Dear R help group
I've been working on this plot for a while now and now getting around to the
minor adjusments. I would like to be able to put a border and background
fill around the legend in this plot.
I understand the legend 'bty' should do this have this capablity but not
sure how the
- This mail is in HTML. Some elements may be ommited in plain text. -
Meu nome é Frank P, eu trabalhei com o Banco SNS em Holland.I habe Você
contactado por conta de parente falecido com os nossos Banj e que foi deixado
sem pretensões.
Por favor, me envia um email com um número de telefone e
Andra Isan andra_isan at yahoo.com writes:
Hi All,
In order to compare two different logistic regressions,
I think I need to compare them based on their BIC
values, but I am not sure if the smaller BIC would mean a better
model or the reverse is true?
Thanks a lot,Andra
Smaller
Hi
Re: [R] two question about plot
The help for boxplot offers suggestions for both those things. You may
be
particularly interested in:
names: group labels which will be printed under each boxplot. Can
be a character vector or an expression (see plotmath).
add:
On 09/01/2011 09:02 PM, baptiste auguie wrote:
Hi,
Are you after this?
last_plot() + opts(aspect.ratio=1)
Even better (I once got correct by Hadley for using aspect.ratio, but
this was plotting spatial data...)
last_plot() + coord_equal()
cheers,
Paul
Also, see
Dear Dennis,
I wouldl ike to thank you for your answer.
That did it work.
how can I know remove the space between the axis labels and the image (blue-red
area)?
Right now things look like are floating a bit. I want the x,y labels to reduce
the overall gap.
I would like to thank you in advance
Thank you both for your replies. I've tried it with a small sample of the
data and it works perfectly. I have no idea yet how it works but I will
spend some time to figure it out.
Thank you!
Thomas
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Hello,
I am having the following error while calling an R script through PHP.
/usr/local/bin/R: line 227: /kk/Programs/R-2.13.0/etc/ldpaths: Permission
denied
ERROR: R_HOME ('/kk/Programs/R-2.13.0') not found
I had compiled R from source and placed the generated R shell script in
/usr/local/bin.
Estimation is realized by MLE, estimators are asymptotically normal
Try this reg-vglm(...)
p.value-1-pnorm(abs(coef(reg)/sqrt(diag(vcov(reg)
Justin BEM
BP 1917 Yaoundé
Tél (237) 76043774
De : suuz suuz_b...@hotmail.com
À : r-help@r-project.org
Envoyé le
Dear R-users
I'm running a maximum likelihood procedure using the spg package. I'd like
to save some output produced in each iteration to a file, but if I put the
capture.output() within the function I get the following message; Error in
spg(par = startval, fn = loglik, gr = NULL, method = 3,
Dear mailing list,
I would like to use the optim() command in order to maximize the logged
likelihood of the following function, where p is the parameter of interest and
should be constrained between 0 and positive infinity.
y = 1/2 * ((te - x)/(te - tc))^p
x and y are given by
x - c(5.18,
Dear R users and hydrological/environmental community,
I'm glad to announce that a major (and recommended) update for the
packages hydroTSM and hydroGOF are now available on CRAN:
-) hydroTSM: http://cran.r-project.org/web/packages/hydroTSM/
-) hydroGOF:
Thanks for your response, that does work, however, it is still not quite what
want. I would like to tell betareg what the mean is (in my case, 0.5) and
force it to use that value. Is this possible?
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On 11-09-02 5:24 AM, Kristian Lind wrote:
Dear R-users
I'm running a maximum likelihood procedure using the spg package. I'd like
to save some output produced in each iteration to a file, but if I put the
capture.output() within the function I get the following message; Error in
spg(par =
i would suggest that if you want to use R that you get a 64-bit version with
24GB of memory to start. if your data is a numeric matrix, you will need 8GB
for a single copy.
Do you really need it all in memory at once, or can you partition the problem?
Can you use a database to access the
Yes, that's true. It works as it's supposed to now.
Thank you.
Kristian Lind
2011/9/2 Duncan Murdoch murdoch.dun...@gmail.com
On 11-09-02 5:24 AM, Kristian Lind wrote:
Dear R-users
I'm running a maximum likelihood procedure using the spg package. I'd like
to save some output produced in
Hi All!
