Dear all:
well, what Duncan has suggested would work in principle. However, the
dependencies of QRMlib as contained in the archive have been deprecated and the
package maintainer (cc'ed to this email directly) is pretty close to a
re-release of his package on CRAN, whereby primarily the
On Feb 27, 2012 at 9:30pm Joyous Fisher wrote:
Q: is there a way to do princomp or another method where every row has at
least one missing column?
You have several options. Try function nipals in packages ade4 and plspm.
Also look at package pcaMethods (on Bioconductor), where you will find a
Hey there,
I have the following problem:
I need to carry out a nonparametric conditional density estimation -- pretty
much what can be achieved using the np package. However, I need to be able
to use the output of the np package in such a way that i get a continuous
function as output, which
Please assist me with all the tests including risk factor analysis i can
use to analyse the enclosed database established from a questionnaire survey
to test for the prevalence of tuberculosis in humans .
Thank you
Nonty
--
View this message in context:
hi,
I know how to use the for loop function like:
for(i in 1:ncol(mat)){
mat[i]-b[i,2]
}
but, in this case
r1-b[1,1]
r2-b[2,1]
r3-b[3,1]
r4-b[4,1]
*
*
*
r3002-b[3002,1]
r3003-b[3003,1]
- must make vectors
how should I make a efficient code for that?
Is there anything in R
Hi All
I am trying to use the unlist() in R to a list variable. The following
statements are within a function.
{
denominator - sqrt(s1 / res.em1$n + s2 / res.em2$n)
returnValue - l2 / (denominator + 11)
attr(returnValue,numerator) - l2
attr(returnValue,denominator) - denominator
On Mon, Feb 27, 2012 at 11:04:13PM -0800, nontokozo mhlanga wrote:
Please assist me with all the tests including risk factor analysis i can
use to analyse the enclosed database established from a questionnaire survey
to test for the prevalence of tuberculosis in humans .
That's quite a
Hi,
There is not anything really like macro functions in R (and honestly,
that is probably a good thing). Most the times I have seen people
generating thousands of variables (vectors in your case) it is due to
a lack of understanding how lists can be utilized to simplify code.
If you give us
Hi Aparna,
Can you please post a reproducible example? It is difficult to
provide much concrete help without having testStatistics. One thing
you might try is looking at:
str(testStatistics[numerator,])
is it actually a list? If it is not (most likely given the error) and
it is supposed to
Hi Aparna,
If you have not defined testStatistics in your code, but it is a
function, then either you defined it at some earlier point and it was
never removed, or some other package you loaded defines it.
You could try using your code in a clean R session (make sure that an
old workspace is not
Hi
Dear who may concern,
I am sending you this email to ask if I could have someone help me to
write this script I am interested in.
I am requesting the script for performing variance-stabilizing
normalizations on the R language. Could you please help me on that?
If I am asking a wrong
Apologies for the phrasing of the question.
I've sorted the problem (thanks Bert Gunter) by using the curly brackets {}
as below (using a simplified version of my real model). I hope this helps
someone else!
Jeff
---
data
Alpha ip X
1 0.7106967
Dear James,
Basically you just need the values (y) and the positions (in your case it
would be the index of the times series). The chromosome argument does not
apply to your case so it can be a vector of ones.
If the positions are at the same distance between (equally spaced) then the
model will
On Mon, Feb 27, 2012 at 10:17 PM, knavero knav...@gmail.com wrote:
Here's my code:
http://pastebin.com/0yRxEVtm
The important parts are uncommented and should be easy to find using the
link above. For the following line of code, I plan on looking for a way to
offset it up 7 rows so that the
TY very much for your setdiffDF(). It does the job perfectly.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Petr Savicky
Sent: lundi 27 février 2012 20:41
To: r-help@r-project.org
Subject: Re: [R]
Hi all!
My question is how to colour pixels by z value in persp plot in raster package.
