[EMAIL PROTECTED] wrote:
The data set has a number of variables each of which is classified into two
groups.
For each variable of each group, I need to create a histogram. All the
histograms are to be lined up into a file that looks like
group1 group2
Variable
I`ve read all the manuals and still couln`t find what is the difference
between the stacked and side-by-side barplots ? Could you explain me ?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
laba diena [EMAIL PROTECTED] writes:
I`ve read all the manuals and still couln`t find what is the difference
between the stacked and side-by-side barplots ? Could you explain me ?
Did you try
par(ask=TRUE)
example(barplot)
--
O__ Peter Dalgaard Øster Farimagsgade 5,
First, produce two barplots for comparison:
par(mfrow=c(2,1) )
barplot(VADeaths,beside=TRUE)
barplot(VADeaths)
The same information is in both plots; in the top, it is displayed as
5 separate bars for each group, and in the stacked plot it is shown as
5 separate regions in each of the four
On Thu, 2006-10-12 at 10:57 -0400, Leeds, Mark (IED) wrote:
you shouldn't need it. Kalmanlike() ( spelling ) I think is in the base
package and there is atleast
One constributed package and probably many others that do kalman
filtering but I can't recall the names of them.
Check out the
I thought at first that you could use a weighted sample (the sample
function) but, you can't since it doesn't take proper account of
replacement if you try that.
You can use the list approach, but through the power of R, you don't
need a lot of loops to do it...
I can't speak for the
I have a data file with 3 columns and I need to take only 2, how to do that
?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
I'm not quite sure what you mean, but if you are wanting to select
columns of a data frame, have a look at
help([)
David
On 13/10/06, laba diena [EMAIL PROTECTED] wrote:
I have a data file with 3 columns and I need to take only 2, how to do that
?
[[alternative HTML version
First, here's the specific bug I have. Later I'll say why I care.
ls(zappo)
Error in try(name) : object zappo not found
# good.
f = function(zappo) { function(y) zappo + y }
g = f(1)
g(1)
[1] 2
formals(g)
$y
formals(g)$y
formals(g)$y = 2
g
function (y = 2)
zappo + y
g(1)
Ah, it's fixed in 2.4.0. I'll work around it.
-Alex
On 13 Oct 2006, at 11:19, Alex Brown wrote:
First, here's the specific bug I have. Later I'll say why I care.
ls(zappo)
Error in try(name) : object zappo not found
# good.
f = function(zappo) { function(y) zappo + y }
g = f(1)
g(1)
Dear Sir or Madam,
I'm wondering if there is any routine or argument in the function 'glm.fit'
that makes it handle NA's. The function 'glm' can handle NA's but I can't
make make it work (or find anything written on this in the help files) with
'glm.fit'.
Is it even possible in'glm.fit'? How?
Dear R Users,
Suppose comparing two non-normal distributions is our interest. Like
distribution of financial time series, they are negative skewed with fat tail.
Which test can better help and in which pachage? ( For example in
goodness-of-fit) Kolmogorov-Smirnov test has its own
Dear R-help,
I posted this on 4 Oct but got no response (I wasn't even told to go away and
do some more background reading ;) ). I am reposting it in the, perhaps, vain
hope that someone with knowledge of the subject will reply, if only to point me
in a different direction to which I am now
Dear useRs,
I'm trying to create a barplot like so:
x=matrix(1:10,2,5)
barplot(x,leg=c(left,right),besid=T)
The legend is placed in default position topright, however the data are
plotted there too. I tried controlling the legend position by adding
x=topleft but this results in an error that x
Dear all,
I would like to multiply two matrixes with the different dimension column
by column. Let make an example:
If I have two matrixes X and Yas follow:
X- matrix(1:12, nrow=4, ncol=3, dimnames=list(c(A,B,C,D),
c(stage1,stage2,stage3)))
Y- matrix(1:28, nrow=4, ncol=7,
Frank Harrell wrote:
[...]
