OK, I thought I was being explicit enough, but here goes. The out of the first
five rows of mu dataframe is:
Time hour lev.morgan lev.lock2 lev.lock1 flow direction
velocity bearingvelcat
1 2009-07-06 15:00:00 15 3.266 3.274 3.240 1710.6180.28
4.352
In virtually every case where people think they need gobs of similarly-named
variables, what they really need are vectors or data frames. I strongly
recommend reading the Introduction to R document again to learn how to create
them and access individual elements and subsets of elements.Data fram
Better moral: read the basic documentation. You missed matrix
indexing (indexing an array with a matrix subscript), and that is
described in 'An Introduction to R', section 5.3
E.g.
s$p <- match(s$prefix, colnames(y))
y[cbind(s$idx, s$p)] <- s$val
(You are trying to use different types of in
example:
> y=(rbind(c(TRUE,TRUE,TRUE),c(
FALSE,FALSE,FALSE)))
> y
[,1] [,2] [,3]
[1,] TRUE TRUE TRUE
[2,] FALSE FALSE FALSE
> as.numeric(y)
[1] 1 0 1 0 1 0
I am trying to make some important matrixes become nuemric (1 or 0) but they
change their dimensions.. anyone know how to easily fi
Hello All,
?environmentName shows how to get the environment name. But I don't
see how to set the name in the first place. Could you tell me where to
look for the function that set the environment name?
--
Tom
__
R-help@r-project.org mailing list
http
Hi All,
The last line if the following code returns the error right below this
paragraph. Essentially, I use the operator %:% to retrieve a variable
in a nested frame. Then I want to use the same operator (with '<-') to
change its value. I'm not sure if this is possible in R with %:%
operator. Wo
Dear All,
I want to connnect R with oracle. I am using RODBC package for this
connection. Following is my code to connect
library(RODBC)
channel <- odbcConnect("Vikrant")
I am getting following error.
1: In odbcDriverConnect("DSN=Vikrant") :
[RODBC] ERROR: state NA000, code 12288, message [M
example:
> y=(rbind(c(TRUE,TRUE,TRUE),c(FALSE,FALSE,FALSE)))
> y
[,1] [,2] [,3]
[1,] TRUE TRUE TRUE
[2,] FALSE FALSE FALSE
> as.numeric(y)
[1] 1 0 1 0 1 0
I am trying to make some important matrixes become nuemric (1 or 0) but they
change their dimensions.. anyone know how to easily fix
Check out your timezone. Did something change on that day?
You omitted the time, so it is trying to make sense of
"1982-01-01 00:00"
According to
http://en.wikipedia.org/wiki/Singapore_Standard_Time
that time did not exist in that time zone (which I am guessing is the
one your computer is se
Some time ago someone asked for things that make R difficult to
master. Here I want to record one R behavior that took me off-guard
yesterday. Moral of the story: vectorialize, but don't exaggerate.
Hope it helps
mario
### A very simple data frame
tc <- textConn
On Wed, 14 Jul 2010, Meissner, Tony (DFW) wrote:
I have a dataframe that contains time values in the form of
-mm-dd hh:mm i.e. 2010-07-14 13:00. When I convert this to
numeric via tvec <- as.numeric(Time) I get a number that is in
seconds. So far so good. When I then divide the numeric
I have a dataframe that contains time values in the form of -mm-dd hh:mm
i.e. 2010-07-14 13:00. When I convert this to numeric via tvec <-
as.numeric(Time) I get a number that is in seconds. So far so good. When I
then divide the numeric value by the number of seconds in a day eg
tvec/(
Hi,
I am trying to call a C function, that I wrote to parse a flat file, into
R. The argument that will go into this function is an input file that I need
to parse and write the desired output in an output file. I used some hit
and trial approach but i keep on getting the "file not found" or
"seg
Greetings to all --
As always, I am grateful for the help of the R community. The generosity of
other R users never ceases to impress me. Just to recap the responses I
have had to this question, in case they can help anyone else down the line:
Robert LaBudde suggested Applied Survey Data Analys
I have a set of labels arranged along a timeframe in a. Each label has
a timestamp and marks a state until the next label. The dataframe a
contains 5 such timestamps and 5 associated labels. This means, on a
continious scale between 1-100, there are 5 markers. E.g. 'abc' marks
the timestampls betwe
Hi all,
Let's say we have M x N matrix, which represents N time series, each having
M observations in order of time.
