Greetings:
I am forming an R Group on LinkedIn.Com for job seekers and potential
employers. Please consider joining.
http://www.linkedin.com/e/gis/77616/0AEFE3574537
Regards,
Ajit
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On Thu, 27 Mar 2008, Tribo Laboy wrote:
I realized that not everyone has Matlab and that basically the issue
is purely how to deal with the returned data in R, so I have revised
my example code and made it easier to copy-paste and run:
Only for those with matlab! The rest of us have little
Oh please don't recommend misuse of @ to those already confused.
@ is for accessing slots in S4 objects. This 'works' because they happen
to be stored as attributes. See the help page (and the warning that it
does no checking - we may change that).
Similarly,
plt$title - My Title
works
Dear all;
I want to know if I can make a two way MANONA , Manova, and Anovas
with randonization using the R program, if possible, How I can make
my matrix for perform this tests.
Thanks
Catalina Lopez-Ospina
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Could someone please tell me if this book is still worth buying and using or
is it outdated and not very practical for use in R.
Thank you.
--
View this message in context:
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Sent from the R help mailing list
Hi!
In this case you are better off moving away from qplot (which does
various substitute tricks to assemble the named variables in a data
frame) to a more explicit form of ggplot:
pl2[[obs]] - ggplot(cData, aes_string(x=x3, y=obs)) + geom_point()
+ opts(title = obs)
Ok, I will try that,
So am I to understand that the only realy _correct_ and _recommended_
way of accessing the attributes is through
attr(someobject, attributename) ?
Regards,
TL
On Thu, Mar 27, 2008 at 4:16 PM, Prof Brian Ripley
[EMAIL PROTECTED] wrote:
Oh please don't recommend misuse of @ to those already
On Wed, 26 Mar 2008, glenn andrews wrote:
I am using the non-linear least squares routine in R -- nls. I have a
dataset where the nls routine outputs tight confidence intervals on the
2 parameters I am solving for.
nls() does not ouptut confidence intervals, so what precisely did you do?
I
Hello,
Thanks for the input. I thought it wasn't much fun to talk to myself
on the public forums. ;-)
Trying the line you suggested generated and error:
as.data.frame(drop(labpcimport))
Error in data.frame(Maker = list(HP, HP, Sony, DELL, whitebox, :
arguments imply differing number of
Now, how is it that I can access the contents of a named list by
dynamically computed name?
To go back to my previous example I have a list and I know the names.
Now I want do something with that named data in a loop.
lst - list(x = 1:3, y = 4:6, z = 7:9)
nm -names(lst)
nm
[1] x y z
I can
At 05:06 PM 3/26/2008, Ted Harding wrote:
On 26-Mar-08 21:26:59, Ala' Jaouni wrote:
X1,X2,X3,X4 should have independent distributions. They should be
between 0 and 1 and all add up to 1. Is this still possible with
Robert's method?
Thanks
I don't think so. A whileago you wrote
The
Thanks, this really solved my problem.
Actually, I do have a good introduction to R - a book co-authored by
some W.N. Venables and some B.D. Ripley, colloquially called 'MASS' is
on my desk. I find it really very helpful. Still, as it is a book on
statistics, some details on R are only
I have packaged the above posted code as a function and I am posting
it here in case someonw would find it useful in the future.
--function begin -
readMat2df - function(readfiledata, datastr){
tmpdata - readfiledata[[datastr]] # use the string contained
I realized that not everyone has Matlab and that basically the
issue
is purely how to deal with the returned data in R, so I have
revised
my example code and made it easier to copy-paste and run:
#Make a data frame in R
Maker - factor(c(HP, HP, Sony, DELL, whitebox,
Dear R developers,
i would like to start R with a *.RData argument under Linux.
Something like R -f /home/user/workspace.RData
Is this possible?
Thanks in advance for any answers.
