Hi,
Actually, I have two steps, the first would be to import a csv file
(read.csv). There I would need to have the same number of decimals as
there are on the original file.
Then I would run some analyses and probably export the results, e.g.
ANOVA table (write.csv). Here too I would like to
Hi All,
I am completely new to R. I have the below data and want to create a
conditional variable say Prof.H as such that it equals 1 if Prof is 50 and
0 otherwise and create a scatter plot of Value and Dim conditional on the
new variable.
Area Value Dim Prof
1 1 145.1 9.5
Is this what you want:
x
[1] 1225221868 1225221906 1225221906 1225230997 1225231000 1225231003
1225231152 1225231348 1225231351
[10] 1225231400
as.POSIXct(x,origin='1970-1-1',tz='GMT')
[1] 2008-10-28 19:24:28 GMT 2008-10-28 19:25:06 GMT 2008-10-28
19:25:06 GMT
[4] 2008-10-28 21:56:37 GMT
To the R color experts:
I need to detect if a chosen background color (as hex e.g. #910322) is
light or dark.
If it is dark I need to ovelay it with light text and vice versa.
Thus I would like to implement the following pseudo code:
if (brightness(color) somevalue) textcolor= dark else
Hi all,
I thought I'd just point out, to those not having yet seen this, that
today there was a classification challenge posted for astronomy.
The web-site is http://www.hep.anl.gov/SNchallenge/
[I have nothing to do with this project so don't ask me any details!]
Basically the idea behind is
On Fri, 29 Jan 2010, Mark Heckmann wrote:
To the R color experts:
I need to detect if a chosen background color (as hex e.g. #910322) is
light or dark.
If it is dark I need to ovelay it with light text and vice versa.
You can use the colorspace package for that. hex2RGB() converts a hex
I had downloaded and installed a number of packages, successfully, when
I ran into some problem with maptools: It would eat up CPU and most of
all memory.
I rebooted, and tried again, only running the terminal after the reboot;
with the same result:
sp2WB text html latex example
sp2tmap text
Hello,
I read the help as well as the examples, but I can not figure out why
the following code does not produce the *given* row names, x and y:
x - 1:20
y - 21:40
rbind(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
y=cbind(N=length(y), M=mean(y), SD=sd(y))
)
Could you please help?
Thank
This is what I tried:
num.vec - c(12.34,56.78,90.12,34.56)
names(num.vec)-c(first,second,third,fourth)
num.vec
first second third fourth
12.34 56.78 90.12 34.56
seq-(1:4)
num.mat-rbind(num.vec,seq)
num.mat
first second third fourth
Hi,
A small (but annoying) problem with RMySQL library. When a connection is
closed, it echoes 'TRUE' to the console. Like this:
R mysqlCloseConnection(con)
[1] TRUE
The real problem comes when I use RMySQL with Sweave, since the TRUE echo gets
into my final Sweave document. Even when I
Try this:
invisible(mysqlCloseConnection(con))
On Fri, Jan 29, 2010 at 9:10 AM, Stefan Petersson
stefan.peters...@inizio.se wrote:
Hi,
A small (but annoying) problem with RMySQL library. When a connection is
closed, it echoes 'TRUE' to the console. Like this:
R mysqlCloseConnection(con)
This gives what you want:
rbind.data.frame(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
y=cbind(N=length(y), M=mean(y), SD=sd(y))
)
On Fri, Jan 29, 2010 at 8:49 AM, soeren.vo...@eawag.ch wrote:
Hello,
I read the help as well as the examples, but I can not figure out why the
following
Hi!
