孟欣 lm_mengxin at 163.com writes:
v1-c(a,b,c,d)
v2-c(a,b,e)
v3-c(a,f,g)
I want to get the intersection of v1,v2,v3,ie a
How can I do then?
What I know is only for 2 vectors via intersect function,
but don't know how to
deal with multiple
vectors.
Reduce(intersect, list(v1 =
On Wed, Feb 1, 2012 at 11:57 PM, HC hca...@yahoo.co.in wrote:
Hi All,
I have a very (very) large tab-delimited text file without headers. There
are only 8 columns and millions of rows. I want to make numerous pieces of
this file by sub-setting it for individual stations. Station is given as
I have some time data and which is in seconds
time -c( 126230400 126252000 126273600 126295200 126316800 126338400)
now I wanted to convert this time to Y M D H M S format
I have tried following codes but it does not give me the out put in Y M D
H M S
time_t1 - as.POSIXlt(time,
On 02.02.2012 06:37, Ginata86 wrote:
I am using the method to sink the output. However, it can only suppress
'user's initial regression estimate
' and still display the following sentence
' Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27'
You have been told to use
Dear all,
Before my question, I wish to all of you my very best wishes for 2012.
I'm using summary.formula to make table. I have something like this :
s1-summary(fdh~cup5+cup6+schef+cpro1+stratify(id2),data=dat,na.include=F)
the output give the marginal row named overall, but is it possible to
Works for me.
Uwe Ligges
On 02.02.2012 02:09, Nick Matzke wrote:
Hi,
Until recently I was using the knncat classifier function of knncat on
an old computer (2.12, Mac OS X 10.4), and everything worked great.
However, now that I have updated to R 2.14.1 (on Mac OS X 10.7), knncat
seems broken.
On 30.01.2012 20:26, Ajay Askoolum wrote:
When I plot, the plot's user interface offers me a choice:
File | Copy to the Clipboard | as a Bitmap.
What is the equivalent code for achieving this but without the plot interface
becoming visible?
For something *equivalent*, see ?dev.copy.
On 01.02.2012 03:37, Matyas Sustik wrote:
Prof Brian Ripley wrote:
'library' in R has a different meaning: I've altered the subject to be
more accurate 'libR'.
This is what R CMD SHLIB is for: it does all this for you in a portable way.
But if you want to DIY, you can use R CMD config to
On 28.01.2012 05:43, Melissa Patrician wrote:
Hi,
Please excuse my inexperience, but I am just learning R (this is my very
first day programming in R) and having a really hard time figuring out
how to do the following:
I have a matrix that is 1000 row by 6 columns (named 'table.combos') and
On Thu, Feb 02, 2012 at 01:55:07PM +0800, 孟欣 wrote:
v1-c(a,b,c,d)
v2-c(a,b,e)
v3-c(a,f,g)
I want to get the intersection of v1,v2,v3,ie a
How can I do then?
What I know is only for 2 vectors via intersect function,but don't know how
to deal with multiple vectors.
Hi.
Set
On Thu, Feb 02, 2012 at 01:55:07PM +0800, 孟欣 wrote:
v1-c(a,b,c,d)
v2-c(a,b,e)
v3-c(a,f,g)
I want to get the intersection of v1,v2,v3,ie a
How can I do then?
What I know is only for 2 vectors via intersect function,but don't know how
to deal with multiple vectors.
Hi.
Try the
What is the best way to write out comma separated data, as a program
is running (rather than waiting to the end using write.csv)? At the
moment I'm doing this, but I guess it's not the most efficient. The
data is in a column in the matrix postcount, and I'm using a loop to
write out each
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 - c(9,5,1,5)
q2 - c(9,2,1,5)
w1 - c(9,4,4,4,5)
w1 -
Hey Chris,
I would take advantage from the apply function:
apply(cbind(q1,q2),1,function(x)any((x[1]==w1)(x[2]==w2)))
Regards
PF
On Thu, Feb 2, 2012 at 12:55 PM, Chris82 rubenba...@gmx.de wrote:
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have
On 02/02/2012 11:40 AM, Thomas wrote:
What is the best way to write out comma separated data, as a program
is running (rather than waiting to the end using write.csv)? At the
moment I'm doing this, but I guess it's not the most efficient. The
data is in a column in the matrix postcount, and
Correct me if I'm wrong, but I think that write.csv() doesn't have an
append argument; write.table() does though.