How do we find maximal number of linear combinations of these N time series,
their mutual correlation has to be less than certain pre-specified
constraints.
That's to say, we woul
It looks like you want to use a matrix, then you can do proper indexing.
Creating the names will just lead to unnecessary complexity and confusion.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
>
Hi all,
I'm trying out R's optimization functions (nlm() and optim()) in order
to obtain parameter estimates for a combination of distributions. I've got
my likelihood function, and optim() is happy to provide estimates (nlm()
less so), but I'm concerned that uneven parameter scaling may be
To follow up: if you tried to join this SIG but couldn't, please try again.
There was a configuration error but it has been corrected. The link to join is
below. Cheers!
-Original Message-
From: John Williams [mailto:john.willi...@otago.ac.nz]
Sent: Thursday, July 08, 2010 5:55 PM
To:
Hello,
I am a very new R user and not a statistician so please excuse any over
explanation, I'm just trying to be as clear as possible.
I have performed a cross correlation of two time series (my columns) in a
single
data setusing:
ccf(ts(A[rows,columnX]),(A[rows,columnY]), lag=NULL, type="cor
Hi,
I have a question about the parameter C (cost) in svm function in e1071. I
thought larger C is prone to overfitting than smaller C, and hence leads to
more support vectors. However, using the Wisconsin breast cancer example on
the link:
http://planatscher.net/svmtut/svmtut.html
I found that th
Oops, sorry! Here it is:
R version 2.11.1 (2010-05-31)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_Singapore.1252 LC_CTYPE=English_Singapore.1252
LC_MONETARY=English_Singapore.1252 LC_NUMERIC=C
[5] LC_TIME=English_Singapore.1252
attached base packages:
[1] grDevices datasets splines graph
Do you care to share your sessionInfo() , as the Posting Guide asks?
I cannot reproduce on:
R version 2.11.1 (2010-05-31)
i486-pc-linux-gnu
On 07/13/2010 09:07 PM, Ian Seow wrote:
Hi, I'm encountering a strange error in POSIXlt... anyone got a clue on
this?
as.POSIXlt("1982-01-01")
Error i
Hi, I'm encountering a strange error in POSIXlt... anyone got a clue on
this?
> as.POSIXlt("1982-01-01")
Error in as.POSIXlt.character("1982-01-01") :
character string is not in a standard unambiguous format
> as.POSIXlt("1992-01-01")
[1] "1992-01-01"
> as.POSIXlt("1972-01-01")
[1] "1972-01-01"
Hi, I am encountering a strange error in POSIXlt... anyone got a clue?
> as.POSIXlt("1982-01-01")
Error in as.POSIXlt.character("1982-01-01") :
character string is not in a standard unambiguous format
> as.POSIXlt("1992-01-01")
[1] "1992-01-01"
> as.POSIXlt("1972-01-01")
[1] "1972-01-01"
> as.PO
Some of us enter these things with a keyboard! (Don't know what the
rest of you do.) That takes time, and command completion won't always
pick out the right one immediately.
On 07/13/10 18:44, Bert Gunter wrote:
> Rich:
>
> I suspect it's a cultural relic of the "storage is scarce" era that beg
Rich:
I suspect it's a cultural relic of the "storage is scarce" era that begat
"C" and "R" in the first place. Cutesiness. Culture is hard to change, but I
think you make a very salient point!
