With floating point numbers I'm seeing 'cut' putting values in the wrong
bands. An example below places 0.3 in (0.3,0.6] i.e. 0.3 > 0.3.
> x = 1:5*.1
> x
[1] 0.1 0.2 0.3 0.4 0.5
> cut(x, br=c(0,.3,.6))
[1] (0,0.3] (0,0.3] (0.3,0.6] (0.3,0.6] (0.3,0.6]
Levels: (0,0.3] (0.3,0.6]
I'm sure this i
Hi, useRs-
I have been building a set of functions over time and now my R_GlobalEnv
becomes
too crowded.
I would like to put all my functions under the same namespace/environment.
Ideally,
I would like to call them using "env::func" as in C++.
The following code almost do the jobs I want - e
Hi,
I am trying to create a diagram similar to Figures 4.4 (page 125) and 5.2 (page
159) in Stock and Watson2009) Introduction to Econometrics Brief Edition. The
diagrams display the conditional probability distributions of y at different
values of x along the regression line.
I would be gra
Hi Alain,
2009/6/19 Alain Forget
[...]
> Hi Ishwor,
>
> Ugh, lots of roadblocks. Do you think there's any chance that I could just
> install the Windows version of R onto a Windows Mobile device? I'm currently
> trying to set up a Windows Mobile 6.1 emulator, so I may actually find out
> soon.
2009/6/19 Alain Forget
> > > Hell R-list,
> > >
> > > At the cost of sounding far-fetched and almost incredulous, I would
> > > like to know if any R user is remotely considering the use of R on
> > > Mobile devices, and Android in particular.
> >
> > First, rewrite R in Java. Good luck with tha
Hi Alain,
2009/6/19 Alain Forget
> > > Hell R-list,
> > >
> > > At the cost of sounding far-fetched and almost incredulous, I would
> > > like to know if any R user is remotely considering the use of R on
> > > Mobile devices, and Android in particular.
> >
> > First, rewrite R in Java. Good lu
> > Hell R-list,
> >
> > At the cost of sounding far-fetched and almost incredulous, I would
> > like to know if any R user is remotely considering the use of R on
> > Mobile devices, and Android in particular.
>
> First, rewrite R in Java. Good luck with that!
>
> http://code.google.com/android/k
I have haven't neen following this thread but:
1. if RGoogleDocs_0.2-2.tar.gz is a source distribution (as
opposed to built source) then the first line renames it so
that its not the same name as the built file about to be created.
The second line detars it into the RGoogleDocs directory. The thi
Try this:
gsub("^M{1}", "MOLE", names(data))
On Thu, Jun 18, 2009 at 8:24 PM, Mark Na wrote:
> Dear R-helpers,
>
> I would like to adapt the following code
>
> > names(data)<-sub("M","MOLE",names(data))
>
> which changes any occurrence of "M" (in my variable names) to "MOLE"
>
> such that it ON
What do you mean by "cd the.directory.containing.RGoogleDocs"
Do you mean the directory where I downloaded the RGoogleDocs_0.2-2.tar.gz
to? Or do you mean that I must create a directory called RGoogleDocs under
Library and then change to that directory?
Farrel Buchinsky
Google Voice Tel: (412) 567-
On 18-Jun-09 23:24:39, Mark Na wrote:
> Dear R-helpers,
> I would like to adapt the following code
>
>> names(data)<-sub("M","MOLE",names(data))
>
> which changes any occurrence of "M" (in my variable names) to "MOLE"
> such that it ONLY operates on the first character of each variable
> name, i.
Thanks so much for the invaluable pointers folks!
I just also wanted to note that my definition of statistics also
includes data-mining, generic data-analysis, etc. , i.e. the
statistics in the broad sense.
Any more thoughts?
Thanks a lot!
On Thu, Jun 18, 2009 at 4:05 PM, Richard M. Heiberger w
Dear R-helpers,
I would like to adapt the following code
> names(data)<-sub("M","MOLE",names(data))
which changes any occurrence of "M" (in my variable names) to "MOLE"
such that it ONLY operates on the first character of each variable
name, i.e. M will only be changed to MOLE if it's the first
On Thursday 18 June 2009, Jonathan Greenberg wrote:
> Rers:
>
> What is the preferred library/function for doing stratified random
> sampling from a dataset, given I want to control the number of samples
> (rather than the proportion of samples) per strata? Thanks!