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View this message in context:
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Hi R people,
In lme package, I'm searching to find equivalent formula of:
aov(frt~consistency*length*context+Error(subject/(consistency*length*context),data=agE1B4)
I try this :
lme(fixed=frt~consistency*length*context, random=~1|subject, data=agE1B4)
but I did not obtain same F value with
Hi,
This may be a stupid question, but how are the std. error values returned
by lm() calculated? For example
summary(lm)
Call:
lm(formula = log10(moments[2, 1:10]) ~ log10(L_vals[1:10]))
Residuals:
Min 1Q Median 3QMax
-0.0052534 -0.0019473 0.0006785
Sorry to come back on callNextMethod, I am still not very confident about it.
Consideres the following (there is a lot of code, but very simple with
almost only some cat) :
--
setClass(A,representation(a=numeric))
setValidity(A,function(object){cat( * Valid A *\n);TRUE})
Hello,
is there any function in the package snow to check for a really running
cluster?
The function checkCluster only checks the variable cl. And the variable
is still available after stopping the cluster!
( a simple solution would be deleting the cluster variable cl in the
function
Type summary.lm and read the code.
Best,
Uwe Ligges
Nick Chorley wrote:
Hi,
This may be a stupid question, but how are the std. error values returned
by lm() calculated? For example
summary(lm)
Call:
lm(formula = log10(moments[2, 1:10]) ~ log10(L_vals[1:10]))
Residuals:
Reinstall Matrix.
Uwe Ligges
Guy Forrester wrote:
Dear all,
I an running R on a Windows 2000 machine (1.5Gb RAM) and am trying to load
the lme4 package, however, whenever I attempt to load the library lme4 I
get the following error message
I have installed lme4 and Matrix locally from
On Thu, 27 Mar 2008, Bio7 wrote:
Dear R developers,
i would like to start R with a *.RData argument under Linux.
Something like R -f /home/user/workspace.RData
Is this possible?
$ R
...
load(/home/user/workspace.RData)
or put that line in your .Rprofile -- see ?Startup.
(My guess is
That is a *Perl* error, and my guess is that you haven't set your
environment variables correctly. R was not even started.
RSPerl is an Omegahat package, and as the Omegahat mailing lists were down
(last time I looked) I suggest you contact the author. The R posting
guide asks that non-R
Jamie Ledingham wrote:
Dear all, I have a problem with a loop, if anyone has any knowledge on
these things I would appreciate some comments. The code below is
designed to allow me to extract the top record of the data frame, and
them remove rows from the data frame which have an index close
Maybe something like:
f - function(x,y) ifelse(sign(y[x])==-1,return(y[x-1]), return(y[x]))
test[,3] - sapply(seq_along(test[,3]),f,test[,3])
Ravi S. Shankar wrote:
Hi R,
I have a dataframe with
dim(test)
[1] 435150 4
class(test)
[1] data.frame
In the third column every time
Oops, missed the first line of the example:
storm.data-data.frame(meas.index=10001:2,cumrain=runif(1,-10,10)+10)
You can generalize the function to whatever column names you have like this:
find.max.rain-function(raindata,howmany,raincol) {
# a lazy way of getting the same structure
This is to announce that versio 2.0 of the TeachingDemos package is now
on CRAN (and hopefully soon to a mirror near you).
This is a major upgrade of the package, some of the improvements
include:
The package now uses an NAMESPACE so it is easier to load just the parts
that you need.
Some of
Hello Vincent,
On Thu, Mar 27, 2008 at 11:25 AM, Vincent Alcouffe [EMAIL PROTECTED] wrote:
I have 60 Macintoshs on Mac OS X 10.4.11 for learn R to studient
of University and i want to Install New Packages. I click on the button
access list Mirrors and it propose me a list of mirrors.
Hi,
Im using lme to calculate a mixed factors ANOVA according to:
px_anova = anova(lme(dep~music*time*group, random = ~1|id, data = px_data))
where
dep is a threshold,
time is a repeated measures variable (2 levels)
group is a between subjects variable (2 levels)
id is a random factor (subject
Hello.
