Hi Greg
Thanks for responding.
When I scale stackloss and compute the fitted values of the scaled data frame,
the values are not the same:
A-data.frame(scale(stackloss))
lm- lm(stack.loss ~ . -1, qr=T, data=A)
summary(lm)
fitted.values(lm)
Or should I not be comparing the two?
Greg
Hi
r-help-boun...@r-project.org napsal dne 24.02.2009 06:23:23:
Hi R users,
I have a question. How can I use for loop to do pair comparisons. For
example,
x-c(1,2,3)
result-matrix(data=NA, nrow=choose(3,2), ncol=1)
for(i in 1: length(x))
+{ result[i,]-ifelse(x[i] x[i+1],
There are 4 sets of array, each have 2 dimension data:x,y.
And these data comes from bkde.
I tried to plot them in one figure, but details of some lines, like
peak and where it locate, have diminished since one set of array have
fairly high y value.And another array have lots of y value near zero,
It has been solved
On Feb 24, 5:05 pm, yk shiyuankun.deb...@gmail.com wrote:
There are 4 sets of array, each have 2 dimension data:x,y.
And these data comes from bkde.
I tried to plot them in one figure, but details of some lines, like
peak and where it locate, have diminished since one set
Hi
r-help-boun...@r-project.org napsal dne 23.02.2009 18:31:32:
right now I have a vector of about 1000 points. I'd like to iterate
through
each of these points and and test if it is greater than a certain value
and
if not, throw it out.
x=vector
y=empty vector
j=0
for i
Hi
If you had issued axis break into CRAN fearch facility you would have
find several options e.g. plotrix package.
Also look at xlim or ylim and consider logarithmic axes
regards
Petr
r-help-boun...@r-project.org napsal dne 24.02.2009 10:05:26:
There are 4 sets of array, each have 2
Hi Maureen,
There is a specialist mailing list for mixed-models, where you are likely to
get a well-informed answer to this, quite possibly from Prof. Bates himself
(the author of lmer).
https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
Regards, Mark.
PS: R Hompage - Mailing Lists for
One possible first step could be:
ifelse(outer(x, x, ''), 'Big', 'Small')
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
Chunhao Tu wrote:
Hi R users,
I have a question. How can I use for loop
Hi all,
I am using the fMultivar package for calculating probabilities of bivariate
normal distribution. I use the manual's example to understand what is going on,
but let's take it for smaller dimensions of x and y:
## Bivariate Normal Density:
x = c(0.3,10)
y = c(-10,0.2)
X = grid2d(x,y)
z
Patrick Burns wrote:
One possible first step could be:
ifelse(outer(x, x, ''), 'Big', 'Small')
Second step: use 'lower.tri'.
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
Chunhao Tu wrote:
Hi,
Thank you for the reply and suggestions.
I have two questions?
1) If I want to use log, it seems that I have to take log from both sides of
the model which will lead to lm(log(q)~log(-depth)). What is tehdifference
between this syntax and lm(log(q) ~ I(-depth))?
2) How can I calculate the
Hi,
I have a data set (weight) that does not follow the Gaussian (Normal)
distribution. However, I have to transform the data before applying the
Gaussian distribution. I used this syntax and used log(weight) as:
posJy.model-glm(log(weight) ~ factor(pos),
family=gaussian(link='identity'),
Hi
r-help-boun...@r-project.org napsal dne 24.02.2009 11:31:22:
Hi,
Thank you for the reply and suggestions.
I have two questions?
1) If I want to use log, it seems that I have to take log from both
sides of
the model which will lead to lm(log(q)~log(-depth)). What is
tehdifference
Petr Pikal
petr.pi...@precheza.cz
724008364, 581252140, 581252257
r-help-boun...@r-project.org napsal dne 24.02.2009 11:46:07:
Hi,
I have a data set (weight) that does not follow the Gaussian (Normal)
distribution. However, I have to transform the data before applying the
Gaussian
On Tue, Feb 24, 2009 at 02:46:07AM -0800, Saeed Ahmadi wrote:
I have a data set (weight) that does not follow the Gaussian (Normal)
distribution. However, I have to transform the data before applying the
Gaussian distribution. I used this syntax and used log(weight) as:
Hi Peter,
You are totally right and it was a miscalculating and misunderstanding from
me.