>
> --j
Hi Jonathan!
Check
Alan Izenman suggests:
I have lots of places worth checking out for him. It means a lot
of reading.
Probably the first (and best) place to start is the set of Springer
books entitled "Breakthroughs in Statistics," which was edited by Kotz &
Johnson. There' are three (3) volume
The first time I did it I had no idea how I did it. Yesterday I struggled
and tried every combination to get it to work and eventually it worked. Once
again I do not know what I did to get it to work. Now today I am trying to
install a version that I downloaded today. And once again I am banging my
Rers:
What is the preferred library/function for doing stratified random
sampling from a dataset, given I want to control the number of samples
(rather than the proportion of samples) per strata? Thanks!
--j
--
Jonathan A. Greenberg, PhD
Postdoctoral Scholar
Center for Spatial Technolog
You're absolutely right, Marc.
Thanks
> From: marc_schwa...@me.com
> To: jholt...@gmail.com
> Date: Thu, 18 Jun 2009 15:23:06 -0500
> CC: r-help@r-project.org; alexandre_geor...@hotmail.com
> Subject: Re: [R] Random number datasets help
>
> Quite true Jim, however I focused on his request a
On Jun 18, 2009, at 6:09 PM, bram wrote:
Hi all,
how can I effectively add new objects to a list of which the
eventual length is unknown?
I have been using something like
mylist[[length(mylist)+1]] = newobject
but I wonder if there is something better.
this question might look familiar.
On Jun 18, 2009, at 5:50 PM, Marc Schwartz wrote:
On Jun 18, 2009, at 4:36 PM, Jack Luo wrote:
Hi,
I am trying to use glm to fit my data, wondering if there is a easy
way to
fit a glm without typing all the explanatory variable names. For
example, if
I have 100 explanatory variables x1,
Perhaps it was this:
https://stat.ethz.ch/pipermail/r-help/2005-January/063530.html
On Thu, Jun 18, 2009 at 6:09 PM, bram wrote:
> Hi all,
>
> how can I effectively add new objects to a list of which the eventual length
> is unknown?
>
> I have been using something like
>
> mylist[[length(mylist)
This does not use an apply:
unlist(mget(thels, .GlobalEnv))
On Thu, Jun 18, 2009 at 5:25 PM, Carl Witthoft wrote:
> Let's say I have, for some reason, a bunch of scalars (i.e. single-valued
> variables) and I want to merge them all into a single vector of values. Can
> someone recommend a bette
Hi all,
how can I effectively add new objects to a list of which the eventual
length is unknown?
I have been using something like
mylist[[length(mylist)+1]] = newobject
but I wonder if there is something better.
this question might look familiar. It is the same question as has been
asked h
Several folks pointed out the problem is most likely that a column of
data is being read in as a factor.
I prefer to solve this by setting as.is=TRUE as one of the arguments to
read.table() (unless I am mis-remembering and it needs to be set to
FALSE :-) ). The point is to tell the read funct
Hi,
do.call doesn't help, but the solution provided by Henrique D. is much
cleaner than my hack:
> sapply(ls(patt='foo'),get)
Thanks to all.
Carl
Rolf Turner wrote:
On 19/06/2009, at 9:25 AM, Carl Witthoft wrote:
Let's say I have, for some reason, a bunch of scalars (i.e.
single-valu
> In revising my book Regression Modeling Strategies for a second edition, I
> am seeking a dataset for exemplifying multiple regression using least
> squares. Ideally the dataset would have 5-40 variables and 40-1
> independent observations, and would generate significant interest for a wide
On Jun 18, 2009, at 4:36 PM, Jack Luo wrote:
Hi,
I am trying to use glm to fit my data, wondering if there is a easy
way to
fit a glm without typing all the explanatory variable names. For
example, if
I have 100 explanatory variables x1, x2, ..., x100 and response
variable is
y, I don't
Michael wrote:
Hi all,
I apologize for this off-topic question but I need your help -- I know
there are lots of experts here.