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
Le 02/02/12
Dear all,
How to add /*vertical*/ lines above bar graph to display p-values (
between pairs of points )?
Regards
ML
--
Mohamed Lajnef,IE INSERM U955 eq 15#
P?le de Psychiatrie#
H?pital CHENEVIER #
40, rue Mesly
How can I pass from position in length inside a matrix to position in dim ?
a=matrix(c(1:999),nrow=9)
which(a==87)#position in length 1:length(a)
87
which(a==87,arr.ind=TRUE) #position in dim
row col
[1,] 6 10
__
R-help@r-project.org
On Feb 2, 2012, at 6:55 AM, Chris82 wrote:
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 -
On Thu, Feb 02, 2012 at 02:08:37PM +0100, Ana wrote:
How can I pass from position in length inside a matrix to position in dim ?
a=matrix(c(1:999),nrow=9)
which(a==87)#position in length 1:length(a)
87
which(a==87,arr.ind=TRUE) #position in dim
row col
[1,] 6 10
Hi.
Thanks for the advice,
df-read.table(infile, sep=, skip = 1, header=TRUE)
is indeed much cleaner from the outset (and was my usual way to it).
I was unaware that
readLines(infile, n=1)
could get me the first line without reading the whole file again.
But I do need to get my head around
Hi
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 - c(9,5,1,5)
q2 - c(9,2,1,5)
w1 -
Also take a look at the arrayInd() function which is what's used by
which() internally for the arr.ind = TRUE case.
Michael
On Thu, Feb 2, 2012 at 8:25 AM, Petr Savicky savi...@cs.cas.cz wrote:
On Thu, Feb 02, 2012 at 02:08:37PM +0100, Ana wrote:
How can I pass from position in length inside a
I believe connections were designed to do this as efficiently as
possible by keeping the i/o path open rather than reopening it each
time like write.table(append = TRUE) would do, though I may be wrong
on the details: see ?connections. Prof Ripley has a good article about
them in R News 1.1 --
Hi all,
I'm struggling a bit to get pgfSweave to lazyload objects when compiling
a .Rnw file for a second time. Caching works fine except that for every
run all objects get cached again and again. I've used cacheSweave which
works fine; all cached objects from code-chunks with option cache =
I don't think it can be removed, a message like this has been coming out for
several years and there may be a good reason why it is there.
Your best bet is probably to approach the package maintainer with a suggestion
to alter the code.
Regards
Søren
-Oprindelig meddelelse-
Fra:
The example here puzzles me. It seems like the operator doesn't work as
expected.
l - 0.6
u - seq(0.4, 0.7, 0.1)
u
[1] 0.4 0.5 0.6 0.7
mygrid - expand.grid(l = l, u = u)
mygrid
l u
1 0.6 0.4
2 0.6 0.5
3 0.6 0.6
4 0.6 0.7
mygridcollapsed - mygrid[mygrid$l mygrid$u, ]
Hi,
I 'd like to know how to retrieve a column name of a data frame. For
instance :
df = data.frame(c1=c('a','b'),c2=c(1,2))
df
c1 c2
1 a 1
2 b 2
I would like to retrieve the column name which value is 2 (here, the column
is c2)
thanks for your help
--
View this message in context:
Hi,
I would like to know, if there's any R function, which allows
calculation of probability values (0,1) from multivariate normal densities.
I would be grateful for any output.