-- Bert
Bert Gunter
Genentech Nonclinical Statistics
> -Original Message-
> From: r-help-bou
this isn't a whole lot different, but if x is a data.frame (and not a
matrix), you could also try this:
names(x)[which(names(x)=="oldname")]= "newname"
--
View this message in context:
http://r.789695.n4.nabble.com/Rename-column-or-row-names-tp2288122p2288183.html
Sent from the R help mailing
Here is another way of going about it (presuming we are correct about what
you are after):
i=1:10
j=1:10
unlist(lapply(i, function(x) paste("X", x, j, sep="_")))
Tyler
--
View this message in context:
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Sent from
I have a series of animals that are being radio-tracked. I have written
some code that goes through each animal and calculates the total distance
that each animal is traveling. I will need to run these data on a regular
basis and want to write the result to a .php webpage file so that the page
up
Henrique is correct; entering duplicated(df) will return an index of TRUE or
FALSE for every row. TRUE indicates a duplicated row.
df[duplicated(df),] # shows which rows are repeated
df[-duplicated(df),] # shows which rows are unique
--
View this message in context:
http://r.789695.n4.nabble.
Suppose I want create variables with indexes in their names, e.g., X_1_1,
X_1_2, X_1_3, ..., X_1_10, X_2_1, X_2_2, X_2_3, .. X_2_10,..., X_10_1,
X_10_2, ... X_10_10. It looks like I need to use 2 indexes I and J so I is
looped from 1 to 10, and J is looped from 1 to 10. But I don't know how to
See ?duplicated
On Tue, Jul 13, 2010 at 7:42 PM, david hilton shanabrook <
davidshanabr...@me.com> wrote:
> I wrote something to check for duplicate rows in a data frame, but it is
> too inefficient. Is there a way to do this without the nested loops?
>
> This code correctly indicates rows 1-7,
I have been examining the Mann-Whitney test closely. And there are two
features of the R implementation that puzzles me. The test statistic is
reported as "W" and depends on the order of the arguments to the function.
> x <- c(1,3,5,7,9)
> y <- x-1
> x
[1] 1 3 5 7 9
> y
[1] 0 2 4 6 8
> wilcox.t
Maybe I misunderstand your problem. For my understanding, it's quite
simple.I hope it can help.
id <- vector()
for(i in 1:10){
for(j in 1:10){
id<-append(id,paste("X",i,j,sep="_"))
}
}
Wenbo Mu
On Tue, Jul 13, 2010 at 5:44 PM, He, Yulei wrote:
> Hi, there:
>
Hi everyone,
I have another "Random Forest" package question:
- my (presumably incorrect) understanding of the varImpPlot is that it
should plot the "% increase in MSE" and "IncNodePurity" exactly as can be
found from the "importance" section of the model results.
- However, th
Hello All,
Suppose that x is a data.frame. I want to change the colname 'oldname'
to 'newname'. I have the following code. Could anybody let me know if
there is any better way to change a column name?
colnames(x)[grep('oldname', colnames(x))]='newname'
--
Tom
__
I wrote something to check for duplicate rows in a data frame, but it is too
inefficient. Is there a way to do this without the nested loops?
This code correctly indicates rows 1-7, 1-8, 2-9 and 7-8 are duplicates.
> m <- matrix(c(1,1,1,1,1, 2,2,2,2,2, 6,6,6,6,6, 3,3,3,3,3, 4,4,4,4,4,
> 5,5,
Hi, there:
Suppose I want create variables with indexes in their names, e.g., X_1_1,
X_1_2, X_1_3, ..., X_1_10, X_2_1, X_2_2, X_2_3, .. X_2_10,..., X_10_1, X_10_2,
... X_10_10. It looks like I need to use 2 indexes I and J so I is looped from
1 to 10, and J is looped from 1 to 10. But I don't
Dear all R-users,
When I install the package "lme4" in R-2.11.1, and use it to run the
simulation on the Linux system, it appears the following problem.
Attaching package: 'lme4'
The following object(s) are masked from 'package:stats':
   AIC
Error in names(argNew)[1] <- names(formals(new))
Hi,
I want to use the neural network package AMORE and I don't find in the
documentation the weight decay option.
Could someone tell if it is possible to add a regularization parameter (also
known as a weight decay) to the training method.