As a lover and student of statistics, I am thinking of building a tree
of various branches of statistics and keeping track of the greatest
historical inventions/discover
Hi,
I am trying to use glm to fit my data, wondering if there is a easy way to
fit a glm without typing all the explanatory variable names. For example, if
I have 100 explanatory variables x1, x2, ..., x100 and response variable is
y, I don't want to do something like
glm1 <- glm(y ~ x1 + x2 + ...
On 19/06/2009, at 9:25 AM, Carl Witthoft wrote:
Let's say I have, for some reason, a bunch of scalars (i.e.
single-valued variables) and I want to merge them all into a single
vector of values. Can someone recommend a better function, or simpler
way, to do so than the following?
Suppose my sc
You can get the values using get:
sapply(ls(patt = "foo"), get)
On Thu, Jun 18, 2009 at 6:25 PM, Carl Witthoft wrote:
> Let's say I have, for some reason, a bunch of scalars (i.e. single-valued
> variables) and I want to merge them all into a single vector of values. Can
> someone recommend a
Thank you very much, it's what I was looking for.
--
View this message in context:
http://www.nabble.com/Print-column-headers-of-summary-tp24094930p24099660.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
Let's say I have, for some reason, a bunch of scalars (i.e.
single-valued variables) and I want to merge them all into a single
vector of values. Can someone recommend a better function, or simpler
way, to do so than the following?
Suppose my scalars' names are foo1, foo2, foo3, foo1high, foo
On Jun 18, 2009, at 3:10 PM, Guilhem Faure wrote:
Hello,
I would like to build rectangles in a plot and use color and
different type
of hatching for filling rectangles. I don't find the way to draw
hatchings.
I'm thinking to build segment by segment inside each rectangle but
I'm sure
tha
A simple way to do it would be:
mat<-mat[,c(2,1)]
Slightly more fancy (for any number of columns):
mat<-mat[,dim(mat)[2]:1]
I'm sure there are prettier ways to do it.
-Jim
On Jun 18, 2009, at 4:14 PM, RON70 wrote:
Hi,
Suppose I have following dataset :
mat <- matrix(rnorm(100), 50)
Now
Dear Group:
In revising my book Regression Modeling Strategies for a second edition,
I am seeking a dataset for exemplifying multiple regression using least
squares. Ideally the dataset would have 5-40 variables and 40-1
independent observations, and would generate significant interest fo
Chapter 10 "A Twentieth-Century Tour of Categorical Data Analysis" in
Agresti's smaller book with this title "An Introduction to Categorical Data
Analysis includes this history.
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
mat[ , 2:1]
On Jun 18, 2009, at 4:14 PM, RON70 wrote:
Hi,
Suppose I have following dataset :
mat <- matrix(rnorm(100), 50)
Now I want to put 2nd column in the place of 1st and 1st column in
the place
of 2nd. Is there any "quick" way to do that?
Thanks and regards,
--
View this message
Quite true Jim, however I focused on his request and not his code,
presuming that he did not realize what he was doing as a consequence
of the nested loops.
Perhaps Alexandre can provide clarification?
Regards,
Marc
On Jun 18, 2009, at 2:46 PM, jim holtman wrote:
That is not what his 'for
Tena koe Guilhem
Perhaps
?rect
would help ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Guilhem Faure
> Sent: Friday, 19 June 2009 7:11 a.m.
> To: r-help@r-project.org
> Subject: [R] Hatched symbols
> Could anybody give me some pointers about existing books/articles
> about the greatest inventions/discoveries in statistics? And topic
> list?
You could have a look at
http://www.amazon.com/Lady-Tasting-Tea-Statistics-Revolutionized/dp/0805071342/ref=sr_1_1?ie=UTF8&s=books&qid=1245356064&sr=8-1
Hi,
Suppose I have following dataset :
mat <- matrix(rnorm(100), 50)
Now I want to put 2nd column in the place of 1st and 1st column in the place
of 2nd. Is there any "quick" way to do that?