Regarding the R-squared calculation of non linear model (question 2), is
there any way to do that?
Regards
Saeed
Petr Pikal wrote:
Hi
r-help-boun...@r-project.org napsal dne 24.02.2009 11:31:22:
Dear R helpers
How to generate random numbers for
(a) Generalized logistic distribution
(b) Generalized normal distribution
(c) Pearson Type III distribution
(d) Kappa
Thanks in advance
Maithili
__
R-help@r-project.org mailing list
In order ot see if any of these syntaxes work or not, I made the QQ plot.
Attached files are showing the QQ plots made when log was inside the model
formulation and outside teh model formulation. You can see that when
transformation were done outside, then I have a general better QQ plot
I have written the codes below. Unfortunately, the ticks generated by
matplot A and matplot B overlap with each other. Does anyone have some
idea to avoid it automatically?
split_y=2.0e-4
layout(matrix(c(1,2),2,1,byrow=T),height=c(1,3))
par(mar=c(0.5,4.5,1.5,1.5))
x_row=max(row(x_m))
Hi Patrick,
Thank you for your hint. I got it works.
x-1:3
result-ifelse(test=outer(X=x,Y=x,FUN=), yes=Big,
+ ifelse(test=outer(X=x,Y=x,FUN=), yes=Small, Equal))
result
[,1][,2][,3]
[1,] Equal Small Small
[2,] Big Equal Small
[3,] Big Big Equal
See the lmom package by J.R.M Hosking
or
lmomco package by W.H.Asquith
-wha
On Feb 24, 2009, at 5:46 AM, Maithili Shiva wrote:
Dear R helpers
How to generate random numbers for
(a) Generalized logistic distribution
(b) Generalized normal distribution
(c) Pearson Type III distribution
Hello everybody!
I'm using Perl regular Expression for find pattern in my data set.
The pattern is: NaQxy, where a=E, F, G or H and xy != 29. I have tried this:
pattern - ^N[E-H]Q[0-9]{2,2}
index - grep(pattern, X, perl=T) #where X is my vector
But the problem is the xy should not be 29. How
Dear List,
I registered for the useR conference in Rennes today; half an hour after the
confirmation I received a first requested newsletter from a company selling a
product named Inference for R.
This coincidence might be spurious. Or not, depending on frequency.
Dieter
On 24/02/2009 8:06 AM, Dieter Menne wrote:
Dear List,
I registered for the useR conference in Rennes today; half an hour after the
confirmation I received a first requested newsletter from a company selling a
product named Inference for R.
This coincidence might be spurious. Or not, depending
Dieter Menne wrote:
Dear List,
I registered for the useR conference in Rennes today; half an hour after the
confirmation I received a first requested newsletter from a company selling
a
product named Inference for R.
This coincidence might be spurious. Or not, depending on frequency.
I'm looking for a way to create bar charts with nested groups levels.
I have data that is seperated to this factors:
1. Day part - day/night.
2. Season - summer/winter.
3. habitat - Cover/Open
I have created in Excel the following chart which I would like to create in
R:
Do it in two steps:
x - c(NEQ23, NHQ29, NGQ00, NFQ123)
pat - N[E-H]Q[0-9]{2}
bad - N[E-H]Q29
all - grep(pat, x, perl=TRUE)
x29 - grep(bad, x, perl=TRUE)
setdiff(all, x29)
[1] 1 3 4
On Tue, Feb 24, 2009 at 7:57 AM, Katrine Damgaard
katrine.damga...@kunnskapssenteret.no wrote:
Hello
Hi All.
