On 09/10/2010 01:03 AM, David Winsemius wrote:
On Sep 9, 2010, at 6:34 PM, Gosse, Michelle wrote:
Greetings,
I am using R version 2.11.1 on a Dell computer, via a VMware
connection to a remote server. My browser version is IE
8.0.6001.18702 and the OS is some corporate version of
Dear all,
I'm using R function glmmPQL in MASS package for generalized linear mixed
model considering the temporal correlations in random effect. There are 1825
observations in my data, in which the random effect is called Date, and there
are five levels in Date, each repeats 365 times.
When
It is indeed a negative value for sigma that causes the issue.
You can check this by inserting this line
if(sigma = 0 ) cat(Negative sigma=,sigma,\n)
after the line
mu - x %*% beta
in function llk.mar
Negative values for sigma can be avoided with the use of a transformation
I have two questions:
(1) How do you 'create' an 2 x 2 table in R using say an Odd ratio of 3 or
even 0.5
(2) If I have several 2 x 2 tables, how can I 'implement' dependence in the
tables with say 25 of the Tables having an odds ratio of 1 and 75 of the
tables having an odds ratio of 4?
Jim
Hello D,
Thanks for sharing your technique, nice work :)
I hope the solution the people here are helping with will make it both
cheaper and simpler for people with less CSS expreince.
p.s: thank you for the kinds words regarding R-bloggers.com
Best,
Tal
Contact
Hi
I had similar issue and it was solved by setting proxy in browser
preferences to
do not use proxy server for 127.0.0.1
and it helped.
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.09.2010 08:18:53:
On 09/10/2010 01:03 AM, David Winsemius wrote:
On Sep 9, 2010, at 6:34
Hi,
I got two questions :
1st Question
         a=S
         b=data.frame(S=3)
         do.call(`-`,list(do.call(`$`,list(b,S)),5))
= How can I put new values on S column having the column name as a variable ?
2 nd Question
      a=S
    Â
hi,
I have created a hierarchical clustering program which reads in data and
clusters them.
The next part of the problem is to visualize the results.
I would like to know how i could import my cluster results into the R hclust
object so that i can visualise using the ggobi library. Or if there
Hi
r-help-boun...@r-project.org napsal dne 10.09.2010 10:05:37:
Hi,
I got two questions :
1st Question
         a=S
         b=data.frame(S=3)
         do.call(`-`,list(do.call(`$`,list(b,S)),5))
= How can I put new values on S column having the
On Thu, 2010-09-09 at 23:40 -0400, John Sorkin wrote:
Bert,
I appreciate you comments, and I have read Doug Bates writing about p
values in mixed effects regression. It is precisely because I read
Doug's material that I asked how are we to interpret the estimates
rather than how can we
Hello,
I'm trying to do bar plot where 'sex' will be the category axis and
'occupation' will represent the bars and the clusters will represent
the mean 'income'.
sex occupation income
1 female j 12
2male b34
3male j
Yes. A white cube and all lights off to start with give me what I want.
Many thanks!
J
Dr James Foadi PhD
Membrane Protein Laboratory (MPL)
Diamond Light Source Ltd
Diamond House
Harewell Science and Innovation Campus
Chilton, Didcot
Oxfordshire OX11 0DE
Email: james.fo...@diamond.ac.uk
Hi,
You can leverage read.table using a textConnection:
txt - x,y,z,a,b,c,dda,b,c,d,e,f,gd
con - textConnection( gsub( d, \\\n, txt ) )
read.table( con, sep = , )
V1 V2 V3 V4 V5 V6 V7
1 x y z a b c d
2 a b c d e f g
close( con )
Romain
Le 10/09/10 06:41, rajesh j a écrit :
Have a look at the ggplot2 package
install.packages(ggplot2)
library(ggplot2)
ggplot(data, aes(x = sex, y = income, fill = occupation)) +
geom_bar(position = dodge)
Have a look at http://had.co.nz/ggplot2/ for more information and
examples.