29.01.2010 12:49, soeren.vo...@eawag.ch wrote:
Hello,
I read the help as well as the examples, but I can not figure out why
the following code does not produce the *given* row names, x and y:
x - 1:20
y - 21:40
rbind(
x=cbind(N=length(x), M=mean(x), SD=sd(x)),
Tru this:
xyplot(Value ~ Dim | ifelse(Prof 50, 1, 0), data = x)
On Fri, Jan 29, 2010 at 5:25 AM, Brima adamsteve2...@yahoo.com wrote:
Hi All,
I am completely new to R. I have the below data and want to create a
conditional variable say Prof.H as such that it equals 1 if Prof is 50 and
0
On Fri, 2010-01-29 at 18:56 +0800, Uwe Dippel wrote:
This is what I tried:
num.vec - c(12.34,56.78,90.12,34.56)
names(num.vec)-c(first,second,third,fourth)
num.vec
first second third fourth
12.34 56.78 90.12 34.56
seq-(1:4)
Hi,
I am using the FracSim library to simulate a time series. However, the
simulate function ignores my attempt to set the RNG seed I need for
reproducible research. The published docs and google have not yielded an
answer, so any help greatly received.
Thanks,
Selwyn
## Example code snippet
Stefan Petersson wrote:
Hi,
A small (but annoying) problem with RMySQL library. When a connection is
closed, it echoes 'TRUE' to the console. Like this:
R mysqlCloseConnection(con)
[1] TRUE
The real problem comes when I use RMySQL with Sweave, since the TRUE echo gets into my
final Sweave
S: Works! Thanx...
2010-01-29 Henrique Dallazuanna wrote:
Try this:
invisible(mysqlCloseConnection(con))
On Fri, Jan 29, 2010 at 9:10 AM, Stefan Petersson
stefan.peters...@inizio.se wrote:
Hi,
A small (but annoying) problem with RMySQL library. When a connection is
closed, it echoes
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Uwe Dippel
Sent: Freitag, 29. Januar 2010 11:57
To: r-help@r-project.org
Subject: [R] Explanation w.r.t. rbind, please!
This is what I tried:
num.vec -
Hi,
How can we vectorize multiple for loops?
E.g. how do you vectorize this:
for (i in 1:10){
for(j in 1:25){
for(k in 1:19){
x[i,j,k]=i*k-j
}
}
}
THanks
KC
__
R-help@r-project.org mailing list
ComRades,
How do you put a superscript on a scalable delimiter?
I want to put 'b' as the power of the expression in the following plot:
t - 1:25
K - 0.2
y - ((1-exp(-K*t))/(1-exp(-K*t)*exp(K)))^3
plot(t,y,l,main=K=0.2, b=3)
text(15,5,expression(bgroup((,frac(1-e^-Kt,1-e^-Kt*e^K),
Plotmath
Readers,
I am trying to use the zoo package with an array of data:
file1:
hh:mm:ss1
hh:mm:ss2
hh:mm:ss3
hh:mm:ss4
file2:
hh:mm:ss11 55
hh:mm:ss22 66
hh:mm:ss33 77
hh:mm:ss44 88
I wanted to merge these data set
Try this:
aperm(outer(i %o% k, j, FUN = -), c(1, 3, 2))
On Fri, Jan 29, 2010 at 9:58 AM, Kohleth Chia kohl...@gmail.com wrote:
Hi,
How can we vectorize multiple for loops?
E.g. how do you vectorize this:
for (i in 1:10){
for(j in 1:25){
for(k in 1:19){
x[i,j,k]=i*k-j
}
}
}
Hi All,
Does the step function work in this model? I tried to run the
following model but no result obtained. The computer is hanging and I
killed the job several times. Below is the code.
library(survival)
m.fit=clogit(y~x1+x2+x3+x4, data=ftest)
summary(m.fit)
final- step(m.fit)
Thanks in
On 29.01.2010 13:03, Rubén Roa wrote:
ComRades,
How do you put a superscript on a scalable delimiter?
I want to put 'b' as the power of the expression in the following plot:
t- 1:25
K- 0.2
y- ((1-exp(-K*t))/(1-exp(-K*t)*exp(K)))^3
plot(t,y,l,main=K=0.2, b=3)
See FAQ #1 of the zoo faq:
vignette(zoo-faq)
On Fri, Jan 29, 2010 at 7:12 AM, e-letter inp...@gmail.com wrote:
Readers,
I am trying to use the zoo package with an array of data:
file1:
hh:mm:ss 1
hh:mm:ss 2
hh:mm:ss 3
hh:mm:ss 4
file2:
hh:mm:ss 11
On Thu, Jan 28, 2010 at 2:32 PM, Philipp Rappold
philipp.rapp...@gmail.com wrote:
Dear Prof. Broström,
Dear R-mailinglist,
first of all thanks a lot for your great effort to incorporate time-varying
covariates into aftreg. It works like a charm so far and I'll update you
with detailled
Dear all,
I need to solve the following Lyapunov Matrix equation:
C=ACA' + B,
with A and B given square symmetric matrices. Does anyone knows of a
package that can solve the lyapunov matrix equation in R? Or even a
C/Fortran implementation? I did not find one on netlib.