Here is an example:
x - seq(-1.95, 1.95, length = 30)
y - seq(-1.95, 1.95, length = 35)
z - outer(x, y, function(a,b) a*b^2)
r1 - raster(nrows=35, ncols=30, xmn=0, xmx=30, ymn = 0, ymx = 35)
r1[] - c(z)
Hi,
Ok, I understand what you mean.
I wanted to get sorted data group by cluster in output ...
But I have to do it myself using kres$cluster
thanks for your help
--
View this message in context:
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Sent from
Thanks for the reply guys. Well, I've tried the following command after
installing the package dependancies including timeSeries:
install.packages(file_name, type = source, repos = NULL)
Ans still no luck...I get the following error message:
Installing package(s) into ‘C:/Program
Hi Josh
Kindly find below the code as it is executed:
test.functional.t - function(res.em1,res.em2,mint,maxt,se.m=0,points=300)
{
at - seq(mint,maxt,length.out=points)
by - at[2] - at[1]
mu1 - spline(x=res.em1$tau,y=res.em1$eta,xout=at,method=natural)$y
l2 - functional.norm(mu1,mu2,by=by)
Hi there!
I´m a noob when it comes to R and I´m using it to run statisc analysis.
With the code for ARIMA below I´m getting this error: Error in
solve.default(res$hessian * n.used) :Lapack routine dgesv: system is
exactly singular
The code is:
s.ts - ts(x[,7], start = 2004, fre=12)
Hi, I'm wondering if you can help me, this is a really simple query but I
keep getting confused. I have run a GLM to see how boldness varies over
time following a particular treatment. The results are as follows...
Call: glm(formula = boldtwentyfour ~ treatment + boldcontrol)
Deviance
h-array()
h
[1] NA
append(h,9)
[1] NA 9
but what it append.
h
[1] NA
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Sent from the R help mailing list archive at Nabble.com.
But it won't be hard at allyou can likely get what you need using
the tapply() function (or ave)
Michael
On Tue, Feb 28, 2012 at 4:33 AM, ikuzar raz...@hotmail.fr wrote:
Hi,
Ok, I understand what you mean.
I wanted to get sorted data group by cluster in output ...
But I have to do it
i open a table
table-read.table(file.choose(),skip=1)
at sometime i forget which file i open,so i want file name/file path
--
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Sent from the R help mailing list archive at
Actually, what I really want to do is that,
annual productivity data(2011)
firts date 2011-01-01 2011-01-02
2011-01-03 2011-01-04
You need to reassign the value of append back to h -- in more
technical terms, this is a pass-by-value rather than pass-by-reference
behavior:
h - append(h, 9)
This is not likely to be efficient in production code, however.
Michael
On Tue, Feb 28, 2012 at 7:15 AM, sagarnikam123
Hello All,
My algorithm as follows;
y - c(1,1,1,0,0,1,0,1,0,0)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t - matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out - vector(list, )
for(i in 1:10) {
t.s - t[sample(n,n,replace=T),]
y.s - t.s[,1]
x.s - t.s[,2]
z.s - y.s+x.s
out[[i]]
Use this line instead:
table - read.table(fileIchoose - file.choose(), skip = 1)
This will have the side effect of creating fileIchoose as a variable
containing a file path.
Michael
On Tue, Feb 28, 2012 at 8:34 AM, sagarnikam123 sagarnikam...@gmail.com wrote:
i open a table
As stated, you need to install the *deprecated* dependencies of QRMlib as shown
in its DESCRIPTION as well as the reverse dependent *deprecated* packages.
These can still be fetched from R-Forge (Rmetrics project). The package
'timeSeries' will become a dependency of the to be re-released
'append' returns a value that you have assign back to the object you
want, in this case 'h'
h - append(h, 9)
On Tue, Feb 28, 2012 at 7:15 AM, sagarnikam123 sagarnikam...@gmail.com wrote:
h-array()
h
[1] NA
append(h,9)
[1] NA 9
but what it append.
h
[1] NA
--
View this
Hi
Difficult to without knowing what objects you are operating your
functions.