I have made many normal boxplots where I have added a new boxplot to
an existing one. When I have done this, I have used the at command to
move the boxplots a bit so that they could fit next to eachother, like
this:
boxplot(data.., at = number_of_categories-0.15)
boxplot(data..,
Hi,
Im using lme to calculate a mixed factors ANOVA according to:
px_anova = anova(lme(dep~music*time*group, random = ~1|id, data = px_data))
where
dep is a threshold,
time is a repeated measures variable (2 levels)
group is a between subjects variable (2 levels)
id is a random factor (subject
Dear list,
I have a dataset containing values obtained from two different instruments (x
and y).
I want to generate 5 samples from normal distribution for each instrument based
on
their means and standard deviations. The problem is values from both
instruments are
non-negative, so if
Ok, I will try that, thanks. BTW, where is this aes_string option
documented, sounds useful? How could I do the same thing with facetting?
If I want to save something like . ~ groupVar as a string in a
variable, could I pass it with facet_string to ggplot?
It's a see also from ?aes (or at
xydf - data.frame(x = 1:5, y = 11:15)
plt - ggplot(data = xydf, aes(x = x,y = y)) + geom_point()
attributes(plt)
Now we can change the title:
plt$title - My Title
plt
This is rather poorly documented, but the preferred way of setting the
title is now:
p + opts(title = My Title)
Thanks to David and Zaihra for their help.
Besides plotting the difference between 2 ecdfs, I also would like it to
*look like* a step function. Any suggestion?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of David Winsemius
Sent: Friday, March 21, 2008
On Thu, 27 Mar 2008, Tom Cohen wrote:
Dear list,
I have a dataset containing values obtained from two different
instruments (x and y). I want to generate 5 samples from normal
distribution for each instrument based on their means and standard
deviations. The problem is values from both
Please chk out the url below it might be of some help for plotting step
function or *look like* of step function.
On Thu, 27 Mar 2008 09:03:58 -0400 wrote:
Thanks to David and Zaihra for their help.
Besides plotting the difference between 2 ecdfs, I also would like it to
Thanks for this.
I was afraid someone was going to say this ...
Does this mean the only way of getting this to run faster is by
moving to C code?
The cases I'm thinking of applying this in have dimensions of A that
are much larger than
the example, eg n by n by T where n has a max of 10 or so
It could be almost anything. You need to supply some
information about what you are doing and your system
It would be a good idea to supply the information
that sessionInfo() gives.
--- Huilin Chen [EMAIL PROTECTED] wrote:
Hi
When I copy a block of code to R console, the code
would be
Yes, indeed. The help page says that @ extracts the contents of a slot in
S4 objects. But you mention below that this 'works' for S3 objects because
S4 slots are stored as attributes. Doesn't this mean that @ is currently
implemented to access attributes of objects in general (attributes of S3
Dear all,
I have a piece of code along the lines of
f - function(x) {
clipname - LK # but is in real determined based on info in data.frame x
# other manipulations
return( list(clipname=list()))
}
My intention is to do
out - f(dat)
and then (in this example) having/getting
out$LK
Hi all
One suggestion, tranforme the x
0x11 Tranforme x1=exp(u1)/(exp(u1)+exp(u2)+exp(u3)+1)
0x21 Tranforme x2=exp(u2)/(exp(u1)+exp(u2)+exp(u3)+1)
0x31 Tranforme x3=exp(u3)/(exp(u1)+exp(u2)+exp(u3)+1)
0x41 Tranforme x4= 1/(exp(u1)+exp(u2)+exp(u3)+1)
x1+x2+x3+x4=1
Now
Try this:
foo - function(x)
{
clipname - LK
out - list()
out[[clipname]] - rnorm(5)
return(out)
}
On 27/03/2008, Paul Lemmens [EMAIL PROTECTED] wrote:
Dear all,
I have a piece of code along the lines of
f - function(x) {
clipname - LK # but is in real determined based on info in
Brian Ripley's suggestions of truncated normal and
log-normal are of course resources for ensuring that
you get positive simulated results.
However, I think youre real problem (having looked at
the numbers you quote) is that you should not be thinking
of using a normal distribution, or anything
Hello,
Im reading Data out of a Database.