Thank you.
Dear all,
I am trying to put together in a package some functions that maybe of
interest to other. A particular function of the package uses zgges
from Lapack with a .Fortran call. Since R does not load all symbols
from liblapack, I load all the liblapack symbols (lazyload, it is)
with a
Assuming my documentation is correct, my version shows faq 1 to refer
to duplicate times but if file2 is:
01:01:0111 55
01:01:0422 66
01:01:0733 77
01:01:1044 88
I cannot see what is duplicate? If I create two new files:
file3:
01:01:01
prem_R mtechprem at gmail.com writes:
I heard form my friend there is a way to run R in system hard disk space not
in the RAM .By that we may not run out of memory and have problem attached
with the same.Someone could help me in this.Thanks.
Your question is posed extremely vaguely.
Try this:
y - c(v1, v2, v3, v4)
rowf - gl(2, 1, 16)
colf - gl(2, 8, 16, labels=1:0)
dat - data.frame(y)
aggregate(dat[1], list(Row=rowf, Col=colf), mean)
-Peter Ehlers
Jack Siegrist wrote:
I have a data set of many rows and many columns in which both the rows and
the columns have associated
Hi, R Users
I find a problem in extracting the R-squared and P-value from the lm results
described below (in Italic),
*Residual standard error: 2.25 on 17 degrees of freedom*
*Multiple R-squared: 0.001069, Adjusted R-squared: -0.05769 *
*F-statistic: 0.01819 on 1 and 17 DF, p-value: 0.8943 *
Perhaps some variation of
SetTextContrastColor - function(color)
{
ifelse( mean(col2rgb(color)) 127, black, white)
}
Mark Heckmann wrote:
To the R color experts:
I need to detect if a chosen background color (as hex e.g. #910322) is
light or dark.
If it is dark I need to ovelay
On 1/29/2010 9:04 AM, wenjun zheng wrote:
Hi, R Users
I find a problem in extracting the R-squared and P-value from the lm results
described below (in Italic),
*Residual standard error: 2.25 on 17 degrees of freedom*
*Multiple R-squared: 0.001069, Adjusted R-squared: -0.05769 *
Peter, I decided to directly remove the NA's but when I tested it yes I wrote
fromLast = TRUE so I might have checked erroneously sorry and thanks again I
am going to keep this in mind because I might need it at some point, I have
many cases where I need to handle missing values differently.
x - 1:10
y - 2 + 1.5 * rnorm(10, x, 2)
m - lm(y ~ x)
summary(m)$r.squared
[1] 0.6056889
anova(m)$'Pr(F)'
[1] 0.0080142NA
Components of the summary() and anova() methods of lm() can be extracted.
See
names(summary(m))
names(anova(m))
to see the components one can extract.
HTH,
vikrant wrote:
Yes I do have zeros in my data. But I m not able to understand y inclusion of
zeros results in error messages, because range for x in weibull distribution
is x=0. Can you please clarify this doubt?
Well, that's a matter of definition (and is a problem if
the shape parameter is
Thanks, I get it.
Wenjun, ZHENG
2010/1/29 Dennis Murphy djmu...@gmail.com
x - 1:10
y - 2 + 1.5 * rnorm(10, x, 2)
m - lm(y ~ x)
summary(m)$r.squared
[1] 0.6056889
anova(m)$'Pr(F)'
[1] 0.0080142NA
Components of the summary() and anova() methods of lm() can be extracted.
