On Fri, 1 Mar 2013, Frank Harrell wrote:
Thank you Bill. A temporary re-arrangement of the formula will allow me to
do the usual subset= na.action= processing afterwards. Nice idea. I don't
need the dot notation very often for this application.
That's what the Formula package provides. It
Hi everyone,
Imagine that I have a data frame with four columns:
data-
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
I want to replace the zeros in columns a:b for NA only for the rows in
which column d are zero. So
a b c
On 13-03-01 8:35 PM, C W wrote:
Thanks, everyone, I will definitely avoid it. Is there any tips on naming
variables? I've seen the Google R style guider and Hadley R style guide.
Name them in ways that are meaningful to you. R standard function names
are famous for not following any naming
you want to replace all rows where the 4th column is zero.. (data[ , 4 ]
== 0)
and you want to perform that replacement in the first column..
so try
data[ data[ , 4 ] == 0 , 1 ] - NA
On Sat, Mar 2, 2013 at 5:26 AM, Camilo Mora cm...@dal.ca wrote:
Hi everyone,
Imagine that I have a data
Hello,
Try
# for columns a.b it's 1:2, not 1:3
data[data[,4] == 0, 1:3] - NA # columns a, b and c
Hope this helps,
Rui Barradas
Em 02-03-2013 10:26, Camilo Mora escreveu:
Hi everyone,
Imagine that I have a data frame with four columns:
data-
a b c d
0 1 1
Achim this is perfect. I had not seen Formula before. Thanks for writing
it!
Frank
Achim Zeileis-4 wrote
On Fri, 1 Mar 2013, Frank Harrell wrote:
Thank you Bill. A temporary re-arrangement of the formula will allow me
to
do the usual subset= na.action= processing afterwards. Nice idea.
Hi Gabor,
This is not for a regression function but for a major update I'm working on
for the summary.formula function in the Hmisc package. So I need to handle
several data types in the formula.
Thanks
Frank
Gabor Grothendieck wrote
Gabor Grothendieck wrote
On Fri, Mar 1, 2013 at 7:16 PM,
On 13-03-01 12:57 PM, Duncan Murdoch wrote:
On 01/03/2013 11:20 AM, William Dunlap wrote:
A core R function that fails with odd names is reformulate():
reformulate(c(P/E, % Growth), response=+-)
Error in parse(text = termtext) : text:1:16: unexpected input
1: response ~ P/E+%
Dear all,
knitr writes a comment sign in front of each line of console output. This
extremely useful, especially for beginners who are starting to write scripts.
One could just compose their script in the console (getting immediate output)
and then copy the whole chunk into their .r-file.
I
HI,
Not sure I understand it correctly,
data1-read.table(text=
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
,sep=,header=TRUE)
data2- data1
data3- data1
If i follow this logic for the 1st and 2nd columns,
data1[ data1[ , 4 ] == 0 , 1 ] - NA
I would like to have a lattice conditioning ( | var ) variable have
expression() as values because I want panel labels to be able to use
plotmath notation for subscripts, etc. lattice barks at this. Does anyone
know of a trick workaround? An attempted example program is below. Thanks
-Frank
Whoops - these 2 lines should have been omitted from the program:
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
Frank Harrell wrote
I would like to have a lattice conditioning ( | var ) variable have
expression() as values because I want panel labels
Hello,
Arun has a point, I forgot to check the element to be replaced for a
zero. The following code will do that.
dat - read.table(text =
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
, header = TRUE)
dat - data.matrix(dat)
str(dat)
?options
Change the prompt option:
options(prompt = # ) ## is I think what you want.
-- Bert
On Sat, Mar 2, 2013 at 5:35 AM, Marcus Kriele mkri...@me.com wrote:
Dear all,
knitr writes a comment sign in front of each line of console output. This
extremely useful, especially for beginners
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in print.anova().
[Testing this requires the latest version of vcdExtra, from R-Forge
**|install.packages(vcdExtra, repos=http://R-Forge.R-project.org;)|**
]
Try the following to see if it does what you want:
## init
con - textConnection(output, w)
options(echo=FALSE)
sink(con)
addTaskCallback( function(expr, out, err, vis) {
sink()
close(con)
if(vis) {
cat(paste(#, output, collapse=\n), \n)
}
con - textConnection(output, w)
sink(con)
TRUE
})
3+4
Le samedi 02 mars 2013 à 12:37 -0500, Michael Friendly a écrit :
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in print.anova().
[Testing this requires the latest version of vcdExtra, from R-Forge
Hi Michael,
Take a look at ?printCoefmat.
I hope this helps,
John
On Sat, 02 Mar 2013 12:37:36 -0500
Michael Friendly frien...@yorku.ca wrote:
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in
In reference to [1], how would you solve the following regression
problem:
Given observations (X_i,Y_i) with known respective error distributions
(e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters
a and b which maximize the Likelihood of
Y = a*X + b
Taking the example
Hello,
please log me out from the conference. I tried to contact the webmaster but
the only e-mail I reached was this one. Probably fault on my side in the
registration.
Thanks,
Petr Suvorov
[[alternative HTML version deleted]]
__
Thanks Greg, this is great! Regards, Marcus
On 2013-03-02, at 19:06 , Greg Snow wrote:
Try the following to see if it does what you want:
## init
con - textConnection(output, w)
options(echo=FALSE)
sink(con)
addTaskCallback( function(expr, out, err, vis) {
sink()
Some packages have to be installed on new R version, 2.15.3
but I have 2.15.2
If I don't want to go through all download, and uninstall and install
procedures, is there any simple R command that could handle this? Thanks.