Thank you Brian. It seems that no matter what is the right answer, the
answer currently returned on my system is clearly wrong. lowess()$y
should be constrained to be within range(y).
Really? That assertion is offered without proof and I am pretty sure is
Dear Sir/ Madam,
I would like to ask you about mantel test in R. In ade4 package, if I
want to use mantel.rtest, I get error massage Object of class to dist
expected. As I already have two dissimilarity matrices, shall I again
compute distance measure using this function?
If not, could you
On Fri, 2006-10-13 at 12:51 +0200, Fredrik Thuring wrote:
Dear Sir or Madam,
I'm wondering if there is any routine or argument in the function 'glm.fit'
that makes it handle NA's. The function 'glm' can handle NA's but I can't
make make it work (or find anything written on this in the help
For example :
x=matrix(1:10,2,5)
barplot(x,besid=T)
legend(topleft, c(left,right), density= c(0,1000))
2006/10/13, Ingmar Visser [EMAIL PROTECTED]:
Dear useRs,
I'm trying to create a barplot like so:
x=matrix(1:10,2,5)
barplot(x,leg=c(left,right),besid=T)
The legend is placed in
Thanks, this could work!
However, the legend does not reproduce the color/shading used in the
original
barplot, are those available somehow?
Best, Ingmar
From: David Hajage [EMAIL PROTECTED]
Date: Fri, 13 Oct 2006 14:11:21 +0200
To: Ingmar Visser [EMAIL PROTECTED]
Cc: R-help@stat.math.ethz.ch
Dear R helper,
does anyone have an idea on why R.2.4.0 draws the surface for the two
command lines below and the next time it renders the error message below
for exactly the same command lines:
norm.cop - normalCopula(0.5)
persp(norm.cop, dcopula)
Error in ceiling(length.out) : Non-numeric
On Fri, 2006-10-13 at 14:07 +0200, Hossein Moradi wrote:
Dear Sir/ Madam,
I would like to ask you about mantel test in R. In ade4 package, if I
want to use mantel.rtest, I get error massage Object of class to dist
expected. As I already have two dissimilarity matrices, shall I again
Hello
My data looks ugly in a normal histogramm. How can I create a histogramm with a
Y-axis in log-scale?
Thanks for your help!
David Graf
--
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Hi
R newbie here :)
I need to plot 3 barplots in the same axis, something like
|
| ___
| | | _| | _| | _
| _ | || | _ | || | _ | || |
| | || || || || || || || || |
-+-
I f I understand you correctly, I don't think that your model is doing
what you think it is.
Look at the model.matrix. Consider a toy example:
x - 1:10
y - factor(letters[1:2])
dd - expand.grid(x, y)
dd$resp - rnorm(20)
model.matrix(~Var2/Var1+1, dd)
(Intercept) Var2b Var2a:Var1
If you are just modifying an S3 method in a package you may not need to reinsert
the method into the package since UseMethod first looks into the
caller environment
for methods anyways and only second does it look for methods in the
package. Thus:
HTML.data.frame - R2HTML:::HTML.data.frame
Prof Brian Ripley wrote:
Frank Harrell wrote:
[...]
Thank you Brian. It seems that no matter what is the right answer, the
answer currently returned on my system is clearly wrong. lowess()$y
should be constrained to be within range(y).
Really? That assertion is offered without proof
Here are two ways:
1. Using inner from:
http://tolstoy.newcastle.edu.au/R/help/05/04/3709.html
try:
array(inner(t(X), Y, *), c(4, 21))
2. using model.matrix get all terms and interactions and eliminate the
non-interactions:
model.matrix(~ X * Y - X - Y - 1)
On
Majid Iravani wrote:
I would like to multiply two matrixes with the different dimension
column by column. Let make an example: If I have two matrixes X
and Yas follow:
X- matrix(1:12, nrow=4, ncol=3, dimnames=list(c(A,B,C,D),
c(stage1,stage2,stage3)))
Y- matrix(1:28, nrow=4, ncol=7,
apply(Y, 2, function(y)list(y*X))
On 13 Oct 2006, at 12:33, Majid Iravani wrote:
Dear all,
I would like to multiply two matrixes with the different dimension
column
by column. Let make an example:
If I have two matrixes X and Yas follow:
X- matrix(1:12, nrow=4, ncol=3,
Actually, I'm a beginner too :-) It's just this list is so helpful, and I was
helped so many times, that I am trying to bring my small contribution :-)
How about:
#_
Y1=X*Y[,1]
Y=cbind(Y,Y1)
Y
site1 site2 site3 site4 site5 site6 site7 stage1 stage2
Gavin, you are right. I thought he just wanted a routine. My bad.