Cheers,
MG
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R-help@r-project.org mailing list
Hi,
I have the following data.frame:
data.frame(x = c(1:10), y = rnorm(10,2,1), label = rep(c('a', 'b', 'c', 'd',
'e'),2))
in this data.frame there is a label-variable containing strings.
Each string is represented two times.
Now I would like to have the mean of the corresponding x (and
The Sys.setenv(NOAWT=TRUE) code indeed solved my problem which was excatly
what Julien described.
The key is you have to deactivate AWT BEFORE loading RWeka/Snowball. If I do
so it will fire a few warning messages but that should not affect anything.
I am running the lsa package which requires
colnames(df)[2]
Michael
On Thu, Feb 2, 2012 at 10:31 AM, ikuzar raz...@hotmail.fr wrote:
Hi,
I 'd like to know how to retrieve a column name of a data frame. For
instance :
df = data.frame(c1=c('a','b'),c2=c(1,2))
df
c1 c2
1 a 1
2 b 2
I would like to retrieve the column name
There are *many* ways, but here's two:
df = data.frame(x = c(1:10), y = rnorm(10,2,1), label = rep(c('a',
'b', 'c', 'd', 'e'),2))
with(df, ave(x, label)) # Returns the correct value in each spot
(useful if you want to add a group-mean column to df
with(df, tapply(x, label, mean)) # Probably
You need to back up a bit to see the root cause of
the problem, which is that seq()'s calculations necessarily
involve some roundoff error (since it works with 52 binary
digits of precision):
u - seq(from=0.4, to=0.7, by=0.1)
u - c(0.4, 0.5, 0.6, 0.7)
[1] 0.00e+00 0.00e+00
I'm not sure what you mean probability values but you can get a
multivariate normal density from
library(mvtnorm)
? dmvnorm
The same package also provides a pmvnorm but one has to be slightly
more comfortable handling CDFs in the multivariate case.
Michael
2012/2/2 Michał Góralski
This is R FAQ 7.31, about machine representation of
floating point numbers.
mygrid$u[3] - mygrid$l[3]
[1] 1.110223e-16
So
mygrid$l[3] mygrid$u[3]
is true, though the difference is very, very small and due solely
to the limitations of computers.
Sarah
On Thu, Feb 2, 2012 at 5:00 AM, Jonas Hal
It's likely an infelicity of floating point representations (R FAQ
7.31) but admittedly, not a case I would have expected to present
itself.
If you want it to work out as expected, try this:
l - 0.6
u - seq(0.4, 0.7, 0.1)
l.int - (6L) / 10
u.int - seq(4, 7) / 10
l u
l.int u.int
Michael
On
colnames( df )[2]
[1] c2
On Thursday 02 February 2012 07:31:33 ikuzar wrote:
Hi,
I 'd like to know how to retrieve a column name of a data frame. For
instance :
df = data.frame(c1=c('a','b'),c2=c(1,2))
df
c1 c2
1 a 1
2 b 2
I would like to retrieve the column name
I would like to fit the following model:
logit(p_{ij}) = \mu + a_i + b_j
wherea_i ~ N(0, \sigma_a^2) , b_j ~ N(0, \sigma_b^2) and \sigma_a
= \sigma_b.
Is it possible to fit a model with such a constraint on the variance
components in glmer?
--
View this message in context:
Sorry, it was not clear:
my program have to return column name corresponding to a value, for example
'b' (so, the corresponding column is c1)
How to retrieve c1 ?
Thanks
--
View this message in context:
Recently I've run into memory problems while using data.frames for a
reasonably large dataset. I've solved those problems using arrays, and
that has provoked me to do a few benchmarks. I would like to share the
results.
Let us start with the data. There are N subjects classified into G
groups.