Thanks and regards,
--
View this message in context:
http://www.nabble.com/Altering-columns-tp2409959
agresti's book ( forget which one. there are two ) has an appendix about the
history of categorical data that i remember being quite interesting. that's
the only one i know of.
On Jun 18, 2009, Michael wrote:
Hi all,
I apologize for this off-topic question but I need your
For the mean() example, I believe this should work (untested)
for (var in names(Dataset)) print( mean( Dataset[[var]] , na.rm=TRUE ) )
or
for (var in names(Dataset)) print( mean( Dataset[,var] , na.rm=TRUE ) )
But it's harder with lm, glm, and friends. For them I think maybe you
can do i
Ben Bolker ufl.edu> writes:
> > In the Sweave output for summary for several types
> > of model objects and also for the comparison of models
> > with anova, I find that that the display of the call(s)
> > or formula does not obey the width option, even with
> > keep.source=TRUE set, so that a lon
That is not what his 'for' loops are doing. He is iterating through all
combinations and would have created 784. So his problem statement did not
match the code that he sent.
On Thu, Jun 18, 2009 at 2:12 PM, Marc Schwartz wrote:
> Alexandre did say 28 datasets, not 784 (28 * 28)
>
> Thus,
On Jun 18, 2009, at 3:23 PM, Anke Konrad wrote:
Hi all,
I am currently trying to compare different plant occurrence
prediction maps generated in R and exported into GRASS. One of these
maps was generated from a glm fitted to some data, and subsequently
applying this glm model to a wider
Hi all,
I am currently trying to compare different plant occurrence prediction
maps generated in R and exported into GRASS. One of these maps was
generated from a glm fitted to some data, and subsequently applying this
glm model to a wider region using predict.glm. The outcome here was a
prob
Hello,
I would like to build rectangles in a plot and use color and different type
of hatching for filling rectangles. I don't find the way to draw hatchings.
I'm thinking to build segment by segment inside each rectangle but I'm sure
that exists a better way to do that. I didn't find any document
With help from my colleague, I found the problem. After I put another library
dll, zlib1.dll, at the folder, biOps is working again now. Though it's still
amazing to me why biOps worked for about one week before I had zlib1.dll.
I hope message could also help people who have the same problem.
Th
I am calling on the R-help crew once again. The CD4Per for each subject should
be the same for each subject based on a 90 day window either side of the known
value. For example, ID 2931 should have a CD4Per of 36 at 0, 34, and 62
DaysEnrolled. Any ideas how to do this without a loop?
ID
Hi all,
I apologize for this off-topic question but I need your help -- I know
there are lots of experts here.
As a lover and student of statistics, I am thinking of building a tree
of various branches of statistics and keeping track of the greatest
historical inventions/discoveries in statistics
On 18 June 2009 at 14:02, Jim Nemesh wrote:
| Actually, I'm going to take by my comments.
|
| You can use Rscript with multiple -e arguments to do what I was
| looking for:
|
| Rscript -e 'library(broadgap.hapmaptools)' -e 'processAllFiles(dir="/
|
humgen/cnp04/sandbox/steve/1m/hapmap3/genot
On Jun 18, 2009, at 1:10 PM, Alexandre Lockhart wrote:
Hello:
My problem is that I have a data frame of means, and a data frame of
standard deviations which match up to each mean. I have been trying
to create 500 random numbers in a given dataset for each mean/sd
combination, but I am
Dear R-users,
I am trying to modify the mfp() function in the mfp package to model
Weibull survival using fractional polynomials approach. However, I keep
getting into trouble when mfp.fit and other "hidden" functions can't be
found. I did find some of them in Splus but it's getting nowhere.
Hi,
take a look at the following manual pages:
?lm
?longley
In general, you would use Wilkinson-Rogers notation for linear models:
y ~ x + z, etc.
