Hello,
On 6/16/09, Stu @ AGS s...@agstechnet.com wrote:
Error in optim(c(0.66, 0.999, 0.064), pe, NULL, method = L-BFGS-B) :
objective function in optim evaluates to length 6 not 1
skip
pe - function(c) c[1]*x1*x2^c[2]*x3^c[3]
I would suspect a matrix multiplication issue. In order to
On Sun, 14 Jun 2009 22:30:49 -0300 Antonio Olinto
aolint...@bignet.com.br wrote:
AO I would like to receive any suggestion of which is most appropriate
AO for a non-statistician (I am a biologist). Reading only the index I
AO could not evaluate it.
AO reproducible code.
I am an economist but
Dear R-helpers,
I want to make a series of boxplots on several numeric univariates with two
group variables (species and population, population nested in species, and
with population as the X-axis). In order to get a proper order of the
individual populations in X-axis, I need to assign a wanted
Hi,
The way you do it actually renames the factors one after each other (it
replaces the values in the data frame, which is not what you want).
Have a look at this code:
test - data.frame(id=c(1,2,3), fac=c(lv1, lv2, lv3) )
levels(test$fac)
test$fac2 - factor(test$fac, levels=c(lv3, lv2, lv1))
On Tue, Jun 16, 2009 at 03:14:01PM +0800, Mao Jianfeng wrote:
levels(d$population)-c(YXPy01, KMPy01, YLPy01, GSPy02, BCPy01,
LJPy01, GYPt01, YLPd01, CYPd01, CYPd02, CYPd03, BXPd01,
NSPt01)
I'm not at home with factors myself, but maybe this will do the trick for
you:
d$population -
http://biocep-distrib.r-forge.r-project.org/doc.html
Regards
Malcolm
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Hi r-helpers!
I need to save the output of summary() function that Ive
runned like this:
z- lmList(y~x1+x2| x3,
na.action=na.omit,data1,subset=year==1999)
w-summary(z)
The output (w) is something like this:
Call:
Model: y ~ x1 + x2 | x3
Data: data1
Coefficients:
(Intercept)
Hi Mao,
I am confused. And, I want to know how to assign a wanted order to factor
levels, intentionally?
You want ?relevel. Although the documentation leads one to think that it can
only be used to set a reference level, with the other levels being moved
down, presently it can in fact be
Thank you for the reply. I really appreciate it.
I calculated the Scoenfeld residual per event and my results are the
following:
finage race
17 -0.33942334 -2.0722187270.29024804
20 0.394600944 5.303968774 0.517689472
25
Dear list
I'm having trouble with inverting color gradation.
As seen in the rgl example,
library(rgl)
example(rgl.surface)
I understand that I can assign colors to heights for each point.
col - colorlut[ y-ylim[1]+1 ] # assign colors to heights for each point
However, I am now trying to
Dear List,
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3), df$fac, mean)
Thank you,
Stefan
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Hi,
I tend to use a slightly modified version of stats::relevel, (from an
old thread on this list),
relevel =
function (x, ref, ...)
{
lev - levels(x)
if (is.character(ref))
ref - match(ref, lev)
if (any(is.na(ref)))
stop('ref' must be an existing level)
nlev - length(lev)
Mark Difford wrote:
Hi Mao,
I am confused. And, I want to know how to assign a wanted order to factor
levels, intentionally?
You want ?relevel. Although the documentation leads one to think that it can
only be used to set a reference level, with the other levels being moved
down,
Hi,
actullay i am try to intsall Rpy2 (a python interface for R) later, for RPy2
requries R 2.7.0 higher
but default intsall RPy2 will search the system installed R 2.4.0
so i think to replace the system installed R with my R in my home directory
echo $PATH
Hi,
i just add /home/lidaof/R/bin and /home/lidaof/R/lib to the end of the PATH
variable
yet i type R command runs the system installed R 2.4.0...
On Tue, Jun 16, 2009 at 5:48 PM, Daofeng Li lid...@gmail.com wrote:
Hi,
actullay i am try to intsall Rpy2 (a python interface for R) later, for
Dear all,
I've been trying to superscript the '2' in the following command (I don't want
the '^' displayed), but as yet haven't had much luck. I've tried both the paste
and expression commands, but neither have brought me any joy!
mtext(side=2, line=5.5, Monthly Precipitation (mm x
Commenting on this, is there a strong argument against modifying
relevel() to reorder more than one level at a time?