I'd use something like
which(df == b, arr.ind = TRUE)
which, gives the column number in the second spot; this gives you
colnames(df)[which(df == b, arr.ind = TRUE)[2]]
Michael
On Thu, Feb 2, 2012 at 11:00 AM, ikuzar raz...@hotmail.fr wrote:
Sorry, it was not clear:
my program have to
Thank you Duncan, the 2nd instructions worked. The both probably would have
worked, but I had some code (below) that threw an error. It's designed to
automatically set the internet connection to my Windows setting, so R would
know the proxy server used at my worksite. I had seen this
Hi,
I have fixed this. I replaced the = with -. I do not think this
is the most elegant way, so if anyone else has any better ideas they
would be very much apperitiaded.
New lines:
formulaGenotype - test_variable~Genotype + Gender
formulaNull - test_variable~Gender
Cheers,
Hugh
On
Luk: Don't know if this solves your desire for an implementation in R,
but the most general extension of Cohen's kappa for testing agreement that
I'm aware of are the extensions made by using multi-response randomized
block permutation procedures (MRBP) developed by Pual Mielke and Ken
Dear Uwe ,
Thanks for reply
I have tried format function that u suggested (format(time_t1, %Y %m %d %H
%M %S) and I got
format(time_t1, %Y %m %d %H %M %S)
[1] 126230400 126252000 126273600 126295200 126316800 126338400
I think something is not working correct.
--
View this message in
This is very interesting, Brian, thanks. I was starting to wonder if the
reason I can't find an implementation of Kraemer's approach might be that
there are better methods. The key extension of kappa by Kraemer, for my
purposes, was the jacknife estimate to improve on estimated standard errors
If you are willing to use base graphics instead of ggplot2 graphs, then look at
the subplot function in the TeachingDemos package. One of the examples there
shows adding multiple small bar graphs to a map.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
I'm trying to format text on a plot such that the string is vertical but the
letters are horizonal. I tried
text(1,1,label=output, srt=270)
This gives the string rotation I want, but that rotates the entire output so
the letters are also rotated. I've also tried
text(1,1,label=output,
On Thu, Feb 2, 2012 at 3:11 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Feb 1, 2012 at 11:57 PM, HC hca...@yahoo.co.in wrote:
Hi All,
I have a very (very) large tab-delimited text file without headers. There
are only 8 columns and millions of rows. I want to make numerous
One possible solution is to use strsplit to break on each character and then
paste to put in a \n after each character. Then when you plot the text
should be in the format you desire.
x - outputy - unlist(strsplit(x, NULL))p - cat(paste(y, collapse=\n))
plot.new()text(.5, .5, paste(y,
On Thu, Feb 02, 2012 at 10:00:58AM +, Jonas Hal wrote:
The example here puzzles me. It seems like the operator doesn't work as
expected.
l - 0.6
u - seq(0.4, 0.7, 0.1)
u
[1] 0.4 0.5 0.6 0.7
mygrid - expand.grid(l = l, u = u)
mygrid
l u
1 0.6 0.4
2 0.6 0.5
3 0.6 0.6
There are only a few graphics devices that honor the 'crt' setting to
rotate characters differently from the string rotation (postscript is
the only one I know of, and then not always).
For your specific case you could do something like:
text(1,1, paste( unlist(strsplit('output','')),
I apologize for the improperly formatted submission. I had my hotmail set to
plain text instead of rich text.
x - outputy - unlist(strsplit(x, NULL))
plot.new()text(.5, .5, paste(y, collapse=\n))
From: tyler_rin...@hotmail.com
To: israelb...@hotmail.com; r-help@r-project.org
Date: Thu, 2
[cc'ing back to r-help]
On 12-02-02 01:56 PM, Sally Luo wrote:
I tried to adapt your code to my model and got the results as below. I
don't know how to fix the warning messages. It says rearrange the lower
(or upper) bounds to match 'start'.
The warning is overly conservative in this
Dear reader, I'm a student on engineering studies at Silesian University of
Technology in Gliwice in Poland, my field of study is Technology and
Mechanical Engineering on Integrated process of manufacturing systems, also
I held a Bachelor's degree on Automation and Robotics. However I have a view
I've created a linear model and am trying to use the predict function to
predict the outcome of a sports game.