Please find code and the respective lists below. My problem: I specify the
case that lilwin[[p]] is not an NA and want the code found in iwish to be
returned ONLY for that case. Why do I get a list of length 2 (and why is
NULL the first element)? I understand that the code below is
Hi:
Try this:
key1 -
draw.key(list(text=list(levels(Cal_dat$Commodity)),
title=Ore type,
border = TRUE,
background = 'ivory',
points=list(pch=22, cex=1.3, fill=col.pat, col=black)),
draw = FALSE)
key2 -
Dear all,
I would like to know how to merge columns like:
Input file:
V1 V2 V3 V4 V5 V6
1 G A G G G G
2 A A G A A G
Desired output file:
V1 V2 V3
1 G/A G/G G/G
2 A/A G/A A/G
So for every 2 consecutive columns merge their content into one.
Thanks in advance.
Anna Dunietz wrote on 09/02/2011 07:16:45 AM:
Hi All!
Please find code and the respective lists below. My problem: I specify
the
case that lilwin[[p]] is not an NA and want the code found in iwish to
be
returned ONLY for that case. Why do I get a list of length 2 (and why
is
NULL the
I believe when using BIC one needs to compare nested models, i.e. , when
comparing models A and B one must make sure that model A contains all the
parameters of model B and additionally A contains one or more extra parameter
beyond those in B. Further the comparison of BICs requires that models
Hi:
Here's one approach:
d - read.table(textConnection(
V1 V2 V3 V4 V5 V6
1 G A G G G G
2 A A G A A G), header = TRUE, stringsAsFactors = FALSE)
closeAllConnections()
# Create two vectors of variable names, one for odd numbered,
# one for even numbered
vars1 -
On Sep 2, 2011, at 4:14 AM, thomas.chesney wrote:
Thank you both for your replies. I've tried it with a small sample
of the
data and it works perfectly. I have no idea yet how it works but I
will
spend some time to figure it out.
When you get around to putting in the time to figure it
Thank you for your reply, it has been helpful.
Do you know if the parameters estimators are MLE estimators?
One more question:
In my case study I have failures that occured on different objects that have
different age and length, could I use weight to find the estimates of a
weibull law and so
Dear all,
I'm trying to do a multivariate local regression whit the locfit instruction,
such as:Y=m(Z,X,W)+u
However, I have a problem at the moment of calculte the bandwith of the
regression. If I had a univarite local regression model I could use the
instruction regband. However,
Hello!
I am working on a manuscript on sexual dimorphism in an aquatic
invertebrate, where we have estimated sexual dimorphism (SD) for 7 different
traits in four populations (a total of 28 SD-estimates). We have used the
following formula for estimating SD: 100 * (mean male trait value - mean
On Sep 2, 2011, at 8:34 AM, Joao Fadista wrote:
Dear all,
I would like to know how to merge columns like:
Input file:
V1 V2 V3 V4 V5 V6
1 G A G G G G
2 A A G A A G
Looked like an mapply-type problem:
with(dat,
mapply(paste,
list(V1, V3, V5),
On Sep 2, 2011, at 9:30 AM, David Winsemius wrote:
On Sep 2, 2011, at 8:34 AM, Joao Fadista wrote:
Dear all,
I would like to know how to merge columns like:
Input file:
V1 V2 V3 V4 V5 V6
1 G A G G G G
2 A A G A A G
Looked like an mapply-type problem:
with(dat,
I think the following pdf will explain the details of how to use the
optim function.
http://www.unc.edu/~monogan/computing/r/MLE_in_R.pdf
Hope that helps,
Sam
On Fri, Sep 2, 2011 at 7:06 AM, Thiem Alrik th...@sipo.gess.ethz.ch wrote:
Dear mailing list,
I would like to use the optim()
really thxs to David Winsemius..
this websits helps a lot,
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R-help@r-project.org
I have a mixed model with a binomial response, four factor variables and one
random factor.
m1=glmer(nbhf.hour~Season+Diel+Tidal.phase+Tidal.cycle+(1|POD.ID),family=binomial,data
=bl1,control=list(msVerbose=TRUE))
I have really need to try and find a post hoc test for this model and
finding the
ths a lot, david. it helps a lot
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On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information
criteria do not.
--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University
On 2011-09-02 05:16, Anna Dunietz wrote:
Hi All!
Please find code and the respective lists below. My problem: I specify the
case that lilwin[[p]] is not an NA and want the code found in iwish to be
returned ONLY for that case. Why do I get a list of length 2 (and why is
NULL the first
On Fri, 2 Sep 2011, Patrick Breheny wrote:
On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Actually, this is off-topic on this list. But blanket
Inline:
On Fri, Sep 2, 2011 at 8:09 AM, Patrick Breheny patrick.breh...@uky.edu wrote:
On 09/02/2011 08:48 AM, John Sorkin wrote:
I believe when using BIC one needs to compare nested models
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Yes, indeed.