See
Hi
r-help-boun...@r-project.org napsal dne 29.01.2010 15:04:39:
Hi, R Users
I find a problem in extracting the R-squared and P-value from the lm
results
described below (in Italic),
*Residual standard error: 2.25 on 17 degrees of freedom*
*Multiple R-squared: 0.001069, Adjusted
Hello all,
I'm trying to create a 2x2 matrix, 32 times after unlist() so that I can
convert the list to matrix. I've looked through the R archive but
couldn't find the answer. There is what I've done.
f - system(ls *.txt, intern=TRUE)
x - lapply(f, read.table)
x
[[1]]
V1V2
1
If you use tcltk package package, you can do:
as.character(.Tcl(tk_chooseColor))
Best,
Philippe
Greg Snow wrote:
I don't know of any existing palettes that meet your conditions, but here are a
couple of options for interactive exploration of colorsets (this is quick and
dirty, there are
Meyners,Michael,LAUSANNE,AppliedMathematics wrote:
What you (probably) want here is
num.mat [seq,]
num.mat [num.vec]
[1] NA NA NA NA
num.mat[num.vec,]
and so on. You have to use tell R that you want the ROW (that's why the
comma is needed) defined by the NAME seq or num.vec
Hi all
I have a dataframe like I coppied below
ff
a b d
110 5 7
220 4 9
3 3 8 10
4 5 68
5 6 35
67373
74528 9
83 2 8
while I am trying to sort multy coloums at
How to vectorize this for loop and how can I assign result to vector
instead of using print function?
mylist - list(a = letters[1:3], b = LETTERS[1:3], c = c(1, 2, 3))
for (i in seq_along(mylist[[1]])) {
for (j in seq_along(mylist[[2]])) {
print(mylist[[1]][i])
anna wrote:
Hello everyone, I have a vector P and I want to replace each of its
missing values by its next element, for example:
P[i] = NA -- P[i] = P[i+1]
You can also try
P[which(is.na(P))]- P[which(is.na(P))+1]
or avoiding duplicate calculations
index.Pna-which(is.na(P))
Please provide a reproducible examples. You can use the following style:
Lines - 01:01:0111 55
+ 01:01:0422 66
+ 01:01:0733 77
+ 01:01:1044 88
library(zoo)
library(chron)
z - read.zoo(textConnection(Lines), FUN = times)
z
V2 V3
Hi:
The problem, I'm guessing, is that you need to assign each of the matrices
to an object.
There's undoubtedly a slick apply family solution for this (which I want to
see, BTW!),
but here's the brute force method using a loop:
nms - paste('x', 1:32, sep = )
for(i in seq_along(nms))
On Jan 29, 2010, at 9:34 AM, venkata kirankumar wrote:
Hi all
I have a dataframe like I coppied below
ff
a b d
110 5 7
220 4 9
3 3 8 10
4 5 68
5 6 35
67373
74528 9
83
Dear R-ers,
I am using getGEO to download expression data from the Gene Expression
Omnibus. With default settings, when a file is downloaded and parsed,
lots of dotted lines are printed in the terminal, like this:
.. .. .. .. ..
.. ..
On 29/01/2010 10:04 AM, Craig P. Pyrame wrote:
Dear R-ers,
I am using getGEO to download expression data from the Gene Expression
Omnibus. With default settings, when a file is downloaded and parsed,
lots of dotted lines are printed in the terminal, like this:
.. ..
On Jan 29, 2010, at 9:45 AM, Dennis Murphy wrote:
Hi:
The problem, I'm guessing, is that you need to assign each of the
matrices
to an object.
There's undoubtedly a slick apply family solution for this (which I
want to
see, BTW!),
I don't have a method that would assign names but you
Hi,
I am struggling to create a 2 by 2 multiple graphs in one page. I used
par(mfrow=c(2,2)) to divide the screen into 4. In each screen I draw a pie
chart (They are all same).
For example, my data is like this
Concentration value
A1 69
A2
Hi,
I am struggling to create a 2 by 2 multiple graphs in one page. I used
par(mfrow=c(2,2)) to divide the screen into 4. In each screen I draw a
pie chart (They are all same).
For example, my data is like this
Concentration value
A1 69
A2
Hi,
I
am struggling to create a 2 by 2 multiple graphs in one page. I used
par(mfrow=c(2,2)) to divide the screen into 4. In each screen I draw a pie
chart (They are all same).