I get
test.functional.t(1:10,1:10,3,4)
Error in res.em1$eta : $ operator is invalid for atomic vectors
moderated.functional.t(1:10)
Error in testStatistics[numerator, ] : incorrect number of dimensions
And
Hi
My algorithm as follows;
y - c(1,1,1,0,0,1,0,1,0,0)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t - matrix(cbind(y,x), ncol=2)
Do not use t, it is a function for transposing matrix and after you
redefine it you can get nasty surprise in future.
tt - cbind(y,x)
is enough
z = x+y
Hi
Your mail is quite messy. what is ***? How your data are structured?
str(your.data)
Please submit part of your data by
dput(your.data)
and show what the output shall be.
Regards
Petr
Actually, what I really want to do is that,
annual productivity data(2011)
firts date
On Tue, Feb 28, 2012 at 05:59:24AM -0800, helin_susam wrote:
Hello All,
My algorithm as follows;
y - c(1,1,1,0,0,1,0,1,0,0)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t - matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out - vector(list, )
for(i in 1:10) {
t.s -
I think you'll need to install fEcofin from CRAN, and then the following
.tar.gz files from the CRAN archives:
fUtilities
fCalenday
fSeries
And then try installing the QRMlib .tar.gz.
If you're still having problems, try:
C:/Program Files/R/R-2.14.0/bin/x64/R CMD check QRMlib_1.4.5.1.tar.gz
(or
Hi
I''m looking for an Excel Vlookup type function in R.
Example:
list - c(1,2,3,4,5,6,7)
base - c(2.2,3,5.2)
What I want is, for each number in base, the highest value in list,
which is equal to or less than the number in base
So the results would be:
base list
2.2 -- 2
3
For a school course I and a partner developed a GUI in R designed to
enable exploration of data via visualization of hierarchical
clustering and correlation of cluster partitions with external
metadata. The key features were the ability to load in a distance
matrix (most GUI-based clustering
On Tue, Feb 28, 2012 at 09:02:04PM +0530, Priyan Fernando wrote:
Hi
I''m looking for an Excel Vlookup type function in R.
Example:
list - c(1,2,3,4,5,6,7)
base - c(2.2,3,5.2)
What I want is, for each number in base, the highest value in list,
which is equal to or less than the number
On 28-02-2012, at 16:32, Priyan Fernando wrote:
Hi
I''m looking for an Excel Vlookup type function in R.
Example:
list - c(1,2,3,4,5,6,7)
base - c(2.2,3,5.2)
What I want is, for each number in base, the highest value in list,
which is equal to or less than the number in base
So
findInterval returns the index (into list) of what
you want. Use [ to get the numbers at the bottom of
the found intervals:
list - c(1,2,3,4,5,6,7)
base - c(2.2,3,5.2)
findInterval(base, list)
[1] 2 3 5
findInterval(base+100, list+100)
[1] 2 3 5
(list+100)[.Last.value]
[1]
On Tue, Feb 28, 2012 at 09:02:04PM +0530, Priyan Fernando wrote:
Hi
I''m looking for an Excel Vlookup type function in R.
Example:
list - c(1,2,3,4,5,6,7)
base - c(2.2,3,5.2)
What I want is, for each number in base, the highest value in list,
which is equal to or less than the number
According to the QRMlib package pdf,
http://www.mirrorservice.org/sites/lib.stat.cmu.edu/R/CRAN/doc/packages/QRMlib.pdf
the package depennds on methods, fCalendar, fEcofin, mvtnorm,
chron,its,Hmisc. I have installed all of these (some were retrieved from
archive and some were retrieved from
Hi I am using the ggplot2 package for the volcano plot and I am using the
following code for the same:
g = ggplot(data=data, aes(x=data[11], y=-log10(data[12]), colour=threshold))
+
+ geom_point(alpha=0.4, size=1.75) +
+ opts(legend.position = none) +
+ xlim(c(-10, 10)) + ylim(c(0, 15)) +
+
Perhaps, the following does it as well.