Hi there,
I'm trying to apply the Mann Kendall test for trend analysis of a time
series. I have downloaded and installed the package Kendall and
subsequently loaded it into the software.
My time series is a .txt file with 2 columns - column 1 is the year (1985 -
2009) and column 2 is the
Hello,
I'm attempting to return the date (in form '%Y-%m-%d') of the Monday
previous to the current date. For example: since it is 2011-09-02 today, I
would expect 2011-08-29 to be the return value.
I found the following in:
http://www.mail-archive.com/r-help@r-project.org/msg144184.html
Start
On 09/01/2011 10:13 PM, Worik R wrote:
I am starting on a (section of the) project where I need to build a matrix
with on the order of 5 million rows and 200 columns
I am wondering if I can stay in R.
I need to do rollapply type operations on the columns, including some that
will be
On 09/02/2011 11:26 AM, Prof Brian Ripley wrote:
This is wrong. Hypothesis tests rely on nested models; information criteria
do not.
Actually, this is off-topic on this list. But blanket statements are
often themselves untrue: there are hypothesis tests of non-nested
models (most famously due
I don't have the Kendall package at my fingers, but it seems like you have
some deeper trouble with R syntax if you are writing things like
MannKendall(1:27(data),data)
when you know that MannKendall only takes one argument.
Can you verify that your data object actually has a full set of data
On Sep 2, 2011, at 10:35 AM, Ben qant wrote:
Hello,
I'm attempting to return the date (in form '%Y-%m-%d') of the Monday
previous to the current date. For example: since it is 2011-09-02 today, I
would expect 2011-08-29 to be the return value.
I found the following in:
Michael,
Cheers for the input - still learning syntax as I go - very new to the use of
programming language and having to self teach as part of my PhD.
The input data was fine - it was the call on the R website which was confusing
me - by defining the location of the actual data in the second
I didn't sort out the issue in my email below but here is a (not very
R'ish) solution:
pm = function(x) {
+ for(i in 1:7){
+ if(format(as.Date(Sys.Date()-i),'%w') == 1){
+ d = Sys.Date() - i;
+ }
+ }
+ d
+ }
pm(Sys.Date())
[1] 2011-08-29
On Fri, Sep 2, 2011 at 9:35 AM, Ben
Oh OK, missed that.
Here is a solution using base: (already posted)
I didn't sort out the issue in my email below but here is a (not very
R'ish) solution:
pm = function(x) {
+ for(i in 1:7){
+ if(format(as.Date(Sys.Date()-
i),'%w') == 1){
+ d = Sys.Date() - i;
+ }
+ }
+ d
I'd suggest that you look into using a time series class to organize your
data rather than just keeping levels and times next to each other. This will
also make it alot easier to work with your data in specific time-series sort
of ways.
I really like the xts class and you can get to it with
On Fri, Sep 2, 2011 at 9:59 AM, Marc Schwartz marc_schwa...@me.com wrote:
On Sep 2, 2011, at 10:35 AM, Ben qant wrote:
Hello,
I'm attempting to return the date (in form '%Y-%m-%d') of the Monday
previous to the current date. For example: since it is 2011-09-02 today,
I
would expect
Hello,
I need to calculate a moving average and an exponentially weighted moving
average over a fairly large data set (500K rows).
Doing this in a for loop works nicely, but is slow.
ewma - data$col[1]
N - dim(data)[1]
for(i in 2:N){
data$ewma - alpha * data$ewma[i-1] + (1-alpha) *
Have you looked at SMA/EMA from the TTR package? That's a pretty quick
implementation.
runmean from caTools is even better for the SMA but I don't think there's an
easy way to turn that into an EWMA.
Hope this helps,
Michael Weylandt
On Fri, Sep 2, 2011 at 12:43 PM, Noah Silverman
Dear All,
I will be confronted (relatively soon) with the following problem:
given a set of known statistical indicators {s_i} , i=1,2...N for a N
countries I would like to be able to do some data clustering i.e.
determining the best way to partition the N countries according to their
known
On Fri, Sep 2, 2011 at 11:47 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
Have you looked at SMA/EMA from the TTR package? That's a pretty quick
implementation.
runmean from caTools is even better for the SMA but I don't think there's an
easy way to turn that into an EWMA.