I have four explanatory variables a,b,c,d. where a,b relate to the home
team and c,d relate to the away team.
i'd like to know the probability that the home team wins (assuming no
Is it possible to apply a kernel smoothing regression whose estimator or
indeed the confidence intervals cannot take negative values or values
greater than 1?
Best regards,
Ioanna
--
View this message in context:
Hello,
I want to perform a permanova where the first factor called Treatment
has four levels. The second factor involves sampling the same research
plots for four consecutive years, hence the repeated measurements.
I have been able to use the adonis function from the package vegan to
run this
It works for me as well so there's something funny on your end: please
run the following *verbatim* (in a vanilla R session):
sink(ForRHelp.txt)
print(sessionInfo())
cat(\n)
print(.Platform)
time -as.POSIXct(c( 126230400, 126252000, 126273600),
origin=2005-01-01, tz=GMT)
print(time)
Hi.
I am looking for a function in R for computing the Fiedler vector of a graph
(the eigenvector associated with the second smallest eigenvalue of the
Laplacian of the graph). Alternatively, I am searching for an efficient method
to compute just few eigenvalues/vectors of a matrix (the
Caveat: Both of the following claims are subject to verification by
true experts, which I am not. But I believe:
1. If all values being smoothed are positive, then the smoother must
be also. If there are negative values, this is no longer true, and
your question needs much more detail to get an
On 02-02-2012, at 19:23, R. Michael Weylandt wrote:
It works for me as well so there's something funny on your end: please
run the following *verbatim* (in a vanilla R session):
sink(ForRHelp.txt)
print(sessionInfo())
cat(\n)
print(.Platform)
time -as.POSIXct(c( 126230400, 126252000,
Disclaimer: I am a lawyer so this all should be verified elsewhere,
but best I understand it (and would welcome verification by someone
who knows more about this):
The R-Project (broadly taken) is an extensive collection of packages +
a core interpreter. The interpreter, the base packages, and
[cc'ing back to r-help again -- I do this so the answers can be
archived and viewed by others]
On 12-02-02 02:41 PM, Sally Luo wrote:
Prof. Bolker,
Thanks for your quick reply and detailed explanation.
I also ran the unrestricted model using glmfit -
glm(y~x1+x2+x3+x4+x5+x6+x7+x8,
On 02-02-2012, at 21:10, Berend Hasselman wrote:
On 02-02-2012, at 19:23, R. Michael Weylandt wrote:
It works for me as well so there's something funny on your end: please
run the following *verbatim* (in a vanilla R session):
sink(ForRHelp.txt)
print(sessionInfo())
cat(\n)
Hi,
the following Code demonstrates an possibly Error in R
(or you can explain me, why this happens, thanks in advance)
Code:
#
testClass - function( stackData= c())
{
list(
write= function( ...)
{
sChain=
for( s in c( stackData, ...))
{
Hi Paul! Thanks a lot!
I tried downloading the Rdisop file and encountered this error:
Error: package ‘Rdisop’ is not installed for 'arch=x64'
I tried downloading directly from the source using R and got this error:
Error in file(filename, r, encoding = encoding) :
cannot open the
I also tried downloading the JDK version of Java and received this new error
when running it:
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: dirname(this$RuntimeLib)
error: a character vector argument expected
Error: package/namespace load failed for ‘rJava’
--
View
push (sorry ;-))
--
View this message in context:
http://r.789695.n4.nabble.com/Splitting-up-large-set-of-survey-data-into-categories-tp4323327p4352611.html
Sent from the R help mailing list archive at Nabble.com.
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Hi R-helpers,
TukeyHSD() works for models fitted with aov(), but could anyone point
me to a function that performs a similar post hoc test for models
fitted with lm() or glm()?
Thanks in advance,
Mark
__
R-help@r-project.org mailing list
Wow. Thanks very much for pointing that out - I never would have guessed it was
deliberate that + and - were reversed!
