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
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 defined.
[2]
adolfpf wrote:
How do I group my data in dolf the same way the data Orthodont are
grouped.
show(dolf)
distance age Subjectt Sex
16.83679 22.01 F1 F
26.63245 23.04 F1 F
3 11.58730 39.26 M2 M
I know that many sample in that excellent book use
Greetings
In a lattice barchart (lattice_0.19-26, R 2.12.1), I am trying to control the
order in which packets are displayed (in other words, which packet goes in
which panel), and the order in which bars are displayed in each panel. I tried
the simple idea of changing the levels of the
Hi everyone,
I determined the presence of three types parasites in a passerine bird
over two years. I would like to create a bar chart that shows the
proportion infected on the y and year/parasite on the x such that each
type of parasite is grouped together (single label) and a bar for each
As an alternative, you could try the HWExact function in the GWASExactHW
package. (Haven't tried it myself though ...).
Brad
- Original Message -
From: Jim Silverton jim.silver...@gmail.com
To: r-help@r-project.org
Sent: Tuesday, 12 July, 2011 8:33:54 PM
Subject: Re: [R] Hardy
Hi,
par(mfrow = c(2,2))
will create a 2x2 window that I can use to plot 4 diferent figures in:
[plot1 plot2]
[plot3 plot4]
But how can do 3 so that the bottom spans the width of the upper two:
[plot1 plot1]
[p l o t 3]
Is this possible in R?
--
View this message in context:
bbolker wrote:
What about
library(pscl)
example(hurdle)
-2*logLik(fm_hnb2)
?
Should this not be -2*(logLik(M) - logLik(S)) where M is the current model
and S is the saturated or full model? In which case, is it safe to assume
that L(S) = 0? Or have I got my deviances crossed?
@Dimitri: I tried to enter it as numeric and still got the same outcome. I
still wonder if there is any way to get the same result from both programs.
@David, Bert: Yes, I found that the gender coefficient is R is exactly twice
that of the one from SPSS. Need to study on parametrization.
Thanks,
All the examples in 'nlme' are in Grouped Data: distance ~ age | Subject
format.
How do I group my data in dolf the same way the data Orthodont are
grouped.
show(dolf)
distance age Subjectt Sex
16.83679 22.01 F1 F
26.63245 23.04 F1 F
3 11.58730 39.26 M2 M
I finally got the same result by converting gender variable as numeric, and
standardize it.
I guess SPSS automatically doing the same thing when doing analysis.
But, it still is not clear to me how I can interpret standardized
categorical (dummy coded) variable.
I'd rather stick to use R.
Thanks
Hi,
Try looking at ?layout. Here is a simple example:
layout(matrix(c(1, 2, 3, 3), 2, byrow = TRUE))
plot(1:10); plot(11:20); plot(21:40)
Cheers,
Josh
On Tue, Jul 19, 2011 at 4:07 PM, DrCJones matthias.godd...@gmail.com wrote:
Hi,
par(mfrow = c(2,2))
will create a 2x2 window that I can
DrCJones wrote:
But how can do 3 so that the bottom spans the width of the upper two:
[plot1 plot1]
[p l o t 3]
?layout
for standard graphics (plot..), but that's what you are referring to. For
trellis, you must use other methods.
Dieter
--
View this message in context:
Hi all,
I intend to use R for some image manipulations/analysis. It's relatively
straightforward to read and manipulate image files in R, but I also would like
to store the resulting image (after manipulation) as a bitmap image with the
same resolution as the original image. Here is a
-- Forwarded message --
Date: Mon, 18 Jul 2011 10:17:00 -0700 (PDT)
From: KHOFF kuph...@gmail.com
To: r-help@r-project.org
Subject: [R] cforest - keep.forest = false option?
Hi,
I'm very new to R. I am most interested in the variable importance
measures
that result from
You may want to try out the beta (!!!) version of a new RWinEdt
available from http://www.statistik.tu-dortmund.de/~ligges/RWinEdt
Please use R-2.13.1 or later to try it out (2.13.0 won't work with
Windows 7 and RWinEdt).
