x - read.table(textConnection( V1 V2
1 1 160.54%
2 1 201.59%
3 1 18.45%
4 1 179.03%
5 1 274.37%
6 1 0.00%
7 1 24.52%
8 1 39.17%
9 3 43.72%
10 1 53.06%
11 1 64.97%
12 1 79.84%
13 1 98.08%
14 1 115.32%
15 1 127.96%
16 1 155.38%
17 1 157.25%
18 1 193.17%
[adapted repost of question
http://tolstoy.newcastle.edu.au/R/e4/help/08/03/6260.html]
Dear R community,
by default, text in the strips of a trellis plot is centered in the
strip.
Is there a way to have the text left-aligned?
For example:
library(lattice)
test - data.frame(x=rnorm(100),
Hi Solomon,
On Tue, May 13, 2008 at 05:53:44PM -0400, Messing, Solomon O. wrote:
Hi Gabor,
Thank you for your help, and thanks for making the excellent igraph
package. The function below seems not generate an edge list that works
for my data. I coerced a my data from a data frame using
Hi,
I have two dataframes one with 144 rows and 160 columns (SDF1) and one with
12 rows and 160 columns (SDF2).
Now I'm trying to divide rows 1:12 with SDF2, rows 13:24 with SDF2, rows
25:36 with SDF 2, .
In S-Plus the following code works fine:
DFS = SDF1[1:144,1:60] /
Please stay on the list.
On Tue, May 13, 2008 at 06:05:15PM -0400, Messing, Solomon O. wrote:
Gabor,
By the way, this seems to work:
I'm a bit lost. So now you're converting your data frame
to a matrix? Why? Or you're doing the two-mode to one-mode
conversion here? It does not seem so to
RT == Rolf Turner [EMAIL PROTECTED]
on Wed, 14 May 2008 13:12:02 +1200 writes:
RT I have written a function which reads data from files using read.delim
RT ().
RT The names of these files are complicated and are built using arguments
RT to the wrapper function. Since the
Hi, I wanted to make sure you were all aware of this upcoming event. There is
one remaining spring seminar in Predictive Analytics, on June 5-6 in New
York City. This is intensive training for managers, marketers, and IT
people who need to make sense of customer data to predict buying behavior,
Oops, there where some typos / thinkos in my post,
in the more sophisticated tryCatch() example.
I append the full message with corrected example.
--
Martin
---
RT == Rolf Turner [EMAIL PROTECTED]
on Wed, 14 May 2008 13:12:02
On 14-May-08 09:11:31, Achim Zeileis wrote:
On Wed, 14 May 2008, Mike Ryckman wrote:
Hello,
I am trying to run a negative binomial regression model in R and
can't get the standard errors to match the output I get from the
Stata nbreg command. I've tried a few different options but
haven't
Hello,
I am trying to run a negative binomial regression model in R and can't get
the
standard errors to match the output I get from the Stata nbreg command. I've
tried a few different options but haven't had much luck. The closest I've
found
is:
gamlss(formula, family = NBI, sigma.formula = ~
On Wed, 14 May 2008, Mike Ryckman wrote:
Hello,
I am trying to run a negative binomial regression model in R and can't get
the
standard errors to match the output I get from the Stata nbreg command. I've
tried a few different options but haven't had much luck. The closest I've
found
is:
Dear Frank
Frank E Harrell Jr [EMAIL PROTECTED] wrote:
A few observations.
1. With minimal overfitting, rcorr.cens(predict(fit), Y) gives a good
standard error for Dxy = 2*(C-.5) and bootstrapping isn't very necessary
2. If you bootstrap use the nonparametric bootstrap percentile method
Dear Nathan,
How is you C function test defined? What type of argument is it
expecting? The mappings between R and C type of arguments you can pass
is explained in:
http://cran.r-project.org/doc/manuals/R-exts.html#Interface-functions-_002eC-and-_002eFortran
which also provides some examples.
PhGr == Philippe Grosjean [EMAIL PROTECTED]
on Tue, 13 May 2008 16:10:15 +0200 writes:
PhGr Hello,
PhGr I did this bechmark test. Perhaps is it a good oppotunity to rewrite
it
PhGr and make it compatible with R 2.7.0, David?