#v+
rs - dbGetQuery(con,SELECT * ... )
attach(rs)
#v-
There ist a colum I convert into Time.
#v+
zeit-strptime(datum,format=%Y-%m-%d %H:%M:%S);
class(zeit)
[1] POSIXt POSIXlt
#v-
1.
A plot(zeit,money) plots the Data.
All i see on the x-achis are
On 3/26/08, Agustin Lobo [EMAIL PROTECTED] wrote:
Dear list,
Is there any way of making barplots as a Trellis graphic?
Yes, the function to use is 'barchart'.
-Deepayan
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Hi Henrique,
On Thu, Mar 27, 2008 at 2:52 PM, Henrique Dallazuanna [EMAIL PROTECTED] wrote:
Try this:
foo - function(x)
{
clipname - LK
out - list()
out[[clipname]] - rnorm(5)
return(out)
}
That easy ... Hadn't thought of it. But now I have a refinement in foo()
foo - function(x)
On Thu, 27 Mar 2008, Christos Hatzis wrote:
Yes, indeed. The help page says that @ extracts the contents of a slot in
S4 objects. But you mention below that this 'works' for S3 objects because
S4 slots are stored as attributes. Doesn't this mean that @ is currently
implemented to access
Hello, my name is Alfonso. I'm trying to apply bootstrap to a database.
This database present as variables: cohort, age, length, an group. The variable
group is unique for each age*cohort, at this way, if we have 18 cohorts and we
study individuals with an age between 2 and 10 years (9
HI,
I don't understand why you're using lapply.
Please provide a example of your 'x' data.frame
str(x)
On 27/03/2008, Paul Lemmens [EMAIL PROTECTED] wrote:
Hi Henrique,
On Thu, Mar 27, 2008 at 2:52 PM, Henrique Dallazuanna [EMAIL PROTECTED]
wrote:
Try this:
foo - function(x)
Not sure if this is sufficient but if you always open your RData
file in the same directory this may be good enough:
mkdir ~/tmp
cd ~/tmp
R
x - 33
q() # answer y to save in .RData
Now whenever you open R in ~/tmp it will load .RData containing x.
You must be in ~/tmp and the file must be called
You normally want to represent your times as POSIXct rather than
POSIXlt so try
xct - as.POSIXct(xlt)
or you may want to use chron. Read read more about this in
R News 4/1 which has info on dates and times.
Use the axis or Axis command to create custom axes.
?axis
?Axis
You may also want to
Hi,
Sorry yes, I forgot to give comment about the lapply().
foo - function(x) {
out - list()
lapply(x$clipno, function(c) {
clipname - x$clipname
# stuff
out[[clipname]] - rnorm(5)
})
return(out)
}
So x is a dataframe with the data from one subject
Here is one approach to drawing simple cylinders at specified locations, the
height, width, and angle should all be numbers between 0 and 1 and you may want
to rewrite the ms.cylinder function below so that the arguments are more
meaningful relative to the input data that you want to use:
On Thu, 27 Mar 2008, [EMAIL PROTECTED] wrote:
Brian Ripley's suggestions of truncated normal and
log-normal are of course resources for ensuring that
you get positive simulated results.
I was answering the question asked!
Wanting to simulate from the fitted mean and sd does *not* mean you
On Thu, 27 Mar 2008, Robert A LaBudde wrote:
At 05:06 PM 3/26/2008, Ted Harding wrote:
On 26-Mar-08 21:26:59, Ala' Jaouni wrote:
X1,X2,X3,X4 should have independent distributions. They should be
between 0 and 1 and all add up to 1. Is this still possible with
Robert's method?
Thanks
I
As Gabor said, your zeit should be a POSIXct object, not a POSIXlt object.
Then, for example:
zeit - Sys.time() + 150*runif(10)
val - rnorm(10)
plot(zeit,val,xaxt='n')
axis.POSIXct(1,zeit, format='%Y-%m-%d %H:%M')
Or better (perhaps) yet:
plot(zeit,val,xaxt='n')
Can someone explain me this spec.pgram effect?