(d - data.frame(x1=letters[2*1:4 - 1], x2=letters[2*1:4]))
c(t(d))
Alemtsehai
Hello,
I am looking for a way to transform an array into a list (or a string).
My array has two columns 1 and 2, and I would like to create a list of
the
values.
Folks:
Graphics afficianados might enjoy this link (although conditional
probability is also mentioned!):
http://news.yahoo.com/blogs/signal/republican-nomination-extends-march-candidates-face-weakening-odds-152347629.html
Pay attention to the graph. Never thought I'd see the phrase lowess
Hi everyone,
Im very new with R and I would like to change the size text (names) in the ylab
in a dotplot. I have checked many webpages and R sites, but I have not found
any help.
This is the same structure of the plot in R graph Gallery:
In command prompt, I direct to the following directory:
C:\Program Files\R\R-2.14.1\bin
Now I enter:
R CMD check my_local_folder\QRMlib_1.4.5.1.tar.gz
The following appears:
* using log directory 'C:/Program Files/R/R-2.14.1/bin/QRMlib.Rcheck'
* using R version 2.14.1 (2011-12-22)
* using
Dear Petr Pikal and Petr Savicky thank you for your replies..
If the y vector contains different elements my algorithm gives this result;
y - c(1,2,3,4,5,6,7,8,9,10)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t - matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out -
Next, you should check C:/Program
Files/R/R-2.14.1/bin/QRMlib.Rcheck/00install.out for details on the error.
DT54321 wrote
In command prompt, I direct to the following directory:
C:\Program Files\R\R-2.14.1\bin
Now I enter:
R CMD check my_local_folder\QRMlib_1.4.5.1.tar.gz
The
I really appreciate to your attention.
I want to change the data form A into B
http://r.789695.n4.nabble.com/file/n4428807/1.png
http://r.789695.n4.nabble.com/file/n4428807/2.png
In fact, there are 1095 column in data A.
Specifically, variable publish_day is when some book in the same
I installed fEcofin, fUtilities, fCalenday, and fSeries from the CRAN
archives and was able to install and load QRMlib.
Which errors does R CMD check QRMlib_1.4.5.1.tar.gz show?
DT54321 wrote
According to the QRMlib package pdf,
Hello,
Is it just a columns names issue?
Try (DF is your data, 'A' or other data.frame)
# colNames - colnames(DF)
colNames - c(name, publish day, 2011-01-01, 2011-01-02,
2011-01-03, 2011-01-04, 2011-01-05)
x - as.Date(colNames[-(1:2)])
x
[1] 2011-01-01 2011-01-02 2011-01-03 2011-01-04
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote:
Dear Petr Pikal and Petr Savicky thank you for your replies..
If the y vector contains different elements my algorithm gives this result;
y - c(1,2,3,4,5,6,7,8,9,10)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t -
There seems to be a clash of character sets so I can't see just what
you're trying to do, but I think you'll find out how to adjust the
text size with a scales list. Check out how it's used in the help for
xyplot.
HTH
On Tue, 28-Feb-2012 at 06:05PM +, Jose Bustos Melo wrote:
| Hi
Hello,
I am trying to paste together a formula to use in the mob function of
party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN.
I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the
conditional part of the equation using update().
Here's the test
On Tue, Feb 28, 2012 at 08:50:45AM -0800, helin_susam wrote:
Dear Petr Pikal and Petr Savicky thank you for your replies..