I'll not really rewrite it (to make it nice in my
Martin Maechler wrote:
PhGr == Philippe Grosjean [EMAIL PROTECTED]
on Tue, 13 May 2008 16:10:15 +0200 writes:
PhGr Hello,
PhGr I did this bechmark test. Perhaps is it a good oppotunity to rewrite it
PhGr and make it compatible with R 2.7.0, David?
I'll not really rewrite it
Hi R,
I have the data frames, data1, data2data50. Now I want to put all of
these in a single list. But,
list(data1, data2,.data50) is very big to write. How do I then
do it?
Thanks,
Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
How do I proceed from this stage?
paste(data,1:10,sep=,collapse=,)
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL
Hallo All,
I have difficulties understanding how factors work in R. Suppose a have
data in the panel form below. I would to compute a correlation coefficient
(actually apply a different function of two time series) in the V variable
between members of the two sexes in each city over time. How
Does
mget(paste(data, 1:50, sep = ), envir = globalenv())
give what you want?
hth,
Matthias
Shubha Vishwanath Karanth wrote:
How do I proceed from this stage?
paste(data,1:10,sep=,collapse=,)
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo *
MM == Martin Maechler [EMAIL PROTECTED]
on Wed, 14 May 2008 12:05:00 +0200 writes:
PhGr == Philippe Grosjean [EMAIL PROTECTED]
on Tue, 13 May 2008 16:10:15 +0200 writes:
PhGr Hello,
PhGr I did this bechmark test. Perhaps is it a good oppotunity to rewrite
it
PhGr and
Hi
I am not sure what you want to do as you speak about using different
function for different part of your data.
head(data)
city year sex V
11 1975 F 25.3044
21 1975 M 16.5711
31 1976 M 16.6072
41 1976 F 24.2841
51 1977 M 14.8838
61 1977 F 24.8124
On May 14, 2008, at 3:47 AM, RINNER Heinrich wrote:
[adapted repost of question
http://tolstoy.newcastle.edu.au/R/e4/help/08/03/6260.html]
Dear R community,
by default, text in the strips of a trellis plot is centered in the
strip.
Is there a way to have the text left-aligned?
For example:
Dear R Users!
Thanks very much!
I really don't know why the scale function didn't come to my mind when
writing my silly question.
But anyway thanks to all of you.
Yukihiro Ishii
2-3-28 Tsurumakiminami, Hadano, 250-0002 Japan
+81463691922
__
jiho.han wrote:
hello, useRs~
suppose i have a matrix as follows:
itemcategory sub-category
A 1 11
B 1 12
C 1 12
D 2 21
E 2 22
i like to draw
Hello,
Thank you for your replies. I think that information largely focuses on
robust
standard errors... maybe I wasn't clear. By default Stata parameterizes the
dispersion. From the stata help files it is calculated as 1 + alpha*exp(xb +
offset).
As I understand it, this process should happen
Try this:
DFS - lapply(split(seq(1, 144, by = 1), rep(1:12, each = 12)),
function(x)SDF1[x,]/SDF2)
On Wed, May 14, 2008 at 5:02 AM, Bert Jacobs [EMAIL PROTECTED] wrote:
Hi,
I have two dataframes one with 144 rows and 160 columns (SDF1) and one
with
12 rows and 160 columns (SDF2).
Now
Oops, you will need:
library(prettyR)
to run the barhier function.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
Hi all,
How could I draw a plot with two y axes? For example, if I need too plot two
curves in the same graphic, which are measured in different scales.
Thanks a lot in advance,
Caio
[[alternative HTML version deleted]]
__
On Wed, 14 May 2008, Mike Ryckman wrote:
Hello,
Thank you for your replies. I think that information largely focuses on
robust standard errors... maybe I wasn't clear.
No, you were not and, as pointed out previously (and described in the
posting guide), a reproducible example would clearly
plot(...)
par(new=TRUE)
plot(...,axes=FALSE)
On Wed, May 14, 2008 at 8:11 AM, Caio Azevedo [EMAIL PROTECTED] wrote:
Hi all,
How could I draw a plot with two y axes? For example, if I need too plot
two
curves in the same graphic, which are measured in different scales.