For future reference for anyone else similarly confused by this departure from
time zone and mathematic convention, here's the relevant part of
The R multcomp package provides one general approach to multiplicity
correction. For general contrasts in lm and glm, the rms package's ols and
Glm functions make this even easier to use.
Frank
Mark Na wrote
Hi R-helpers,
TukeyHSD() works for models fitted with aov(), but could anyone
The glht function in the multcomp package is what you are looking for.
There are additional examples in the ?MMC help file in the HH package.
Rich
On Thu, Feb 2, 2012 at 3:42 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
TukeyHSD() works for models fitted with aov(), but could anyone
Thank you Richard and Frank for your very quick and helpful replies.
Cheers, Mark
On Thu, Feb 2, 2012 at 2:58 PM, Frank Harrell f.harr...@vanderbilt.edu wrote:
The R multcomp package provides one general approach to multiplicity
correction. For general contrasts in lm and glm, the rms
I can reproduce it but I do not know why this happens.
FWIW, I tried the knitr package and it worked well except that you
have to write cache=TRUE or FALSE instead of true/false.
library(knitr)
knit('test_pgf.Rnw')
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web:
Hello all,
I was wondering if there is an R function to do the following:
[*] log(pnorm(x)-pnorm(y)), where xy.
I don't want all the area under the natural log of the normal pdf less than
x, I only want the area between y and x.
I am aware of the ability to specify log.p=TRUE, which gives me
Hi michael, thanks, but here is more explanations of my questions to have more
help, (also pls have a look at the data below):
Â
Three questions to give more concrete help:
i) Is your data set stored as a matrix or a data.frame
My data is in a data frame
ii) What are you trying to get the mean
Hi michael, thanks, but here is more explanations of my questions to have more
help, (also pls have a look at the data below):
Â
Three questions to give more concrete help:
i) Is your data set stored as a matrix or a data.frame
My data is in a data frame
ii) What are you trying to get the mean
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of uday
Sent: Thursday, February 02, 2012 8:57 AM
To: r-help@r-project.org
Subject: Re: [R] time conversion from second to Y M D H M S format
Dear Uwe ,
Thanks for reply
I have
I suggest you read the GNU license included in the source code and on the CRAN
website. The essence is that you are free to use it, and to change it, but if
you pass your changes on to anyone else, you have to make the source code of
those changes available to those whom you give it to. Most
I have some elements in a vector with extraneous information (e.g. file
name and sample IDs) that I'd like to strip from every element.
For example, I would like SPI1.S1.str1.P3.sample.tif
SPI1.S1.STR2.P1.sample.tif to read SPI1.S1.str1.P3 SPI1.S1.STR2.P1.
Will someone help me with the syntax
In the example you gave, all that has to be done is
replace .sample.tif at the end of the string with , which
is easy.
avec - c(SPI1.S1.str1.P3.sample.tif, SPI1.S1.STR2.P1.sample.tif)
gsub(\\.sample\\.tif$, , avec)
[1] SPI1.S1.str1.P3 SPI1.S1.STR2.P1
If your real data are more complex, we
Oh, perfect. I was running
gsub(.sample.tif, , avec).
your change
gsub(\\.sample\\.tif$, , avec)
did it.
Thanks Sarah
*Ben Caldwell*
On Thu, Feb 2, 2012 at 1:48 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
In the example you gave, all that has to be done is
replace .sample.tif at the
On Thu, Feb 02, 2012 at 01:18:42PM -0800, justin jarvis wrote:
Hello all,
I was wondering if there is an R function to do the following:
[*] log(pnorm(x)-pnorm(y)), where xy.
I don't want all the area under the natural log of the normal pdf less than
x, I only want the area between y and
I thought some of you might be amused by this.
In my non-work time, I'm an avid weaver and teacher of weaving. I'm
working on a project involving creating many detailed weaving
patterns, so I wrote R code to automate it.
