"Don't do that."
Use $ notation to refer to elements of lists/data frames explicitly.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Hello,
I am using Dr. Harrell's design package to make a nomogram. I was able to
make a beautiful one. However, I want to change 5-year survival probability
to 5-year failure probability.
I couldn’t get hazard rate from Hazard(f1) because I used cph for the model.
Here is my code:
f1 <- cph
sorry the code for nomogram
par(cex=0.8)
nom<- nomogram(f1, conf.int=F,
fun=list(surv5, surv10), funlabel=c('5-Year Survival Probability', '10-Year
Survival Probability' ), lp=F,
fun.at=c(at.surv, at.surv),label.every=1, force.label=FALSE, cex.axis=0.8,
verbose=TRUE, cex.var=0.8)
Thanks
Min
It is from attatching and detaching data frames and moving them between
functions, there is no library calls in the loop
Michael Weylandt wrote
>
> What does your code look like that you see such messages repeatedly? I
> only see those message upon loading a package: maybe move the
> library()
Hi
Please let me know if we can store the linear model object in the data
base and retrive the object and output from them
Data<- read.csv("C:/FE and RE.csv")
Formula="Y~X2+X3+X4
lmobject = lm(formula=Formula,data=Data)
can i store the lm object in the database and and is it possible to retri
What does your error distribution look like? Have you tried any time series
analysis on your data? Your information is too little for us to give any
concrete directions.
Regards,
Indrajit
From: rch4
To: r-help@r-project.org
Sent: Wednesday, November 30, 201
Agreed, nobody has deemed it "default", but is there any other such package
that you can think of for this purpose.
Regards,
From: Uwe Ligges
Cc: "r-help@r-project.org" ; rch4
Sent: Tuesday, November 29, 2011 7:33 PM
Subject: Re: [R] Negative exponential f
Dear All,
I have one questions about *Matrix U* which using to obtain copula
parameter. For example when *Matrix U* (*n-*by-2) contains data
transformed to the unit hypercube by parametric estimates of their marginal
cumulative distribution functions.
Did R have build function to obtain this *Ma
What does your code look like that you see such messages repeatedly? I
only see those message upon loading a package: maybe move the
library() call outside the loop.
Also, the suppressPackageStartupMessages() function can hide them, but
you should first figure out for yourself why they are there.
Hello,
I'm trying to run a mlogit regression on my data, and have been unsuccessful
so far. The data I am working with consist of many observations of how
people react when given a certain number. I have just 2 data points per
observation: a number (there are ~300 different possible numbers) and t
So I am receiving messages in a loop
"The following object(s) are masked from..."
I tried
triedoptions(warn=-1) and options(show.error.message:=FALSE) but it still
tells me
"The following object(s) are masked from..."
repeatedly, and my problem is that it masks the print statement from my
fu
Thank you for your suggestions. We are currently working with an the nls
function, but are having trouble getting a curve that fits our data. We have
tried using the following formula:
mod1 = nls(time~a*exp(b*(date)), start=list(a=4.178,b=-1.18))
av=mdate
bv=predict(mod1,list(mdate=av))
length(av)
sirilkt hotmail.com> writes:
>
> Hi All,
>
> When we run the command : summary ( newmod<-gam(Dlq~ formula,family,,data) )
>
> in R, the output would the effect of smoothness in R.
>
> As of now to calculate the probability I am following the below approach:
>
> 1) Run the plot of the GAM
Oh, and your message:
In log(LikeGi(l, i, par[1], par[2])) : NaNs produced
means your LikeGi is returning something negative. Can't take the log of it...
On Tue, Nov 29, 2011 at 8:09 PM, Florent D. wrote:
> I also think your last write-up for LogLiketot (using a single
> argument "par") is the
I also think your last write-up for LogLiketot (using a single
argument "par") is the correct approach if you want to feed it to
optim().
So now you have a problem with log(LikeGi(l, i, par[1], par[2])) for
some values of par[1] and par[2].
Where is LikeGi coming from? a package or is it your ow
Hi,
IMHO, if you will be focussing mostly in incorporating C/C++ in your
own packages you build, instead of wanting to go into other mixed R/C
packages to change them up, I'd go with C++ together with the Rcpp
package.
Rcpp makes lots of things easier on your end, that having been said,
you shoul
Dear List,
Id like to modify the R rpart package source code to add a new split
criterion. Im familiar with R, but not at all with C or C++.
I understand C and C++ are quite different, and I dont have the time to
learn both, so my questions are (i) which one should I learn for the
specific tas
On Tue, Nov 29, 2011 at 6:55 PM, Gyanendra Pokharel
wrote:
> yes, log of negative number is undefined and R also do the same and produces
> NaNs. Here I want to reject the value of exp(-alpha*d^(-beta)) when greater
> than 1, and want to run the loop otherwise.
> Thanks
Then just add another if()
yes, log of negative number is undefined and R also do the same and
produces NaNs. Here I want to reject the value of exp(-alpha*d^(-beta))
when greater than 1, and want to run the loop otherwise.
Thanks
On Tue, Nov 29, 2011 at 6:48 PM, Sarah Goslee wrote:
> > Here p[i] <- 1 - exp(-alpha*d^(-beta
> Here p[i] <- 1 - exp(-alpha*d^(-beta))> so, log(p[i]) produces NaNs when
> exp(-alpha*d^(-beta)) is greater than 1.> How can I remove it.After
> generating the out put we can omit it, but the> problem is different.