SMA
Joshua,
Thanks for the tip.
I need to roll my own code on this. But perhaps I can borrow some code from
the package you mentioned.
Is the package just performing the loop, but in a faster language?
--
Noah Silverman
UCLA Department of Statistics
8117 Math Sciences Building #8208
Los
There is a recent blog post by Dirk Eddelbeutel on how to do something
similar using his Rcpp package and C++, with massive time improvements.
http://dirk.eddelbuettel.com/blog/
On 9/2/2011 12:43 PM, Noah Silverman wrote:
Hello,
I need to calculate a moving average and an exponentially
On Fri, Sep 2, 2011 at 12:06 PM, Noah Silverman noahsilver...@ucla.edu wrote:
Joshua,
Thanks for the tip.
I need to roll my own code on this. But perhaps I can borrow some code
from the package you mentioned.
Is the package just performing the loop, but in a faster language?
As I said,
Along the lines of one of Jim's suggestions, if you have some
basic MySQL knowledge check out the RMySQL package. I use it to
convert / partition a matrix similar to yours to R objects and it
works fine.
Hope this helps,
A.
On Fri, 2 Sep 2011 06:33:13 -0400
Jim Holtman jholt...@gmail.com
Look at the function daisy in the package cluster.
require(cluster)
?daisy
Jean
Lorenzo Isella wrote on 09/02/2011 11:50:04 AM:
Dear All,
I will be confronted (relatively soon) with the following problem:
given a set of known statistical indicators {s_i} , i=1,2...N for a N
countries I
Thanks Joshua,
I really like the example given in the blog post that Abhijit pointed me to.
Doing it in C++ using the Inline seems like an easy way to get a massive
improvement in speed without the hassle of writing a package.
I'm working on coding that now.
--
Noah Silverman
UCLA Department
On Sep 1, 2011, at 21:11 , R. Michael Weylandt wrote:
Dropping all occurences of a factor does not drop that level. This actually
turns out to be much more useful than it first might appear, but if you
really need to get around it, it can be done.
...most expediently by using factor(), as
The 'filter' function should be able
to do what you want efficiently.
On 02/09/2011 18:06, Noah Silverman wrote:
Joshua,
Thanks for the tip.
I need to roll my own code on this. But perhaps I can borrow some code from
the package you mentioned.
Is the package just performing the loop, but
Dear everyone,
I am new to R, and I am looking at doing text classification on a huge
collection of documents (500,000) which are distributed among 300 classes
(so basically, this is my training data). Would someone please be kind
enough to let me know about the R packages to use and their
Greetings -
I am working on some data that contain chemical names with air concentrations,
and I am creating a data frame with date/time and each chemical having its own
column. However, these are organic chemicals (e.g. 1-butene,
2,3,4-trimethylbenzene etc). The package I am going to be using
On Sep 2, 2011, at 3:13 PM, Durant, James T. (ATSDR/DTEM/PRMSB) wrote:
Greetings -
I am working on some data that contain chemical names with air
concentrations, and I am creating a data frame with date/time and
each chemical having its own column. However, these are organic
chemicals
?make.names perhaps.
On Fri, Sep 2, 2011 at 4:13 PM, Durant, James T. (ATSDR/DTEM/PRMSB)
h...@cdc.gov wrote:
Greetings -
I am working on some data that contain chemical names with air
concentrations, and I am creating a data frame with date/time and each
chemical having its own column.
possibly with unique = TRUE:
make.names(c(', '))
[1] X. X.
make.names(c(', '), unique=TRUE)
[1] X. X..1
Spencer
On 9/2/2011 12:28 PM, Gustavo Carvalho wrote:
?make.names perhaps.
On Fri, Sep 2, 2011 at 4:13 PM, Durant, James T. (ATSDR/DTEM/PRMSB)
h...@cdc.gov wrote:
Hello
Please see the example below
class(testX)
[1] matrix
class(testX[1,])
[1] numeric
Why not matrix? What am I missing here? Is there a way to keep the same
class?
The reason for the question is that I want to implement a k-step ahead
prediction for my own routines
Try this:
predict(fit10,testX[1,,drop = FALSE])
On Fri, Sep 2, 2011 at 5:05 PM, Eduardo M. A. M.Mendes
emammen...@gmail.com wrote:
Hello
Please see the example below
class(testX)
[1] matrix
class(testX[1,])
[1] numeric
Why not matrix? What am I missing here? Is there a way
On Sep 2, 2011, at 3:05 PM, Eduardo M. A. M.Mendes wrote:
Hello
Please see the example below
class(testX)
[1] matrix
class(testX[1,])
[1] numeric
Why not matrix? What am I missing here? Is there a way to keep the same
class?