If the y vector contains different elements my algorithm gives this result;
y - c(1,2,3,4,5,6,7,8,9,10)
x - c(1,0,0,1,1,0,0,1,1,0)
n - length(x)
t -
On Feb 28, 2012, at 4:16 AM, Hans Ekbrand wrote:
On Mon, Feb 27, 2012 at 11:04:13PM -0800, nontokozo mhlanga wrote:
Please assist me with all the tests including risk factor analysis
i can
use to analyse the enclosed database established from a
questionnaire survey
to test for the
Hey,
I just googled my error and many things came up. I followed the leads and
read the ?read.delim page; I tried changing header = TRUE, and row.names =
TRUE-- but I've still been having trouble fixing it, so I would greatly
appreciate any help you can provide. Here is my code:
rm(list=ls())
Your code seems to be screwed up. There are no (small) size
limitations on formulas (afaik).
The following worked fine for me:
x - paste(x,1:100,sep=,collapse = + )
part1 - paste(y,x,sep= ~ )
part2 - paste(f,1:50, sep=,collapse = + )
fml - formula(paste(part1,part2,sep= | ))
Cheers,
Bert
On
On Feb 28, 2012, at 4:34 AM, Omphalodes Verna wrote:
Hi all!
My question is how to colour pixels by z value in persp plot in
raster package. Here is an example:
x - seq(-1.95, 1.95, length = 30)
y - seq(-1.95, 1.95, length = 35)
z - outer(x, y, function(a,b) a*b^2)
r1 - raster(nrows=35,
My basic worry is that the GC does not work properly,
i.e., the unreachable data is never collected.
* Bert Gunter thagre.ore...@trar.pbz [2012-02-27 14:35:14 -0800]:
This appears to be the sort of query that (with apologies to other R
gurus) only Brian Ripley or Luke Tierney could figure
On Tue, Feb 28, 2012 at 11:57 AM, Sam Steingold s...@gnu.org wrote:
My basic worry is that the GC does not work properly,
i.e., the unreachable data is never collected.
Highly unlikely. Such basic inner R code has been well tested over 20
years. I believe that you merely don't understand the
On Feb 28, 2012, at 12:10 PM, mrzung wrote:
I really appreciate to your attention.
I want to change the data form A into B
http://r.789695.n4.nabble.com/file/n4428807/1.png
http://r.789695.n4.nabble.com/file/n4428807/2.png
Please use text copied versions of dput(a) and dput(b) rather
Look into environments that may be stored
with your data. object.size(obj) does not
report on the size of the environment(s)
associated with obj. E.g.,
f - function(n) {
+d - data.frame(y=rnorm(n), x1=rnorm(n), x2=rnorm(n))
+terms(data=d, y~.)
+ }
z - f(1e6)
On 28/02/2012 2:46 PM, Bert Gunter wrote:
Your code seems to be screwed up. There are no (small) size
limitations on formulas (afaik).
The following worked fine for me:
x- paste(x,1:100,sep=,collapse = + )
part1- paste(y,x,sep= ~ )
part2- paste(f,1:50, sep=,collapse = + )
fml-
Unfortunately, some data I need to work with was delivered in a rather messy
Excel file. I want to import into R and clean up some things so that I can do
my analysis. Pulling in a CSV from Excel is the easy part.
My current challenge is dealing with some text mixed in the values.
i.e.
First of all when reading in the CSV file, use 'as.is = TRUE' to
prevent the changing to factors.
Now that things are character in that column, you can use some pattern
expressions (gsub, regex, ...) to search for and change your data.
E.g.,
sub(.*, 0, yourCol)
should do it for you.
On Tue,
-Original Message-
From: Noah Silverman
Sent: Tuesday, February 28, 2012 3:27 PM
To: r-help
Subject: [R] Cleaning up messy Excel data
Unfortunately, some data I need to work with was delivered in a rather messy
Excel file. I want to import into R and clean up some things so that I can
That's exactly what I need.
Thank You!!
--
Noah Silverman
UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095
On Feb 28, 2012, at 1:42 PM, jim holtman wrote:
First of all when reading in the CSV file, use 'as.is = TRUE' to
prevent the changing to factors.