I started a topic a while back (recursive relevel,
https://stat.ethz.ch/pipermail/r-help/2009-January/184397.html) and I've
happily used the proposed change since then by
Thanks to Luke Tiernay and some experimenting I found out some issues. I
don't claim this infomation is complete, but it may be helpful for anyone
experimenting with SNOW on Linux:
- environment variables PATH and R_SNOW_LIB need to be set on master and
slaves.
(manually or permanent in
Dear Henrique,
I think that R is not actually the best statistical tool to model MS-VAR.
Indeed, the package msvar only allow a simple specification of the model.
One tool I have ever used is on Ox with the package MSVAR built by Krolzig.
This package allow a large variety of model
Stefan Uhmann wrote:
Dear List,
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3), df$fac, mean)
because
length(cbind(df$var1, df$var2, df$var3))
# 9
length(df$fac)
On Tue, 16-Jun-2009 at 05:48PM +0800, Daofeng Li wrote:
| Hi,
|
| actullay i am try to intsall Rpy2 (a python interface for R) later, for RPy2
| requries R 2.7.0 higher
But as I mentioned earlier, R-2.9.0 is current and R-2.9.1 will be
released next week. It's nearly always best to use a
On Tue, Jun 16, 2009 at 10:01:00AM +, Steve Murray wrote:
I've been trying to superscript the '2' in the following command (I
don't want the '^' displayed), but as yet haven't had much luck.
I've tried both the paste and expression commands, but neither have
brought me any joy!
Setting methods for groups (compare in this case) does not work
properly. Once one method is set ,redefining it or even removing does
not change the behavior:
Can anyone suggest how to refresh methods table for some particular
generic? (in the above case Compare group generic).
For now
On Tue, Jun 16, 2009 at 5:16 AM, Stefan Uhmann stefan.uhm...@googlemail.com
wrote:
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3), df$fac, mean)
Because tapply is defined for
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 12:45:04:
Stefan Uhmann wrote:
Dear List,
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3), df$fac, mean)
because
Stavros Macrakis wrote:
On Tue, Jun 16, 2009 at 5:16 AM, Stefan Uhmann stefan.uhm...@googlemail.com
wrote:
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3),
Hi All,
There are several replies to the question below, but I think there must
exist a better way of doing so.
I just want to check whether all the elements of a vector are same. My
vector has one million elements and it is highly likely that there are
distinct elements in the first few
For the sake of brevity, I like to use this trick,
plot(0, 0)
mtext(~Monthly Precipitation (mm x *10^2*/month))
HTH,
baptiste
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Petr PIKAL wrote:
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 12:45:04:
Stefan Uhmann wrote:
Dear List,
why does this not work?
df - data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
fac = c('A', 'A', 'B'))
tapply(cbind(df$var1, df$var2, df$var3),
Just check that the first (or any other element) is equal to all the rest:
x = c(1,2,rep(1,1000)) # 10,000,000
system.time(print(all(x[1] == x)))
[1] FALSE
user system elapsed
0.180.000.19
This was for 10M entries.
On Tue, Jun 16, 2009 at 7:42 AM, utkarshsinghal
Steve Murray wrote:
Dear R Users,
I am trying to plot a barchart with a line graph superimposed (using
par(new=TRUE)). There are 12 bars and 12 corresponding points for the line
graph. This is fine, except that I'm encountering two problems:
1) The position of the points (of the line graph)
Take advantage of a 20% discount on the most recent R books from Chapman
Hall/CRC!
We are pleased to offer our latest books on R at a 20% discount through our new
website. To take advantage of this offer, simply visit
http://www.crcpress.com/, choose your titles and insert code 281DW in the
Hi,
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 long formula will overshoot
the margins in
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character vectors, it gets more than double.
In my modeling, which is already highly
I was using older version of R (installed early). I install new version of R
(R.2.9.0) but i could not find package xlsReadWrite to read Excel file.
As others have pointed out (thanks) you can find it here:
http://treetron.googlepages.com/.
It runs fine in 2.9.0.
Is there any alternatives to
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly since it is
having to loop to check the equality of each character in each element.