HTH,
Thierry
On 09/10/2010 01:35 AM, Steve Murray wrote:
Dear all,
I have a barplot upon which I hope to superimpose horizontal lines extending
across the width of each bar. I am able to partly achieve this through the
following set of commands:
positions- barplot(bar_values, col=grey)
par(new=TRUE)
On Sep 10, 2010, at 11:38 , Tanvir Khan wrote:
Hello,
I'm trying to do bar plot where 'sex' will be the category axis and
'occupation' will represent the bars and the clusters will represent
the mean 'income'.
sex occupation income
1 female j 12
2male
On 09/10/2010 01:07 AM, Filoche wrote:
Hi everyone.
I'm trying to break the y axis on a plot. For instance, I have 2 series
(points and a loess). Since the loess is a continuous set of points, it
passes in the break section. However, with gap.plot I cant plot the loess
because of this (I got
On 09/10/2010 07:38 PM, Tanvir Khan wrote:
Hello,
I'm trying to do bar plot where 'sex' will be the category axis and
'occupation' will represent the bars and the clusters will represent
the mean 'income'.
sex occupation income
1 female j 12
2male b
Dear all,
I am looking for ways to compute standardized logistic regression coefficients.
I found papers describing at least 6 different ways to standardize logistic
regression coefficients. I also found a very old (Thu May 12 21:50:36 CEST
2005) suggestion by Frank E Harrell (one of the
Dear all,
I am looking for ways to compute standardized logistic regression coefficients.
Before asking for your time I did some homework. I found papers describing at
least 6 different ways to standardize logistic regression coefficients. I also
found a very old (Thu May 12 21:50:36 CEST
There are two z-scores reported in the summary: Naive z and Robust z.
pvalue=2*min(pnorm(z-score), 1-pnorm(z-score)) # two-sided test
--
View this message in context:
http://r.789695.n4.nabble.com/gee-p-values-tp2533835p2534302.html
Sent from the R help mailing list archive at Nabble.com.
Hi
I have a question regarding an output of a binomial lmer-model.
The model is as follows:
lmer(y~diet * day * female + (day|female),family=binomial)
The corresponding output is:
Generalized linear mixed model fit by the Laplace approximation
Formula: y ~ diet * day * female + (day | female)
Peng,
If the answer were as simple as you suggest, I would expect that gee would
automatically produce the p values. Since gee does not produce the values, I
fear that the computation may be more complex, or perhaps computing p values
from gee may be controversial. Do you know which, if either
windows Vista
R 2.10.1
What is the difference (or differences) between lme and lmer? Both appear to
perform mixed effects regression analyses.
Thanks
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Windows Vista
R 2.10.1
Does the following use of groupedData and lme produce an analysis with both
random intercept and slope, or only random slope?
zz-groupedData(y~time | Subject,data=data.frame(data),
labels = list( x = Time,
y = y ),
units =
plyr is a set of tools for a common set of problems: you need to
__split__ up a big data structure into homogeneous pieces, __apply__ a
function to each piece and then __combine__ all the results back
together. For example, you might want to:
* fit the same model each patient subsets of a data
Reshape2 is a reboot of the reshape package. It's been over five years
since the first release of the package, and in that time I've learned
a tremendous amount about R programming, and how to work with data in
R. Reshape2 uses that knowledge to make a new package for reshaping
data that is much
Hi Josh,
Thanks for your reply. I gave a reply yesterday but found that it was not
posted.
I managed to plot the bar pot and overlay points.
The problem I am facing now is the spread of Y scale. The values I am
plotting
in Y scale are very close. so they look pretty flat. (lowest value 7.5 and
John Sorkin jsor...@grecc.umaryland.edu 10/09/2010 13:21:09
What is the difference (or differences) between lme and lmer? Both
appear to perform mixed effects regression analyses.