[[alternative HTML version deleted]]
HI Utpal,
Alight, I will look into it. I was under the impression that this is what you
wanted:
dat1- structure(list(V1 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), V2 = c(1L,
1L, 1L, 1L, 1L, 1L, 0L), V3 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L),
V4 = c(1L, 1L, 0L, 1L, 1L, 1L, 1L), V5 = c(1L, 1L, 1L, 1L,
On Sat, Mar 2, 2013 at 1:49 PM, capricy gao capri...@yahoo.com wrote:
Some packages have to be installed on new R version, 2.15.3
but I have 2.15.2
If I don't want to go through all download, and uninstall and install
procedures, is there any simple R command that could handle this? Thanks.
xYplot has many options that are passed to panel.xYplot. Did you read the
documentation? You can suppress labels, use an automatically generated
Key() function to control where you want keys, and use several other
options. Start with label.curve=FALSE and go from there.
Frank
bwr87 wrote
I'm
Frank,
As you probably realize, the problem is not with lattice but with
unique (actually: .Internal(unique, ...)):
unique(expression('a','b','b'))
generates your error.
Lattice needs a factor variable for the panels, and it tries to
use unique() on your 'vn' to do that.
For example, if
Duncan's comment may not qualify as a fortune, but it did make
me chuckle.
Peter Ehlers
On 2013-03-02 03:01, Duncan Murdoch wrote:
On 13-03-01 8:35 PM, C W wrote:
[...snip...]
pie is a function, but all it does is draw pie charts, so who cares if
you mask it? :-).
Duncan Murdoch
There's a no homework policy in R-help.
Rui Barradas
Em 02-03-2013 18:28, Cedric Sodhi escreveu:
In reference to [1], how would you solve the following regression
problem:
Given observations (X_i,Y_i) with known respective error distributions
(e_X_i,e_Y_i) (say, 0-mean Gaussian with known
Greetings,
I have been exploring the use of the caret package to conduct some plsda
modeling. Previously, I have come across methods that result in a R2 and
Q2 for the model. Using the 'iris' data set, I wanted to see if I could
accomplish this with the caret package. I use the following code:
Perhaps it would have been clearer that this is no homework if I
hadn't forgotten to say what [1] is. Sorry for that.
[1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225
(This is no homework but genuinely adresses the problem that R to my
knowledge does not have models for error in
Hello R-Help,
I want to be able to read in a raster image, plot it with grid.raster
or rasterImage and save the image with one pixel per a pixel element
from my array. Saved preferably in a common image format.
The real goal of my question is to eventually read in images with text
on them,
I have discovered on of my errors. The timematrix was unnecessary and an
unfortunate habit I brought from another package. The following provides
the same R2 values as it should, however, I still don't know how to
retrieve Q2 values. Any insight would again be appreciated:
library(caret)
Based on your comments in the (not-a-)bug report, I *think* this might help:
quanttrader.info/public/betterHedgeRatios.pdf
or more generally, the idea of total least squares regression.
Cheers,
MW
On Sat, Mar 2, 2013 at 9:55 PM, Cedric Sodhi man...@gmx.net wrote:
Perhaps it would have been
Dear all,
I am very grateful that Wolfgang Viechtbauer implemented the standardised mean
change for dependent
groups. I was playing around a bit today, and I am not sure if I understand the
SMCR procedure correctly. The documentation states that sd1i and sd2i are
needed, but it seems to me
Well *I* think it should be a fortune!
cheers,
Rolf
On 03/03/2013 10:30 AM, Peter Ehlers wrote:
Duncan's comment may not qualify as a fortune, but it did make
me chuckle.
Peter Ehlers
On 2013-03-02 03:01, Duncan Murdoch wrote:
On 13-03-01 8:35 PM, C W wrote:
[...snip...]
Hello,
Like you say, apparently R doesn't have models for error in variables.
But R packages might have.
library(sos)
findFn('errors-in-variables')
Some look promising. Hope you find something.
Rui Barradas
Em 02-03-2013 21:55, Cedric Sodhi escreveu:
Perhaps it would have been clearer that
On Mar 2, 2013, at 1:55 PM, Cedric Sodhi wrote:
Perhaps it would have been clearer that this is no homework if I
hadn't forgotten to say what [1] is. Sorry for that.
[1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225
(This is no homework but genuinely adresses the problem
Hi,
I'm trying to set up R to run a simulation of two populations in which every
3.5 days, the initial value of one of the populations is reset to 1.5. I'm
simulation an experiment we did in which we fed Daphnia populations twice a
week with algae, so I want the initial value of the algal
just seeing this.
when you say y1 and y2, are you asking about 2 differnt models in an effort to
model select? i need more info, i might be able to help. please supply a bit
more info.
~n
On Feb 25, 2013, at 4:39 AM, Ali A. Bromideh wrote:
Dear Sir/Madam,
I apologize for any
also, kruschke at indiana has some info on this, both online and youtube. (if
homework.) if not, more infor will be helpful.
~n
On Feb 25, 2013, at 9:41 AM, Bert Gunter wrote:
Homework? We don't do homework here.
If not, search (e.g. via google -- R hierarchical Bayes -- or some such).
cdouglass wrote
Hello all,
Totally new to this and I'm just doing a frequency distribution analysis
on T-shirt sales by size. I have a .csv with 60 orders. I read in the
data using read.csv. If I look at the summary() or table() of the data it
looks fine, except that the shirt sizes are
Charles,
You should not be treating the classes as numeric (is virginica really
three times setosa?). Q^2 and/or R^2 are not appropriate for classification.
Max
On Sat, Mar 2, 2013 at 5:21 PM, Charles Determan Jr deter...@umn.eduwrote:
I have discovered on of my errors. The timematrix was
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