Now
* William Dunlap jqha...@gvopb.pbz [2012-02-28 20:19:06 +]:
Look into environments that may be stored with your data.
thanks, but I see nothing like that:
for (n in ls(all.names = TRUE)) {
o - get(n)
print(object.size(o), units=Kb)
e - environment(o)
if (!identical(e,NULL)
You need to walk through the objects, checking for
environments on each component or attribute of an
object. You also have to look at the parent.env
of each environment found. E.g.,
f - function(n) {
+ d - data.frame(y = rnorm(n), x = rnorm(n))
+ lm(y ~ poly(x, 4), data=d)
+ }
z
Just replace that value with zero. If you provide some reproducible
code I could probably give you a solution.
?dput
good luck,
Stephen
On 02/28/2012 03:27 PM, Noah Silverman wrote:
Unfortunately, some data I need to work with was delivered in a rather messy
Excel file. I want to import
Hello,
I can get the median for each factor, but I'd like another column to go
with each factor. The nm column is a long name for the lvls column. So
unique work except for the order can get messed up.
Example:
x =
aggregate(val~lvls+nm,data=x,FUN='median')
On Tue, Feb 28, 2012 at 4:43 PM, Ben quant ccqu...@gmail.com wrote:
Hello,
I can get the median for each factor, but I'd like another column to go
with each factor. The nm column is a long name for the lvls column. So
unique work except for the
Excellent! I wonder why I haven't seen aggregate before.
Thanks!
ben
On Tue, Feb 28, 2012 at 4:51 PM, ilai ke...@math.montana.edu wrote:
aggregate(val~lvls+nm,data=x,FUN='median')
On Tue, Feb 28, 2012 at 4:43 PM, Ben quant ccqu...@gmail.com wrote:
Hello,
I can get the median for
What is initialCuratedDF? I'm not seeing it in any standard location
so there's no way to debug it...
Also see inline.
Michael
On Tue, Feb 28, 2012 at 2:41 PM, Ben Ganzfried ben.ganzfr...@gmail.com wrote:
Hey,
I just googled my error and many things came up. I followed the leads and
read
Thanks Gabor. Answering all my noob questions haha. Interesting strategy to
adjust the input raw data as opposed to Excel where one offsets the output
data instead. I'll remember that.
--
View this message in context:
Error: could not find function hier.part
things I have tried:
1. reinstall R (lastest version, on windows XP)
2. install package gtools
3. include:
library(gtools)
require(gtools)
4. how I call this function:
hier.part(c$Y, xdata, fam = gaussian, gof = Rsqu)
5.when I try to check what's in the
Yes, I did. But my Windows did not know what program to open it with...
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__
Hello,
I have posted this question earlier and got a lot of helps. Appreciate all
those helps. After intensive learning and researching, this is to summurize
what I have now.
Target: Install the package Rcplex
OS: windows 7 64bit
The Story:
Installed the Rtool, and setted the path to all
Dear Petr Savicky,
Actually, this is based on jackknife after bootstrap algorithm. In summary,
I have a data set, and I want to compute some values by using this
algorithm.
Firstly, using bootstrap, I create some bootstrap re-samples. This step O.K.
Then, for each data point within these
Hi all,
I am trying to analyze some graphs in R. Given a weighted directed network
with 7 nodes. I want to find a path from node 1 to node 7, so the sum of all
the edges on the path is the smallest comparing to all the other paths. I am
wondering if I can do this in igraph. After I identify the
I have a dataset that does not include native scores, but only serial
quantile rankings for a set of units.
Clearly these observations are dependent (in that you can't alter one
observation without also altering others).
Are there methods for dealing with quantile dependent variables. My atempt
Dear List,
I would be pleased if someone can help me with the following issue:
I'm about to plot two time series in one plot via ts.plot which looks like:
ts.plot(series1, series2, main=main, xlab=xlab, ylab=ylab, col=c(green,
red, blue), lwd=2)
The problem is, that R
Hi all,
I´m new using Access. I see that many things that you can do on Access you
can do on CRAN R but not on contrary.