From a user's point of view:
- lme only accepts nested random effect; lmer handles crossed random
effects
- lme has
John Sorkin jsorkin at grecc.umaryland.edu writes:
windows Vista
R 2.10.1
What is the difference (or differences) between lme and lmer? Both appear to
perform mixed effects
regression analyses.
Thanks
John
in a nutshell:
lmer is newer, much faster, handles crossed random
Hi,
On Fri, Sep 10, 2010 at 4:05 AM, omerle ome...@laposte.net wrote:
Hi,
I got two questions :
1st Question
a=S
b=data.frame(S=3)
do.call(`-`,list(do.call(`$`,list(b,S)),5))
I think there is some confusion here. Why are you setting a equal to
S but
Dirk E. has properly focussed the discussion on measurement rather than opinion. I'll add
the issue of the human time taken to convert, and more importantly debug, interfaced code.
That too could be measured, but we rarely see human hours to code/debug/test reported.
Moreover, I'll mention the
Hi, I'm having a bit of trouble using SSOAP to send requests containing
complex types. It seems as though processWSDL does not generate functions
for converting the generated types into SOAP requests, but it does generate
them for converting **from** SOAP requests to the complex types.
The error
Message du 10/09/10 14:53
De : Ista Zahn
A : omerle
Copie à : r-help@r-project.org
Objet : Re: [R] Data.frames : difference between x$a and x[, a] ? - How set
new values on x$a with a as variable ?
Hi,
On Fri, Sep 10, 2010 at 4:05 AM, omerle wrote:
Hi,
I got two questions :
Dear all,
I would like to separate the gbm C++ code from any R dependencies, so that
it could be compiled into a standalone module.
I am wondering if anyone has already done this and could provide me with
some pointers/help ?
Thanks!
Markus
[[alternative HTML version deleted]]
On Sep 9, 2010, at 10:57 AM, David Winsemius wrote:
On Sep 9, 2010, at 6:50 AM, Jan private wrote:
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves the
On Fri, Sep 10, 2010 at 9:22 AM, omerle ome...@laposte.net wrote:
Message du 10/09/10 14:53
De : Ista Zahn
A : omerle
Copie à : r-help@r-project.org
Objet : Re: [R] Data.frames : difference between x$a and x[, a] ? - How
set new values on x$a with a as variable ?
Hi,
On Fri, Sep 10,
I'm having trouble parsing this. What exactly do you want to do?
1 - Put a list as an element of a data.frame. That's quite convenient for my
pricing function.
I think this is a really bad idea. data.frames are not meant to be
used in this way. Why not use a list of lists?
It can be very
Le 9/10/2010 15:37, Ista Zahn a écrit :
On Fri, Sep 10, 2010 at 9:22 AM, omerleome...@laposte.net wrote:
Message du 10/09/10 14:53
De : Ista Zahn
A : omerle
Copie à : r-help@r-project.org
Objet : Re: [R] Data.frames : difference between x$a and x[, a] ? - How
set new values on x$a with
Dear all, I am perplexed when trying to get the same results using
pairwise.t.test and t.test.
I'm using examples in the ISwR library,
attach(red.cell.folate)
I can get the same result for pairwise.t.test and t.test when I set the
variances to be non-equal, but not when they are assumed to be
Dear,
When I try to to execute the following command, R don't read all lines (reads
only 57658 lines when the file has 814125 lines):
dados2-read.table(C:\\Documents and Settings\\mgoncalves\\Desktop\\Tábua
Hi Mamum,
You can look at the ylim argument to plot(). It lets you control the
limits; however, in your example graph, part of the issue is that the
bars have a much higher value, so you could not change ylim too much
(looks like in your example graph you could set it to something like
ylim =
On Sep 10, 2010, at 9:42 AM, Hadley Wickham wrote:
I'm having trouble parsing this. What exactly do you want to do?
1 - Put a list as an element of a data.frame. That's quite
convenient for my pricing function.