Best,
Uwe Ligges
On 18.07.2011 23:16, Soyeon Kim wrote:
Dear All,
Hi All,
This is not really an R question but a statistical one. If someone could
either give me the brief explanation or point me to a reference that might
help, I'd appreciate it.
I want to estimate the mean of a log-normal distribution, given the (log
scale normal) parameters mu and sigma
Melinda,
Any chance you could send me the data that goes with this, offline, (if
you can, I'll only use the data for investigating this issue, of
course)? I can reproduce something going wrong with the edf computation
for fixed smooths in gamm4, when there are random effects, but I can't
get
Dear list,
When I try to run the Ansari-Bradley test on two long vectors I obtain a
warning and the p-value is NA:
set.seed(12)
x - rnorm(10)
y - rnorm(10)
ansari.test(x,y)
Ansari-Bradley test
data: x and y
AB = 5002890779, p-value = NA
alternative hypothesis: true ratio
Hello all,
I am new to xtable. I have several datasets in the form of matrices.
Consider the following two simple datasets which are 2 x 3 matrices. The
rows in both matrices have the same meaning. For example the first row of
both matrices are variable 1 and the second row of both matrices are
Hi all,
How can I calculate the mean from several imputed data sets with the package
mice?
I know you can estimate regression parameters with, for example, lm and
subsequently pool those parameters to get a point estimate using functions
included in mice. But if I want to calculate the mean value
Hi Jeff,
One way to graph the differences between the two years for the first set of
data is via barchart(), a function equivalent to barplot in the lattice
package.
Please check if with this portion of code (and with your data) the graph you
get is quite self-explanatory.
I don't think the OP specified an operating system, but...
A few weeks ago I had a closely analogous problem, seeking files 'menus.txt'
in subdirectories 'etc' but not from other subdirectories;
'/etc/menus.txt'. I made this post
http://tolstoy.newcastle.edu.au/R/e14/help/11/06/5685.html, but
At 03:56 20/07/2011, Joshua Wiley wrote:
On Mon, Jul 18, 2011 at 10:48 AM, a.me...@yahoo.co.uk
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
At 19.07.2011 18:50 -0700, Spencer Graves wrote:
On 7/19/2011 4:04 PM, Bert Gunter wrote:
On Tue, Jul 19, 2011 at 3:45 PM, David
Winsemiusdwinsem...@comcast.net wrote:
On Jul 19, 2011, at 6:29 PM, J. wrote:
Thanks for the answer.
#
However, I am still curious about
Hi,
Instead of hist you can use functions histogram() or densityplot() in
lattice package:
require(lattice)
histogram(
~ length | factor(sex), data=dene
)
densityplot(
~ length, data=dene,
groups=factor(sex),
auto.key = list(space =
On 07/20/2011 01:24 AM, Sally_roman wrote:
In my first post is example data.
Hi Sally,
I was going to suggest:
fish-read.table(fish.dat,header=TRUE)
controlfish-
rbind(fish$pounds[fish$net==Control fish$type == kept],
fish$pounds[fish$net==Control fish$type == discard])
expfish-
On 20/07/11 21:08, Simon Knapp wrote:
Hi All,
This is not really an R question but a statistical one. If someone could
either give me the brief explanation or point me to a reference that might
help, I'd appreciate it.
I want to estimate the mean of a log-normal distribution, given the (log
On 07/20/2011 01:56 PM, Stratford, Jeffrey wrote:
Hi everyone,
I determined the presence of three types parasites in a passerine bird
over two years. I would like to create a bar chart that shows the
proportion infected on the y and year/parasite on the x such that each
type of parasite is
On 20/07/11 11:07, DrCJones wrote:
Hi,
par(mfrow = c(2,2))
will create a 2x2 window that I can use to plot 4 diferent figures in:
[plot1 plot2]
[plot3 plot4]
But how can do 3 so that the bottom spans the width of the upper two:
[plot1 plot1]
[p l o t 3]
Is this possible in R?
In R
2011/7/19 Prof Brian Ripley rip...@stats.ox.ac.uk:
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
Sarah,
You can try this
mean(sapply(1:n.imp, function(x) complete(imp,x)$y))
Weidong Gu
On Wed, Jul 20, 2011 at 6:05 AM, Sarah s1327...@student.rug.nl wrote:
Hi all,
How can I calculate the mean from several imputed data sets with the package
mice?
