> However, I am still curious about which result I should use? The result
>> from
>> R or the one from SPSS?
>
> It is becoming apparent tha
Pei -
A file pattern can't contain a directory separator, but it's
easy to search for one outside the context of list.files. I think
grep('B/file2.txt',list.files(path = routeStr, all.files = TRUE,
full.names = TRUE, recursive = TRUE),value=TRUE)
should give y
On Jul 19, 2011, at 6:29 PM, J. wrote:
Thanks for the answer.
However, I am still curious about which result I should use? The
result from
R or the one from SPSS?
It is becoming apparent that you do not know how to use the results
from either system. The progress of science would be saf
Hi, all:
My folders are organized in such a way:
root
branch1
---A
---file1.txt
---file2.txt
---B
---file1.txt
---file2.txt
branch2
---A
---file1.txt
I don't think SPSS does anything with the variables you enter there.
Have you entered it as numeric?
Have you entered gender as numeric in R?
On Tue, Jul 19, 2011 at 6:11 PM, Bert Gunter wrote:
> Answer: Contrasts, i.e. the parameterization of the categorical variable(s)
> df.
>
> ?contrasts may
Thanks for the answer.
However, I am still curious about which result I should use? The result from
R or the one from SPSS?
Why the results from two programs are different?
Jay
--
View this message in context:
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Answer: Contrasts, i.e. the parameterization of the categorical variable(s) df.
?contrasts may be of some help, but you really need to do some
background studying of the linear models principles involved. Googling
may provide tutorials. Also searching the mail archives, e.g.:
https://stat.ethz.ch
Hi, I am trying to do a simple multiple regression analysis that has one
nominal variable (gender) and three numeric variables as independent
variables and one numeric variable as dependent variable.
So, I got a formula like this:
summary(out.3 <- lm(scale(DV) ~ gender + scale(IV.1) + scale(IV.2)
Hi R-Help!
I am trying to find a nicer way of extracting all the "complete" diagonals
of a matrix. I am working with very large matrices that have many more rows
than columns. I want to be able to extract each of the diagonals that are
as long as the number of columns in the matrix. I have writ
Hello,
I'm running mixed models in GAMM4 with 2 (non-nested) random intercepts and
I want to include a spline term for one of my exposure variables. However,
when I include a spline term, I always get reported degrees of freedom of
less than 1, even when I know that my spline is using more than 1
Hi All,
I am working on CGH datasets.
When I am running clustering, it shows me this error.
d <- dist(mydata, method = "euclidean")
Error in vector("double", length) :
cannot allocate vector of length 1010723280
If anyone can help me, I will really appreciate.
Thanks,
Kinnari
--
View this me
strptime("04-MAY-11 1428",format="%d-%b-%y %H%M")
[1] "2011-05-04 14:28:00"
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600
Sarah et. al:
On Tue, Jul 19, 2011 at 1:56 PM, Sarah Goslee wrote:
> In the more general case, that approach is prone to machine precision
> error (FAQ 7.31).
>
> Here's a clunky but safer alternative:
>
Perhaps ?zapsmall .
However, I would agree with your sentiments that it may depend on
conte
Hi everyone,
I was wondering if there was a simple way to assign a color to a cell
based on value in a data frame and return the cell as an image? For
example, if the value is >1, then blue, if between 1 and 2, green, if
between 2 and 3, yellow, etc.
I tried using a heatmap function but I wa
On Wed, Jul 20, 2011 at 5:42 AM, AO_Statistics wrote:
>
> Terry Therneau-2 wrote:
>>
>> This query of "why do SAS and S give different answers for Cox models"
>> comes
>> up every so often. The two most common reasons are that
>> a. they are using different options for the ties
>> b.
On Jul 19, 2011, at 4:18 PM, Peter Lomas wrote:
Hi Richard,
As others have said, try to use the "apply" functions rather than
loops.
There is also an apply function for lists, see ?lapply. This is
much more
efficient.
Actually the "apply" functions are not "more efficient" in the usual
I dare the conjecture that if you had written the code, you would know how to
do this. This suggests that you are asking us to do your homework, which is
not the purpose of this list. A simple inclusion of the code in a for or
while loop and storing the estimated parameters with the index of the
it
Thanks a lot, Sarah.
I assume, if the values against which I am comparing are REALLY zero
("0") - then even the first one (mean(testvec[testvec != 0])) should
work, right?