Imagine you have a large block diagonal matrix. I'd like to replace
the zeros in this matrix with small random (runif) numbers. Any ideas
for a simple and efficient way to do this?
Best regards,
Rick DeShon
__
R-help@r-project.org mailing
Berwin A Turlach wrote:
G'day Dimitris,
On Tue, 24 Feb 2009 11:19:15 +0100
Dimitris Rizopoulos d.rizopou...@erasmusmc.nl wrote:
in my opinion the point of the whole discussion could be summarized
by the question, what is a design flaw? This is totally subjective,
and it happens almost
I received it too without the conference registration.
2009/2/24 Wacek Kusnierczyk waclaw.marcin.kusnierc...@idi.ntnu.no:
Dieter Menne wrote:
Dear List,
I registered for the useR conference in Rennes today; half an hour after the
confirmation I received a first requested newsletter from a
say 'mat' is your matrix, then you could try something like (untested):
ind - mat == 0
mat[ind] - runif(sum(ind))
I hope it helps.
Best,
Dimitris
Rick DeShon wrote:
Hi All.
Imagine you have a large block diagonal matrix. I'd like to replace
the zeros in this matrix with small random
The same company caused a complaint about a year ago
https://stat.ethz.ch/pipermail/r-help/2008-March/157423.html
The mailing company they are using (iContact.com) claims to have a tough
antispam policy. So does everyone, of course.
-thomas
Thomas Lumley Assoc.
Hi everyone,
I am currently using the function optim() to maximize/minimize functions and
I would like to see more output of the optimization procedure, in particular
the numerical gradient of the parameter vector during each iteration.
The documentation of optim() describes that the trace
I got the same spam message today and I havent signed up for anything except
this forum mailing list.
The software they are trying to sell doesnt seem to cover any new ground
anyway.
Simon.
- Original Message -
From: Thomas Lumley tlum...@u.washington.edu
To: ronggui
Dear R-list members,
I have a data file with thousands of lines (cases), where each line
contains the values of several variables. I would like to separate
these lines in small groups, with each group followed by a blank
line, to ease the visual inspection of the data in some situations.
I am
Dear all,
When using odfWeave, I get the following error:
odfWeave(file.in,file.out)
Copying Example2.odt
Setting wd to /tmp/Rtmp8ekeDC/odfWeave24145003519
Unzipping ODF file using unzip -o Example2.odt
Archive: Example2.odt
End-of-central-directory signature not found. Either this
Fox, Gordon gfox at cas.usf.edu writes:
Example: 40 and 80 have these factors: c(1,2,2,2,5) and c(1,2,2,2,2,5).
We can use match() to get the common factors c(1,2,2,2,5). What we want
to be able to get from this is the list of all the possible products,
which should be the concatenation of
Hi,
If you look at the source code for optim() in the optim.c file, you will see
the following lines for BFGS:
if (trace (iter % nREPORT == 0))
Rprintf(iter%4d value %f\n, iter, f);
This means that BFGS does not output gradient values when you trace the
iterations. Let us
Hi,
As a number of people suggested, I downloaded the e071 package and am using
the function svm for my classification problem. I now want to take the
classifier outside
the R environment, for example to a C program, and use it.
I know that the classifer is of the form
WTX+b where W is defined
Dear list,
Sorry for bothering you with a pure odfSweave question, but I just ran
into a problem that I cannot find the cause of.
Anyonse seen this before? This file used to work, but not anymore.
Would apreciate all the help I could get.
/Fredrik
Hi Ravi,
Thanks for your great suggestion, it does exactly what I need as it provides
more insight into what is going on in the 'black box'. In addition, it's
much faster than optim(). I will use this function in the future.
Kind Regards,
Shimrit
On Tue, Feb 24, 2009 at 2:33 PM, Ravi
Hi I have managed to do a paired t-test with a data set
i have 5 colums of data im dealing with
GENE SampA SampB SampC SampVehicle
ctcc 859 na145 24
gtcg 45 5 54 69
and so on but they are much larger
How does one control the size and type of data symbols in pairs()? I
am trying to use the little dot (as in type=.) with absolutely no
success.