Some nice examples here:
http://data.princeton.edu/R/linearmodels.html
Cheers,
Dylan
On Thu, Jun 18, 2009 at 11:04 AM, Martin
Batholdy wrote:
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Henrique Dallazuanna
> Sent: Thursday, June 18, 2009 10:52 AM
> To: Alexandre Lockhart
> Cc: r-help@r-project.org
> Subject: Re: [R] Random number datasets help
>
> Try this also:
Alexandre did say 28 datasets, not 784 (28 * 28)
Thus, either:
mapply(rnorm, n = 500, mean = a1, sd = a2)
or
apply(cbind(a1, a2), 1, function(x) rnorm(500, x[1], x[2]))
HTH,
Marc Schwartz
On Jun 18, 2009, at 12:51 PM, Henrique Dallazuanna wrote:
Try this also:
a <- expand.grid(a
hi,
How can I compute a regression between x and y adding a covariate z?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commente
Actually, I'm going to take by my comments.
You can use Rscript with multiple -e arguments to do what I was
looking for:
Rscript -e 'library(broadgap.hapmaptools)' -e 'processAllFiles(dir="/
humgen/cnp04/sandbox/steve/1m/hapmap3/genotypes/ftp.hapmap.org/genotypes/2008-07_phaseIII/hapmap_form
On Thu, Jun 18, 2009 at 1:54 PM, Jim Nemesh wrote:
> I did look at ?Rscript. My question there was:
>>>
>>> Unfortunately, it looks like each expression is evaluated separately.
>>> Is there a way to evaluate multiple expressions in the same
>>> environment to make the above work?
>
> I could writ
I did look at ?Rscript. My question there was:
Unfortunately, it looks like each expression is evaluated separately.
Is there a way to evaluate multiple expressions in the same
environment to make the above work?
I could write an Rscript client file, but it's still poor CLI support
compared
Try this also:
a <- expand.grid(a1, a2)
x <- mapply(rnorm, n = 500, mean = a[,1], sd = a[,2])
On Thu, Jun 18, 2009 at 2:10 PM, Alexandre Lockhart <
alexandre_geor...@hotmail.com> wrote:
>
> Hello:
>
> My problem is that I have a data frame of means, and a data frame of
> standard deviations whic
See the getopt R package and have a look at ?Rscript
Also on Windows see Rscript.bat and #Rscript.bat in
http://batchfiles.googlecode.com
and on Linux see:
http://dirk.eddelbuettel.com/code/littler.html
On Thu, Jun 18, 2009 at 1:25 PM, Jim Nemesh wrote:
> I develop quite a bit of R code that I t
try this:
>
a1<-c(178.07,178.28,178.08,177.74,177.04,178.17,177.58,57.71,59.6,60.92,59.48,59.32,61.59,59.94,28.9,29.82,30.73,25.68,27.93,28.98,29.76,123.48,127.27,127.8,127.2,127.13,126.71,125.5)
>
a2<-c(1.69,1.3,1,.18,1.53,1.31,1.35,1.83,1.56,1.12,.74,1.48,1.67,1.53,.95,.87,0.03,1.12,1.95,1.22,1.
approx() almost does what you want, but you have to patch
up its output to account for a possible initial run of 0's in the
input:
> x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0)
> y <- approx(seq_along(x)[x!=0], x[x!=0], xout=seq_along(x),
method="const", f=0, rule=2)$y
> y
[1] -1 -1 -1 -1 -1 -1 1 -1
I develop quite a bit of R code that I tend to distribute, or let
other people embed in their software. One of the things I'd really
like to do is find a way to load an R library I've developed, and call
one function with arguments.
Currently, I achieve that by building a separate R client
The zoo package has na.locf which replaces NAs with the last non-NA.
So, first replace 0's with NA's, apply na.locf and then replace NAs with 0's.
> library(zoo)
> x.na <- replace(x, x == 0, NA)
> x0 <- na.locf(x.na, na.rm = FALSE)
> replace(x0, is.na(x0), 0)
[1] 0 -1 -1 -1 -1 -1 1 1 1 1
On
Hi,
I installed biOps on my XP, and installed the required dll packages,
jpeg62.dll, libfftw3-3.dll, and libtiff3.dll, as suggested by this forum.