-Original Message-
From: Gavin Simpson [mailto:[EMAIL PROTECTED]
Sent: Friday, October 13, 2006 5:00 AM
To: Leeds, Mark (IED)
Cc: Malini Subramanian; R-help@stat.math.ethz.ch
Subject: Re: [R] C code for KalmnaLike
On Thu,
As there is no function normalCopula in R 2.4.0, your example is
incomplete. Please see the footer of this message (and perhaps contact
the maintainer of the package involved).
Please don't send messages repeatedly: if you don't get an answer the
first time study the posting guide and work
Hello,
I am trying to insert a lot of data into a table using windows R (2.3.1)
and a mysql database via RODBC.
First I read a file with read.csv and then form sql insert statements
for each row and execute the insert query one row at a time. See the
loop below.
This turns out to be very slow.
On Thu, 2006-10-12 at 23:08 -0300, Andre Nathan wrote:
Hi
R newbie here :)
I need to plot 3 barplots in the same axis, something like
|
| ___
| | | _| | _| | _
| _ | || | _ | || | _ | || |
| | || || || ||
When updating to the very last version of RODBC under freebsd 6.1 the
errors below pop up but RODBC compiles till the end and, it seems, to
work properly.
What are those errors about?
Vittorio
..
checking for suffix of
executables...
checking for
Hi
How do I generate all ways of ordering sets of indistinguishable items?
suppose I have two A's, two B's and a C.
Then I want
AABBC
AABCB
AACBC
ABABC
. . .snip...
BBAAC
. . .snip...
CBBAA
[there are 5!/(2!*2!) = 30 arrangements. Note AABBC != BBAAC]
How do I do this?
--
Robin Hankin
Hi all,
I have the following -newbye- problem.
Inside R, I am trying to process a file and creating from it many files.
The file is organized in different columns, the second containing a code. I
want to create as output objects, which contain only entries in a certain code
range, and whose
e.g.:
barplot(x,col=c(lightgrey,darkgrey),besid=T)
legend(topleft,c(left,right),fill=c(lightgrey,darkgrey))
try:
?legend
and
example(legend)
for documentation!!!
Ingmar Visser schrieb:
Thanks, this could work!
However, the legend does not reproduce the color/shading used in the
original
I suggest you look at one of the guides:
http://cran.r-project.org/other-docs.html
before answering questions like this to the mailing list... please read
the posting guide!
laba diena schrieb:
I have a data file with 3 columns and I need to take only 2, how to do that
?
[[alternative
Hi,
I have some similar problems. Some times ago this problem dont there existed.
Look this simple example:
Y - c(0,0,0,1,4,8,16)
X - c(1,2,3,4,5,6,7)
m - glm(Y~X,family=gaussian(link=log))
Error in eval(expr, envir, enclos) : cannot find valid starting values: please
specify some
m -
Hi everyone,
I need some help to produce a random bistochastic matrix,
that is a squared matrix of positive real numbers e_ij, with sum(e_i)=1
and sum(e_j)=1.
Thanks
Florent Bresson
___
Thank you, Alex! That's exactly what I was looking to do. I'm going to
remove the loops and use your apply function approach. Best regards and
much thanks, brian
On 10/13/06, Alex Brown [EMAIL PROTECTED] wrote:
I thought at first that you could use a weighted sample (the sample
function)
On Fri, 2006-10-13 at 09:09 -0400, Bill Szkotnicki wrote:
Hello,
I am trying to insert a lot of data into a table using windows R (2.3.1)
and a mysql database via RODBC.