Dimitri
On Tue, Jul 19, 2011 at 4:56 PM, Sarah Goslee wrote:
> In the more general case, that approach is prone to machine pr
Tena koe Michael
The help file for strptime suggests you should be using %b (three letter month)
rather than %m (decimal number month).
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of mdkz...@aol.com
Please read the posting guide (requires a self-contained example of code) and
consult the help pages before posting. If you type ?predict.lm the help page
clearly states that the argument 'newdata' takes "[a]n optional data frame
in which to look for variables with which to predict..."
x1<-rnorm(1
I am a novice with network fuctions! I have been exploring the network
function in the statnet package, but haven't been able to figure out
how to hold vertices in position while varying edge features. Can
anyone advise on whether this is possible, and if so, how to do it?
Thanks!
--
Matthew Bakke
In the more general case, that approach is prone to machine precision
error (FAQ 7.31).
Here's a clunky but safer alternative:
> set.seed(1234)
> testvec <- sample(0:10, 100, replace=TRUE)
> mean(testvec)
[1] 4.31
> mean(testvec[testvec != 0])
[1] 4.842697
> mean(testvec[!sapply(testvec, function
You can do it by subsetting or indexing
r<-c(0,0,0,rnorm(10,10,5))
> mean(r)
[1] 8.052215
> mean(r[r!=0])
[1] 10.46788
Weidong Gu
On Tue, Jul 19, 2011 at 4:36 PM, Dimitri Liakhovitski
wrote:
> Sorry if it's been discussed before - don't seem to find it.
> I'd like to calculate a mean while ign
Dear R Experts:
I am trying to convert a date and time character field to timeDate where the
month is presented as three letters, such as "JUN" for June, etc.
This is an example of the full character field:
"04-MAY-11 1428"
What is the proper format syntax?
I've tried
timeDate("04-MAY
Sorry if it's been discussed before - don't seem to find it.
I'd like to calculate a mean while ignoring zeros.
"mean" doesn't seem to have an option for that.
Any other function/package that could do it?
Thanks for a pointer!
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
Thanks a lot, Roger!
Dimitri
On Tue, Jul 19, 2011 at 11:00 AM, Bos, Roger wrote:
> Yes. I have it running on a win server 2008 machine with no problems.
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Dimitri Liakhovitski
>
Hi Richard,
As others have said, try to use the "apply" functions rather than loops.
There is also an apply function for lists, see ?lapply. This is much more
efficient. I also like writing my own functions. For example:
f <- function(x) {
x^2
}
Which can then be used by:
> f(2)
[1] 4
Thi
On Jul 19, 2011, at 4:03 PM, RichardLang wrote:
Thanks for your advice!
I found another method to solve my problem (on page 20 of the manual
=) ).
Here's my code
In r-help most of the readers and most of the regular responders are
reading this as a mailing list posting not on Nabble
Thanks for your advice!
I found another method to solve my problem (on page 20 of the manual =) ).
Here's my code
# Set n and m to whatever you want
n = 25
m = 10
# Build a random vector (here with exponential distribution) with lenght of
n*m
x = rexp(n*m,1)
# Set up a factor in order to group
Dear list members,
I'm trying to use DCC-GARCH model to estimate the correlation. I have
downloeaded ccgarch packeage but can't understand some argument in the
formula.
dcc.estimation(inia, iniA, iniB, ini.dcc, dvar, model, method="BFGS",
gradient=1, message=1)
which is on R.Help
I understand oth
On Jul 19, 2011, at 2:43 PM, RichardLang wrote:
Hi everyone!
I'm trying to teach myself R in order to do some data analysis. I'm a
mathematics student and (only) familiar with matlab and latex. I'm
working
trough the "official" introduction to R at the moment, while
simultaneously
solving
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of RichardLang
> Sent: Tuesday, July 19, 2011 11:44 AM
> To: r-help@r-project.org
> Subject: [R] calculating the mean of a random matrix (by row) and some
> general questions
>
> Hi
Dear list, I am currently writing up some of my R models in a more
formal sense for a paper, and I am having trouble with the notation.
Although this isn't really an 'R' question, it should help me to
understand a bit better what I am actually doing when fitting my
models!
Using the analysis of co
Abraham Mathew thisorthat.com> writes:
>
> I'm trying to develop a stacked bar plot in R with ggplot2.
