On Sun, 31 Jul 2011, Iasonas Lamprianou wrote:
Thanks
Pscl seems to be a sensible option.
I have the counts variable with the name N. This variable can only
take values bigger than zero!
I have two explanatory variables with the names type and diam
but when I run
hpm - hurdle(n ~
oaxacamatt wrote:
I am calculating two t-test values for each of many files then save it
to file calculate another set and append, repeat.
You did not tell use what you want to do with the data in the file. If you
just want a copy of the output, bracketing with sink(file) and sink() can be
Hi Matt,
I assume that you want a tabular text file of the results. Since I
don't know what your tempA and tempB are I'll steal some examples from
?t.test
t.example.1 - t.test(1:10,y=c(7:20))
t.example.2 - t.test(1:10,y=c(7:20, 200))
Now looking at ?dump, the first argument needs to be
Dear R users,
I'm trying to fit a set an ODE to an experimental time series. In the
attachment you find the R code I wrote using modFit and modCost of FME
package and the file of the time series.
When I run summary(Fit) I obtain this error message, and the values of the
parameters are equal to
marcel wrote:
I have a figure with a lattice plot and a basic plot. Is there a way to
select the color and line width of the surrounding boxes for each of
these? I could not find any documentation on this.
Thanks for providing a nice self-contained example. There was nothing wrong
Rosario Garcia Gil-2 wrote:
I have a problem on keeping the format when I export a matrix file with
the write.table() function.
When I import the data volcano from rgl package it looks like this in R:
data[1:5,]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
(I'm replying to your original post because your follow-up omits the
context.)
The K-S test is designed for continuous distributions. You have far
too many zeros in your data to get anything reasonable out of the
test. For your data, the K-S statistic is the difference in the
(e)cdfs at zero.
Hi,
sorry for repeating the question but this is kind of important to me and i
don't know whom should i ask.
So as noted before when I do a parameter fit to the beta distr i get:
fitdist(vectNorm,beta);
Fitting of the distribution ' beta ' by maximum likelihood
Parameters:
estimate
Thank you, it works! And the estimates (as well as the standard errors) seem to
be more reasonable now, compared to the normal Poisson model. Thank you.
However, I tried to find a manual (although I did manage to find the paper
published in the Journal of Statistical Software. For example, how
On Aug 1, 2011, at 10:33 , baxy77 wrote:
Hi,
sorry for repeating the question but this is kind of important to me and i
don't know whom should i ask.
So as noted before when I do a parameter fit to the beta distr i get:
fitdist(vectNorm,beta);
Fitting of the distribution ' beta '
On 2011-08-01 01:33, baxy77 wrote:
Hi,
sorry for repeating the question but this is kind of important to me and i
don't know whom should i ask.
So as noted before when I do a parameter fit to the beta distr i get:
fitdist(vectNorm,beta);
Fitting of the distribution ' beta ' by maximum
Dear Peter,
The spatial taskview lists a number of interpolation methods [1]. Some
of those support spatio-temporal interpolation. For example gstat
supports spatio-temporal kriging [2,3,4].
regards,
Paul
[1] http://cran.r-project.org/web/views/Spatial.html
[2]
yes it is the fitdistrplus package. Sorry form not mentioning it earlier.
Usually i do those things but this time it somehow slipped my mind , sorry
and
Thank you both!
--
View this message in context:
http://r.789695.n4.nabble.com/Beta-fit-returns-NaNs-tp3709139p3709277.html
Sent from the R
On 08/01/2011 04:52 AM, Cheryl Johnson wrote:
Hello,
I have two plots on the same screen. I use the command par(mfrow=c(1,2)) in
order to do this. When I try to make a legend for both plots, it only puts
the legend in the plot on the right side. If I would like a legend that is
outside of both
Wir sind bis am 20. August in den Ferien und werden keine e-mails beantworten.