Thanks a lot in
Hi
so basically you want to
1. select one city
2. compute a correlation coeficient between value V for males and
females
3. repeat this for each city
As I am not an expert in statistics I could be off track but maybe
cor(do.call(cbind,split(data$V, interaction(data$city,
Caio Azevedo cnaberdl at gmail.com writes:
How could I draw a plot with two y axes? For example, if I need too plot two
curves in the same graphic, which are measured in different scales.
http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-base:2yaxes
Gabor Grothendieck wrote:
Is it possible to use the console from within tcltk?
Not on Windows RGui, not without deeper magic, anyway (your examples
work quite happily in a terminal on Linux). If you can figure out how to
wire a Tcl channel to the R console, then I suppose it could be made to
Hello all,
I'm using R2.7.0 (on Windows 2000) and I'm trying do run a robust
regression on following model structure:
model = Y ~ x1*x2 / (x3 + x4 + x5 +x6)
where x1 and x2 are both factors (either 1 or 0) and x3.x6 are numeric.
The error code I get when running rlm(as.formula(model),
DAVID ARTETA GARCIA wrote:
Dear Frank
Frank E Harrell Jr [EMAIL PROTECTED] wrote:
A few observations.
1. With minimal overfitting, rcorr.cens(predict(fit), Y) gives a good
standard error for Dxy = 2*(C-.5) and bootstrapping isn't very necessary
2. If you bootstrap use the nonparametric
Hi Henrique,
I think I understand your formula, but your final result is a 12x160
data.frame, while it should be a 144x160 data.frame
divide rows 1:12 with SDF2 (=12 rows)
divide rows 13:24 with SDF2 (=12 rows)
divide rows 25:36 with SDF2 (=12 rows)
divide rows 133:144 with SDF2 (=12
SDF3 - SDF1/SDF2[ rep(1:12,12), ]
-Original Message-
From: Bert Jacobs [EMAIL PROTECTED]
To: 'Henrique Dallazuanna' [EMAIL PROTECTED]
Cc: r-help@r-project.org r-help@r-project.org
Sent: 5/14/08 7:41 AM
Subject: Re: [R] Dividing Two Dataframes
Hi Henrique,
I think I understand your
Dear Bert,
Another option is to replicate SDF2 so the number of rows matches with those of
SDF1.
SDF1 - data.frame(matrix(rnorm(144 * 2), nrow = 144))
SDF2 - data.frame(matrix(rnorm(12 * 2), nrow = 12))
SDF1 / SDF2 #yields the error
SDF2a - do.call(rbind, rep(list(SDF2), nrow(SDF1) /
http://ion.researchsystems.com/cgi-bin/ion-p
I would like a continuous wavelet transform. I have downloaded wavethresh,
Rwave, and waveslim. I would like an output very similar to the above
website. any suggestions?
--
Let's not spend our time and resources thinking about things that are so
Christophe LOOTS Christophe.Loots at ifremer.fr writes:
Hi,
I've two fitted models, one binomial model with presence-absence data
that predicts probability of presence and one gaussian model (normal or
log-normal abundances).
I would like to evaluate these models not on their
Dear useRs:
I'm not sure if it's the correct place to ask but I'll try it out. I've been
reading about how to perform Principal Component Analysis (PCA) in
microarrays (see [1]) and there's something that I don't get it. Basically
it's related with performing PCA over data sets which number of
I am trying to use rgenoud and snow with an external model and file I/O.
Unlike the typical application of rgenoud and snow, I need to run an
external executable and do pre- and post-processing of input and output
files for each parameter set generated by genoud(). I'm hoping that
someone can
I assume by 'console' you mean the Rgui window. That is not where stdout
is connected -- it is not a file. As the CHANGES file for 2.7.0 says
o When Rgui is launched from a terminal, C output from ill-formed
packages that write to stdout or stderr rather than use
Rprintf
Dear R community,
I wrote a small program using for loop but it does not make cycles.