>
> My data:
>
> conv = c(10, 4.76, 17.14, 25, 26.47, 37.5, 20.83, 25.53, 32.5, 16.7, 27.33)
> click = c(20, 42, 35, 28, 34, 48, 48, 47, 40, 30, 30)
> date = c("July 7", "July 8", "July 9", "
Hello,
I have some data that exhibits a negative binomial distribution and also
spatial structure. I would like to create a model that accounts for both.
However, instead of locations, I have a distance matrix (cost matrix)
describing the spatial relationships among the locations. I have tried usi
Hi everyone!
I'm trying to teach myself R in order to do some data analysis. I'm a
mathematics student and (only) familiar with matlab and latex. I'm working
trough the "official" introduction to R at the moment, while simultaneously
solving some exercises I found in the web. Before I post my (pro
On Jul 19, 2011, at 2:36 PM, Jeff Newmiller wrote:
?predict
Use data.frame() to generate input vectors (newdata) for which you
want predicted values.
The OP probably needs to use expand.grid to generate the spanning
combinations of x values. He will in addition need to include values
f
?predict
Use data.frame() to generate input vectors (newdata) for which you want
predicted values.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Eng
People who participate in this list via email are unlikely to have
your example data, or even to have any idea what you are currently
referring to.
Please leave enough of the previous messages in your reply to the list
to provide context, and include all necessary information. Don't
assume that ev
On Tue, Jul 19, 2011 at 11:31 AM, kev946 wrote:
> I'm using this to plot the data with success, but am unable to figure out how
> to get the H:M:S timestamp included with their respective Dates. Any
> suggestions?
>
> xts(Dataset[,-1],as.Date(Dataset[,1],"%Y-%m-%d"))
>
This is not a plot command.
I'm trying to develop a stacked bar plot in R with ggplot2.
My data:
conv = c(10, 4.76, 17.14, 25, 26.47, 37.5, 20.83, 25.53, 32.5, 16.7, 27.33)
click = c(20, 42, 35, 28, 34, 48, 48, 47, 40, 30, 30)
date = c("July 7", "July 8", "July 9", "July 10", "July 11", "July 12",
"July 13",
"July 14", "Jul
Here is my model with interaction terms and control variables (I changed
variables names for easy read):
reg1 <- lm(y ~ x1*x2*x3 +control1 + control2 + control3)
x1 ranges from 0 to 6; x2 from 0 to 5; and x3 from 0 to 4. All three are
discrete ordinal variables; but I will treat them as continu
I'm using this to plot the data with success, but am unable to figure out how
to get the H:M:S timestamp included with their respective Dates. Any
suggestions?
xts(Dataset[,-1],as.Date(Dataset[,1],"%Y-%m-%d"))
--
View this message in context:
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I had the same problem with the code of the Wikipedia in a 64-bit Windows 7.
The gif works fine if I use the executable located in bin\x64
instead of bin, which produces the ugly gif
eg, in a cmd: \bin\x64\Rscript.exe example.r
I realised the solution using the --verbose flag
Sorry about my eng
In my first post is example data.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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https://stat.ethz.ch/
Terry Therneau-2 wrote:
>
> This query of "why do SAS and S give different answers for Cox models"
> comes
> up every so often. The two most common reasons are that
> a. they are using different options for the ties
> b. the SAS and S data sets are slightly different.
> You have bot
On Jul 19, 2011, at 11:58 AM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
] On Behalf Of Daniel Malter
Sent: Tuesday, July 19, 2011 1:51 AM
To: r-help@r-project.org
Subject: Re: [R] Centering data frame by factor
On 7/18/2011 9:23 PM, Sigrid wrote:
Hi
I apologize for not providing reproducible codes more clearly, and I hope
this will be more understandable.
I have 14 lines (7 per facet that I would like to add). I will provide you
with six of the lines from the data as that should enough data to work
wi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Daniel Malter
> Sent: Tuesday, July 19, 2011 1:51 AM
> To: r-help@r-project.org
> Subject: Re: [R] Centering data frame by factor
>
>
> P1-tapply(P1,Experiment,mean)[Experiment
If I understand you correctly,
> I would like to export the esvr.pred object to a file so that I can
> draw a graph of it against my original data in other software that I'm
> using.