First I read a file with read.csv and then form sql insert statements
for each row and execute the insert query one row
Large for loops are slow. Try to avoid them using apply, sapply, etc.
I've made the paste statements a lot shorter by using collapse. See
?paste for more info.
Append.SQL - function(x, channel){
sql=INSERT INTO logger (time, v1, v2, v3, v4, v5, v6, v7, v8, v9,
v10)
--- Ingmar Visser [EMAIL PROTECTED] wrote:
Thanks, this could work!
However, the legend does not reproduce the
color/shading used in the
original
barplot, are those available somehow?
Best, Ingmar
?legend
Try
x=matrix(1:10,2,5)
barplot(x,besid=T, col=c(red,blue))
legend(topleft,
On Fri, 2006-10-13 at 13:33 +0200, David Graf wrote:
Hello
My data looks ugly in a normal histogramm. How can I create a
histogramm with a Y-axis in log-scale?
Thanks for your help!
David Graf
I'm not sure that you want to use a log scale here, but may be better
served by log
Hi. I'm attempting to fit a logistic/binomial model so I can determine
the influence of landscape on the probability that a box gets used by a
bird. I've looked at a few sources (MASS text, Dalgaard, Fox and
google) and the examples are almost always based on tabular predictor
variables. My
On Fri, 2006-10-13 at 14:59 +0100, Vittorio wrote:
When updating to the very last version of RODBC under freebsd 6.1 the
errors below pop up but RODBC compiles till the end and, it seems, to
work properly.
What are those errors about?
I don't know what adverse effect this may have in RODBC.
Is there a reason why the data have to be inserted 1 row at a time?
Is it possible to insert the entire table at once?
sqlSave perhaps.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bill Szkotnicki
Sent: Friday, October 13, 2006 9:09 AM
To:
--- laba diena [EMAIL PROTECTED] wrote:
I have a data file with 3 columns and I need to take
only 2, how to do that
Have a look at the manual?
An Introduction to R
2.7 Index vectors; selecting and modifying subsets of
a data set
__
Dear all,
I would like to use R-comands within Perl-Scripts. How can I do this?
Yours
Torsten
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear R users,
we are trying to do some parallel computing using library(snow).
In particular we have a cluster with 3 nodes
cl - makeCluster(3, type = MPI)
3 slaves are spawned successfully. 0 failed.
and we want to compute the function op_mat (see below) first with the
master and then
comments in line
Roy Sanderson wrote:
Dear R-help
I have two separate experiments, one a repeated-measures design, the other
a split-plot. In a standard ANOVA I have usually undertaken a
multiple-comparison test on a significant factor with e.g TukeyHSD, but as
I understand it such a
I am trying to insert a lot of data into a table using windows R (2.3.1)
and a mysql database via RODBC.
First I read a file with read.csv and then form sql insert statements
for each row and execute the insert query one row at a time. See the
loop below.
This turns out to be very slow.
Can
--- Andre Nathan [EMAIL PROTECTED] wrote:
Hi
R newbie here :)
I need to plot 3 barplots in the same axis,
something like
|
| ___
| | | _| | _| | _
| _ | || | _ | || | _ | || |
| | || || || || || ||
Hi Robin,
This approach first generates all combinations and then eliminates the
non-feasible ones. It should work fine for smallish vectors but might not
scale well for larger vectors. Hopefully it gives you what you need for
this problem.
xx - c(A,A,B,B,C)
yy - 1:length(xx)
zz -
On Fri, 2006-10-13 at 09:28 -0500, Jeffrey Stratford wrote:
Hi. I'm attempting to fit a logistic/binomial model so I can determine
the influence of landscape on the probability that a box gets used by a
bird. I've looked at a few sources (MASS text, Dalgaard, Fox and
google) and the examples
Jeffrey Stratford said the following on 10/13/2006 9:28 AM:
Hi. I'm attempting to fit a logistic/binomial model so I can determine
the influence of landscape on the probability that a box gets used by a
bird. I've looked at a few sources (MASS text, Dalgaard, Fox and
google) and the
Torsten,
your message is a bit terse.