Then it worked perfectly for about one week till this afternoon. When I
tried to load biOps again, I got the following error message again
Error in i
use na.locf in zoo:
> x
[1] 0 -1 -1 -1 0 0 1 -1 1 0
> # replace 0 with NA so na.locf works
> is.na(x) <- x == 0
> x
[1] NA -1 -1 -1 NA NA 1 -1 1 NA
> na.locf(x, na.rm=FALSE)
[1] NA -1 -1 -1 -1 -1 1 -1 1 1
>
You can then go back and replace NAs with 0
On Thu, Jun 18, 2009 at 12:47
Hello:
My problem is that I have a data frame of means, and a data frame of standard
deviations which match up to each mean. I have been trying to create 500
random numbers in a given dataset for each mean/sd combination, but I am only
able to generate the last value in each data set to creat
On 18 June 2009 at 09:36, Bert Gunter wrote:
| -- or Chapter 4 in S PROGRAMMING? (you'll need to determine if it's "reader
| friendly")
+1
It helped me a lot too back in the day. But I am wondering if there are good
current alternatives. Hadley: if you could, please send a summary back to
the
I think I remember reading that some time back and finding it
confusing because it described the ideal implementation in S, rather
than the actual implementation in R. I'll look at it again.
Hadley
On Thu, Jun 18, 2009 at 11:36 AM, Bert Gunter wrote:
> -- or Chapter 4 in S PROGRAMMING? (you'll n
This code works for me,
subset(merge(x1, x2, by = c("category", "id"), all = T), select=-id)
category rnorm.10..5..2. rnorm.10..8..2.
1 1 NA7.810377
2 16.4932819.002411
3 16.4932818.397378
4 12.757321
Folks,
If I have a vector such as the following:
x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0)
and I want to replace the zeroes by the nearest non-zero number to the
left, is there a more elegant way to do this than the following loop?
y <- x
for (i in 2 : length(x))
{
if (y[i] == 0) {
Greg is absolutely right: Generally base and grid graphics(which is what
lattice plots are based on) don't play well together, and gridBase is meant
to address this.
However...
The problem here is specifically that the print.trellis (or synonymous
plot.trellis) function that is silently invoked to
-- or Chapter 4 in S PROGRAMMING? (you'll need to determine if it's "reader
friendly")
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Thursday, June 18, 2009
Of course, closed form hessian is great, but I would still check it using
deriv3 or hessian() from numDeriv.
It should be noted that hessian() is so accurate that it can be a surrogate for
the exact hessian. It is computationally demanding, but it is tolerable when
you are computing it only
Not a suggestion on documentation, but to ease S3 coding and hiding
some of the details, you might want to consider the R.methodsS3
package. It basically provides some S3 code generators and makes it
clear what class of objects a method is dispatched on. From the
example code:
setMethodS3("foo",
There is a section on Object Orientation in MASS (I have 2nd ed).
On Thu, Jun 18, 2009 at 12:06 PM, Hadley Wickham wrote:
> Hi all,
>
> Do you know of any good resources for learning how S3 works? I've
> some how become familiar with it by reading many small pieces, but now
> that I'm teaching it
Greetings:
I created an R group on Linkedin. It has a discussion board but a bit less
about programming and more about networking including jobs but also relevant
web sites. Please join us.
Regards,
Ajit
http://www.linkedin.com/groups?about=&gid=77616&trk=anet_ug_grppro
http://www.linkedin.c
Gilad wrote:
Hi all,
How do I add an asterisk representing a significant difference to a barplot?
Cheers,
Gil
Bar plots are dreadful ways to present information (see
http://biostat.mc.vanderbilt.edu/DynamitePlots). Don't make them worse
by adding arbitrary declarations of 'significance.
Hi all,
Do you know of any good resources for learning how S3 works? I've
some how become familiar with it by reading many small pieces, but now
that I'm teaching it to students I'm wondering if there are any good
resources that describe it completely, especially in a reader-friendly
way. So far
On Jun 18, 2009, at 10:05 AM, nmset wrote:
Hello,
This is a newbee question. I would simply like to write in a text
file the
headers of the summary function along with the computed data.
write(summary(x), file="out.txt")
gives only the data.
I have not found a solution on the forum.
Tha
Hi,
perhaps that is what you want?
> df <- as.data.frame(matrix(runif(100),ncol=10))
> form <- as.formula(paste(names(df)[length(df)], "~ ."))