My data: Dataframes: a2, a1, b0 and b1. Vector: d
I would like to get b1 for each of i., i.e. totally 11. However, the program
gives me b1 only for the last i =11.
d-as.vector(levels(a2$combin2))
for
Dear all,
I have a vector generated using the function strptime:
my.dt
[1] 2004-04-19 08:35:00 W. Europe Daylight Time 2004-04-19 09:35:00 W.
Europe Daylight Time 2004-04-19 11:35:00 W. Europe Daylight Time
[4] 2004-04-19 13:35:00 W. Europe Daylight Time 2004-04-20 07:50:00 W.
Europe Daylight
Hi R,
Suppose
l=c(1,1,1,2,2,1,1,1)
k[-which(k==1)]
[1] 2 2
k[-which(k==2)]
[1] 1 1 1 1 1 1
But,
k[-which(k==3)]
numeric(0)
I do not want this numeric(0), instead the whole k itself should be my
result... How do I do this?
Thanks,
Shubha
This e-mail may
Hello all,
I was wondering if there is a package in which someone would have created a
quantile function that handles vectors of class Date.
In advance thanks.
David Gouache
ARVALIS - Institut du végétal
Station de La Minière
78280 Guyancourt
Tel: 01.30.12.96.22 / Port: 06.86.08.94.32
k - c(1,1,1,2,2,1,1,1)
k[k != 1]
[1] 2 2
k[k != 2]
[1] 1 1 1 1 1 1
k[k != 3]
[1] 1 1 1 2 2 1 1 1
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Shubha
Vishwanath Karanth
Sent: Wednesday, May 14, 2008 11:16 AM
To: [EMAIL
On 5/14/2008 11:16 AM, Shubha Vishwanath Karanth wrote:
Hi R,
Suppose
l=c(1,1,1,2,2,1,1,1)
k[-which(k==1)]
[1] 2 2
k[-which(k==2)]
[1] 1 1 1 1 1 1
But,
k[-which(k==3)]
numeric(0)
I do not want this numeric(0), instead the whole k itself should be my
result...
l[!l==3]
On 14 May 2008, at 17:16, Shubha Vishwanath Karanth wrote:
Hi R,
Suppose
l=c(1,1,1,2,2,1,1,1)
k[-which(k==1)]
[1] 2 2
k[-which(k==2)]
[1] 1 1 1 1 1 1
But,
k[-which(k==3)]
numeric(0)
I do not want this numeric(0), instead the whole k itself should be my
Thanks. I've updated to the latest R on the mirror I used and
puts stdout 12
from tcl does indeed produce output from Rterm on Windows but gives an
error message from Rgui on Windows (both on Windows Vista SP1).
- this works
C:\tmp2Rterm
R version 2.7.0 Patched (2008-05-11
On Wed, 14 May 2008, GOUACHE David wrote:
Hello all,
I was wondering if there is a package in which someone would have
created a quantile function that handles vectors of class Date.
What is the median of today and tomorrow, expressed as a day?
Quantiles of discrete distributions are not
On Wed, 14 May 2008, Lukas Rode wrote:
Dear all,
I have a vector generated using the function strptime:
my.dt
[1] 2004-04-19 08:35:00 W. Europe Daylight Time 2004-04-19 09:35:00 W.
Europe Daylight Time 2004-04-19 11:35:00 W. Europe Daylight Time
[4] 2004-04-19 13:35:00 W. Europe Daylight
On Wed, 14 May 2008, Jorge Ivan Velez wrote:
Dear useRs:
I'm not sure if it's the correct place to ask but I'll try it out. I've been
reading about how to perform Principal Component Analysis (PCA) in
microarrays (see [1]) and there's something that I don't get it. Basically
it's related with
Hi,
I am trying to split a string into 2 separate strings
in 2 lines. So, tried with
sprintf(%s \n %s, ID, Name)
But R prints: [1] ID \n Name instead of
ID
Name
Is it something I am missing here?
Thanks in advance.