I know you can estimate regression
On Jul 20, 2011, at 1:34 AM, Daniel Malter wrote:
snipped
requests, except that you were referring to SAS and had heard that R
does
not like loops. (This is factually wrong. But R can be slow looping).
Where did you hear this? Can you cites any references?
--
David Winsemius, MD
West
Let me expand a bit on Thomas's answer.
Looking more closely at your data set you have the following:
death time group 0group 1
1.5 0/413/13
3 0/4 5/5
8 4/4 0
At time 1.5 group 1 had 13 deaths out of
*Is there an implementation of this in R? Thanks.
*
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear all,
Coercing a logical vector to a numeric one is easy. The as.numeric function is
used. However what do we use when we have a matrix or an array?
Sumona
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Thomas Lumley-2 wrote:
[...]
The warning and error messages are correct here. Look at the point
estimate. It's a log hazard ratio of about 20 in one case and about
-20 in the other case. The true partial maximum likelihood estimator
is infinite. The estimated standard errors are
I am analyzing a dataset on the effects of six pesticides on population
growth rate of a predatory mite. The response variable is the population
growth rate of the mite expressed as ln(Nfinal/Nstarting) of the mite,
where N final the population of the mite at the end of the experiment
and N
Dear all,
I am using np package in order to estimate a model with Klein and Spady
estimator. To estimate the model I use
KS - npindexbw (xdat=X, ydat=Y, bandwidth.compute=TRUE,
method=kleinspady, optim.maxit=10^3, ckertype=epanechnikov, ckerorder=2)
and to estimate beta hats standard errors I
On Tue, Jul 19, 2011 at 4:19 PM, J. seoulseoulse...@gmail.com wrote:
@Dimitri: I tried to enter it as numeric and still got the same outcome. I
still wonder if there is any way to get the same result from both programs.
There is. ?C ?contrasts
But of course you must do your homework to
On 20/07/2011 7:14 AM, Mitra, Sumona wrote:
Dear all,
Coercing a logical vector to a numeric one is easy. The as.numeric function is
used. However what do we use when we have a matrix or an array?
Usually you don't need to do anything: if a local is used in
arithmetic, the values are
On Jul 20, 2011, at 7:14 AM, Mitra, Sumona wrote:
Dear all,
Coercing a logical vector to a numeric one is easy. The as.numeric
function is used. However what do we use when we have a matrix or an
array?
Here's one way:
m -matrix(c(1, 2, 3, 4),2)
m
[,1] [,2]
[1,] 1 3
[2,] 2 4
Hi all,
Suppose I have a matrix of logical value:
x-matrix(c(TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE),nrow=3)
I would like to change the value of FALSE to 0 and TRUE to 1. An
obvious way to do it is :
y-as.numeric(x)
However this method doesn't keep the dim of x. I also need to copy
Thanks for the reply.
write.table(esvr.pred) worked - I got data out that's been scaled back
to its original range of values.
2011/7/19, Bert Gunter gunter.ber...@gene.com:
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
untested because I don't have access to your data, but this should work.
b13.NEW - b13[, c(Gesamt, Wasser, Boden, Luft, Abwasser,
Gefährliche Abfälle, nicht gefährliche Abfälle)]
Geophagus wrote:
*Hi @ all,
I have a question concerning the possibilty of grouping the columns of a
*Hi @ all,
I have a question concerning the possibilty of grouping the columns of a
matrix.
R groups the columns alphabetically.
What can I do to group the columns in my specifications?
The script is the following:*
#R-Skript: Anzahl xyz
#Quelldatei einlesen
b-read.csv2(Z:/int/xyz.csv,
How about this:
x-matrix(c(TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE),nrow=3)
x
[,1] [,2] [,3]
[1,] TRUE FALSE TRUE
[2,] FALSE TRUE FALSE
[3,] TRUE FALSE TRUE
ifelse(x, 1, 0)
[,1] [,2] [,3]
[1,]101
[2,]010
[3,]101
Sarah
On Wed, Jul
On Jul 20, 2011, at 10:42 AM, Geophagus wrote:
*Hi @ all,
I have a question concerning the possibilty of grouping the columns
of a
matrix.