This
Dear All,
I am running a Ubuntu 8.04 System and trying to install the
Design-package. Hmisc is already installed, all fortran compilers and a
Tex-Package are also on board. I searched the net and the help list for
analogue threads, but didn't find any.
Whenever I run the
Hi
I want to use anova-pca method, but I don't know what package I use to.
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?rev
On Jun 16, 2009, at 5:13 AM, Naoki Irie wrote:
Dear list
I'm having trouble with inverting color gradation.
As seen in the rgl example,
library(rgl)
example(rgl.surface)
I understand that I can assign colors to heights for each point.
col - colorlut[ y-ylim[1]+1 ] # assign colors
Dear list,
I use Sweave almost exclusively for writing papers, and I have become
quite spoiled by the excellent xtable export facilities. Has anybody
written an xtable method for the Anova function in CAR, or has anybody
used a different set of functions to import Anova results into
a table in
utkarshsinghal wrote:
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character vectors, it gets more than double.
In my
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 14:31:21:
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
With numbers you may speed it a bit
Hi all
I have an irregular zoo series, where the time index looks like the
following:
head(time(l.zoo))
[1] 2009-06-15 01:44:20.802 GMT 2009-06-15 01:44:20.812 GMT 2009-06-15
01:44:20.837 GMT 2009-06-15 01:44:20.848 GMT 2009-06-15 06:00:01.320
GMT
[6] 2009-06-15 06:00:01.330 GMT
Thanks for your response!
No, my basic equation does not use matrices at all. It takes scalar values and
returns a scalar.
What I am trying to accomplish is to find the best-fit coefficients to the
equation as follows:
y ~ c1 * x1 * x2^c2 * x3^c3
where y, x1, x2, and x3 are observed data and
-- begin included message
Thank you for the reply. I really appreciate it.
I calculated the Scoenfeld residual per event and my results are the
following:
finage race
17 -0.33942334 -2.0722187270.29024804
20 0.394600944
[correcting a stupid error in my previous post]
testTwoStages - function(x, y, head.stop = 100){
if(!isTRUE(all(head(x, head.stop) == head(y, head.stop
{
print(paste(quick test returned FALSE))
return(FALSE)
} else {
full.test = isTRUE(all(tail(x, length(x) - head.stop) == tail(y,
Not in the direction I was looking, but your function fff really speed
up the process, and using it along with fff3 is fine for me.
Thanks
Petr PIKAL wrote:
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 14:31:21:
Hi Jim,
What you are saying is correct. Although, my computer
Hello,
I want to generate data set from Cox PH model with gamma frailty effects.
theta(parameter for frailty distribution)=2
beta=1.5
n=300
cluster size=30
number of clusters=10
I think I should first generate u from Gamma(Theta,theta) and then using
this theta I could not decide how I should
Thnaks to both of you, with a combined effort of the Wiki and your code I
have now managed to get teh result I was after.
When I have a better understanding of what the code is doing I shall make an
effort to contribut to the Wiki!
Cheers,
Kenny
Kenny Larsen wrote:
Hi All,
I have
Dear Maria,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Maria Wolters
Sent: June-16-09 6:31 AM
To: r-help@r-project.org
Subject: [R] Output of Anova (CAR package) in Sweave
Dear list,
I use Sweave almost exclusively
Not to fix your fundamental problem, but if you don't mind using the
next-to-last versions of the packages you can install them quite easily
as debian/Ubuntu packages. They are named r-cran-hmisc and
r-cran-design. You can use sudo apt-get install ... or the adept package
manager etc.
Frank
See R News 4/1.
On Tue, Jun 16, 2009 at 9:04 AM, rory.wins...@gmail.com wrote:
Hi all
I have an irregular zoo series, where the time index looks like the
following:
head(time(l.zoo))
[1] 2009-06-15 01:44:20.802 GMT 2009-06-15 01:44:20.812 GMT 2009-06-15
01:44:20.837 GMT 2009-06-15
Dear R-helpers,
Very small amount of outliers can greatly affect the mean and many other
statistic of a numeric variable. So, usually we must deal with the outliers
properly in the process of data analysis. Here, I want to replace outliers
with the group median of the variable. But, I can not
Thanks Gabor - I'll check it out.