Best regards,
Ezhil
__
Dear Brian and Gabor,
When the Rcmdr console.output option is set to TRUE, the Rcmdr interface
does something very much like this using cat() for R output, which it
captures. The relevant code is in the doItAndPrint() function in the Rcmdr
sources.
I hope this helps,
John
-Original
Dear all,
I have several datasets and I want to generate pdf plots from them.
I also want to generate automatically the names of the files. They are
country-specific and the element mycurrentdata[1,1] contains this
information.
So what I do is something like this:
A Ezhil wrote:
Hi,
I am trying to split a string into 2 separate strings
in 2 lines. So, tried with
sprintf(%s \n %s, ID, Name)
But R prints: [1] ID \n Name instead of
ID
Name
Is it something I am missing here?
Yes, in order to respect teh control character you have to cat() it as
Maybe another basic question but...
Is is possible to accomodate more than a single (independent, not resulting
from arranjment of layout=c()) lattice graph in a single win.graph() device?
Thanks in advance,
PS Maybe duplicated
Paulo
De: [EMAIL PROTECTED]
Hello friends!!
I have two questions, and I would like that you could answer me!!!
I have created a plot as
plot(range(10,-10),range(10,-10),col=blue,col.axis=blue,col.lab=blue,col.main=blue,col.sub=blue);
1º) I want that the square of plot and the lines that indicates the value
Hi,
My function test is properly declared in the header file of the code:
void test(Model *);
and we just discovered yesterday we should be using .Call instead ,of .C.
However we still can't get it to work. We arent trying to pass a primitive
type from R but a complex one. We are trying to
GOUACHE David D.GOUACHE at arvalisinstitutduvegetal.fr writes:
I was wondering if there is a package in which someone would have
created a quantile function that handles vectors of class Date.
Since dates are just decorated numbers, you can convert in both way.
Written the long way:
dt =
Hello,
I am trying to run a nonparametric comparison of multiple datasets (A-E).
After importing my csv file I sorted the variable value by the letter class
(representing height along a streambank).
a=Z[Letter==A]
b=Z[Letter==B]
c=Z[Letter==C]
d=Z[Letter==D]
e=Z[Letter==E]
Next I tested
Hello,
I am new to R, can anyone give me an idea of how R handle a large dataset
(e.g. couple of Gbytes)? Thanks a lot!
Best,
Mingjun
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Using the heatmap.2 function, I am trying to generate a heatmap of a
2 column x 500 row matrix of numeric values. I would like the 1st
column of the matrix sorted from the highest to the lowest values -
so that the colors reflected in the first column of the heatmap (top
to bottom) go from
Susanne Komhard susanne.komhard at ise.fraunhofer.de writes:
I'm using R2.7.0 (on Windows 2000) and I'm trying do run a robust
regression on following model structure:
model = Y ~ x1*x2 / (x3 + x4 + x5 +x6)
where x1 and x2 are both factors (either 1 or 0) and x3.x6 are numeric.
The
Date: Wed, 14 May 2008 12:06:39 -0400
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [R] strip white in character strings
Dear all,
I have several datasets and I want to generate pdf plots from them.
I also want to generate automatically the names of the files. They are
Thank you very much. It works !
Ezhil
--- Uwe Ligges [EMAIL PROTECTED]
wrote:
A Ezhil wrote:
Hi,
I am trying to split a string into 2 separate
strings
in 2 lines. So, tried with
sprintf(%s \n %s, ID, Name)
But R prints: [1] ID \n Name instead of
ID
Name
Is it
You got exactly what you were asking for: the value the last time through
the loop. If you are trying to capture all 11 values, then you might
consider a list:
d-as.vector(levels(a2$combin2))
a1 - b1 - vector('list',11)
for (i in 1:11){
a1[[i]] -a2[a2$combin2%in%d[i],]
a = United Kingdom
gsub( , , a, fixed=TRUE )
[1] UnitedKingdom
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Jeremiah Rounds
Sent: Wednesday, May 14, 2008 12:48 PM
To: '[EMAIL PROTECTED]'
Subject: Re: [R] strip white in character strings
Hi Stephen,
Your link doesn't work. In any case, check out the wavCWT function in
the wmtsa package.