Code:
period.6-c(0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10
,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10)
period.5-c(0,0,0,0,0,10,0,0,0,0,10,0,0,0,0,0,0,10,0,0,0,0,10,0,0,0,0,0,0,10
Not at present. The next release should include something along these
lines.
luke
On Thu, 27 Mar 2008, Markus Schmidberger wrote:
Hello,
is there any function in the package snow to check for a really running
cluster?
The function checkCluster only checks the variable cl. And the
Hi,
I'm new to R and want to use it to analyse a number of data files (e.g.
100).
I want to read in multiple files in a loop in a specific order (e.g. 1 to
100) and was hoping to do something like:
for (r in 1:100) {
d1 -read.table(r.anl)
for (r2 in 1:100) {
d2 -read.table(r2.anl)
cc
On Thu, 27 Mar 2008, Nuno Prista wrote:
Can someone explain me this spec.pgram effect?
Code:
period.6-c(0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10
,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10,0,0,0,0,0,10)
Hi All,
I would like to compute the simple finite-difference approximation to the
gradient of a scalar function of a large number of variables (on the order
of 1000). Although a one-time computation using the following function
grad() is fast and simple enough, the overhead for repeated
On Thu, Mar 27, 2008 at 4:40 PM, Zu Thur Yew [EMAIL PROTECTED] wrote:
for (r in 1:100) {
d1 -read.table(r.anl)
read.table(paste(r,.anl, sep=))
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Thanks Prof Brian for your suggestion.
I should know that for right-skewed data,
one should generate the samples from a lognormal.
My problem is that x and y are two instruments that were thought to
be measured the same thing but somehow show a wide confidence interval
of the difference
[EMAIL PROTECTED] wrote:
Sorry to come back on callNextMethod, I am still not very confident
about it.
Consideres the following (there is a lot of code, but very simple with
almost only some cat) :
--
setClass(A,representation(a=numeric))
But, your lapply is:
lapply(x$clipno, function(c){
clipname - x$clipname
out[[clipname]] - rnorm(5)
})
So, you don't use 'c' argument.
On 27/03/2008, Paul Lemmens [EMAIL PROTECTED] wrote:
Hi,
Sorry yes, I forgot to give comment about the
Here is as solution that calculates derivatives using central differences by
appropriately embedding the vectors:
grad.1
function(x, fn) {
x - sort(x)
x.e - head(embed(x, 2), -1)
y.e - embed(fn(x), 3)
hh - abs(diff(x.e[1, ]))
y.e - apply(y.e, 1, function(z) (z[1] - z[3])/(2 * hh))
Dear R Help,
is someone going to write a R/S language lexer for the Pygments Python syntax
highlighter http://pygments.org/? As it is used now by Trac, Django, or the
Python documentation tool Sphinx, the R community can apply it in Python-based
Wikis like Moinmoin and others.
Hans Werner
If 'myfunc' is a vector function and can be vectorized in R, then it
is even faster to use the following:
grad.vec - function(x, fn, ..., eps = sqrt(.Machine$double.neg.eps)){
x1 - x + eps * pmax(abs(x), 1)
x2 - x - eps * pmax(abs(x), 1)
(fn(x1, ...) - fn(x2, ...)) / (x1 - x2)
}
I am using the two functions hclust and plot to draw a dendrogram. The
result is a vertical dendrogram , but I prefer a horizontal one.
Need help.
This is my script:
dd1=hclust(d1, method=ward)
plot(dd1)
[[alternative HTML version deleted]]
Thank you, Dimitris Christos.
Yes, myfunc is a scalar function that needs to be minimized over a
high-dimensional parameter space. I was afraid that there might be no
better way, apart from coding in C. Thanks, Dimitris, for confirming my
fear!
Best regards,
Ravi.
On Thu, 27 Mar 2008, Ravi Varadhan wrote:
Thank you, Dimitris Christos.
Yes, myfunc is a scalar function that needs to be minimized over a
high-dimensional parameter space. I was afraid that there might be no
better way, apart from coding in C. Thanks, Dimitris, for confirming my
fear!