My question is: Is there any manual with examples comparing how to do data
base analysis on access and making the same on CRAN R?
Imagine I want to compare two columns Name
It would seem that hier.part is not in the gtools packagedo you
have some reference that suggests otherwise? I don't know the
function, but I might suggest you check the hier.part package (also on
CRAN).
Michael
On Tue, Feb 28, 2012 at 4:27 PM, haiyan tucsonaug...@gmail.com wrote:
Error:
Please quote context.
You can open it with any plain text editor. (i.e., notepad)
Michael
On Tue, Feb 28, 2012 at 2:42 PM, DT54321 deepan.tailo...@gmail.com wrote:
Yes, I did. But my Windows did not know what program to open it with...
--
View this message in context:
The previously posted code contains bugs. The code below should work:
x: a vector of data
y: Group indicator (if id.col=TRUE); data of the second group (if
id.col=FALSE). If y is the group indicator it MUST take 0 or 1 to indicate
the groups, and x must contain the data for both groups.
Hi Josh, Petr
I checked that testStatistics is a function but it is not defined in the
code anywhere else. But if I am able to remove it and give the input by my
own to the function which calls testStatistics, it might work.
The hurdle here for me is that the input is a set* of values and each
Dear All,
I've been searching relevant topics about replacing values, none seemed to
be applicable to me...
I have a file with many many varieties, and want to replace some of them
into different names.
I tried various of ways, still don't know how to do that most efficiently..
Here is part of
Hello,
I am trying to use a for loop to name objects in each iteraction. As in the
following example (which doesn't work quite well)
my_list-c(A,B,C,D,E,F)
for(i in c(1:length(my_list))){
url- http://finance.yahoo.com;
doc = htmlTreeParse(url, useInternalNodes = T)
tab_nodes =
Dear Oscar,
Â
I am extremely grateful to your help and detailed explanation of the use of
RJaCGH package.
But, when runing the sample codes you listed, another issue I am a little
confused is as following:
After runing summary(res), I have got the estimation of the random matrix Beta:
David thanks!
Maybe is better ''orginal'' example from raster package:
r - raster(system.file(external/test.grd, package=raster))
persp(r)
class(r)
It is not a problem to colour plot by z value in perspective plot (persp) in
package graphic - there is an example in help files:
par(bg =
Hannah,
If Gen is a factor you can simply build the new factor on top of it:
dataframe$Gen- factor( c('Wynda' , 'A_2' , 'B_1' , 'Wynda' , 'Wynda'
, 'OP1_5')[Gen] )
Just make sure the replacement labels are in the same order as levels(Gen).
Cheers
On Tue, Feb 28, 2012 at 8:39 PM, hannahmaohuang
On Tue, Feb 28, 2012 at 12:57 PM, Jochem Schuster
jochem.schus...@web.de wrote:
ts.plot(series1, series2, main=main, xlab=xlab, ylab=ylab, col=c(green,
red, blue), lwd=2)
What I've tried before is deleting the X axes via gpars=list(xaxt=n) in
the ts.plot-code. But after that I was
Hi.
You can do something like this:
df - read.table(textConnection(
GenRep
A_1 1
A_1 2
A_2 1
A_2 2
B_1 1
B_1 2
B_3 1
B_3 2
OP1_1 1
OP1_1 2
OP1_5 1
OP1_5 2),header=TRUE,stringsAsFactors=FALSE)
str(df)
#adding a new column
cbind(df,ncol =
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Hi,
today i wrote a function in R of the type:
index.refraction - function(Temp,Press, RH, CO2)
When i try to plug a negative number in Temp, i got this type of error:
n - index.refraction(Temp= -40,100,80,CO2)
Messages d'avis :
1: In Ops.ordered(left, right) : '-' is not meaningful for
Hello to the members of the list,
I am using the 'spectrum' function from 'stats' package, to calculate
the squared coherency between two time series. The function
'plot.spec.coherency' provides information for the coverage probability
for the normal distribution.
It seems that the
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