Julian
stephen sefick wrote:
http://ion.researchsystems.com/cgi-bin/ion-p
I would like a continuous wavelet transform. I have downloaded wavethresh,
Rwave, and waveslim. I would like an
I believe this is determined by how much memory your computer has, not
particularly by R itself.
Mingjun Huang wrote:
Hello,
I am new to R, can anyone give me an idea of how R handle a large dataset
(e.g. couple of Gbytes)? Thanks a lot!
Best,
Mingjun
It depends on what you mean by striping the white space. Here is an example
of how it might be done:
x - c(United Kingdom, Europe, United States, a
longer string)
gsub( +, _, x)
[1] United_Kingdom Europe United_States a_longer_string
On Wed, May 14, 2008 at 12:06 PM,
Dear all,
I would like to say thanks for the help of Mark Leeds (off-list),
Jeremiah Rounds (off-list), and Jim Holtman.
I will use the solution Mark and Jim provided using the gsub function.
A bit embarrassing for me that I did not remember the function since I
used it myself awhile ago.
Use logical subscripts rather than which:
k - c(1,1,1,2,2,1,1,1)
k[ k != 1 ]
[1] 2 2
k[ k != 2 ]
[1] 1 1 1 1 1 1
k[ k != 3 ]
[1] 1 1 1 2 2 1 1 1
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
R-Fellow Travellers:
Asked from the perspective of a software development amateur ...
Suppose one wishes to create data structures that have both data and
metadata -- e.g. a data.frame and details about where, when, and by whom the
data were gathered (specifics not important here).
1. Is there
Thank you everyone for your help. I tried the following. Please let me know if
I can use this method.
df1$V2-as.numeric(gsub([%],, df1[,2]))
xbin - binning(df1$V2,df1$V1,nbins=12)
barplot(xbin$sums,xbin$x,xlab=V2,border = TRUE,ylab=V1,col=red)
Thanks,
Sachin
- Original Message
From:
The key difference is inheritance.
Objects consisting of data frames with attributes, including a class attribute
such as c(myclass, data.frame), can inherit data frame methods but a list
with a data frame component will require entirely new methods to be constructed
for everything.
On Wed,
Try this:
k=c(1,1,1,2,2,1,1,1)
k[(k!=1)]
[1] 2 2
k[(k!=2)]
[1] 1 1 1 1 1 1
k[(k!=3)]
[1] 1 1 1 2 2 1 1 1
Julian
Shubha Vishwanath Karanth wrote:
Hi R,
Suppose
l=c(1,1,1,2,2,1,1,1)
k[-which(k==1)]
[1] 2 2
k[-which(k==2)]
[1] 1 1 1 1 1 1
But,
k[-which(k==3)]
Dear List Members,
I have encountered two problems when using the step function to
select models. To better illustrate the problems, an R image
(step.add1.test.RData)
which includes the objects needed to run the code (step.add1.test.R) can be
found at
Hello All,
I have a covariance matrix, generated by read.table, and cov:
co-cov(read.table(c:/r.x))
XYZ
X 0.0012517684 0.0002765438 0.0007887114
Y 0.0002765438 0.0002570286 0.0002117336
Z 0.0007887114 0.0002117336 0.0009168750
And a weight vector
Hello Abhijit,
When I try that, I get:
diag(w)%*%co%*%diag(w)
Error in diag(w) %*% co : requires numeric matrix/vector arguments
Dan Stanger
Eaton Vance Management
200 State Street
Boston, MA 02109
617 598 8261
-Original Message-
From: Abhijit Dasgupta [mailto:[EMAIL PROTECTED]
Sent:
try this:
V - var(matrix(rnorm(100*3), 100, 3))
w - c(0.5, 0.3, 0.2)
V * (w %o% w)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
Using R 2.6.2, say I have the following list of lists, comb:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
So that all names for the lowest level list are common. How can I most
Try this:
do.call(rbind, lapply(comb, '[', 'a'))
On Wed, May 14, 2008 at 4:40 PM, [EMAIL PROTECTED]
wrote:
Using R 2.6.2, say I have the following list of lists, comb:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1
Hi Phil,
That solved the problem.