> lm(form,data=df)
Call:
lm(formula = form, data = df)
Coefficients:
(Intercept) V1 V2 V3 V4
-1.367 -
The word "merge" in the context of R suggests the use of the merge()
function, but I don't think that's the right tool for what you want.
The merge() function is for relational database type merges, which
for your data would have a many to many merge. Not good.
In terms of the R language, you'
Thanks! I did not look at the output of str(df) closely. Since y is
defined as a character variable when df is created (but stored as a
factor), it looks like str(df) is sorting the factors, at least when it
is displayed to the screen.
Jude
-Original Message-
From: David Winsemius [mailto
Look at the gridBase package.
--
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-
> project.org] On Behalf Of Titus v. d. Malsburg
> Sent: Thu
It's not a solution. Unfortunately data.matrix is no different with
respect to factors than other functions. Note what str(df) produced
for df$y.
--
David.
On Jun 18, 2009, at 10:59 AM,
wrote:
David Winsemius' solution:
> apply(data.matrix(df), 1, I)
[,1] [,2] [,3] [,4] [,5] [,6]
Hello,
This is a newbee question. I would simply like to write in a text file the
headers of the summary function along with the computed data.
write(summary(x), file="out.txt")
gives only the data.
I have not found a solution on the forum.
Thank you for any help.
--
View this message in con
Hello!
I want to estimate strength and correlation of RandomForest, but in package
"randomForest" there is not an interface to get it. I think I must to change
the source code. Is there any advise?
Thanks,
Li
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If I understand correctly, you can try something about like this:
x <- c(1, 4, 5, 10)
bp <- barplot(x, ylim = c(0, max(x) + 2))
text(bp, x,
labels = symnum(x, cutpoints = c(1, 5, 10),
symbols = c("*", "**")), pos = 3)
On Thu, Jun 18, 2009 at 12:40 PM, Gilad wrote:
> Hi all,
>
> How
David Winsemius' solution:
> apply(data.matrix(df), 1, I)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
x12345678910
y13456789 10 2
For y and [,2] above the value is 3. Why is the value not 2?
It look
Hi!
I have a dataset with some 300+ variables and 2000+ records. I'd like to grind
through a bunch of analyses on the variables by using a script, but can't
figure out how to refer to variable names properly. For some of the simpler
stuff I use various "apply" functions, but for others (like t-tes
Thank you. One question, though. In the case where I have closed form
formulæ for the Hessian, or the functions are parseable by "deriv3", would
it be better to use that than the approximation?
--Avraham
"
Where did you get the idea that Helvetica was a Windows font family
name? (It is a Linotype trademark font name.) It is rarely seen on
Windows, so attempts to use it almost always end up with a font
substitution, often to Arial which is already the default family in
win.metafile(). That will
Here is a way of determining where the dups are:
> vec <- scan(textConnection(' "STAT1" "STAT1" "STAT1" "STAT1" "GAPDH"
"GAPDH" "GAPDH"
+
+ "ACTB" "ACTB" "ACTB" "DDR1" "RFC2" "HSPA6" "PAX8"
+ "GUCA1A" "UBE1L" "THRA" "PTPN21" "CCL5" "CYP2E1" "STAT1"
+ "THRA" "PAX8"'), what=
On Jun 18, 2009, at 5:01 AM, alessia matano wrote:
Dear all,
I 'm implementing the koenker procedure for quantile fixed effects.
Who he? Wuz 'dat?
I would like also to apply weights to the procedure, so that to give
more weight to the observation that better represent my original
sample (
Hi all,
How do I add an asterisk representing a significant difference to a barplot?
Cheers,
Gil
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On Jun 18, 2009, at 9:28 AM, njhuang86 wrote:
Hi all,
Suppose I have a vector like this:
[1] "STAT1" "STAT1" "STAT1" "STAT1" "GAPDH" "GAPDH" "GAPDH"
"ACTB"
"ACTB"
[10] "ACTB" "DDR1" "RFC2" "HSPA6" "PAX8" "GUCA1A" "UBE1L"
"THRA"
"PTPN21"
[19] "CCL5" "CYP2E1" "STAT1"
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