For example, my data is like this
Concentration value
A1 69
A2
Duncan Murdoch wrote:
getGEO is a Bioconductor package. Feel free to insult the helpful
people there if you want to alienate them, but please don't crosspost
here.
Duncan Murdoch
Duncan,
Do you consider this sort of output from a function helpful in any way?
I don't, and I can't see
thanks Berend, that's a very smart to do it :)
-
Anna Lippel
--
View this message in context:
http://n4.nabble.com/Error-on-using-lag-function-tp1399935p1415584.html
Sent from the R help mailing list archive at Nabble.com.
__
Original Message
Subject:Re: [R] Suppress output from getGEO
Date: Fri, 29 Jan 2010 10:29:54 -0500
From: Duncan Murdoch murd...@stats.uwo.ca
To: Craig P. Pyrame crap...@gmail.com
References: 4b62f906.90...@gmail.com 4b62fb32.4040...@stats.uwo.ca
Below is a 'pretty-poor' version of a color picker I wrote using the
code on the color chart page I cited. It's slightly 'pretty' in terms
of layout, but 'poor' in terms of usability and flexibility. Still,
it is more useful to me than the nearly-null set of alternatives
I have found in R.
¿Did you do your homework before posting? ¿Did you read the posting guide?
I do not know of anything in R, but searching the web gives a lot of info,
there is a good article in wikipedia, and there are a lot of papers accessible.
Code, which probably can be used with R, can be found on the
Hello,
Can you tell me what R function to use to do a two-sample chi-squared
test? I want to see if two distributions are significantly different
from each other, and I don't specify the theoretical distribution of
either. For example, I have the following fake count data:
x -
Hi Michael,
have a look at colors.plot(T) from the epitools-package (and perhaps at
colorbrewer.display() as well). Maybe this suits you?
hth
Michael Friendly schrieb:
I'm looking for a scheme to generate a default color palette for
plotting points, lines and text (on a white or transparent
Hello everyone, I have the following matrix
[,1] [,2] [,3] [,4]
[1,] 0.002809706 0.0063856960 0.0063856960 0.011749681
[2,] 0.004893124 0.0023118418 -0.0005122951 -0.014646465
[3,] 0.003547897 0.0063355297 0.0030410542 0.011403953
[4,]
Hi everybody,
To run some statistical tests from the package WRS (from Rand R Wilcox),
I need to store my data in a list, which fac2list() from this package
does very well.
But I would like to do it in a loop for each numerical variable. It
would be easier!
For now, I have the loop with the
On Jan 29, 2010, at 11:11 AM, eric lee wrote:
Hello,
Can you tell me what R function to use to do a two-sample chi-squared
test? I want to see if two distributions are significantly different
from each other, and I don't specify the theoretical distribution of
either. For example, I have
There is a col2grey (and col2gray) function in the TeachingDemos package that
use a common algorithm to convert colors to grey based on perceived lightness,
that may work for you on deciding the color.
For placing text on colored backgrounds, look at the shadowtext function (also
in
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition
Hi,
Would this do as an alternative syntax?
g1 - quote(1/Tm)
mat - list(0, bquote(f1*s1*.(g1)))
vals - data.frame(f1=1, s1=.5, Tm=2)
sapply(mat, eval, vals)
HTH,
baptiste
On 29 January 2010 17:51, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out
Hello,
You could do something along the following lines:
sapply( 1:ncol( my.matrix ),
function( i )
my.foo( my.matrix[,i], my.parm[i]
)
Best regards,
Carlos J. Gil Bellosta
http://www.datanalytics.com
anna wrote:
Hello everyone, I have the following matrix
Hmm
I *think* this will work, but may break in a further sub routine.
It certainly works in this example, but my expression vector is used in
many scenarios and it will take a while to check them all.
For instance, I take the derivative of each element with respect to each
variable using
Hello,
You may problably need to create the lagged vars yourself and use them
as input for the NN.