Hello all,
I know I'm not making friends with this, but: I absolutely see the point
in dual-(or more!)-y-axis plots! I find them quite informative, and I
see them often. In Earth-Sciences (and I very generously include
atmospheric sciences here, as Johannes has given an example of a
I know this comes up, but I didn't see my exact issue in the archives. I
have variables in a dataframe that need to be recoded. Here is what I'm
dealing with
I have a factor called aa
class(aa)
[1] factor
table(aa)
aa
*0123ABCDLNT
00
If I understand, you can try this:
levels(x)[is.na(as.numeric(levels(x)))] - 0
On 27/03/2008, Doran, Harold [EMAIL PROTECTED] wrote:
I know this comes up, but I didn't see my exact issue in the archives. I
have variables in a dataframe that need to be recoded. Here is what I'm
dealing with
Perfect. My headache is gone. Thanks.
-Original Message-
From: Henrique Dallazuanna [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 27, 2008 12:50 PM
To: Doran, Harold
Cc: r-help@r-project.org
Subject: Re: [R] Recode factors
If I understand, you can try this:
Thanks, it was a matter of reshaping the data matrix as I usually have
it, ie:
datos -
data.frame(x=abs(round(rnorm(100,10,5))),y=abs(round(rnorm(100,2,1))),f=factor(round(runif(100,1,3
to become:
datos2 -
On Mar 27, 2008, at 1:47 PM, Agustin Lobo wrote:
Thanks, it was a matter of reshaping the data matrix as I usually have
it, ie:
datos -
data.frame(x=abs(round(rnorm(100,10,5))),y=abs(round(rnorm
(100,2,1))),f=factor(round(runif(100,1,3
to become:
datos2 -
Pierre Legendre has developed a beta version of a new redundancy analysis
package called RdaTest that is available on his web page at the Universit® de
Montréal. The test example that is included with the package is based on the
example provided in his book (Numerical Ecology, Chapter 11
Thanks for the nice tip.
I also consider to use the clipboard to transfer a command with the
file path which then can be pasted easily in the shell and executed.
Gabor Grothendieck wrote:
Not sure if this is sufficient but if you always open your RData
file in the same directory this
Another suggestion is:
blah = as.character(aa)
blah=gsub('[a-z]','0',blah,ignore.case=T)
aa = as.factor(blah)
I've found changing factors to characters rather than numeric is
generally safer.
Abhijit
Doran, Harold wrote:
Perfect. My headache is gone. Thanks.
-Original Message-
hi,
library(lattice)
x-c(-1,-1,-1,0,0,0,0,1,1,2)
rng-range(x)
histogram(x,endpoints=c(rng[1]-0.5,rng[2]+0.5),nint=length(unique(x)))
instead of contiguous bins, i'd like spaces between them to indicate
that the data is discrete, or even better, line segments positioned at
integer values.
i
Hi !
I am familiar with the rdaTest package.
It works perfectly find for me. Can you give me more details on the
problems you have so I can be of greater help !
Have a nice day !
Guillaume Blanchet
PhD student
Stanfield, Les (MNR) a écrit :
Pierre Legendre has developed a beta version of a
OK,
I really dread to ask that much more that I know some discussion about
p-values and if they are relevant for regressions were already on the list. I
know to get p-val of regression coefficients - this is not a problem. But
unfortunately one editor of a journal where i would like to
James == James Root [EMAIL PROTECTED] writes:
Is there a way to run all paired t-tests where a paired
t-test is run for every possible combination?
Sounds like pairwise.t.test is the sort of thing you are looking
for...
Mike
__
Hi all,
I would like to know if it is posible by, someway, to get colMeans from
a data.frame with numeric as well as character data, dispersed all over
the object. Note that I would like to get colMeans neglecting character
data.
I am really in need of some function proceeding in that way
You might consider something based on the concept of:
y - table( x)
plot( as.numeric(names(y)) , y, type='h')
-Don
At 2:59 PM -0400 3/27/08, [EMAIL PROTECTED] wrote:
hi,
library(lattice)
x-c(-1,-1,-1,0,0,0,0,1,1,2)
rng-range(x)
Thanks!