>
you cannot do this. You can export **data**, but of course any R
"object" is either a binary or text (via dput)
I also noticed, after sending the code that the modification does only work
when the legend is positioned at the bottom of the figure region , and the
function crashes when no title is provided.
The modification I applied to the code was intended to be more the solving of a
particular problem th
Hi,
I'm using R 2.12.0 on Windows XP.
I've used the e1071 package to tune a Support Vector Regression object
and I've created the SVR object:
> epsilon.svr <- svm(C8R004 ~.,data = rain_flow.train, scale = T, type =
> "eps-regression",
+ kernel = "radial", cost = 0.9, epsilon=0.55,tolerance=0.00
As Sarah requested, could you at least read the posting guide and provide us
with some sample data?
--
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A. T. Still University of Health Sciences
800 W. Jefferson St
creamers rdeft.nhs.uk> writes:
>
> Thanks David...I am trying to plot out data for various consultants by
> specialty - each specialty has a varying number of consultants - each
> consultant a varying number of data pointsI found direct access of the
> elements of the dataframe was the only
On Jul 19, 2011, at 10:38 AM, Duncan Murdoch wrote:
On 11-07-19 8:16 AM, loic wrote:
As suggested by David, I applied some modifications to the code of
the legend
function so that the legend box size adapts to the number of line
jumps in
the legend title.
Attached the modified code of the
Dear Help-list, I have solved the problem by simply deleting the erroneous "0"
in the "CR core - CR EC" contrast and deleting the unnecessary command
"test=adjusted(summarytype = "single-step")". Regards,B. Jessop
> From: deel...@hotmail.com
> To: r-help@r-project.org
> Date: Sun, 17 Jul 2011
Thanks David...I am trying to plot out data for various consultants by
specialty - each specialty has a varying number of consultants - each
consultant a varying number of data pointsI found direct access of the
elements of the dataframe was the only way to plot this type of variation,
otherwis
On 11-07-19 9:42 AM, John Minter wrote:
Duncan, thanks for your work. I could not get the workaround you suggested
Rterm.exe --no-restore --slave -e utils::Sweave("file.Rnw")
to work under 2.13.1 and so gave up and installed the 2.14.0 development
build. I can verify that
R CMD Sweave file.Rnw
On 11-07-19 8:16 AM, loic wrote:
As suggested by David, I applied some modifications to the code of the legend
function so that the legend box size adapts to the number of line jumps in
the legend title.
Attached the modified code of the function.
No code was attached.
In case you didn't, cou
Duncan, thanks for your work. I could not get the workaround you suggested
Rterm.exe --no-restore --slave -e utils::Sweave("file.Rnw")
to work under 2.13.1 and so gave up and installed the 2.14.0 development
build. I can verify that
R CMD Sweave file.Rnw
works properly on the 2.14.0 development
On Jul 19, 2011, at 4:50 AM, Daniel Malter wrote:
P1-tapply(P1,Experiment,mean)[Experiment]
Another way would be with ave(), but I discovered that it does not
accept subsidiary arguments and does not issue warnings either, so
this works:
> with(dfrm, ave(P1, Experiment, FUN=function(x
Grand merci! Will try!
Kind regards,
Alex
2011/7/19 Paul Hiemstra :
> e you
__
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 commen
On Tue, Jul 19, 2011 at 12:30 AM, Joshua Wiley wrote:
[snip] I guess that I must have a data frame to plot a histogram.
>>>
>>> Not at all!
>>>
>>> ## a *vector* of 100 million observation
>>> x <- rnorm(10^8)
>>> ## a histogram for it (see attached for the result from my system)
>>> hist(x)
On Jul 19, 2011, at 5:40 AM, creamers wrote:
Hi.I am relatively new to R but was quite pleased with myself at
having
generated a series of lattice plots as PDFs. I was very surprised when
plotting these out as jpegs (or png or tiff) that the strip title
information above each lattice plot
Perfect! Made my day!
--
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Sent from the R help mailing list archive at Nabble.com.
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https://stat.ethz.c
As suggested by David, I applied some modifications to the code of the legend
function so that the legend box size adapts to the number of line jumps in
the legend title.
Attached the modified code of the function.
Thanks to David
Regards,
Loïc
Wageningen University
http://r.789695.n4.nabble.c
Hi.I am relatively new to R but was quite pleased with myself at having
generated a series of lattice plots as PDFs. I was very surprised when
plotting these out as jpegs (or png or tiff) that the strip title
information above each lattice plot vanished. The pdf was fine. Has anybody
any ideas?