Bei dringenden Fällen melden Sie sich bei Stefanie von Felten
steffi.vonfel...@oikostat.ch
We are on vacation until 20. August. In urgent cases, please contact Stefanie
von Felten steffi.vonfel...@oikostat.ch
Dear All,
I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried using
: ylab = expression(paste(st / , plain(km)^2, / day)) to see if I can
at least get the unit before thinking about the power of 10 (10^-8).
However, ylab = expression(paste(st / , plain(km)^2, / day)) didn't
Bumping this one up because the 'before.plot.new' solution turned out
to be sub-optimal after all.
It should be possible to do this with a before.plot.new hook, right?
Yes, sure, if you treat the first and last plot separately.
It turns out that the before.plot.new hook does not is not
On 07/31/2011 05:57 PM, r student wrote:
I'm wondering if anyone can give some basic advice about how to approach a
specific task in R.
I'm new to R but have used SAS for many years, and while I can muscle
through a lot of the code details, I'm unsure of a few things.
Specific questions:
Hi
Hi,
I have a dataframe that I imported from a .txt file by:
skogTemp - read.delim2(Skogaryd_shoot_data.txt, header=TRUE,
fill=TRUE)
and the data are factors, how can avoid factors from the beginning?
Although
the file contains both characters and numbers.
You have got an answer
On 2011-08-01 03:32, ogbos okike wrote:
Dear All,
I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried using
: ylab = expression(paste(st / , plain(km)^2, / day)) to see if I can
at least get the unit before thinking about the power of 10 (10^-8).
However, ylab =
Hello,
I am looking for some help with this question: how could I test structural
breaks in a instrumental variables´s model?
For example, I was trying to do something with my model with three time
series.
tax_ivreg - ivreg(l_y ~ l_x2 + l_x1+ dl_y | lag(l_x2, -1)+lag(l_x2, -2)+
lag(l_x1,
On Aug 1, 2011, at 3:41 AM, Paola Lecca wrote:
Dear R users,
I'm trying to fit a set an ODE to an experimental time series. In the
attachment you find the R code I wrote using modFit and modCost of FME
package and the file of the time series.
This is getting a bit tiresome. None of the
* Apologies for multiple posting *
I attached to my previous e-mail a .r file, and it was not permitted by the
rules of the mailing lis. Again, please receive my sincere apologies for
this.
I re-send again the e-mail with .txt attachemnt in the hope someone an help
me to solve my problem.
I'm
If you are not going to be using factors, then you can keep everything
a character (if there are non-numerics in a column) by adding
'as.is=TRUE' as a parameter on the 'rad.table' functions.
On Mon, Aug 1, 2011 at 7:55 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
Hi,
I have a dataframe
On Jul 31, 2011, at 7:54 PM, Rosario Garcia Gil wrote:
Hello
I have a problem on keeping the format when I export a matrix file
with the write.table() function.
The quick answer is ... don't do that. Use save() if you want to
preserve the attributes of an R object. And that especially
If you are just exporting it so you can read it back into R later, it
is better to use save/load since it keep the data in the internal
format so it will look the same.
Can you describe what you are going to be doing with the data that you
'export'; that might help us come up with a solution to
Can you define better exactly what you what to do with the data. I
would suggest that you keep each of the outputs (objects) of the test
in a 'list' that way you can access each one and do what you need.
You can also 'save' the list and later 'load' it into another session.
On Sun, Jul 31, 2011
Am Sonntag, den 31.07.2011, 23:32 -0500 schrieb R. Michael Weylandt :
Glad to help -- I haven't taken a look at Dennis' solution (which may be far
better than mine), but if you do want to keep going down the path outlined
below you might consider the following:
I will try Dennis’ solution
Hi Peter,
Many thanks. It worked.
Regards
Ogbos
On 1 August 2011 14:05, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-08-01 03:32, ogbos okike wrote:
Dear All,
I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried
using
: ylab = expression(paste(st / , plain(km)^2, /
Greetings all,
Thanks for all your help so far.
Let me give a better idea of what I am doing. I have hundreds of
files that I need to plow thru with a t-test and correlation test.