Actually I just realised I can also do what I am looking for in a
ridiculously simple manner (as the data I have is intra-day):
aggregate(l.zoo, hours(index(l.zoo)), mean)
Cheers
-- Rory
On Jun 16, 2009 2:46pm, Gabor Grothendieck ggrothendi...@gmail.com
Hi Cecilia,
Could you send us a reproducible example?
cheers
milton
On Tue, Jun 16, 2009 at 4:29 AM, Cecilia Carmo cecilia.ca...@ua.pt wrote:
Hi r-helpers!
I need to save the output of summary() function that Ive runned like this:
z- lmList(y~x1+x2| x3,
This is a sure way to get a biased variance estimate.
Instead, use a robust dispersion (scale) estimator such as Gini's mean
difference (average absolute difference between any two observations).
The median is a robust location estimator. There are others. If your
ultimate goal is a
Hi All,
I'm trying to format the y-axis in an xyplot to show numbers with a comma
separating the thousands but I'm not able to do it using formatNum(y,
big.mark=,). This is what I have:
library(lattice)
year-c(2003,2004,2005,2006,2007,2008,2003,2004,2005,2006,2007,2008)
Many thanks to John Fox, who replied on-list,
and to David Hajage, who replied to me
off list. David suggested this quick hack of the
print.anova.mlm function, which I am sharing with his
permission
# From print.Anova.mlm
xtable.Anova.mlm - function (x, ...) {
test - x$test
repeated -
Hello!
I am starting to use the lattice package. I generated an xyplot
conditioned on a factor that has three levels: hence I get three plots
in three panels spaces and one is left empty. I would like to add a
plot to the empty panel space. Is it possible?
Thank you
Antonio,
I just got the Cowpertwait and Metcalfe book (released today in the U.S.) and I
think it's what you're looking for. It has good, practical examples and lots of
sample code. I looked at Cryer's book, but it had too much on time series
theory for my purposes; Cowpertcalfe helpfully
Hi all,
As of now, I have a 15x8 matrix (name is asdf). The first seven columns
contain numbers while the last column contains a string. The class of each
column is character. When I use the plot function to display a scatter
plot between any of the two columns, ie. plot(asdf[, 1], asdf[, 2])
I wonder whether R provides an interface to access miRecords data.
Particularly, I am looking for extracting humans miRNA and target genes
sequences.
All such information is stored in there in a set of structured web site pages
(http://mirecords.umn.edu/miRecords)
I would greatly appreciate
Hello!
I am trying to write a function with vector and data.frame parameters that
uses the sum() function and values from the rows of the data.frame.
I need to pass this function as a parameter to optim().
My starting point is:
observs - data.frame(y, x1, x2, x3)
Fn - function(par,
On Tue, 16 Jun 2009, jim holtman wrote:
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly since it is
having to loop to check the
plot(asdf[,1:7])
On Tue, Jun 16, 2009 at 10:58 AM, njhuang86njhuan...@yahoo.com wrote:
Hi all,
As of now, I have a 15x8 matrix (name is asdf). The first seven columns
contain numbers while the last column contains a string. The class of each
column is character. When I use the plot function
At 10:57 16/06/2009, Daofeng Li wrote:
Hi,
i just add /home/lidaof/R/bin and /home/lidaof/R/lib to the end of the PATH
variable
yet i type R command runs the system installed R 2.4.0...
[snip]
On Sat, 13-Jun-2009 at 03:43PM +1000, bill.venab...@csiro.au wrote:
| You need to adjust your
Why not just use rowSums?
Ronggui
2009/6/16 Stu @ AGS s...@agstechnet.com:
Hello!
I am trying to write a function with vector and data.frame parameters that
uses the sum() function and values from the rows of the data.frame.
I need to pass this function as a parameter to optim().
My
On Tue, 16 Jun 2009, Prof Brian Ripley wrote:
On Tue, 16 Jun 2009, jim holtman wrote:
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly
I'm trying to identify the positions of all genes within a specific
chromosomal region using biomart. When using the current biomart
database I'm able to do this without issue. However, I need to use
build 36 of the mouse genome which was last included in ensembl mart
46. I selected this
Looks like a BioConductor question.
On Jun 16, 2009, at 11:05 AM, mau...@alice.it wrote:
I wonder whether R provides an interface to access miRecords data.