R groups the columns alphabetically.
What can I do to group the columns in my specifications?
Dear Earth Eater;
You can create a factor whose levels
Two further methods:
x+0
[,1] [,2] [,3]
[1,]101
[2,]010
[3,]101
mode(x)- numeric; x
[,1] [,2] [,3]
[1,]101
[2,]010
[3,]101
On Jul 20, 2011, at 11:27 AM, Sarah Goslee wrote:
How about this:
Yes. If you want a program that does some of what Mplus does, use the
lavaan package or the sem package.
Dimitri Liakhovitski dimitri.liakhovit...@gmail.com
Sent by: r-help-boun...@r-project.org
07/18/2011 05:36 PM
To
r-help r-help@r-project.org
cc
Subject
[R] Using Mplus via R
Hi,
I am not sure about correct, but R stores logical values TRUE/FALSE
as 1/0 already so simply changing the mode would suffice:
mode(x) - numeric
alternately
x + 0
HTH,
Josh
On Wed, Jul 20, 2011 at 8:16 AM, Julian TszKin Chan cjul...@bu.edu wrote:
Hi all,
Suppose I have a matrix of
what is exactly difference between gee, geese, and ordgee
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
A series of package updates is on CRAN (or in the process to get there).
Already available from CRAN are:
- signal: Since I took over maintainership years ago, I have not
invested the required amount of time into this package - until this
spring and now the package got a NAMESPACE and dozens
It is easier and more straightforward than any of the suggestions
so far. Simply multiply the matrix by 1, or add 0 to it:
x-matrix(c(TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE,TRUE),nrow=3)
1*x
# [,1] [,2] [,3]
# [1,]101
# [2,]010
# [3,]101
-Original Message-
From: Peter Lomas [mailto:peter.lo...@ucalgary.ca]
Sent: Tuesday, July 19, 2011 6:42 PM
To: Nordlund, Dan (DSHS/RDA)
Cc: r-help@r-project.org
Subject: Re: [R] Taking all complete diagonals of a matrix
Thanks very much to everyone who replied. Peter got me on my
EIP(Enterprise Integration Patterns), ESB( Enterprise service bus) are
commonly used terms utilized with Message oriented middleware.
Apache ActiveMQ implements message queues and message topics. The Rjms
package brings the capabilities of messaging to R.
The description states:
This package uses
A new version of irtoys is / will be available on CRAN. Two minor bugs have
been fixed. One of these is more interesting: the previous code did not
anticipate negative estimates for discriminations. What I did not know is that,
unlike ICL or Bilog, ltm does not constrain discriminations to be
On Jul 20, 2011, at 03:42 , Peter Lomas wrote:
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
Dear all, my question is not directly related to R, however I believe that
experts here would not mind anything to have a look on my problem.
Please consider a symmetric matrix and it's eigen values:
set.seed(1)
mat - matrix(rnorm(36), 6)
mat - mat %*% t(mat) # symmetric matrix
mat
Hi
Is it possible to use grofit to get the AIC of several (e.g. two) growth
models and compare both these and model parameters? All I can get it to do
so far is return parameters for a single model.
Cheers
Roland
[[alternative HTML version deleted]]
Thanks for the assistance
How does
strptime(04-MAY-11 1428,format=%d-%b-%y %H%M)
differ from
timeDate(04-MAY-11 1428,format=%d-%b-%y %H%M)
?
-Original Message-
strptime(04-MAY-11 1428,format=%d-%b-%y %H%M)
[1] 2011-05-04 14:28:00
--
Clint BowmanINTERNET:
On Jul 20, 2011, at 3:38 PM, mdkz...@aol.com wrote:
Thanks for the assistance
How does
strptime(04-MAY-11 1428,format=%d-%b-%y %H%M)
differ from
timeDate(04-MAY-11 1428,format=%d-%b-%y %H%M)
?
Wouldn't one need to know where you got this non-base function?