BTW, 'tempA' and tempB' are simply columns of numbers from a gene-chip
experiment that spits out dna 'amounts'. So I
Hello,
I was wondering if someone knows the formula used by the function lm to compute
the t-values.
I am trying to implement a linear regression myself. Assuming that I have K
variables, and N observations, the formula I am using is:
For the k-th variable, t-value= b_k/sigma_k
With
On Aug 1, 2011, at 9:27 AM, Samuel Le wrote:
Hello,
I was wondering if someone knows the formula used by the function lm
to compute the t-values.
I am trying to implement a linear regression myself. Assuming that I
have K variables, and N observations, the formula I am using is:
For
On Aug 1, 2011, at 15:27 , Samuel Le wrote:
Hello,
I was wondering if someone knows the formula used by the function lm to
compute the t-values.
I am trying to implement a linear regression myself. Assuming that I have K
variables, and N observations, the formula I am using
Yes! Would you mind filing an issue so I dont forget?
Hadley
On Friday, July 29, 2011, Stavros Macrakis macra...@alum.mit.edu wrote:
Perfect! Thanks!
By the way, I see that, unlike base rbind, it does not work for vectors
and lists:
rbind(c(a=1),c(b=2)) =
-Original Message-
[mailto:r-help-boun...@r-project.org] On Behalf Of Samuel Le
Subject: [R] formula used by R to compute the t-values in a
linear regression
I was wondering if someone knows the formula used by the
function lm to compute the t-values.
Typing
summary.lm
I
Exactly.
My formula holds only for k=1, this is how I generated it.
Do you have any references concerning the rather more careful algorithms?
Thanks,
Samuel
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: 01 August 2011 14:45
To: Samuel Le
Cc:
Yes, that's what I was looking for.
Many thanks,
Samuel
-Original Message-
From: S Ellison [mailto:s.elli...@lgcgroup.com]
Sent: 01 August 2011 15:16
To: Samuel Le; r-h...@stat.math.ethz.ch
Subject: RE: formula used by R to compute the t-values in a linear regression
-Original
Dear R community,
I have a general question regarding indexing in multidiemensional arrays.
Imagine I have a three dimensional array and I only want to extract on
vector along a single dimension from it:
data- array(rnorm(64),dim=c(4,4,4))
result - data[1,1,]
If I want to extract
Thanks a lot for the help.
Actually, I am using a mac which (R for Mac OS X GUI 1.40-devel Leopard
build 32-bit (5751)) but I think I can find access on windows 7 64-bit. What
I am trying to do is a maximization through grid search (because I am not
sure that any of the optim() methods works
Dear group,
I experience some problems with gam() function after R update to version 2.13.1
The function in both gam and mgcv packages stopped to work. Before, with the
same code I used, everything was fine.
The function from gam package yields following warning:
Residual degrees of freedom are
Dear group,
I experience some problems with gam() function after R update to version 2.13.1
The function in both gam and mgcv packages stopped to work. Before, with the
same code I used, everything was fine.
The function from gam package yields following warning:
Residual degrees of freedom
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The points
are
distinguished by a third variable which is categorical with 10 levels. I have
been plotting x against y and using gray scales to distinguish the level of the
categorical variable for each point. It looks ok
Dear Mailing-list
I used hclust to make a dendrogram of 2613 leafs. I also have a list
with the names of certain labels which are of interest and I would like
to visualize their appearance within the dendrogram. I found an example
how to use dendrapply to colour the labels but the problem is that
Dear all,
I have been attempting to use multiple imputation (MI) to handle missing data
in my study. I use the mice package in R for this. The deeper I get into this
process, the more I realize I first need to understand some basic concepts
which I hope you can help me with.
For example, let
Dear all,
this must have been a temporary problem. In this case I assume that the
build cycle did not finish in time, i.e., binaries were synced to the
staging area although not all were built.
best,
stefan
On 07/31/2011 05:52 PM, David Winsemius wrote:
On Jul 31, 2011, at 11:26 AM,
On 11-08-01 5:38 AM, Jannis wrote:
Dear R community,
I have a general question regarding indexing in multidiemensional arrays.