Is it possible to alter the gradient descent rule for
I've got a couple of more changes that I want to make to my plot, and I
can't figure things out. Thanks for all the help.
I'm using this R script
library(ggplot2)
library(lattice)
# Generate 50 data sets of size 100 and assign them to a list object
low <- 1
n <- 50
#Load data from file
for(i in
On 13-Jul-10 22:21:29, Melissa Peters wrote:
> Hello,
>
> 1. I have a number of snow depth measurements (centimeters depth) that
> were acquired in a row (linear transect).
>
> A: I am trying to plot the probability density function (PDF) of these
> points. When I use the "dnorm" command and lin
On Tue, Jul 13, 2010 at 4:21 PM, eugen pircalabelu
wrote:
> Hello R-users,
> Checking the archives, I recently came across this topic:
> "export tables to Excel files"
> (http://r.789695.n4.nabble.com/export-tables-to-Excel-files-td1565679.html#a1565679),
> and the following interesting reference
Hello,
1. I have a number of snow depth measurements (centimeters depth) that were
acquired in a row (linear transect).
A: I am trying to plot the probability density function (PDF) of these points.
When I use the "dnorm" command and line type = points, the PDF is displayed
correctly. When I
If I am correctly understanding what Eugen is trying to do, WriteXLS() won't
get him there. WriteXLS() will enable you to label/name the worksheets (tabs)
but not allow you to precede the actual data frame rows and columns on the
sheet with a title or label.
I suspect that you may have to look
Check the WriteXLS package, I think it does that and also saves
each R object on a different excel sheet.
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
- Original Message
> From: eugen pircalabelu
> To: R-help
Although there are a number of documents describing library development, in
which compiling and using C or FORTRAN source code is discussed, such
documents are too detailed and complicated for my limited needs (i.e. I
can't follow them!). I am looking for a worked example (i.e. the series of
comm
Hi:
My head is spinning with this latex doc so hopefully after I align my tables to
the left of the page
my headache are going to be over. I always use:
\hspace*{-0.1in}
to move my figures horizontally to the left margin of the page but the table
below doesn't move at all,
but instead it gets sid
It's not really clear what the structure of your data is, nor how exactly you
are sampling it (you need to provide an example), but you could try adding
this to the end of your sample command (if the result is a data.frame or
matrix): [,-1]
Tyler
--
View this message in context:
http://r.789695
On Tue, Jul 13, 2010 at 3:58 PM, Megh Dal wrote:
> Dear all,
>
> Please forgive me if there is a duplicate post; my previous mail perhaps
> didnt reach the list...
>
> Let say I have following time series
>
> library(zoo)
>> dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1)
For example: When I take a sample of 5 from 10 columns, I want to exclude
from the sample column 1 from being chosen. How do I do this?
--
View this message in context:
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Sent from the R h
http://r.789695.n4.nabble.com/file/n2287976/BBMM_BUG.RData BBMM_BUG.RData
I've been constructing utilization distributions for 63 animal movement
trajectories using a Brownian bridge movement model with BBMM. I started
noticing after running the routine numerous times at different resolutions
t
which(!is.na(dat1))
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://s
Hello R-users,
Checking the archives, I recently came across this topic:
"export tables to Excel files"
(http://r.789695.n4.nabble.com/export-tables-to-Excel-files-td1565679.html#a1565679),
and the following interesting references have been proposed:
http://rwiki.sciviews.org/doku.php?id=tips:da
Megh,
I don't know whether this is the best way, but it works:
> seq(1,length(dat1))[!is.na(dat1)]
[1] 1 2 4 5 6 9 10
Jonathan
On Tue, Jul 13, 2010 at 1:58 PM, Megh Dal wrote:
> Dear all,
>
> Please forgive me if there is a duplicate post; my previous mail perhaps
> didnt reach the lis
Dear all,
Please forgive me if there is a duplicate post; my previous mail perhaps didnt
reach the list...