I think this is a really bad idea. data.frames are not meant to be
used in this
Dear All,
I'm using R for Mac OS X Cocoa GUI R version 2.11.1.
I can save contents of my console window by using command + s, but I
would like to do same thing using R commands.
My question is can I save the contents automatically by using R editor
with some R commands.
Thank you.
Nobu
Dear all,
I am struggling to modify the axis labels/ticks in a panel provided to
xyplot.
To begin with, I do not know the equivalent of the xaxt=n directive for
panels that would set the stage for no default x axis being drawn.
My goal is to draw ticks and custom formatted labels at certain hours
Hello,
I have some Data, that I analyse using a regression tree (by using rpart).
Now I would like to save this tree to a file for later use, and load it (not
necessarily in the same workspace). It would be most preferrable, if I could
just save some parameters (later probably to a database)
On Sep 10, 2010, at 16:01 , Jabez Wilson wrote:
Dear all, I am perplexed when trying to get the same results using
pairwise.t.test and t.test.
I'm using examples in the ISwR library,
attach(red.cell.folate)
I can get the same result for pairwise.t.test and t.test when I set the
On Sep 10, 2010, at 9:01 AM, Jabez Wilson wrote:
Dear all, I am perplexed when trying to get the same results using
pairwise.t.test and t.test.
I'm using examples in the ISwR library,
attach(red.cell.folate)
I can get the same result for pairwise.t.test and t.test when I set the
On 10/09/2010 10:03 AM, Marcelo Estácio wrote:
Dear,
When I try to to execute the following command, R don't read all lines (reads
only 57658 lines when the file has 814125 lines):
dados2-read.table(C:\\Documents and Settings\\mgoncalves\\Desktop\\Tábua
I think this is a really bad idea. data.frames are not meant to be
used in this way. Why not use a list of lists?
It can be very convenient, but I suspect the original poster is
confused about the different between vectors and lists.
I wouldn't be surprised if someone were confused, since
Well, let's see if the following helps or just adds to the confusion.
First lists are vectors of mode list . But they are general
recursive structures (in fact, completely general).
Second, data frames are lists: each column of a data frame is a
component (member) of the list with the additional
Don't underestimate the importance of the choice of the algorithm you
use. That often makes a huge difference. Also, vectorization is key
in R, and when you use that you're really up there among the top
performing languages. Here is an example from the official R wiki
illustrating my points:
On Fri, Sep 10, 2010 at 10:23 AM, Henrik Bengtsson h...@stat.berkeley.edu
wrote:
Don't underestimate the importance of the choice of the algorithm you
use. That often makes a huge difference. Also, vectorization is key
in R, and when you use that you're really up there among the top
This really depends on why you want to do this and what results you want. If
your main goal is to look at some basic tests, goodness of fit, then the add1
function may do everything you need. If you just want coefficient estimates
then some basic matrix algebra will give those to you.
Thanks a lot, Peter. Excellent book btw.
Jab
--- On Fri, 10/9/10, peter dalgaard pda...@gmail.com wrote:
From: peter dalgaard pda...@gmail.com
Subject: Re: [R] pairwise.t.test vs t.test
To: Jabez Wilson jabez...@yahoo.co.uk
Cc: R-Help r-h...@stat.math.ethz.ch
Date: Friday, 10 September, 2010,
Hello,
I want to make a general routine to draw barplots with numbers plotted
above each bar. See the example below.
I could not place the numbers on the middle of each bar because I
could not calculate the right position of each x-axis tick. axTicks(1)
indicated a unitary step, but it
Dear all,
I've wanted to give a try to the rattle GUI for R. After long struggle
with dependencies I've finally managed to install rattle and all of
its dependencies, although for some of them I've been forced to use
cran2deb. And now all I get is:
library(rattle)
Loading required package: pmml
Thank Jim for your answer.