Here is the pairs call I am using:
pairs(data.frame, panel=function(x,y) {points(x,y); lines(lowess(x,y))})
or even simpler:
pairs(data.frame,
Yes, initially, it didn't work and thanks to one of the examples in the help
file, I found out that I need to set maximize = T...but thanks for your
suggestion anyway.
I mainly work with state space models and I'm currently dealing with a case
where the estimation time is halved (!!!) by spg().
Cheers for that information; I've just registered for the useR meeting
in London and then about 10 minutes later got that same bit of spam
too which made me a wee bit suspicious.
On 24 Feb, 13:39, Thomas Lumley tlum...@u.washington.edu wrote:
The same company caused a complaint about a year
I am using R to access .mdb files (created in Microsoft access 2003) through
RODBC.
I am able to view the Tables and also list their attributes through the
sql. commands. However, when try to acces the primary keys (using
sqlPrimaryKeys /odbcPrimaryKeys) I get a value of '-1' on
On Tue, Feb 24, 2009 at 9:58 AM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
TL == Thomas Lumley tlum...@u.washington.edu
on Tue, 24 Feb 2009 05:39:33 -0800 (PST) writes:
TL The same company caused a complaint about a year ago
TL
Kirk Wythers wrote:
How does one control the size and type of data symbols in pairs()? I am
trying to use the little dot (as in type=.) with absolutely no success.
I guess you mean pch=. and hence
pairs(data.frame, panel=panel.smooth, pch=.)
Uwe Ligges
Here is the pairs call I am
Dear Tania,
Shouldn't that be varIdent(form = ~species.group) instead of
varIdent(form = ~1|species.group)?
Notice that this is untested as you did not provide a self-contained
example (as de posting guide asked you to do).
HTH,
Thierry
PS Next time try the mixed-models mailing list with
Hi there,
I wanted to search some packages related with 'genetics'. I know I can do it on
CRAN webpage. I'm just wondering if there's some function in R could do that.
Debian apt offers one way. It would be perfect if R has some builtin function
like Debian apt.
Appreciate any help.
Xin
Dear All,
I have matrix of size 2500 x 12. I would like to select all the rows if 6 out
12 values are 0. How can I do that in R?
Thanks in advance.
Kind regards,
Ezhil
__
R-help@r-project.org mailing list
The same as what? It is not clear what you are trying to compare.
Is this the comparison that you are looking for?:
A - scale(stackloss)
fit1 - lm( stack.loss ~ . - 1, data=as.data.frame(A))
tmp1 - fitted(fit1)
fit2 - svd(A[,-4])
tmp2 - fit2$u %*% t(fit2$u) %*% A[,4]
all.equal(
one way is:
mat - matrix(rnorm(2500*12), 2500, 12)
ind - rowSums(mat 0) 6
mat[ind, ]
I hope it helps.
Best,
Dimitris
A Ezhil wrote:
Dear All,
I have matrix of size 2500 x 12. I would like to select all the rows if 6 out 12 values are 0. How can I do that in R?
Thanks in advance.