Let say I have following time series
library(zoo)
> dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1)
> dat1[c(3, 7,8)] = NA
> dat1
2010-01-01 2010-01-02 2010-01-03
Raubertas, Richard wrote:
I agree that 'list' is a terrible package name, but only secondarily
because it is a data type. The primary problem is that it is so generic
as to be almost totally uninformative about what the package does.
For some reason package writers seem to prefer maximall
Dear all,
Apologies if this question is bit theoretical and for the longish email.
I am meta-analyzing the coefficients and standard errors from multiple
studies where the raw data is not available.
Each study analyst runs a model that includes an interaction term for,
say, between sex and s
Hello!
I am trying to scale the outliers in a boxplot. I am passing "pars =
list(boxwex=0.1, staplewex=0.1, outwex=0.1)" to the boxplot command. The
boxes are scaled correctly, but the circles (outliers) are not scaled at
all, and thus pretty big compared to the boxes scaled with 0.1.
Am I m
On Tue, Jul 13, 2010 at 9:54 AM, Aaditya Nanduri
wrote:
> Guys, I wrote to the finance mailing list earlier with my questions but was
> directed here.
>
> Sorry for the repeat.
>
> ---
> library(quantmod)
>
> now <- Sys.time()
>
> midnight <- strptime() # < I want to ma
My solution was based on using vectors (which were your original example), now
you are using data frames. The actual result is NA, then you just print
content again (which my code never modified) so you are going to see the full
content data frame.
Try:
content[na.omit(pmatch(searchset$signat
Hi everybody
I am trying to install RJSONIO from source in on Mac OS X 10.5.8. I used the
Package Installer.
The message and sessionInfo is attached below
Can someone help me to understand the error message and maybe give hint
towards solving the problem
thanks in advance
Christiaan
Message:
T
When running the combined code with your suggested line:
content <- data.frame(urls=c(
"http://www.google.com/search?source=ig&hl=en&rlz=&=&q=stuff&aq=f&aqi=g10&aql=&oq=&gs_rfai=CrrIS3VU8TJqcMJHuzASm9qyBBgAAAKoEBU_QsmVh";,
Using this dataframe with quite long column headers, how can I wrap the
text so that the columns are narrower. I was trying to use strwrap without
success. Thanks
reportDF <- structure(list(IDDate = c("3/12/2010", "3/13/2010", "3/14/2010",
"3/15/2010"), FirstRunoftheYear = c("33 (119 ? 119)",
I agree that 'list' is a terrible package name, but only secondarily
because it is a data type. The primary problem is that it is so generic
as to be almost totally uninformative about what the package does.
For some reason package writers seem to prefer maximally uninformative
names for the
You can't try this:
sapply(names(reportDF), toString, width = 10)
abbreviate(names(reportDF))
On Tue, Jul 13, 2010 at 2:43 PM, Felipe Carrillo
wrote:
> Hi:
> Using this dataframe with quite long column headers, how can I wrap the
> text so that the columns are narrower. I was trying to use strw
What is the original intent? The bandwidth:productivity ratio is not
looking encouraging for this problem.
Frank
On 07/13/2010 12:38 PM, schuster wrote:
Hello,
are you trying to pase SAS code (or lightly modified SAS code) and run it in R?
Then you are right: the hard part is parsing the c
N_runs -1 seems a bit of an odd df to choose to calculate the CI for a
mean. To answer your question, I think that t.test() is the easiest
way to get a CI in R. That said, you can use the MS_residuals from
ANOVA to take advantage of variance calculated on groups and pooled.
Something like:
foo <
Hi:
Using this dataframe with quite long column headers, how can I wrap the
text so that the columns are narrower. I was trying to use strwrap without
success. Thanks
reportDF <- structure(list(IDDate = c("3/12/2010", "3/13/2010", "3/14/2010",
"3/15/2010"), FirstRunoftheYear = c("33 (119 ? 119)"
Hello,
are you trying to pase SAS code (or lightly modified SAS code) and run it in R?