Best regards,
Carlos J. Gil Bellosta
http://www.datanalytics.com
http://datanalytics.wordpress.com
CJ Rubio wrote:
For example I have a time series
Q(t) ~ Q(t-1) + Q(t-2) + Q(t-3)
meaning
On Fri, 29 Jan 2010, Ashta wrote:
Hi All,
Does the step function work in this model? I tried to run the
following model but no result obtained. The computer is hanging and I
killed the job several times. Below is the code.
library(survival)
m.fit=clogit(y~x1+x2+x3+x4, data=ftest)
But then I would have to make a loop right?
-
Anna Lippel
--
View this message in context:
http://n4.nabble.com/Applying-a-function-on-each-columns-of-a-matrix-tp1415660p1415743.html
Sent from the R help mailing list archive at Nabble.com.
__
Thanks David Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a 2x2
matrix with subsets, I could do,
y - matrix(xx[c(1:4)], 2, 2).
This returns,
[,1] [,2]
[1,] -27.3 14.4
[2,] 29.0 -38.1
If I do,
y2 - matrix(xx[c(5:8)],2,2)
On Jan 29, 2010, at 12:43 PM, Muhammad Rahiz wrote:
Thanks David Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a
2x2 matrix with subsets, I could do,
y - matrix(xx[c(1:4)], 2, 2).
This returns,
[,1] [,2]
[1,] -27.3 14.4
[2,]
OK, I've got this. The output prints what I want, but I'm not sure if
there will be problems in further analysis because the main idea is to
convert the data from list to matrix. I'm quite concerned with how I
define xx2.
xx - unlist(x) # Unlist from lapply + read.table
a -
On Jan 29, 2010, at 1:07 PM, Muhammad Rahiz wrote:
OK, I've got this. The output prints what I want, but I'm not sure
if there will be problems in further analysis because the main idea
is to convert the data from list to matrix. I'm quite concerned with
how I define xx2.
xx -
See sweep function
On Fri, Jan 29, 2010 at 2:32 PM, anna lippelann...@hotmail.com wrote:
Hello everyone, I have the following matrix
[,1] [,2] [,3] [,4]
[1,] 0.002809706 0.0063856960 0.0063856960 0.011749681
[2,] 0.004893124 0.0023118418
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
proc - function(e, env =
Dear R Users,
Using codetools I obtained the text representation of the parse tree for this
snippet
z=quote({x[1]-2})
showTree(z)
({ (- ([ x 1) 2)) (A)
If I understand correctly, x[1]-2 ought to be [-(x,1,2), so shouldn't i see
({ ( [- x 1 2 ) )
If indeed the parse tree in (A) is
Dear list members,
I'm tryng to write the code in order to calculate the index (expression 2)
published in Ricotta Burrascano 2008 (Preslia 80, pp 61-71).
Specifically, I'm having some problems in extending the index for more than
two observations. Does anyone already write a function for such
On 29/01/2010 2:03 PM, Saptarshi Guha wrote:
Dear R Users,
Using codetools I obtained the text representation of the parse tree for this
snippet
z=quote({x[1]-2})
showTree(z)
({ (- ([ x 1) 2)) (A)
If I understand correctly, x[1]-2 ought to be [-(x,1,2), so shouldn't i see
({ ( [- x 1
I'm currently using r scripts in sweave to grab some data via ODBC,
process it then generate some tables. I'd like to be able to give
someone the files and let them reproduce what I've done. Is there some
way to store the data that is gathered by ODBC so that the second person
can recreate
Hello -- I posted this question yesterday and for some reason the post seems to
be attached to the wrong thread. Also, I extended my test a little and it seems
to indicate the problem is with spm. I would appreciate any help. Thanks.
==
[Env: Win Xp]
Is there any options() setting or call to help() or help.start() that
will allow me to view a help file for
a package built under R 2.10.x under R 2.9.2?
I'm using both R 2.9.2 and R 2.10.1, but prefer the former because the
CHM help is so much easier
to use. Yet, if I install
On 29/01/2010 2:28 PM, Paul wrote:
I'm currently using r scripts in sweave to grab some data via ODBC,
process it then generate some tables. I'd like to be able to give
someone the files and let them reproduce what I've done. Is there some
way to store the data that is gathered by ODBC so
On 29/01/2010 2:34 PM, Michael Friendly wrote:
[Env: Win Xp]
Is there any options() setting or call to help() or help.start() that
will allow me to view a help file for
a package built under R 2.10.x under R 2.9.2?