Thanks,
Dan Stanger
Eaton Vance Management
200 State Street
Boston, MA 02109
617 598 8261
-Original Message-
From: Phil Spector [mailto:[EMAIL PROTECTED]
Sent: Wednesday, May 14, 2008 3:42 PM
To: Dan Stanger
Subject: RE: [R] Newbie question about vector
Try this:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
sapply(comb, [[, a)
data1 data2 data3
1 4 3
# Also, this can be useful:
comb[[c(data2, b)]]
[1] 5
[EMAIL
[EMAIL PROTECTED] said the following on 5/14/2008
12:40 PM:
Using R 2.6.2, say I have the following list of lists, comb:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
So that all
check the following:
sapply(comb, [[, a)
# or
data.frame(
names = names(comb),
value = sapply(comb, [[, a),
row.names = seq_along(comb)
)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address:
Date: Wed, 14 May 2008 15:18:32 -0400
From: [EMAIL PROTECTED]
To: r-help@r-project.org
Subject: [R] Newbie question about vector matrix multiplication
Hello All,
I have a covariance matrix, generated by read.table, and cov:
co-cov(read.table(c:/r.x))
X Y Z
X 0.0012517684
I'm trying to work on a large dataset and after each segment of run, I need
a command to flush the memory. I tried gc() and rm(list=ls()) but they don't
seem to help. gc() does not do anything beside showing the memory usage.
I'm using the package BSgenome from BioC.
Thanks a bunch
--
Regards,
Depends on the RAM in your machine. And in your definition of 'handle'.
You may be able to load a very large dataset into R, but won't be able
to use some functions that require additional memory.
A vague answer to a vague question... :)
Julian
Mingjun Huang wrote:
Hello,
I am new to
On Wed, 14-May-2008 at 09:12AM -0700, ermimi wrote:
|
| Hello friends!!
|
| I have two questions, and I would like that you could answer me!!!
|
| I have created a plot as
|
|
plot(range(10,-10),range(10,-10),col=blue,col.axis=blue,col.lab=blue,col.main=blue,col.sub=blue);
|
|
| 1º) I
Dan,
You need to do
diag(w) %*% as.matrix(co) %*% diag(w)
The reason is that read.table creates a data frame. Although it looks like
a matrix it is not - actually it is a special kind of list. The
as.matrix function will coerce it to a matrix.
By the way, this will generate
On 5/14/2008 3:59 PM, Anh Tran wrote:
I'm trying to work on a large dataset and after each segment of run, I need
a command to flush the memory. I tried gc() and rm(list=ls()) but they don't
seem to help. gc() does not do anything beside showing the memory usage.
How do you know it does
Try
par(fg=blue)
plot(range(10,-10),range(10,-10),col=blue,col.axis=blue,col.lab=blue,col.main=blue,col.sub=blue)
Blay
KATH
ermimi wrote:
Hello friends!!
I have two questions, and I would like that you could answer me!!!
I have created a plot as
Hello,
I am trying to convert this splus line of code to R :
var(outcome[1, ], unbiased = FALSE)
It seems the var function in R doesn't have the unbiased argument. Could
someone help me figure the correct equivalent line in R?
Thank you
--
View this message in context:
Patrick and Blay Thank you very much for help me, I have drawn in blue the
axis
Blay the solution that you give me for start in (-10,-10) and finish in
(10,10) isn´t very well because I want that if my range is (-10,10) the axis
is in -10,10. I need that there isn´t space between the less
See the 'variance' function in ifultools package. If you'd like the
unbiased estimate of the variance, just use 'var' without the extra
argument.
Best,
Erik Iverson
Applejus wrote:
Hello,
I am trying to convert this splus line of code to R :
var(outcome[1, ], unbiased = FALSE)
It seems
Is is possible to accomodate more than a single independent (not resulting
from arranjment of layout=c()) lattice graphs in a single win.graph()
device?
Thanks in advance,
PS Maybe duplicated
Paulo
De: [EMAIL PROTECTED] em nome de Roland Rau
Enviada: qua
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