Zheng, Xin (NIH) [C] wrote:
Hi there,
I wanted to search some packages related with 'genetics'. I know I can do it
on CRAN webpage. I'm just wondering if there's some function in R could do
that. Debian apt offers one way. It would be perfect if R has some builtin
function like Debian
Hi everyone,
I would like to test for the statistical significance(for what it worth ...) in
increasing classification accuracy and kappa statistics from different land
classifications. The classifications were done using other software (like
eCognition and See5), but the results were sampled
Hi,
I'm looking for any information on calculating and using Shapley Values
in a TURF context. A paper was presented at an S-Plus conference about
this:
Conklin M., Lipovetsky S. Modern Marketing Research Combinatorial
Computations:
Shapley Value versus TURF Tools, Proceedings of 1998
What did the data appear like when it was read in? Have you just
tried to read in the lines (readLines) to see if the decompression is
working? Does this compare to what you get if you decompress the file
outside of R? Not exactly sure what you mean by force since the
command it probably
On Tue, Feb 24, 2009 at 10:56 AM, Zheng, Xin (NIH) [C]
zheng...@mail.nih.gov wrote:
Hi there,
I wanted to search some packages related with 'genetics'. I know I can do it
on CRAN webpage. I'm just wondering if there's some function in R could do
that. Debian apt offers one way. It would be
It looks like you found a solution, but if you find yourself in this situation
again using optim, then one approach is to modify your function that you are
optimizing (or write a wrapper for it) to produce the tracing information for
you.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data
Fredrik Karlsson dargosch at gmail.com writes:
Dear list,
Sorry for bothering you with a pure odfSweave question, but I just ran
into a problem that I cannot find the cause of.
Anyonse seen this before? This file used to work, but not anymore.
Would apreciate all the help I could get.
Irina Ursachi irina.ursachi at itwm.fraunhofer.de writes:
Dear all,
When using odfWeave, I get the following error:
odfWeave(file.in,file.out)
Copying Example2.odt
Setting wd to /tmp/Rtmp8ekeDC/odfWeave24145003519
Unzipping ODF file using unzip -o Example2.odt
Archive:
All
Mango Solutions are pleased to announce the first meeting of the London
useR Group.
DateTuesday 31st March
Time4pm to 7pm
Venue The Wall
45 Old Broad St
London
EC2N 1HU
Tel 020 7588 4845
Dear Gordon,
Try also,
unique(apply(expand.grid(commfac,commfac),1,prod))
[1] 1 2 5 4 10 25
HTH,
Jorge
On Mon, Feb 23, 2009 at 9:52 PM, Fox, Gordon g...@cas.usf.edu wrote:
This is a seemingly simple problem - hopefully someone can help.
Problem: we have two integers. We want (1) all
Rob Steele suggested the same thing but I'm not sure I understand how to
implement this exactly. Is there any documentation that you could suggest?
This might be something that could be useful for the future.
Thanks,
Shimrit
On Tue, Feb 24, 2009 at 5:09 PM, Greg Snow greg.s...@imail.org wrote:
Tony Breyal tony.breyal at googlemail.com writes:
Cheers for that information; I've just registered for the useR meeting
in London and then about 10 minutes later got that same bit of spam
too which made me a wee bit suspicious.
Welcome in the Fooled by Randomness society. 2:0 is a bit away from
Hi again,
Looking more into test statistics i realized that maybe i can use the
power.prop.test to see if the difference between the 2 accuracies are zero or
not. Do you have any comments about that? Also, should i considered kappa
statistics also a kind of proportion and use the same test?
Dear all,
I'm using the samr-package to identify significantly differentially expressed
genes in microarray data.
So far, I had no problems, but when I used a large multiclass data set with 327
samples, I obtained the following error/warning message:
Warning message:
Inf
In
I'm growing a large dataframe by composing new rows and then doing
row - compute.new.row.somehow(...)
d - rbind(d,row)
Is this a fast/preferred way?
Cheers,
Alexy
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Here is a simple example (finding the parameters of a normal) that shows one
way to do it:
x - rnorm(25, 100, 5)
tmpfun - function(x) {
steps - matrix( 0:1, nrow=1 )
myfn - function(param) {
print(param)
Try aperm(x, c(2,1,3))
--
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 MarcioRibeiro
Sent: Tuesday, February
Dear R user,
I am working with LOO. Can any one who is working
with leave one out cross validation (LOO) could send me the code?
Thanks in advance
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org
If you know the final size that your matrix will be, it is better to
preallocate the matrix, then insert the rows into the matrix:
mymat - matrix( nrow=100, ncol=10 )
for( i in 1:100 ){
mymat[i, ] - rnorm(10)
}
Even better than this is to use replicate or sapply if you can, they will
Alex Roy wrote:
Dear R user,
I am working with LOO. Can any one who is working
with leave one out cross validation (LOO) could send me the code?