I actually did my own function to plot with the loess. I just calculated the
intersection between the first and second horizontal gap lines with the line
formed by the 2 points before and after the gap. So I can now plot the loess
from both sides of the gap section.
Hi,
I just started using R and need some guidance.
I need to create a time series chart in R, but the problem is the data is
not numeric.
The data is in the following format
Study
A
A
B
B
B
A
C
C
D
Then there is also another column with dates. How can I manipulate this in
order to have
Duncan, thanks for your answer.
I tried this:
con-file(C:\\Documents and Settings\\mgoncalves\\Desktop\\Tábua
IFPD\\200701_02_03_04\\200701_02_03_04.txt,open=rb)
Hi,
Look at the table() function. Here is an example with your data:
dat - read.table(textConnection(
Study
A
A
B
B
B
A
C
C
D), header = TRUE)
closeAllConnections()
table(dat)
Hope that helps,
Josh
On Fri, Sep 10, 2010 at 8:53 AM, dfong df...@medicine.umaryland.edu wrote:
Hi,
I just
I'm actually importing it from a CSV, so I already have that in a table. But
i Can't make a graph with text. I assume I need to do some counting in order
to draw the graph?
Any example of this?
thanks
--
View this message in context:
Hi Everyone,
I have a 2-dim data.matrix(e.g., table1) in which row1 specifies a range of
values. row2 - rown specify the number of times I want to replicate each
corresponding value in row1. I can do this with the following function:
rep(c(table1[1,]),c(table1[X,])) #where X would go from 2 -
Hi,
Yes, the table() function is not to read the data, but to do the
frequency counts of each level. I just included the read.table() part
so that you could copy and paste my code, but I did not include the R
output from table(dat).
table(dat)
dat
A B C D
3 3 2 1
It nicely tallies for you.
Hello,
This is definitely possible with R, there a lots of package to make
good graphics.
However, the easiest way for us to help you is if you give us a small
reproducible example, as you started to with your initial post.
If you have an object in your R session that you'd like help with
you
Hi A,
Here is a little example that I believe does what you want. I am not
quite sure how you want all the output in a new matrix, because as you
repeat each value is the first row varying numbers of times, you will
not have rows of equal length. Although perhaps your data is setup so
that you
I do not follow. Could you please provide a small reproducible example
of what table1 might look like, and what you want as a result?
Surely you don't need a for loop.
alfredo wrote:
Hi Everyone,
I have a 2-dim data.matrix(e.g., table1) in which row1 specifies a range of
values. row2 - rown
try this:
x - matrix(sample(25,25), 5)
x
[,1] [,2] [,3] [,4] [,5]
[1,] 12 24 143 20
[2,] 2175 15 17
[3,] 11 10 22 169
[4,]6 2541 23
[5,]2 198 13 18
# save result in a list
result - lapply(2:nrow(x), function(.row){
+
from the output, I think it's both.
- Original Message
From: John Sorkin jsor...@grecc.umaryland.edu
To: r-help@r-project.org
Sent: Fri, September 10, 2010 5:25:44 AM
Subject: [R] lme, groupedData, random intercept and slope
Windows Vista
R 2.10.1
Does the following use of
Hi:
On Fri, Sep 10, 2010 at 7:16 AM, Markus Loecher markus.loec...@gmail.comwrote:
Dear all,
I am struggling to modify the axis labels/ticks in a panel provided to
xyplot.
To begin with, I do not know the equivalent of the xaxt=n directive for
panels that would set the stage for no default
But as far as I know, profile() seems to be de-activated in the lme4 package.
- Original Message
From: Gavin Simpson gavin.simp...@ucl.ac.uk
To: John Sorkin jsor...@grecc.umaryland.edu
Cc: r-help@r-project.org; Bert Gunter gunter.ber...@gene.com
Sent: Fri, September 10, 2010 2:05:37 AM
On Fri, 2010-09-10 at 09:51 -0700, array chip wrote:
But as far as I know, profile() seems to be de-activated in the lme4 package.
