The current (you give no version information) help file says:
file: A file name. (If a path is given, see 'Note'.)
zipname: The file name (not path) of a 'zip' archive, including the
'.zip' extension if required.
...
This is a helper function for 'help', 'example' and
Sergio Della Franca wrote:
Dear R-Helpers,
I have the following data set(y):
Test_Result #_Test
t 10
f 14
f 25
f NA
f 40
t45
t44
NA
Hi,
I would like to add confidence intervales to an ANCOVA with 2
covariates when using xyplot.
What would be a good way of accomplishing this?
--8---cut here---start-8---
rm(list = ls(all = TRUE))
rm(list = c(ls()))
library(lattice)
## 1. generate data
Hi everybody,
I'm doing a sqlSave() in R, to insert a big data frame of 1 rows.
However, there is problems, since several rows contains quotations marks,
that can leave inserts buggy.
I would like to find a way to add slashes in front of these quotation marks.
Best regards,
Laurent
--
We
A simple question - using the following fishers test it appears that the P
value is significant, but the CI includes 1. Is this result correct?
data.50p10min - matrix(c(16,15, 8, 24),nrow=2)
fisher.test(data.50p10min)
Fisher's Exact Test for Count Data
data: data.50p10min
On 31-Mar-07 13:36:04, Williams Scott wrote:
A simple question - using the following fishers test it appears that
the P value is significant, but the CI includes 1. Is this result
correct?
data.50p10min - matrix(c(16,15, 8, 24),nrow=2)
fisher.test(data.50p10min)
Fisher's Exact
The two-sided test of odds-ratio=1 is not necessarily (nor in
this case) the same thing as an equal-tailed confidence interval: see the
help page comment
Two-sided tests are based on the probabilities of the tables, and
take as 'more extreme' all tables with probabilities less than
I think we occasionally think that it is very easy to get information
because we know how to find the information. This does not mean that other
people know how to find the answer. It is for this reason that questions
appear on the listserver that we might think could be easily found from
G'day Kaltja,
On Sat, 31 Mar 2007 10:19:00 +0300 (EEST)
[EMAIL PROTECTED] wrote:
[...] I did not even now there was R wiki. I couldn't find a link
from the R organization pages to it or am I just blind?
If you talk about CRAN (e.g. http://cran.r-project.org/) then no, but
if you really talk
Hi,
I am using the function lda() from MASS for finding reduced-dimensional
representations of a datset. In reading various texts to compare LDA with
Fisher's LDA approach (including Ripley's Modern Applied Statistics with
S-Plus), it is still unclear to me whether or not they produce the
Try this:
'my dog named Spot and my cat named Kitty fight like cats and dogs'
On 3/31/07, Laurent Valdes [EMAIL PROTECTED] wrote:
Hi everybody,
I'm doing a sqlSave() in R, to insert a big data frame of 1 rows.
However, there is problems, since several rows contains quotations marks,
On Sat, 31 Mar 2007, Andre Jung wrote:
Dear all,
I have three timeseries Uts, Vts, Wts. The relation between the time
series can be expressed as
Uts = x Vts + y Wts + residuals
How would I feed this to lm() to evaluate the unknowns x and y?
If the time series are aligned (and univariate)
SteT == Stephen Tucker [EMAIL PROTECTED]
on Fri, 30 Mar 2007 18:41:39 -0700 (PDT) writes:
[..]
SteT For dates, I usually store them as POSIXct classes
SteT in data frames, but according to Gabor Grothendieck
SteT and Thomas Petzoldt's R Help Desk article
SteT
Hi everyone,
Currently I work for Master degree
in statistics at Garyounis University.
My dissertation on
On 3/31/07, Martin Maechler [EMAIL PROTECTED] wrote:
SteT == Stephen Tucker [EMAIL PROTECTED]
on Fri, 30 Mar 2007 18:41:39 -0700 (PDT) writes:
[..]
SteT For dates, I usually store them as POSIXct classes
SteT in data frames, but according to Gabor Grothendieck
SteT and
Hi team,
I have the data of the form:
a- data.frame(x=c(1,2,1,4,3), y=c(1,2,1,4,3), z=c(1,2,3,4,5))
I need the output of the form
b- data.frame(x=c(1,2,3,4), y=c(1,2,3,4), z=(3,2,5,4) )
As you can see, the Z value contains the maximum for each of the (x,y)
combinations.
I used
c-by(a$z,
Try this:
a
x y z
1 1 1 1
2 2 2 2
3 1 1 3
4 4 4 4
5 3 3 5
# create indices of groups
indices - split(seq(nrow(a)), list(a$x, a$y), drop=TRUE)
combine - lapply(indices, function(.ind){
+ # create a row representing the max of z
+ c(x=a$x[.ind[1]], y=a$y[.ind[1]], z=max(a$z[.ind]))
+
Dear R experts:
I am a bit stymied by how the argument picking-off works in R batch file usage.
$ cat commandArgs.R
cat( Command Line Arguments were , commandArgs(), \n);
$ /usr/bin/R CMD BATCH commandArgs.R --args 1 2 3
$ /usr/bin/R --args 1 CMD BATCH commandArgs.R
... I am now getting into
Hi Gabor and Martin,
Thanks very much for the information. (and Gabor for the Fold() routine
included in original reply)
Regarding changes, I wonder if the behavior of plot() on POSIXct objects
changed also. According to Rnews Vol. 4/1, p. 31,
=
dp - seq(Sys.time(),len=10,by=day)
I think you are right. Plot does seem to have changed and maybe its due to
a change in plot or maybe due to a change in some routine it in turn calls.
Here is another test:
dd - ISOdatetime(2007, 1, 1:3, 0, 0, 0, tz = )
plot(dd, 1:3)
dd.gmt - ISOdatetime(2007, 1, 1:3, 0, 0, 0, tz = GMT)
Try this:
aggregate(a[3], a[1:2], max)
On 3/31/07, Deepak Manohar [EMAIL PROTECTED] wrote:
Hi team,
I have the data of the form:
a- data.frame(x=c(1,2,1,4,3), y=c(1,2,1,4,3), z=c(1,2,3,4,5))
I need the output of the form
b- data.frame(x=c(1,2,3,4), y=c(1,2,3,4), z=(3,2,5,4) )
As
Hello all,
I am trying to figure out an optimal linear model by using stepwise
regression which requires partial f-test, I did some Googling on the
Internet and realised that someone seemed to ask the question before:
Jim Milks [EMAIL PROTECTED] writes:
Dear all:
I have a regression model
Hello everybody,
I'm interesting in evolutionary algorithms. I have tested genetic
algorithms with R but has someone tried with genetic programming? Do
you know, if there are code somewhere written in R.
Best wishes,
Atte Tenkanen
University of Turku, Finland
And what about to read the help page ?anova ...?
When given a sequence of objects, 'anova' tests the models against
one another in the order specified.
Generally you almost never fit a full model (including all possible
interactions etc) - no one can interpret such complicated models.
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