Do you have Omegahat's RSPerl package installed?
If not, visit www.omegahat.org, and have a look at the documentation. (Btw,
you will find a lot of other useful stuff there.)
Sorry that I can't offer more at the moment -- but to know where to start
may be
Hi Christos
thanks for this. Unfortunately, this approach wouldn't work for me
because the real problem is too big for it: I have
letters A-F and two of each, giving
12!/(2^6) ~= 7e6 combinations (borderline feasible)
But in the approach you coded up below, matrix zz would have
6^12 ~= 2e9
Use 'permutations' in 'gtools'
x - permutations(5,5)
y - c('a','a','b','b','c')[x]
dim(y) - dim(x)
unique(y)
On 10/13/06, Robin Hankin [EMAIL PROTECTED] wrote:
Hi
How do I generate all ways of ordering sets of indistinguishable items?
suppose I have two A's, two B's and a C.
Then I
On Fri, 2006-10-13 at 09:28 -0500, Jeffrey Stratford wrote:
Hi. I'm attempting to fit a logistic/binomial model so I can determine
the influence of landscape on the probability that a box gets used by a
bird. I've looked at a few sources (MASS text, Dalgaard, Fox and
google) and the examples
Thanks for the help ... the sqlSave() function was the solution.
The lesson, which has been stated many times before, is to avoid loops
wherever possible!
Bill
# fast RODBC inserting
dat - cbind(as.character(strptime(ti[,2],%d/%m/%y %H:%M:%S
%p)),ti[,3:12])
# you need the as.character to make
There is.
# ls /usr/local/lib/libpcre*
@libpcre.so
libpcre.so.0
Vittorio
Alle 14:39, venerdì 13 ottobre 2006, Jerome Asselin ha scritto:
On Fri, 2006-10-13 at 14:59 +0100, Vittorio wrote:
When updating to the very last version of RODBC under freebsd 6.1 the
errors below pop up but RODBC
clusterCall invokes the same function on all three nodes. You have
basically discovered the communication costs of performing the
calculation in parallel.
You'll get the easiest gains from snow (and other parallel packages in
R) with 'embarrassingly parallel' problems, where the same algorithm is
On Fri, 13 Oct 2006, Marc Schwartz wrote:
On Fri, 2006-10-13 at 13:33 +0200, David Graf wrote:
Hello
My data looks ugly in a normal histogramm. How can I create a
histogramm with a Y-axis in log-scale?
Thanks for your help!
David Graf
There is a log-histogram (called log.hist) in my
I would like to add auxiliary information to the bottom of two strips on
each panel that comes from a table look-up using the values of two
variables that define the panel. For example I might panel on sex and
race, showing 3 randomly chosen time series in each panel and want to
add (n=100)
On Fri, 13 Oct 2006, Michela Cameletti wrote:
Dear R users,
we are trying to do some parallel computing using library(snow).
In particular we have a cluster with 3 nodes
cl - makeCluster(3, type = MPI)
3 slaves are spawned successfully. 0 failed.
and we want to compute the
Why don't you read the whole file in and the use subsetting to get
your ranges instead of reading the file multiple time using 'gawk'.
You can then use 'assign' to create your objects; it would be better
to use a list.
indice - (201:399)
result - list()
x - read.table('base_6_mod')
for (i in
below is a very simple bash script to run Sweave from a cygwin terminal, run
pdflatex on
the generated .tex file, and then view the resulting .pdf output.
i usually use cygwin when i am (forced to be on) Windoze, but i found a few
issues
with paths that this script works around.
pdfview,
I have a user who is currently running R on a desktop system which takes 3
days to run.