--
David Winsemius, MD
West
Hi,
I posted this question on stats.stackexchange.com 3 days ago but the
answer didn't really address my question concerning the speed in
competing risk regression. I hope you don't mind me asking it in this
forum:
I’m doing a registry based study with almost 200 000 observations and
I want to
I have a 5 column matrix like
12 10 8 6 3
10 9 8 7 5
14 NA 4 NA NA NA
15 NA 10 NA 5
...
I want to select the position of the first entry for each row =5
for example, for the first row, I want to select the last element and return
its position as 5;
for th e third row, I want to select the third
On Jul 20, 2011, at 4:23 PM, gallon li wrote:
I have a 5 column matrix like
12 10 8 6 3
10 9 8 7 5
14 NA 4 NA NA NA
15 NA 10 NA 5
...
Probably something along the lines of
aapply(mtx, 1, function(x) { c( x[ which(x = 5)[1] ], # first row are
the values
Tena koe
Assuming your matrix is called yourMatrix, then try
apply(yourMatrix, 1, function(x) which(x=5))
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of gallon li
Sent: Thursday, 21 July 2011 8:23
Hi friends,
I have got a list where each element might have variable number of members.
$`4213`
[1] 214077_x_at
$`164832`
[1] 225996_at 235977_at
$`339010`
[1] NA
$`23410`
[1] 221562_s_at 221913_at 49327_at
$`285386`
[1] 229764_at
$`2099`
[1] 205225_at 211233_x_at 211234_x_at
Hi Jim,
Perhaps somebody else knows a smoother way, but in the past I have
just built my table in R as a matrix then used xtable. Here's what I
would do with your example:
library(xtable)
dataset1 = matrix( c(1,2,3,4, 5, 6 ), 2 , 3)
dataset2 = matrix( c(4,3,5,10, 1, 0), 2, 3)
dataset -
Looping over the rows, as David did below, is one way.
You can also loop over the columns. This can be faster
when nrow(matrix)ncol(matrix). E.g., with the
following 2 functions
f0 - function (x) {
apply(x, 1, function(aRow) which(aRow = 5)[1])
}
f1 - function (x) {
isGood -
On Jul 20, 2011, at 5:04 PM, Vickie S wrote:
Hi friends,
I have got a list where each element might have variable number of
members.
$`4213`
[1] 214077_x_at
$`164832`
[1] 225996_at 235977_at
$`339010`
[1] NA
$`23410`
[1] 221562_s_at 221913_at 49327_at
$`285386`
[1] 229764_at
$`2099`
---BeginMessage---
Vickie, try something like this
# Dummy data
lst - list(This,c(should, work,
just),fine,I,guess...,c(NA,NA))
names(lst) - letters[seq(1,length(lst))]
lst
# Arranging
for (i in 1:length(lst)) {
lst[[i]] - as.matrix(lst[[i]])
rownames(lst[[i]]) - rep(names(lst)[i],
On Wed, Jul 20, 2011 at 12:02 PM, AO_Statistics abouesl...@gmail.com wrote:
Thomas Lumley-2 wrote:
[...]
The warning and error messages are correct here. Look at the point
estimate. It's a log hazard ratio of about 20 in one case and about
-20 in the other case. The true partial maximum
Hi all,
I am facing difficulty on how to use bootstrap sampling and
below is my example of function.
Read a data , use some functions and use iteration to find the solution(
ie, convergence is reached). I want to use bootstrap approach to do it
several times (200 or 300 times) this whole
Apologies, for a very simple question. I forgot how to do it -
although I remember reading about getting a warning in such a
situation.
I have a data frame. It happens to be 10 rows but it could be 11 or 3 or 13...
x-data.frame(a=1:10)
I need to add variable b that is a sequence of 1:3 - repeated
Never mind - found it:
x-data.frame(a=1:10)
x$b-rep(1:3,nrow(x)%/%3,len=nrow(x))
Dimitri
On Wed, Jul 20, 2011 at 6:15 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Apologies, for a very simple question. I forgot how to do it -
although I remember reading about getting a
Or you could take advantage of R's automatic recycling:
x$b - rep(1:3, length=nrow(x))
x
a b
1 1 1
2 2 2
3 3 3
4 4 1
5 5 2
6 6 3
7 7 1
8 8 2
9 9 3
10 10 1
Thanks for providing a simple reproducible example.