Thanks in advance
Alex
I don't think that LOO adequately penalizes for model uncertainty. I
recommend the bootstrap or 50
MarcioRibeiro wrote:
Hi Listers,
Is there a way that I can transpose an array...
Suppose I have the following array...
x-array(c(1,2,3,4),dim=c(1,2,2))
, , 1
[,1] [,2]
[1,]12
, , 2
[,1] [,2]
[1,]34
And I would like to get the following result...
, , 1
[,1]
[1,]
Hi Alex,
Give a look at:
http://search.r-project.org/cgi-bin/namazu.cgi?query=leave+one+outmax=20result=normalsort=scoreidxname=Rhelp02aidxname=functionsidxname=docs
Cheers
miltinho astronauta
brazil
On Tue, Feb 24, 2009 at 3:07 PM, Alex Roy alexroy2...@gmail.com wrote:
Dear R user,
And, since my son asked me and I am basketball ignorant: Why are
basketball scores mostly much too close to equality? The arguments
(loose power when leading)
Ted.Harding at manchester.ac.uk writes:
Or: Once you are in the lead, become much more defensive against
attacking play by
Suppose we want 3 rows and the ith row should have 5 columns of i.
Create a list whose ith component is the ith row and rbind them:
rows - 1:3
do.call(rbind, lapply(rows, rep, 5))
[,1] [,2] [,3] [,4] [,5]
[1,]11111
[2,]22222
[3,]3333
Dear all,
Here is one more way to go though using rep() and then matrix():
rows - 1:3
matrix(rep(rows,5),ncol=5)
[,1] [,2] [,3] [,4] [,5]
[1,]11111
[2,]22222
[3,]33333
HTH,
Jorge
On Tue, Feb 24, 2009 at 1:43 PM, Gabor
One more followup on this: I just added a function shapelist3d() to rgl
(so far only on R-forge, not CRAN) that automates a lot of this. To get
the plot as below, the following code works:
x - rep(1:5, each=25)
y - rep(rep(1:5, each=5), 5)
z - rep(1:5, 25)
shapelist3d(cube3d(), x,y,z,
Thank you Greg! Yes this is the comparison I was looking for.
Greg Snow greg.s...@imail.org 2009/02/24 06:07 PM
The same as what? It is not clear what you are trying to compare.
Is this the comparison that you are looking for?:
A - scale(stackloss)
fit1 - lm( stack.loss ~ . - 1,
Hi,
In the result shown, the District 1 is used as the base category. How to
change to make District 4 as a base category ?
On Mon, Feb 23, 2009 at 11:05 AM, choonhong ang angie.bear...@gmail.comwrote:
I have used the insurance data from R library and I have 2 questions:
I use the following:
Hi, may be simle question, but a do not find it anywhere.
Is there same function like max() ,but giving more results.
max() give 1number-maximum
I need funcion what give p bigest number.
many thanks
--
View this message in context:
Try ?relevel
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of choonhong ang
Sent: Tuesday, February 24, 2009 2:33 PM
To: r-help@r-project.org
Subject: Re: [R] Insurance data in library(MASS)
Hi,
In the result shown, the District
On Tue, 24 Feb 2009 11:36:06 -0800 (PST) (SO, 55 Chs 3175 YOLD)
Peterko lanikpe...@gmail.com wrote:
Hi, may be simle question, but a do not find it anywhere.
Is there same function like max() ,but giving more results.
max() give 1number-maximum
I need funcion what give p bigest number.