We have an Itanium 2 Cluster running HP UX. My system manager has tried to
install R and has sent the following message
INSTALL file says do
./configure
make
./configure fails with ( tail end
read your data frame in all at once and then cut it on x[2] and split
the result, e.g.
split(iris, cut(iris$Sepal.Length, 4:8))
Please provide reproducible code. Without input its not reproducible.
See last line of every message to r-help.
On 10/13/06, Marco Grazzi [EMAIL PROTECTED] wrote:
On Fri, 2006-10-13 at 18:07 +, vittorio wrote:
There is.
# ls /usr/local/lib/libpcre*
@libpcre.so
libpcre.so.0
I don't use freebsd. So I'm not sure how to help. As I hinted before,
I'd first try to reinstall or update the pcre package and make sure that
all its dependencies are
On Fri, 13 Oct 2006, Michela Cameletti wrote:
Dear R users,
we are trying to do some parallel computing using library(snow).
In particular we have a cluster with 3 nodes
cl - makeCluster(3, type = MPI)
3 slaves are spawned successfully. 0 failed.
and we want to compute the
On 13-Oct-06 Robin Hankin wrote:
Hi
How do I generate all ways of ordering sets of indistinguishable
items?
suppose I have two A's, two B's and a C.
Then I want
AABBC
AABCB
AACBC I think you mean AACBB here!
ABABC
. . .snip...
BBAAC
. . .snip...
CBBAA
[there are
On Fri, 13 Oct 2006, Robin Hankin wrote:
Hi
How do I generate all ways of ordering sets of indistinguishable items?
suppose I have two A's, two B's and a C.
Then I want
AABBC
AABCB
AACBC
ABABC
. . .snip...
BBAAC
. . .snip...
CBBAA
[there are 5!/(2!*2!) = 30 arrangements. Note
Folks:
This interesting dicussion brings up an issue of what I have referred to for
some time as safe statistics, by which I mean:
Usually, but not necessarily automated)Statistical procedures that are
guranteed to give either
(a) a reasonable answer; or
(b) Do not give an answer and when
Hello R group,
Given a correlation matrix, I would like to obtain the best subset of
pairs in the matrix of some size n such that the mean of r for that
subset is a maximum compared to any other possible subset of size n.
I've been looking at the deal and subselect packages but they don't
Gavin,
That worked! I went through and I found a few missing cases where I had
. instead of NA - I'm still in SAS mode.
Many thanks!
Jeffrey A. Stratford, Ph.D.
Postdoctoral Associate
331 Funchess Hall
Department of Biological Sciences
Auburn
I've tried to think of an efficient and economical (and therefore
clever) way of doing this for larger problems; but that will have
to wait for another day!
The ruby permutations library
(http://permutation.rubyforge.org/doc/index.html) references The
Algorithm Design Manual, Steven S.
Take a look at the xysplom function in package HH.
You can use it as a model for what you want.
tmp - data.frame(x=rnorm(24),
y=rnorm(24),
a=factor(rep(letters[1:2],12)),
b=factor(rep(LETTERS[1:3], c(8,8,8
xysplom(y ~ x | a*b, data=tmp,
Hello R group,
Given a correlation matrix, I would like to obtain the best subset of
pairs in the matrix of some size n such that the mean of r for that
subset is a maximum compared to any other possible subset of size n.
I've been looking at the deal and subselect packages but they don't
On Windows you could just put this into sweave.bat, say, and then
place that anywhere in your path (or in the current directory):
set infile=%~sdpn1
set infile=%infile:\=/%
cmd Rcmd Sweave %infile%.Rnw
pdflatex %infile%.tex
start %infile%.pdf
On 10/13/06, Thomas Harte [EMAIL PROTECTED] wrote:
Hello R group,
Given a correlation matrix, I would like to obtain the best subset of
pairs in the matrix of some size n such that the mean of r for that
subset is a maximum compared to any other possible subset of size n.