Particularly, I am looking for extracting humans miRNA and target
genes sequences.
All such information is stored in there in a set of
The apply function is the way to iterate over rows or columns of
dataframes or matrices. An example would have made this process easier
testing and I have given up doing that job for
You might try:
apply(observs, 1, function(x) Fn(par, x) )
On Jun 16, 2009, at 11:08 AM, Stu @ AGS wrote:
?data.matrix
Perhaps:
pairs(data.matrix(asdf[ , 1:7]) )
On Jun 16, 2009, at 10:58 AM, njhuang86 wrote:
Hi all,
As of now, I have a 15x8 matrix (name is asdf). The first seven
columns
contain numbers while the last column contains a string. The class
of each
column is character. When
Frank,
Your template is very interesting. The pretty-ifing of the left arrow and
tilde in the input chunk has the unfortunate side-effect of making the code
non-paste-able into R.
I seem to recall that your reports used to include the latex source code as
an appendix. Maybe one option could be
Hi,
yes, that's amazing
follows my .bashrc file
# .bashrc
# User specific aliases and functions
# Source global definitions
if [ -f /etc/bashrc ]; then
. /etc/bashrc
fi
alias rm='rm -i'
if [ ! -d $HOME/trash ]; then
mkdir $HOME/trash
fi
del ( ) { mv $@ $HOME/trash/.; }
Hi, your problem has nothing to do with your specific dataset. Further, it
is - let's say - suboptimal to push a 5 MB dataset through the mailing list.
The posting guides asks to provide an example with self-contained code; this
is what Milton has asked for. Such an example is not out of reach for
Hi everyone,
I experience some problems with adressing of data.frames when I retrieve
some information for geographical position (ypos, xpos) ot of a MySQL
Database and want to perform some simple statistics. The problem is
adressing the dataframes with a construct like
rawdata[c(type)] vs.
Hello,
I would like to create a matrix with one of the columns named
$\delta$. I have also created columns $\beta_1$ , $\beta_2$, etc.
However, it seems like \d is an escape sequence which gets
automatically removed. (Using these names such that they work right in
xtable - latex)
Hello,
On 6/16/09, Stu @ AGS s...@agstechnet.com wrote:
Thanks for your response!
No, my basic equation does not use matrices at all. It takes scalar values
and returns a scalar.
Not quite. Taking the example above, if you run the following:
with(observs , {1*x1*x2^2*x3^3})
[1] 0.000e+00
Liviu,
Thanks for your comments.
With continued study and experimentation, I have discovered the
following:
a. I need to rewrite the function to return a 1x1 as you suggested;
b. it seems that constrOptim() is the most appropriate routine to use
on a
Kevin W wrote:
Frank,
Your template is very interesting. The pretty-ifing of the left arrow
and tilde in the input chunk has the unfortunate side-effect of making
the code non-paste-able into R.
Kevin,
I haven't needed to do that. It's best to make the .Rnw file available
to others, or
I have not figured out why you seem so attached to the c(type)
strategy but try:
numbers -summary(rawdata[ , c(type)] )
On Jun 16, 2009, at 12:06 PM, John Fitzgerald wrote:
Hi everyone,
I experience some problems with adressing of data.frames when I
retrieve
some information for
Hello list,
I wonder if anyone might be able to help me troubleshoot an attempt at
porting some simple Python code to R.
The function below is supposed to take a matrix containing item ratings from
various users and, given a vector containing at least 1 rating and 1 missing
value, employ a
On Fri, Jun 12, 2009 at 10:54 AM, Chosid, David
(FWE)david.cho...@state.ma.us wrote:
I'm wondering if I am dealing with a limitation in lattice. It's
probably due to my own limitations though.
I'm working with a lattice dotplot. The x-axis is set at free. In
one panel, there are only two
Hi guys
Does anyone know if it is possible to index a zoo series by a sequence? For
instance, with the following irregular zoo object, I can calculate the
range of its time-based index:
r - range(index(l.zoo))
r
[1] 2009-06-15 01:44:20.802 GMT 2009-06-15 16:54:24.124 GMT
If I just want to
Patrick,
Thanks for your suggestion!
The R-Inferno was especially useful!! The first chapter had me
chuckling aloud despite the fact that I work alone.