Then you are right: the hard part is parsing the code. I don't believe that's
possible without a custom parser, and even then it's really hard to parse all
the SAS "sub languages" right: data step, macro c
Thank you very much, Gabor!
Dimitri
On Tue, Jul 13, 2010 at 12:25 PM, Gabor Grothendieck
wrote:
> On Tue, Jul 13, 2010 at 11:19 AM, Dimitri Liakhovitski
> wrote:
>> Actually,
>> I realized that my task was a bit more complicated as I have different
>> (let's call them) Markets and the dates repe
i try to show the result of the cluster-analysis (hclust, method=ward) in a
table with following information
first column: height
second column: number of clusters
third column: clustering information
0,041 | 20 | (3)-(5)
0,111 | 19 | (6)-(11)
0,211 | 18 |
Paul wrote:
I am trying to recreate an analysis that has been done by another group
(in SAS I believe). I'm stuck on one part, I think because my stats
knowledge is lacking, and while it's OT, I'm hoping someone here can help.
Given this dataframe;
Well, that will teach me to read the quest
I write about R every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month
of particular interest to readers of r-help.
So, in case you missed them, here are some articles related to R from
the month of June:
ht
Guys, I wrote to the finance mailing list earlier with my questions but was
directed here.
Sorry for the repeat.
---
library(quantmod)
now <- Sys.time()
midnight <- strptime()# < I want to make this a static variable
that will be equal to 12:00:00 am but I dont know
On Tue, Jul 13, 2010 at 11:19 AM, Dimitri Liakhovitski
wrote:
> Actually,
> I realized that my task was a bit more complicated as I have different
> (let's call them) Markets and the dates repeat themselves across
> markets. And the original code from Gabor gives an error - because
> dates repeate
Create a shortcut that targets your batch file and edit its properties to open
minimized, and call the shortcut from R rather than the calling the batch file
directly.
"Gushta, Matthew" wrote:
>R list-
>i have begun using the "MplusAutomation" while piloting a large-scale
>simulation (~200,00
For one definition of random:
ss <- rexp(100)
ss <- ss/sum(ss)
ss <- 5 + round( ss*9500 )
cnt <- 0
while( ( d <- sum(ss) - 1 ) != 0 ) {
tmpid <- sample.int(100,1)
ss[tmpid] <- ss[tmpid] - d
ss[ ss > 500 ] <- 500
ss[ ss < 5 ] <- 5
cnt <- cnt
...
> > Aug;235(2):E97-102). The paper is freely available. The idea is to fit
> > the curves individually and obtain the residual sum of squares and
> > then fit the curves altogether somehow constraining some parameters
> > and then you have another residual sum of squares. Then you can do a
> >
content[na.omit(pmatch(searchset, content,,TRUE))]
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Ralf B
> S
There are several functions in several packages for plotting intervals that
will give you plots much better than the excel one. The RSiteSearch function
or the sos package may help you find those.
But it is also easy to create such plots using just a few lines of R code and
base graphics. Re
Hi Felipe,
See in line below.
On Tue, Jul 13, 2010 at 11:04 AM, Felipe Carrillo
wrote:
> Thanks Izta:
> I see your point, then I should extract the column names when the
> dataset is first read because is a dataframe:
That might work, but it's definitely not how I would do it.
> report <- stru
Good idea Romain, there is quite a bit of type testing in the function
versions of STRING_ELT and CHAR, not to mention the function call
overhead. Since the types are checked explicitly, I believe this
function is safe. All together now...
> system.time(strings[-1] == strings[-1e5])
user syste
If the curves are sufficiently close to sine (cosine) curves and you know the
period, then this can be restructured as a linear model and you can avoid all
the complexities that come with non-linear models. Further, from your
description, it does not sound like you really gain much from using t
Is it possible to download data from password-protected ftp sites? I saw
another thread with instructions for uploading files using RCurl, but I
could not find information for downloading them in the RCurl documentation.