I'm using both R 2.9.2 and R 2.10.1, but prefer the former because the
CHM
Dear Michael,
the help system has been rearranged considerably. It is not possible to
use binary packages prepared under R-2.10.x with earlier versions of R.
The other way round is also not a really good idea.
Note also that the most recent version of TeachingDemos (2.5) uses help
markup
On 29/01/2010 2:55 PM, Tobias Verbeke wrote:
Duncan Murdoch wrote:
On 29/01/2010 2:28 PM, Paul wrote:
I'm currently using r scripts in sweave to grab some data via ODBC,
process it then generate some tables. I'd like to be able to give
someone the files and let them reproduce what I've
On Jan 29, 2010, at 1:28 PM, Paul wrote:
I'm currently using r scripts in sweave to grab some data via ODBC, process
it then generate some tables. I'd like to be able to give someone the files
and let them reproduce what I've done. Is there some way to store the data
that is gathered by
Dear All,
I am intending to build a package (pksmooth) on Linux to work on
Windows. Some c++ functions need c++ libraries (numerical recipes)
from a static library libNR.a.
Building the package on Linux for Linux is allright. However, when
sending the 'pksmooth_1.0.tar.gz' to the online Windows
Hello,
Im trying to combine 3 affybatches (1x hgu133+2 array and 2x hgu133a array)
Im useing this script:
library(matchprobes)
library(affy)
library(AnnotationDbi)
library(hgu133plus2probe)
library(hgu133aprobe)
library(hgu133a.db)
u133p2 = ReadAffy() # reading hgu133 +2 cel file into
Dear All,
I have been attempting to save the output from xpose.VPC as a windows
metafile. When I ran vpc on PSN I had 6 groups and would like to output
a single wmf for each graph. When I set max.plots.per.page=1, xpose
turns on recording (page# in top right corner of output) and prints one
Loris Bennett loris.benn...@fu-berlin.de writes:
Hi,
I have a list of dates like this:
date
2009-12-03
2009-12-11
2009-10-07
2010-01-25
2010-01-05
2009-09-09
2010-01-19
2010-01-25
2009-02-05
2010-01-25
2010-01-27
2010-01-27
...
and am creating a
Hi,
what I would need are 2 vector pairs (x,y) and (x1,y1). x and x1 must have
the same sd. y and y1 should also exhibit the same sd's but different ones
as x and x1. Plotting x,y and x1,y1 should produce a plot with 2 vectors
having a different slope. Plotting both vector pairs in one plot with
Thank you, that worked great!
Sean
Peter Alspach wrote:
Tena koe Sean
I suspect the apply() and merge() functions are working, but they may
not be doing what you expect :-) You could try rbind() and aggregate():
data.frame1$HAD - as.numeric(NA)
data.both - rbind(data.frame1,
Hi
I am working with spgrass6 package and GRASS v.6.2. Everything was
fine until I tryed to read a vector file with readVECT6 (and other
related vector commands, like vInfo). When I ran these commands, the
problem immediately appeared ( Sorry, is not a valid flag ).
Ok, the solution
Dear Mailing List Members,
the problem I've been grappling with für quite some time now is the following:
I have a 100 rows x 200 columns matrix.
data.set - matrix(rnorm(2, 100, 200))
Now I would like to get a vector of length 100 which collects the values from
the following procedure:
On Fri, Jan 29, 2010 at 10:04 AM, Craig P. Pyrame crap...@gmail.com wrote:
Dear R-ers,
I am using getGEO to download expression data from the Gene Expression
Omnibus. With default settings, when a file is downloaded and parsed, lots
of dotted lines are printed in the terminal, like this:
Thanks again, Dennis and Petr!
The solution using the plyr package was perfect:
ddply(data, .(id, mod1), summarize, es = mean(es), mod2 = head(mod2, 1))
Take care,
AC
On Thu, Jan 28, 2010 at 11:26 PM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
r-help-boun...@r-project.org napsal dne
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