Well worth the price!
Thanks!
Stu
-Original Message-
From: Patrick Burns [mailto:pbu...@pburns.seanet.com]
Hi R-helpers,
I'm trying to use this code
pvh_dnv-pvh[sapply(pvh==dnv),]
to make a new dataframe containing the rows from pvh that contain the
value of dnv in ANY column.
But, it's not working. I get this error
Error in match.fun(FUN) : element 1 is empty;
the part of the args list of
See ?window.zoo
Also suggest you use dput to display sample data in posts to r-help.
On Tue, Jun 16, 2009 at 1:19 PM, rory.wins...@gmail.com wrote:
Hi guys
Does anyone know if it is possible to index a zoo series by a sequence? For
instance, with the following irregular zoo object, I can
No example. You are a regular now and examples ARE requested.
Would have thought something along the lines of:
pvh[ unlist( apply(pvh, 1, function(x) dnv %in%
as.character(x) ) ) , ]
Not sure if the unlist or as.character are really needed because ...
no example on which to test.
On
Try this:
pvh[apply(pvh == dnv, 2, any)]
On Tue, Jun 16, 2009 at 2:41 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
I'm trying to use this code
pvh_dnv-pvh[sapply(pvh==dnv),]
to make a new dataframe containing the rows from pvh that contain the
value of dnv in ANY column.
But,
Sorry, it should be:
pvh[apply(pvh == dnv, 1, any),]
On Tue, Jun 16, 2009 at 3:09 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
pvh[apply(pvh == dnv, 2, any)]
On Tue, Jun 16, 2009 at 2:41 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
I'm trying to use this code
Dear Mark,
If I understood correctly, the following should work:
index - apply(pvh, 1, function(x) any(x == dnv) )
pvh_dnv - pvh[index,]
pvh_dnv
HTH,
Jorge
On Tue, Jun 16, 2009 at 1:41 PM, Mark Na mtb...@gmail.com wrote:
Hi R-helpers,
I'm trying to use this code
Hi R-helpers,
I would like to subset my dataframe, keeping only those rows which
satisfy the following conditions:
1) the string dnv is found in at least one column;
2) the value in the column previous to the one dnv is found in is not 0
Here's what my data look like:
POND_ID 2009-05-07
Hi Livia and everyone,
Did you ever get a response on this question from last year (Jan 2008)?
I am also looking for more explanatory documentation on the
ui and ci parameters for the function constrOptim().
The examples provided in the R help and the full
Frank E Harrell Jr wrote:
Dear Group,
I have made significant improvements to our Sweave template, have made
the template self-contained (i.e., you can run it yourself and it will
find the datasets it needs), and have included the output pdf file.
This is at
Backslashes in character strings need to be doubled. Your \b is a
backspace. \\ is a backslash.
http://cran.r-project.org/doc/manuals/R-lang.html#Literal-constants
On 16/06/09 17:12, Stephen J. Barr wrote:
Hello,
I would like to create a matrix with one of the columns named
$\delta$. I
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Mark Na
Sent: Tuesday, June 16, 2009 11:27 AM
To: r-help@r-project.org
Subject: [R] How to subset my dataframe? (a bit tricky)
Hi R-helpers,
I would like to subset my
Try TargetScan, Pictar, miRbase.
These are all useful miRNA databases. Data can be downloaded as cvs or tab
delimited files and parsed in R after that. In fact this may be possible with
the resource you have looked at (although I haven't checked).
Cheers
Iain
--- On Tue, 16/6/09, David
Hello, and thanks in advance.
I have a data.frame from which I want to count observations that occur
on each day and determine the mean and std.error of said counts.
For instance:
x-split(my.df, my.df$julian.days)
Although I'm still in my R learning infancy I am under the impression
that x
Dear all,
Is there a instruction that can help me obtain the coefficient of
determination R^2 after doing linear/nonlinear regression using lm/nls?
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
This easy function you are looking for is tapply. Take a look at the
following example:
day=rep(1:30,each=30)
##There are thirty days
##with thirty obs each
y=rnorm(length(day),mean=2*day,sd=day)
##a dep. variable
##with mean=2*day index no.
##and sd=day
tapply(y,day,length)
##shows no. of obs
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