It is beta software. The lme4a version of the lme4 package might have
had profile re-enabled, IIRC.
G
- Original Message
From: Gavin Simpson
I just wrote up some code for differencing two .RData files or
environments (or one of each). Available from source here:
http://www.maths.lancs.ac.uk/~rowlings/R/Ediff/
In its handiest form, running:
ediff()
will tell you the difference between your working environment and the
.RData file
Thanks for reminding this. So I found lme4a package from Doug's UserR!2010
presentation folder:
http://lme4.r-forge.r-project.org/slides/2010-07-20-Gaithersburg/pkg/
However, after installation, I got the following error message when trying to
load the library:
library(Matrix)
library(Rcpp)
Hi all,
I have to filter a tab-delimited text file like below:
GeneNamesvalue1value2log2(Fold_change)
log2(Fold_change) normalizedSignature(abs(log2(Fold_change)
normalized) 4)
ENSG0209350435-3.81131293562629-4.14357714689656TRUE
ENSG0177133
Hi:
It's not immediately clear what you have in mind, as others have noted, but
here are a couple of ideas that seem as though they may apply to your
problem, as dangerous as it is to play clairvoyant:
I'm using vectors instead of a matrix, but the first vector, val, contains
the values whereas
See this message and the replies to it (and the replies to the replies, etc.):
http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22858.html
In there is a discussion of why you don't really want to do that along with
better alternatives and examples of the improved plots.
--
Gregory (Greg) L.
Duke -
One possibility is to check the help files for the functions
involved to see if there are options to control this behaviour.
For example, the check.names= argument to read.table, or the
quote= argument to write.table. How about
expFC - read.table(test.txt, header=TRUE, sep=\t,
On Fri, 2010-09-10 at 10:23 -0700, array chip wrote:
Thanks for reminding this. So I found lme4a package from Doug's UserR!2010
presentation folder:
http://lme4.r-forge.r-project.org/slides/2010-07-20-Gaithersburg/pkg/
What is wrong with the one on the packages tab of the lme4 project page:
x=1:16
S=summary(x)
S
Min. 1st Qu. MedianMean 3rd Qu.Max.
1.004.758.508.50 12.25 16.00
S[-4]
Min. 1st Qu. Median 3rd Qu.Max.
1.004.758.50 12.25 16.00
par(mfrow=c(1,2))
boxplot(S[-4]) # based on the summarized stats
boxplot(x)
Hello
Imagine I have a vector with ones and zeroes
I write it compactly:
011100101
I need to get a new vector replacing the N ones following the zeroes to
new zeroes.
For example for N = 3
011100101 becomes
For base graphics the bxp function does the actual plotting given the
statistics.
--
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-
Thanks a lot for this incredibly helpful and thorough reply.
I had actually meant to cut out the scales part before sending the email,
very sorry about the confusion, so I was actually executing just
xyplot(Comp ~ time | PC ,data = Xlong, pane1l = WeekhourPanel)
The scales part was a later
Hi Phil,
On 9/10/10 1:45 PM, Phil Spector wrote:
Duke -
One possibility is to check the help files for the functions
involved to see if there are options to control this behaviour.
For example, the check.names= argument to read.table, or the quote=
argument to write.table. How about
is.nan(bd.coerce(as.bdVector(c(1.0, N -Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Brian Diggs
Sent: Thursday, September 09, 2010 12:41 PM
To: David A.
Cc: R-help
Subject: Re: [R] boxplot knowing Q1, Q3, median,upper and
Hi all,
Is it possible to change the filling of the squares used to represent
the colour legend in a bar plot with ggplot?
in this example, fillings are raven black, I'd like them white.
ggplot(diamonds, aes(clarity, colour = cut)) + geom_bar()
Regards
--
-
Benoit Boulinguiez
Hi:
To add to Greg's sound advice, if you want to put the numbers on top of the
bars, why bother with the numerical scale? The entire point of a scale is to
provide a reference for comparing different (sets of) values.
\begin{rant}
And when I see things like this:
dat.bar
VAR1
Are you trying to say that you don't really like barplots?