Details here:
http://stringpage.com/blog/?p=822
If the overlap between R
I want to run the glm () function for my data but instead of using the family
distributions in R, I need the 4P Burr distribution.
Can some please explain how can I go about doing that. Or please provide me
with an example.
I'm new to R.
Eg.
Model1 - glm(Postwt ~ Prewt + Treat +
I'm using bigkmeans in 'biganalytics' to cluster my 60,000 by 600,000 matrix.
I'm using a 8 core Linux VM.
I have register parallel backend with
registerDoMC()
And I checked how many cores registered with
getDoParWorkers()
It returns 8, which is the number of cores I have on my machine.
And I
Brilliant Sarah ! I love seeing such unexpected and creative applications.
I'm not a weaver but am a knitter (and a knotter actually) and have
mused about using R to help design elements of textured knitting
patterns e.g. as seen in single-colour, traditional fisherman's
jumpers from England and
Dear all,
parallel(~iris[1:4], groups = Species, iris, par.settings = simpleTheme(lwd
= c(1,3,1), lty = c(1,1,2), col.line = 1), auto.key = T)
Despite the use of par.settings and simpleTheme, the lines in the key and
graph are not the same. Any suggestions why?
Regards,
Marcin
Well, I have to say, how nice to find a valid use for string theory :-) .
Now that we all know you are in fact the mistress of skulls, guess we
better tread lightly!
Carl
quote
From: Sarah Goslee sarah.goslee_at_gmail.com
Date: Thu, 02 Feb 2012 17:54:04 -0500
I thought some of you might
Hi Gabor,
Thank you very much for your guidance and help.
I could run the following code successfully on a 500 mb test data file. A
snapshot of the data file is attached herewith.
code start***
library(sqldf)
library(RSQLite)
iFile-Test100.txt
con -
On Thu, Feb 2, 2012 at 8:07 PM, HC hca...@yahoo.co.in wrote:
Hi Gabor,
Thank you very much for your guidance and help.
I could run the following code successfully on a 500 mb test data file. A
snapshot of the data file is attached herewith.
code start***
library(sqldf)
Dear R-users,
I have 3 plant populations (fixed). Within each population there is the
same number of âfamiliesâ (random) â the seed progeny of the same plant.
These families were exposed to 2 treatments (fixed) and their response was
measured (mean values for 25 seedlings per family per
Hi.
I was looking for help on how to use Tukey multiple comparison on Type
III SS because I read on Quick R that it is using Type I SS by default.
I am wondering if the use of glht helps.
Thanks for help.
Vera
__
R-help@r-project.org mailing
I don't know what your data look like, but I recently ran into this error
message while using Anova() in the {car} package, and I resolved in by
replacing the categorical predictors in my model with orthogonal contrasts.
I did something along these lines:
fac - factor(c(M,F,M,M,F)) # A
Hello,
I have some data that will be in the form:
structure(list(station = structure(c(20L, 2L, 4L, 19L, 3L, 11L,
1L, 5L, 10L, 12L, 17L, 18L, 6L, 9L, 13L, 16L, 7L, 8L, 15L, 14L
), .Label = c(1, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 2, 3, 4, 5, 6, 7, 8, 9, scope
), class = factor), distance =
Matthew,
My Fault I should have sent you to the current release cycle page. The link was
for the old 2.6 Bioconductor release, sorry about that. :(
Personally the easies way to install any bioconductor package is to open R copy
and paste the following code:
Hello, I am new to R and I am having problems trying to model logistic
population growth with the deSolve package. I would like to run the model
for four populations with the same initial population and carrying capacity
but with different growth rates and put the results into a data frame. When
I
Ok,
I've since found this:
# called previously posted dataset dat
attach(dat)
library(akima)
data.interp - interp(x,y,z)
contour(data.interp)
any idea how to calculate area within specified contour lines?
Thanks
chuck.01 wrote
Hello,
I have some data that will be in the form:
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