I've been looking at the deal and subselect packages but they don't
Dear Deepayan and Sundar,
Thank you so much for your help with this. However, I may have phrased
my problem too specifically, assuming that *in general* I could apply
your response to all Lattice graphics.
What I need is a barchart or vertical dotchart, with error bars, across
three
Using par seems easily put a hist and a density side by side on the same
output window.
I would like to use some features in histogram from Lattice, but how can I put
histogram and densityplot side by side on the same graph?
Thank you
par(mfrow=c(2,1))
hist(y)
plot(density(y))
Jue Wang,
On 10/13/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Using par seems easily put a hist and a density side by side on the same
output window.
I would like to use some features in histogram from Lattice, but how can I
put
histogram and densityplot side by side on the same graph?
Dear Listers:
I happen to have this question in mind, is there a way to evaluate the
correlation between
a continuous variable and a categorical variable (without discretizing
the former)? My intuitive is using lda by considering the latter as
response variable but not sure.
thanks,
weiwei
--
Thanks for the thought in any case Mark. Your right about the brute force.
I'll expand a bit with an example though for the sake of clarity.
Given a correlation matrix of 4 covariates ABCD with distances of:
AB=0.2; AC=0.6; AD=0.3 ; BC=0.9 ; BD=0.8 ; CD=0.7
Find the optimal subset (size n, n
Thanks for the thought in any case Mark. Your right about the brute force.
I'll expand a bit with an example though for the sake of clarity.
Given a correlation matrix of 4 covariates ABCD with distances of:
AB=0.2; AC=0.6; AD=0.3 ; BC=0.9 ; BD=0.8 ; CD=0.7
Find the optimal subset (size n, n
useRs and developeRs-
Apologies for my naivety, but I just couldn't figure out how to open an
old workspace (created using R 2.3.0) using R 2.3.0 and not R 2.4.0
which is currently happening. For all I know this has always been the
case, but I'm having a problem with a function that doesn't work
On 10/13/06, Daniel E. Bunker [EMAIL PROTECTED] wrote:
Dear Deepayan and Sundar,
Thank you so much for your help with this. However, I may have phrased
my problem too specifically, assuming that *in general* I could apply
your response to all Lattice graphics.
What I need is a barchart or
On Fri, 13 Oct 2006 17:15:45 -0400 Weiwei Shi wrote:
Dear Listers:
I happen to have this question in mind, is there a way to evaluate the
correlation between
a continuous variable and a categorical variable (without discretizing
the former)? My intuitive is using lda by considering the
I see.
i think the question is, I did not have a clear idea of the
correlation between them (if I insist no transformation). Otherwise,
for a binary variable case, maybe a simple one-way t-test serves the
purpose if I defined such correlation or dependency as the group mean
difference.
thanks.
Thanks for the thought in any case Mark. Your right about the brute force.
I'll expand a bit with an example though for the sake of clarity.
Given a correlation matrix of 4 covariates ABCD with distances of:
AB=0.2; AC=0.6; AD=0.3 ; BC=0.9 ; BD=0.8 ; CD=0.7
Find the optimal subset (size n, n
Start R-2.3.1 from the icon which is probably still on your desktop.
setwd() to the directory where your CoolStats.RData sits.
load(CoolStats.RData)
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On Fri, 13 Oct 2006 18:17:10 -0400 Weiwei Shi wrote:
I see.
i think the question is, I did not have a clear idea of the
correlation between them (if I insist no transformation). Otherwise,
for a binary variable case, maybe a simple one-way t-test serves the
purpose if I defined such
Reading the list of changes for R version 2.4.0, I was happy to see that the
row names of dataframes can be stored compactly (as the integer n when
row.names(df) is 1:n).
help(row.names) contains this paragraph:
Row names of the form '1:n' for 'n 2' are stored internally in a
compact
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Hi Robin.
When I saw this, I thought expand.grid would do.
But since it is too big and since I sympathize
that C isn't the ideal use of ones time, perhaps
the Combinations package on www.omegahat.org
might be helpful. This provides a
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