At least the OP did not ask for error bars as well. :)
--- On Fri, 9/10/10, Dennis Murphy djmu...@gmail.com wrote:
From: Dennis Murphy djmu...@gmail.com
Subject: Re: [R] adding labels above bars in a barplot
To: Antonio Olinto
Sure, just change the color of the fill.
ggplot(diamonds, aes(clarity, colour = cut)) + geom_bar(fill=white)
-Ista
On Fri, Sep 10, 2010 at 2:24 PM, Benoit Boulinguiez
benoit.boulingu...@ensc-rennes.fr wrote:
Hi all,
Is it possible to change the filling of the squares used to represent the
Dear all,
Is it possible to generate AIC or something equivalent for nonlinear
model estimated based on maximum log likelihood l in R?
I used nls based on least squares to estimate, and therefore I cannot
assess the quality of models with AIC. nlme seems good for only mixed
models and mine is not
On Fri, Sep 10, 2010 at 1:24 PM, Duke duke.li...@gmx.com wrote:
Hi all,
I have to filter a tab-delimited text file like below:
GeneNames value1 value2 log2(Fold_change)
log2(Fold_change) normalized Signature(abs(log2(Fold_change)
normalized) 4)
ENSG0209350 4 35
Do you need a table with an odds ratio exactly equal to 3 (or other value), or
a realistic sample from a population with odds ratio 3 where the sample table
will have a different OR (but the various tables will cluster around the true
value)?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data
Not sure how you handle the ending sequence of '0101'; here is one approach:
x - 011100101
gsub(0111, , x)
[1] 000100101
x
[1] 011100101
For the final one, you could
Error: attempt to apply non-function
In addition: Warning message:
In method(obj, ...) : Invalid object type `GtkFrame'
Rattle timestamp (for the error above): 2010-09-10 17:24:14
Ditto in Debian testing i386 32 bit.
Johan Nyberg
__
My experience is that most medical journals (and probably others as well, but I
am most familiar with the medical journals) have word or page limits on
articles. Diagnostic plots and tests that just show you what you expected and
say that it is ok to use your model are not exciting enough to
Hi:
Try the following:
f - function(x, n) {
r - rle(x)
t1 - cumsum(r$lengths)[r$values == 0L] + 1
repl - as.vector(mapply(seq, t1, t1 + n - 1))
replace(x, repl, 0)
}
f(x, 3)
HTH,
Dennis
On Fri, Sep 10, 2010 at 10:51 AM, skan juanp...@gmail.com wrote:
Hello
Imagine I have a
Odette Gaston odette.gaston at gmail.com writes:
Dear all,
Is it possible to generate AIC or something equivalent for nonlinear
model estimated based on maximum log likelihood l in R?
I used nls based on least squares to estimate, and therefore I cannot
assess the quality of models with
Hi
I'll study your answers.
I could also try
gsub(01, 00, x) N times
but it could be very slow if N is large
In fact when I wrote 10011I mean a vector
1
1
1
1
1
0
0
1
1
not a string, but I wrote it more compactly.
I also could by shifting the elements of the vector one position and
I'm trying to find a more elegant way of doing this. What I'm trying to
accomplish is to count the frequency of letters (major / minor alleles) in a
string grouped by the factor levels in another column of my data frame.
Ex.
DF-data.frame(c(CC, CC, NA, CG, GG, GC), c(L, U, L, U,
L, NA))
Denis.Aydin at unibas.ch writes:
I have a question regarding an output of a binomial lmer-model.
The model is as follows:
lmer(y~diet * day * female + (day|female),family=binomial)
A reproducible example would always be nice.
The corresponding output is:
Generalized linear mixed model
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