The reason for the
Many thanks to all for the reply.
I do apologize for bothering the list with a FAQ but I have to confess that,
although I read Faq in the past, I did not remember to do it again.
Cheers
Ed
-Original Message-
From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Friday, September
Hello, I have a dataframe that looks like this:
a b NA Honduras China NA NA Sudan Japan NA NA Mexico NA Mexico
I would like to replace the NA values in column b with the non-NA values in
column a. I have tried a number of techniques, (if, ifelse) but I must have
the logic wrong.
Thanks
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using optim method L-BFGS-B to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
Dear R users,
When I Save Workspace... and then reopen it, my platform switches
from 64-bit to 32-bit, i.e. the Gui switches between these:
R version 2.13.1 (2011-07-08)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Hi users
I'm student who is struggling with basic R programming. Would you please
help me with this problem.
My english is bad I hope that my question is clear:
I have a matrix in wich there are two colmns( yp, yt)
Yp: predicted values from my model.
yt: true values ( my dependante variable y is
Dear all,
I'm new to frailty model. I have a question on the output from 'survival'
pack. Below is the output. What does gamma1,2,3 refer to? How do I
calculate joint hazard function or marginal hazard function using info
below? Many thanks!
Call:
coxph(formula = surv ~ as.factor(tibia) +
Your data frame didn't come across legibly, try sending it in plain text
using the dput() command.
That said, I'd guess you want something like this:
d[is.na(d$a),a] - d[is.na(d$b),b]
The idea is that is.na(d$a) selects only those rows where column a is NA
and then moves b values into a for
On Sep 2, 2011, at 3:51 PM, Josh Tewksbury wrote:
Hello, I have a dataframe that looks like this:
a b NA Honduras China NA NA Sudan Japan NA NA Mexico NA Mexico
I would like to replace the NA values in column b with the non-NA
values in
column a. I have tried a number of techniques,
Hello All,
I have used logistic regression glm in R and I am evaluating two models both
learned with glm but with different predictors. model1 - glm (Y ~ x4+ x5+ x6+
x7, data = dat, family = binomial(link=logit))model2 - glm (Y~ x1 + x2 +x3 ,
data = dat, family = binomial(link=logit))
and I
You don't say how you re-open it, but I would guess you are double-clicking the
.RData file and letting Windows look up which program to run. If so, you can
either open the 64-bit GUI directly and use File Open Data to open your data,
or you can change which version of the R GUI program is
On Sep 2, 2011, at 7:51 PM, R. Michael Weylandt wrote:
Your data frame didn't come across legibly, try sending it in plain
text
using the dput() command.
That said, I'd guess you want something like this:
d[is.na(d$a),a] - d[is.na(d$b),b]
One of the rare instances where I disagree with
Y'all are both absolutely right -- I just skimmed the problem and wrote an
answer all too briefly that both didn't do what was asked and did what it
did instead entirely incorrectly. The syntax suggested by David's email of
7:52 is what I meant to get at...
My apologies for any confusion this
hi users
I have a data frame in with there are two colomns real values and predicted
ones (for a dichotomic response).
How can i obtain a confusion matrix (miscalssification rat and errors)?
The costs are egal.
Thanks
--
View this message in context:
Dear R experts.
I might be missing something obvious. I have been trying to fix this problem
for some weeks. Please help.
#data
ped - c(rep(1, 4), rep(2, 3), rep(3, 3))
y - rnorm(10, 8, 2)
# variable set 1
M1a - sample (c(1, 2,3), 10, replace= T)
M1b - sample (c(1, 2,3), 10, replace= T)
M1aP1 -
On Sep 2, 2011, at 11:18 PM, Maya Joshi wrote:
Dear R experts.
I might be missing something obvious. I have been trying to fix this
problem
for some weeks. Please help.
#data
ped - c(rep(1, 4), rep(2, 3), rep(3, 3))
y - rnorm(10, 8, 2)
# variable set 1
M1a - sample (c(1, 2,3), 10,
On Fri, 2 Sep 2011, Jeff Newmiller wrote:
You don't say how you re-open it, but I would guess you are
double-clicking the .RData file and letting Windows look up which
program to run. If so, you can either open the 64-bit GUI directly
and use File Open Data to open your data, or you can
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