Sarah
On Wed, Jul 20, 2011 at 6:20 PM, Dimitri Liakhovitski
I didn't see bootstrap steps in your code. One way to modify your codes
for (Ncount in 1:100)
{
b.data-data[sample(1:nrow(data),replace=T),]
y -b.data[,1]
x -b.data[,2]
n - length(x)
... ### make appropriate changes if needed
}
Weidong Gu
On Wed, Jul 20, 2011 at 6:09 PM, Val valkr...@gmail.com
A homework problem?
-- Bert
On Wed, Jul 20, 2011 at 10:06 AM, B. Jonathan B. Jonathan
bkheijonat...@gmail.com wrote:
Dear all, my question is not directly related to R, however I believe that
experts here would not mind anything to have a look on my problem.
Please consider a symmetric matrix
Hi Armin,
Please copy the list on your emails. Providing your matrix A (or some
other reproducible example) would be useful to anyone who wanted to
help you. It is easy to do by copying the output from your console
from running:
dput(A)
This would at least let us try out your code on your
Dear list,
I've been learning how to make a 2x2 paneled dotplot in lattice without any
previous experience using lattice.
my code thusfar is:
nut-read.table(/Users/colinwahl/Desktop/nutsimp_noerror.csv, T, sep=
,)
attach(nut)
nut1-data.frame(Nitrate, Total_Nitrogen, Phosphate,
Hi all,
I have R installed on a box, which is running on a machine with 16 core and
Redhat - Linux. I am handling huge (size of dataset will be 5 GB) dataset.
Lets assume that my data is in the form of structured (multiple) logs. I
access the data by using all.files(). Since by default basic
On 20/07/11 18:56, Dieter Menne wrote:
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
Make this reproducible.
On Wed, Jul 20, 2011 at 6:44 PM, Madana_Babu madana_b...@infosys.com wrote:
Hi all,
I have R installed on a box, which is running on a machine with 16 core and
Redhat - Linux. I am handling huge (size of dataset will be 5 GB) dataset.
Lets assume that my data is in
It is not any homework problem. I just need some pointer. Given that I think
I would be able to carry forward.
Thanks,
On Thu, Jul 21, 2011 at 4:52 AM, Bert Gunter gunter.ber...@gene.com wrote:
A homework problem?
-- Bert
On Wed, Jul 20, 2011 at 10:06 AM, B. Jonathan B. Jonathan
Dear List,
After running a compound symmetric model using gls, I realized that
the default contrasts were not the ones that made the most sense given
the biological relationships among the factor levels. When I either
changed the factor levels to re-arrange the order they occur in the
gls model
Hi,
I attempted to install Rgraphviz but ran into a problem. It requires
graphviz 2.20.3. I have this and installed it but the Windows 7 system
path variable has to be modified to include the path to the graphviz bin
file. How do I do this on Windows 7? It's been a long time since I had
This is off-topic, and extremely easy to find an answer to using Google.
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Research
On Jul 20, 2011, at 11:17 PM, B. Jonathan B. Jonathan wrote:
It is not any homework problem. I just need some pointer. Given that
I think
I would be able to carry forward.
Then what kind of problem _is_ it? You say:
nearest matrix ... using what measure for distance or similarity?
...
Thanks David for your pointer. Here my original matrix is VCV matrix which
is the utmost important matrix in finance. However in reality what happens
is that due to incomplete data. lot of missing values (or some other
problems) that matrix may be unstable like min eigen value is negative or
very
http://mlg.eng.cam.ac.uk/dave/rmbenchmark.php
I haven't ever tried it myself, but online sources suggest that Matlab possibly
gains speed by internally avoiding loops rather than looping faster. What would
stand at the end if this were true, however, is improved end user speed.
Daniel
On Jul 21, 2011, at 1:04 AM, Daniel Malter wrote:
http://mlg.eng.cam.ac.uk/dave/rmbenchmark.php
I haven't ever tried it myself, but online sources suggest that
Matlab possibly gains speed by internally avoiding loops rather than
looping faster. What would stand at the end if this were
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