Yes, but I also want to be able to condition easily hence
histogram(...), not plot
-Original Message-
From: Don MacQueen [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 27, 2008 3:55 PM
To: Rogard, Erwann RD/US/EXT; r-help@r-project.org
Subject: Re: [R] histogram for integer
On 28/03/2008, at 8:42 AM, Michael A. Miller wrote:
James == James Root [EMAIL PROTECTED] writes:
Is there a way to run all paired t-tests where a paired
t-test is run for every possible combination?
Sounds like pairwise.t.test is the sort of thing you are looking
for...
Mike
try this:
dat - data.frame(x = rnorm(10), y = rexp(10), z = letters[1:10])
colMeans(data.matrix(dat[sapply(dat, is.numeric)]))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35,
Hi !!
Stanfield, Les (MNR) a écrit :
Thank you so much for coming to my rescue!:
OK I have been able to open it (but only in the R editor. I assumed that we
had to generate our own files of XX.txt and YY.txt and bring them into R
before the program could run Is this right?
On Thu, Mar 27, 2008 at 3:05 PM, Dimitris Rizopoulos
[EMAIL PROTECTED] wrote:
try this:
dat - data.frame(x = rnorm(10), y = rexp(10), z = letters[1:10])
colMeans(data.matrix(dat[sapply(dat, is.numeric)]))
Alternatively
sapply(dat, mean)
x y z
-0.5260131
I've been struggling to do the following:
After a lengthy computation, I receive an output along the lines of the list
below.
This list has 41 values and is not the end of my computations. I have
another computation to do on the list below, but in this final computation the
list is
I wrote some functions for multiway CANDECOMP, i.e. for least
squares fitting of
a_{i_1\cdots i_m}\approx\sum_{s=1}^p x^1_{i_1s}x^1_{i_1s}\cdots
x^m_{i_ms}
with arrays of arbitrary dimension. Reminded me of the good old APL
days. I could not find this in the archives, but if it's already
summaryBy in the doBy package can do that. The builtin iris data
set has 4 numeric columns and one factor column:
library(doBy)
summaryBy(.~1, iris, fun = mean, keep = TRUE)
Sepal.Length Sepal.Width Petal.Length Petal.Width
1 5.843.0573333.7581.199333
On Thu, Mar
Sorry. I mean
a_{i_1i_2\cdots i_m}\approx\sum_{s=1}^p x^1_{i_1s}x^2_{i_2s}\cdots
x^m_{i_ms}
-- J.
On Mar 27, 2008, at 12:57 , Jan de Leeuw wrote:
I wrote some functions for multiway CANDECOMP, i.e. for least
squares fitting of
a_{i_1\cdots i_m}\approx\sum_{s=1}^p
On Thu, Mar 27, 2008 at 7:01 AM, Aberg Carl [EMAIL PROTECTED] wrote:
Hi,
Im using lme to calculate a mixed factors ANOVA according to:
px_anova = anova(lme(dep~music*time*group, random = ~1|id, data = px_data))
where
dep is a threshold,
time is a repeated measures variable (2 levels)
Ayman,
It is difficult to say without seeing some code, but your output
seems to be not a list in the R sense but a collection of
vectors, each of length 1. The best way to put the values into a
vector probably is to assign them to the elements of the vector
during your computations.
Mike Prager
Hi All,
I need to place lists or vectors within dataframes as single
elements. However when I try this:
df=data.frame(y=1, x=I(list(c(a,b), c(f,c), c(a
df
df[1,'x']=I(c(a,d))
I get this error, even though I am using I():
Error in `[-.data.frame`(`*tmp*`, 1, x, value = c(a, d)) :
I have two data sets with locations, X, Y of houses (df.house) and
habitats(df.habitat), respectively. In each dataset, there are 3
replicates (Repeat). Because each replicate has different locations of
houses and habitats, I would like to plot them in panels. I wrote
something like this:
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