Imagine I have a three dimensional array and I only want to extract on
vector along a single dimension from it:
data- array(rnorm(64),dim=c(4,4,4))
result-
Dear R-help
We are getting an error message `jpeg62.dll missing'.
We are running Windows 7 64-bit, from a Mac using Boot Camp.
Do you know of this error message, and can you give us help trying to
resolve the problem?
many thanks
Rocky
Rocky Hyacinth
Technician
Department of Archaeology
On 11-08-01 5:44 AM, Andrew McCulloch wrote:
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The points are
distinguished by a third variable which is categorical with 10 levels. I have
been plotting x against y and using gray scales to distinguish the level of the
Dear David and Hans- Werner,
Thank you very much for your help. I would like to compare now if a
polynomial or the sinus model fits better. How can I see R-squared or
the F- Statistic for the sinus regression, so as to be able to compare
it with the polynomial model?
Thanks a lot and have a nice
On Aug 1, 2011, at 3:04 AM, Dimitris.Kapetanakis wrote:
Thanks a lot for the help.
Actually, I am using a mac which (R for Mac OS X GUI 1.40-devel
Leopard
build 32-bit (5751)) but I think I can find access on windows 7 64-
bit.
I don't think that was what Holtman was advising. You just
Hi again,
I have tried playing around with the code given to me by Alan and Jim, thank
you for the code but unfortunatelyI can't seem to get either of them to
work... Alans does not work with the sample data and Jims is giving the
error :
Error in radial.grid(labels = labels, label.pos =
On Aug 1, 2011, at 10:50 AM, Duncan Murdoch wrote:
On 11-08-01 5:38 AM, Jannis wrote:
Dear R community,
I have a general question regarding indexing in multidiemensional
arrays.
Imagine I have a three dimensional array and I only want to extract
on
vector along a single dimension
Dear Contributors
thanks for any help you can provide. I searched the threads
but I could not find any query that satisfied my needs.
This is my database:
index time values
13732 27965 DATA.Q211.SUM.Index04/08/11 1.42
13733 27974 DATA.Q211.SUM.Index05/10/11 1.45
plot(1:10, pch=letters[1:10])
On Mon, Aug 1, 2011 at 4:44 AM, Andrew McCulloch amccu...@yahoo.co.ukwrote:
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The
points are
distinguished by a third variable which is categorical with 10 levels. I
have
been plotting x
Hi,
I'm trying to create an abbreviated data file from a larger version. I can
use the subset command to create a value for this data:
dat -subset(raw.data, select=c(SNP, Pvalue))
head (dat)
SNP Pvalue
1 rs11 0.6516
2 rs12 0.3311
3 rs13 0.5615
but when I try to write.table using:
Hello,
I am having a problem with the function matrix. Specifically, when I pass
three arguments (two more being instantiated in the function), I get the
following error message:
Error in matrix(0, 30, 10) :
5 arguments passed to .Internal(matrix) which requires 7
I looked into it, and
On Aug 1, 2011, at 5:01 AM, Przemek Jura wrote:
Dear group,
I experience some problems with gam() function after R update to
version 2.13.1
The function in both gam and mgcv packages stopped to work. Before,
with the same code I used, everything was fine.
Reports like this often turn
IMHO:
On Mon, Aug 1, 2011 at 7:51 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 11-08-01 5:44 AM, Andrew McCulloch wrote:
Hi,
I use R to draw my graphs. I have 100 points on a simple xy-plot. The
points are
distinguished by a third variable which is categorical with 10 levels. I
What do you think about this?
apply(data, 3, '[', indices)
On Mon, Aug 1, 2011 at 4:38 AM, Jannis bt_jan...@yahoo.de wrote:
Dear R community,
I have a general question regarding indexing in multidiemensional arrays.