Dear Peter,
Perhaps:
set.seed(1) # For reproducibility
p-2 # Two biggest values of x
x-rnorm(10)
tail(sort(x),p)
[1] 0.7383247 1.5952808
In a function mode,
foo-function(x,p=2) tail(sort(x),p)
foo(x,p=3)
[1] 0.5757814 0.7383247 1.5952808
HTH,
Jorge
On Tue, Feb 24, 2009 at 2:36
There's probably something built in to R but you can change the values
of the percentiles (p) below to get the value that corresponds to it.
rounding might be problematic also.
temp - c(1,4,8,3,5)
p - 0.8
temp[order(temp)][round(length(temp)*p)]
On Tue, Feb 24, 2009 at 2:36 PM, Peterko
On Tue, 24 Feb 2009, bbbnc wrote:
I'd like to carry out a Monte Carlo simulation test where given data is a
contingency table. I think this is something to do with using rmultinonom(),
but I'm not sure how to code this, to simulate contingency tables. Could
anyone please help with how to use R
Nothing wrong with prior suggestions, but strictly speaking, (fully) sorting
the vector is unnecessary.
y[y quantile(y, 1- p/length(y))]
will do it without the (complete) sort. (But sorting is so efficient anyway,
I don't think you could notice any difference).
-- Bert Gunter
Genentech
One way is to use %in% like below:
df-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
print(df)
df - df[,!(names(df) %in% c('var.a','var.b')),drop=FALSE]
print(df)
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
Dear R-helpers:
I am an R novice and would appreciate answer
Dear R-helpers:
I am an R novice and would appreciate answer to the following question.
Want to delete many variables in a dataframe.
Am able to delete one variable by assigning it as NULL
Have a large number of variables and would like to delete them without using
a for loop.
Is there a
Setting to NULL works only if a single column is selected.
More generally,
df[, !(colnames(df) %in% c(var.b, var.c)), drop=FALSE]
-Christos
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Sean Zhang
Sent: Tuesday, February
Hi list,
I'd like to install an archived version of lmer to compare results with
the current version. I guess one way to do this would be to download the
source, rename the package and then install it. Is there a better
alternative?
--
http://mutualism.williams.edu
signature.asc
Description:
Manuel Morales wrote:
Hi list,
I'd like to install an archived version of lmer to compare results with
the current version. I guess one way to do this would be to download the
source, rename the package and then install it. Is there a better
alternative?
Yes: Install into a different
Dear R Users,
I have three questions.
1) Is there a way to get the output tex file to include
\documentclass{report}
\begin{document}
and
\end{document}
so I can generate a PDF straight away. (I am trying to generate hundreds of
these and don't want to have to manually type this in
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
...Want to delete many variables in a dataframe
df-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
df[,'var.a']-NULL #this works for one single variable
df[,c('var.a','var.b')]-NULL #does not work for multiple variables
Dear R helpers,
When producing a PCA biplot, vectors of environmental variables (as red
arrows with labels) and scores of the observations (black labels
(observation names)) are plotted by default. How can I change the
graphical output? Let's say I would like that the scores are plottet
only
You're right. The help page is somewhat misleading at first read.
?`[-.data.frame` states that (with added emphasis)
value A suitable replacement value: it will be repeated a whole number of
times if necessary and it may be coerced: see the Coercion section. *** If
NULL, deletes the column if
On Tue, Feb 24, 2009 at 3:01 PM, Bert Gunter gunter.ber...@gene.com wrote:
Nothing wrong with prior suggestions, but strictly speaking, (fully) sorting
the vector is unnecessary.
y[y quantile(y, 1- p/length(y))]
will do it without the (complete) sort. (But sorting is so efficient anyway,
I
Well, yes to all -- that's why I pointed this out. Sorting is (depending on
details) = O(n log(n)) while the quantile algorithm uses R's partial sort
option, which I think would be O(n) for a single quantile. For n of a few
thousand or so, the difference shouldn't be noticeable. For large n, of
Stavros Macrakis wrote:
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
...Want to delete many variables in a dataframe
df-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
df[,'var.a']-NULL #this works for one single variable
df[,c('var.a','var.b')]-NULL #does not
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