Did you try the ?getURL function in RCurl? See the `Test the
password
Actually,
I realized that my task was a bit more complicated as I have different
(let's call them) Markets and the dates repeat themselves across
markets. And the original code from Gabor gives an error - because
dates repeate themselves and apparently zoo cannot handle it. So, I
had to do program
"You could try rlm in the MASS package; it doesn't use he resampling
step."
That seems to do the trick. Thank you!
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Dear Clay,
dear list,
I face the same problem when rounding POSIXct objects. Have you (or has
anybody) found an explanation meanwhile, or a way to work around this issue?
Example:
opt<-options(digits.secs=3)
ts1<-as.POSIXct(c("2006-11-01 09:00:00.03", "2006-11-01 09:00:01",
"2006-11-01 09:00:0
Dear nuncio
my internet is connected properly, i am running yast as a superuser , i am
getting the following error
>
>
> *Problem* :
>
> Cannot access installation media
>
> http://download.opensuse.org/repositories/devel:languages:R:patched/openSUSE_11.2
> (Medium 1).
> Check whether the ser
R list-
i have begun using the "MplusAutomation" while piloting a large-scale
simulation (~200,000 replications). since the package takes advantage of the
DOS batch mode available in Mplus, each replication starts and activates a new
instance of a command prompt window. this effectively locks me
I think the easiest way is from calling anova() on your aov class
object. For instance
y <- 1:10
x <- runif(10)
my.aov <- aov(y ~ x)
anova(my.aov)["Residuals", "Sum Sq"]
anova(my.aov)["x", "Pr(>F)"]
You can also extract these values from a call to summary(my.aov), but
that output is a list (even
The high-level concept you need is called "Regular Expressions". R
supports these through several functions, see ?regex .
Ralf B wrote:
Hi all,
I would like to detect all strings in the vector 'content' that
contain the strings from the vector 'search'. Here a code example:
content <- data.f
Thanks Izta:
I see your point, then I should extract the column names when the
dataset is first read because is a dataframe:
report <- structure(list(Date = c("3/12/2010", "3/13/2010", "3/14/2010",
"3/15/2010"), Run1 = c("33 (119 ? 119)", "n (0 ? 0)", "893 (110 ? 146)",
"140 (111 ? 150)"), Run2
On Jul 13, 2010, at 10:35 AM, Luis Borda de Agua wrote:
Hi,
I have used the instruction aov in the following manner:
res <- aov(qwe ~ asd)
when I typed "res" I get:
_
Call:
aov(formula = qwe ~ asd)
Terms:
asd Residuals
Sum of Squares 0.0708704 0.5255957
Deg.
Hi,
I have used the instruction aov in the following manner:
res <- aov(qwe ~ asd)
when I typed "res" I get:
_
Call:
aov(formula = qwe ~ asd)
Terms:
asd Residuals
Sum of Squares 0.0708704 0.5255957
Deg. of Freedom 1 8
Residual standard error:
On Jul 13, 2010, at 10:26 AM, David Winsemius wrote:
On Jul 13, 2010, at 9:24 AM, Josh B wrote:
In my opinion the try and tryCatch commands are written and
documented rather poorly. Thus I am not sure what to program exactly.
Didn't you see the "silent" parameter? Its seems to be documen
On Jul 13, 2010, at 9:24 AM, Josh B wrote:
In my opinion the try and tryCatch commands are written and
documented rather poorly. Thus I am not sure what to program exactly.
Didn't you see the "silent" parameter? Its seems to be documented
fairly clearly to me.
The testing of try at the
Hi Felipe,
The problem has nothing to do with Sweave or \Sexpr. The problem is
that by the time you call \Sexpr report is a matrix, and you cannot
access the column names of a matrix with names(). You need to use
colnames() or convert the matrix to a data.frame.
Perhaps a true useR can write R cod
On Tue, 13 Jul 2010, Rodrigo Aluizio wrote:
Hi List, just to know if the issue is only a problem of mine or if it is a
general issue due to the new MS Office pack. I'm using R 2.11.1 32 bits in a
It's a Microsoft muddle, covered in the RODBC manual for the next
release. Note that R-sig-db is
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