Imagine I have a three dimensional array and I only want to extract on
Hi Margaux,
Check the row.names and col.names arguments of write.table.
See ?write.table
write.table (dat, file = /path/to/my/data.txt, sep = ,
col.names=FALSE, row.names=FALSE)
HTH,
Ivan
Le 8/1/2011 17:18, Margaux Keller a écrit :
Hi,
I'm trying to create an abbreviated data file from
Kindly do not attach questions in a separate document.
Install and read the documentation for the R rms package, and see handouts
at http://biostat.mc.vanderbilt.edu/rms
Frank
sytangping wrote:
Dear R users,
I am a new R user and something stops me when I try to write a academic
article.
Hi Guys,
working on a merge for 2 data frames.
Using the command:
x - merge(annotatedData, UCSCgenes, by.x=names,
by.y=Ensembl.Gene.ID, all.x=TRUE)
names and Ensembl.Gene.ID are columns with similar elements from the x
and y data frames.
annotatedData has 8909 entries, so has x(as expected).
Dear R users,
I am a new R user and something stops me when I try to write a academic
article. I want to make a nomogram to predict the risk of prostate cancer
(PCa) using several factors which have been selected from the Logistic
regression run under the SPSS. Always, a calibration plot is
Try this: had to add extra names to your data since it was not clear
how it was organized. Next time use 'dput' to enclose data.
x - read.table(textConnection( index time key date values
+ 13732 27965 DATA.Q211.SUM.Index04/08/11 1.42
+ 13733 27974 DATA.Q211.SUM.Index
What you see and what the data really is may be two different
things. You should have at least enclosed an 'str' of the two data
frames; even better would be a subset of the data using 'dput'. Most
likely your problem is that your data is not what you 'expect' it to
be.
On Mon, Aug 1, 2011 at
Dan,
If the variables you are merging by are character variables, there may be
subtle differences that you haven't noticed, e.g., capitalization or
spacing. You can look for differences by listing off the unique values:
table(c(annotatedData$names, UCSCgenes$Ensembl.Gene.ID))
Jean
`·.,,
On Aug 1, 2011, at 12:17 PM, world peace wrote:
Hi Guys,
working on a merge for 2 data frames.
Using the command:
x - merge(annotatedData, UCSCgenes, by.x=names,
by.y=Ensembl.Gene.ID, all.x=TRUE)
names and Ensembl.Gene.ID are columns with similar elements from the x
and y data frames.
I've only got a 20 minute layover, but three quick remarks:
1) Do a sanity check on your data size: if you want a million walks of a
thousand steps, that already gets you to a billion integers to store--even at a
very low bound of one byte each, thats already 1GB for the data and you still
the answer was indeed in subtle differences, and 'str' did help.
Problem is solved.
Thanks everybody for comments which was all very useful.
Best,
On Mon, Aug 1, 2011 at 12:25 PM, jim holtman jholt...@gmail.com wrote:
What you see and what the data really is may be two different
things. You
Hi Tina,
That is quite a bit of missingness, especially considering the sample
size is not large to begin with. This would make me treat *any*
result cautiously. That said, if you have a reasonable idea what the
mechanism causing the missingness is or if from additional variables
in your study,
Hello,
I am having a problem with the function matrix. Specifically, when I pass
three arguments (two more being instantiated in the function), I get the
following error message:
Error in matrix(0, 30, 10) :
5 arguments passed to .Internal(matrix) which requires 7
I looked into it, and
Dear all,
I have a very simple question.I have data frame of 50 columns and i want to
insert a column in 30th position.But i do not want to delete that column.Is it
possible to include a column in between, so that new values are in 30th column
and 30 th column is now 31st and 31st is
Robert,
What code did you run to get that error?
Do you get the error if the only code that you run is ...
matrix(0, 30, 10)
You gave three arguments to matrix, which requires none, but can take up
to five.