On 11-07-18 2:16 PM, Nipesh Bajaj wrote:
Hi all, I am trying to understand the R's "environment" concept
however the underlying help files look quite technical to me. Can
experts here provide me some more intuitive ideas behind this concept
like, why it is there, what exactly it is doing in R's a
On 07/19/2011 09:57 AM, Juan Carlos Borrás wrote:
> Ideally you'd have the next two items available:
> - tests that ensure that your code carries out what it should and as it
> should.
> - a coverage analysis tool that reports what parts of your code have
> been and have not been executed by your
Ideally you'd have the next two items available:
- tests that ensure that your code carries out what it should and as it should.
- a coverage analysis tool that reports what parts of your code have
been and have not been executed by your tests above.
Neither of those are mandatory though, but they
On 07/19/2011 04:40 AM, Ana-Maria Pistea wrote:
> >>> B
Hi Ana-Maria,
A quick google for you error message shows that there can be quite a
number of causes for this problem. Therefore it is impossible for us to
help you. Please read the R-help posting guide to improve your question
[1]. The most
Some hints:
list.files() will return the list of files in a directory
readLines() will allow you to load text files as vectors of lines
strsplit() will allow you to break lines into words
c(x,y) concatenates vectors x and y ; x <- c(x,y) appends vector y to x
unique() will allow you to get rid of r
On 2011-07-19 01:27, psombe wrote:
Well yeah it works fine for small data but when i tried the exact same
command with a large data set (abt 167 rows and 4000 columns) it gave me a
different data frame.
either i get the first column as row names and so when i put data[1,1] i
get the the first r
Well yeah it works fine for small data but when i tried the exact same
command with a large data set (abt 167 rows and 4000 columns) it gave me a
different data frame.
either i get the first column as row names and so when i put data[1,1] i
get the the first row second column data (from the origin
Hi,
I would like to center P1 and P2 of the following data frame by the factor
"Experiment", i.e. substruct from each value the average of its experiment,
and keep the original data structure, i.e. the experiment and the group of
each value.
RAW=
data.frame("Experiment"=c(2,2,2,1,1,1),"Group"=c(
Hello everyone,
I'm doing some JGR (a gui frontend for R) development, specifically adding
functionality from tm. In order to enable users to select some text files from
a file dialog, and turn them into a corpus, I need to be able to generate a
corpus using a *SINGLE* text file as a single do
but even this is dubious, since there is no year 0 AD. In Gregorian
and Julian calendars, 1 BC continues directly into 1 AD.
True, but these days we are ruled by ISO 8601:2004, which does define
a year 0 (the year before 1CE aka 1AD). See
http://en.wikipedia.org/wiki/0_(year) .
It seems also
P1-tapply(P1,Experiment,mean)[Experiment]
HTH,
Daniel
ronny wrote:
>
> Hi,
>
> I would like to center P1 and P2 of the following data frame by the factor
> "Experiment", i.e. substruct from each value the average of its
> experiment, and keep the original data structure, i.e. the experiment a
Very Sorry for the instinctive bad comment... I didn't express
correctly what I meant.
I meant that I would have never expected such behaviour, by default
(but may be I am wrong and most of the people need quoting, and hence
has sense to put it by default)
Problem solved,
Philipp guessed right.
I
Can you explain a little more?
I have created a small CSV file following your pattern which looks like this in
a text editor:
A,B,C,D,E
65,68,71,74,77
67,71,75,79,83
69,73,77,81,85
71,77,83,89,95
When I load it into R with
> x <- read.csv( "a.csv" )
I get this which I think is what you would e
I am very happy using the R-Sweave plugin that works with WinEdt 6.0. It
understands not only R but, as the name suggests, Sweave as well.
http://www.winedt.org/Config/modes/R-Sweave.php
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Hi Jim,
Thanks for the reply. But i was looking for percentages in venn diagram.
Regard's
Poornima
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Hi,
I'm a new R user and I'm having trouble with the read.csv command. It
somehow treats the first column as a row name field even though it's not a
row name. there are no missing columns/entries and i'm not sure how to
resolve this.
the format of my data is
A, B, C, D,..(3984 columns)
12,
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