In the function matrix there is a call to .Internal(matrix) which requires
7
Y'know, you aren't likely to get many responses with this kind of request. Why
don't you go read the posting guidelines and come back with:
R version info
Sample data
Actual commands used, so we can reproduce the problem
---
Providing the data will help, but the first thing I noted is that you have more
columns (variables) than rows (cases). PCA will return a maximum of (the number
of columns) or (the number of rows-1) whichever is less. With 84 columns and 66
rows means you can get no more than 65 components. If
x - cbind(x[,1:29], newcolumn, x[,30:ncol(x)])
On Mon, Aug 1, 2011 at 12:59 PM, Bansal, Vikas vikas.ban...@kcl.ac.uk wrote:
Dear all,
I have a very simple question.I have data frame of 50 columns and i want to
insert a column in 30th position.But i do not want to delete that column.Is
it
Doesn't work -- you lose column names.
Try this instead:
yourframe[,30:51] - cbind( newcolumn,yourframe[,30:50])
Adjust column names after via:
names(yourframe) [30:51] - c(newcolname,names(yourframe[30:50])
Cheers,
Bert
On Mon, Aug 1, 2011 at 10:10 AM, Sarah Goslee sarah.gos...@gmail.com
Searching R Graphical Manual (http://www.oga-lab.net/RGM2/, mirror
http://www.oga-lab.net/RGM2/) shows possible candidates in packages circular
(windrose), IDPmisc (plot.rose), climatol (rosavent), openair (windRose),
and oce (as.windrose).
--
David L
Folks:
I consider my reply below rather clumsy: One has to keep track of
index numbers other than that which is inserted and must separately
change column names. Is there as essentially better way to do this,
either via base R or via an R package. I leave it to you to define
essentially better.
Since I didn't get an answer to this question, I'm rephrasing my question in
simpler terms:
I have a dataframe and I want to split it based on the levels of one of its
columns, and apply a function to each section of the data. Output of the
function may be drawing a plot, returning a value,
Not when I do it.
a - data.frame(A=1:10, B=11:20, D=31:40, E=41:50)
a
A B D E
1 1 11 31 41
2 2 12 32 42
3 3 13 33 43
4 4 14 34 44
5 5 15 35 45
6 6 16 36 46
7 7 17 37 47
8 8 18 38 48
9 9 19 39 49
10 10 20 40 50
b - cbind(a[,1:2], C=21:30, a[,3:4])
b
A B C D E
1
Bert,
On Mon, Aug 1, 2011 at 1:17 PM, Bert Gunter gunter.ber...@gene.com wrote:
Doesn't work -- you lose column names.
But I don't lose column names:
x - data.frame(A=1:3, B=1:3, C=1:3, D=1:3, E=1:3)
x
A B C D E
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
newcol - 4:6
cbind(x[,1:2], newcol,
Bert,
On Mon, Aug 1, 2011 at 1:27 PM, Bert Gunter gunter.ber...@gene.com wrote:
Folks:
I consider my reply below rather clumsy: One has to keep track of
index numbers other than that which is inserted and must separately
change column names. Is there as essentially better way to do this,
On Mon, Aug 1, 2011 at 1:43 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Bert,
On Mon, Aug 1, 2011 at 1:27 PM, Bert Gunter gunter.ber...@gene.com wrote:
Folks:
I consider my reply below rather clumsy: One has to keep track of
index numbers other than that which is inserted and must
Merik,
You did get an answer to the question, and it's even included in the material
below.
What doesn't work for you in Ista's suggestion?
id- c(1,1,1,1,1,2,2,2,3,3,3)
month - c(1, 1, 2, 3, 6, 2, 3, 6, 1, 3, 5)
value - c(10, 12, 11, 14, 16, 12, 10, 8, 14, 11, 15)
dat.tmp - data.frame(id,
Thanks Sarah and David.
Yes, but note this:
z - data.frame(a=1:2,b=3:4)
z
a b
1 1 3
2 2 4
newdat - 5:6
cbind(z[,1],newdat,z[,2])
newdat
[1,] 1 5 3
[2,] 2 6 4
cbind.data.frame(z[,1],newdat,z[,2])
z[, 1] newdat z[, 2]
1 1 53
2 2 64
dear all,
i have a quite simple question, i want to fill up a Matrix like done in the
following function,
but the performance is very bad for large dimensions
is there a way to do this like with apply or something similar?
makeMatrix - function(a, b,dim) {
X=matrix(0,ncol=dim,nrow=dim)
See the footer of this and every R-help message.
In particular, that DLL is not used by R itself, so this is probably
something called from a third-party package.
A number of packages used to use that DLL (which is rather out of
date), but no longer, so is your R actually current (the
Making use of the row() and col() functions speeds things up a bit.
makeMatrix2 - function(a, b, dim) {
X - matrix(NA, ncol=dim, nrow=dim)
X - exp( (-1*abs(row(X) - col(X)))/(3*b) )
diag(X) - a
X
}
system.time(makeMatrix(1, 2, 1000))
Yes, even if I only run the command matrix(0,30,10) I get the error. I am
running R with Ubuntu 10.10 (maverick) with R version:
R version 2.13.1 (2011-07-08)
When I check the function matrix, I can see that it is only passing five
arguments to the function .Internal() (shown below).
function
That's interesting. My function matrix() looks like this:
function (data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames = NULL)
{
if (is.object(data) || !is.atomic(data))
data - as.vector(data)
.Internal(matrix(data, nrow, ncol, byrow, dimnames, missing(nrow),
Most certainly you can speed it up:
X - exp(-abs(row(X) - col(X)) / (3*b))
diag(X) - a
should do what you want. This is called
'vectorization' and is discussed lots of
places -- for instance, in the two documents
mentioned below in my signature.
On 01/08/2011 19:12, monk wrote:
dear all,
Actually Sara's method fails if the insertion is after the first or before
the last column:
x - data.frame(A=1:3, B=1:3, C=1:3, D=1:3, E=1:3)
newcol - 4:6
cbind(x[,1], newcol, x[,2:ncol(x)])
x[, 1] newcol B C D E
1 1 4 1 1 1 1
2 2 5 2 2 2 2
3 3 6 3 3 3 3
Hello!
I am trying to identify which ones of a vector of dates are US
holidays. And, ideally, which is which. And I do not know (a-priori)
which dates those should be.
I have, for example:
x-seq(as.Date(2011-01-01),as.Date(2011-12-31),by=day)
(x)
I think chron should help me here - but maybe I
bjmjarrett wrote:
...
rate - function(x){
storage - matrix(nrow=length(x),ncol=1)
ifelse(length(x)==1,storage[1,] - NA,{
storage[1,] - x[1]/max(x)
for(i in 2:length(x)){
p - i-1
storage[i,] - ((x[i] - x[p]) / max(x))
}
On 7/31/2011 6:24 PM, Alexandre Aguiar wrote:
Em Domingo 31 Julho 2011, você escreveu:
My memory is that this question gets asked every few months and one of
the stock answers is to use the function 'package.skeleton' in the
utils package as a starting point.
Got that from docs. And actually
Hi Everyone,
When i try to install a package using
install.packages(agricolae)
--- Please select a CRAN mirror for use in this session ---
|
The cursor keeps blinking i dont get a popup menu to choose a CRAN mirror? Is
it due to my proxy server settings? I tried to echo $http_proxy ,
thanks a lot , that will do the trick
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I have a function to calculate the rate of increase (the difference between
the value and the previous value divided by the total number of eggs in a
year) of egg production over the course of a year:
rate - function(x){
storage - matrix(nrow=length(x),ncol=1)
storage[1,] - x[1] /
It looks like I had been missing an update needed for Ubuntu systems. All I
needed was the following. Thank you.
update.packages(lib.loc = /usr/local/lib/R/site-library)
On Mon, Aug 1, 2011 at 2:39 PM, Jean V Adams jvad...@usgs.gov wrote:
That's interesting. My function matrix() looks like
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