Wait... you're complaining that you can't take the natural log of a negative
n
I have following code:
loglikelihood <- function(alpha,beta= 0.1){
loglh<-0
d<-0
p<-0
k<-NULL
data<-read.table("epidemic.txt",header = TRUE)
attach(data, warn.conflicts = F)
k <-which(inftime==1)
d <- (sqrt((x-x[k])^2+(y-y[k])^2))^(-beta)
p<-1 - exp(-alpha*d)
Please convert to rms. Design is no longer supported. See
http://biostat.mc.vanderbilt.edu/Rrms
Frank
min wrote
>
> sorry the code for nomogram
>
> par(cex=0.8)
> nom<- nomogram(f1, conf.int=F,
> fun=list(surv5, surv10), funlabel=c('5-Year Survival Probability',
> '10-Year Survival Probabilit
Dear R experts---I think I need to figure out how to stop in my error
function without triggering an error again. so, I think I need the
equivalent of C's exit(0) call. Here is what I mean:
$ R CMD BATCH die.R
and die.R is
# in my .Rprofile, but for now in die.R
options(error=function(e) print
Forgot the blog
http://sas-and-r.blogspot.com/
M
On Tue, Nov 29, 2011 at 5:51 PM, R. Michael Weylandt
wrote:
> Also, this blog has implementations of many common subroutines in both
> SAS and R so you can learn "translations" between the two.
>
> Hope it helps,
>
> Michael
>
> On Tue, Nov 2
Also, this blog has implementations of many common subroutines in both
SAS and R so you can learn "translations" between the two.
Hope it helps,
Michael
On Tue, Nov 29, 2011 at 5:35 PM, R. Michael Weylandt
wrote:
> Try something like this:
>
> Suppose your data object is called df:
>
> cbind(pa
Dear R Community,
I'm pleased to announce the availability of 'colorout' on CRAN.
The package colorizes R output when it is running on a terminal
emulator.
Screenshot: http://www.lepem.ufc.br/jaa/R_color_output.png
You can use either 16 colors (8 colors * 2 degrees of intensity)
or 256 colors to
Try something like this:
Suppose your data object is called df:
cbind(paste("sd", 1:NROW(df), sep = ""), df)
If you are just moving from SAS check out the website (and books
associated with) http://r4stats.com -- it's got some tutorials on
moving from other platforms to R.
Michael
On Tue, Nov
rch4 geneseo.edu> writes:
>
> We need help
>
> We are doing a project for a statistical class in and we are looking at
> world record times in different running events over time. We are trying to
> fit the data with a negative exponential but we just cant seem to get a
> function that works
Extension to the complex plane?
-- Bert
Hint: What does the graph of y = log(x) look like?
On Tue, Nov 29, 2011 at 2:02 PM, Gyanendra Pokharel
wrote:
> Hi all I have a function of log defined by y = log(1- exp(-a)), when
> exp(-a) is greater, 1, it produce NaN. How can I remove this in R?
>
>
Marianne Stephan hotmail.com> writes:
> Dear R and statistics experts:
>
> I have data of a behavioral experiment with the aim to
> investigate the effect of a memory task on motor
> learning.
>
> Question:
> I would appreciate help in figuring out a
> possible formula to determine whether m
Hi @ all and sorry for the confusion,
my problem is solved. I also know a corellation is - it was only a
formatting and describing problem. Grant gave me the right advice with the
kruskal-wallis!
Thanks
GeOphagus
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-for-correlatio
I am needing to create a new identifier variable for a data set which has no
ID variable in the original file. I am basically wanting to take the count
of each row and add "sd" in front of it - so it would look like so
sd1
sd2
sd3
sd4
sd5 etc...
I have no idea how to do this. I am a SAS user
I managed to solve the problem myself without using this code.
thx
2011-11-24 12:26 keltezéssel, Kehl Dániel írta:
> Dear Community,
>
> I am trying to write code for the following problem.
> Lets assume we have a beta distribution.
> I know one quantile, lets say, 10% of the mass lies above .8,
Do you mean remove the NaNs? Try na.omit() or complete.cases() or many
other options.
If you mean you want the complex log, try
log(as.complex(1-exp(-a)))
Michael
On Tue, Nov 29, 2011 at 5:02 PM, Gyanendra Pokharel
wrote:
> Hi all I have a function of log defined by y = log(1- exp(-a)), when
>
Hi all I have a function of log defined by y = log(1- exp(-a)), when
exp(-a) is greater, 1, it produce NaN. How can I remove this in R?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listin
On Tue, 29 Nov 2011, Rich Shepard wrote:
Pointers on how to determine why this one variable has some values and
characters rather than as numerics are needed.
Joshua, Marc, David, Bill, Sarah, Bert, et al.:
Thank you all for the insights and ideas. It was a valuable lesson and it
helped me
On Thu, 2011-11-24 at 12:16 +, Edwin Lebrija Trejos wrote:
> Hi, First I should note I am relatively new to R so I would appreciate
> answers that take this into account.
>
> I am trying to perform an MDS ordination using the function “metaMDS”
> of the “vegan” package. I want to ordinate spe
On Thu, 2011-11-24 at 07:57 -0800, B77S wrote:
> Try the daisy() function from the package "cluster", it seems to be able to
> handle NAs and non-dummy coded character variables
>
> metaMDS(daisy(df, metric="gower"))
That won't help the OP as the species scores (the species data, i.e. the
traits
On Wed, 2011-11-23 at 16:02 -0300, Alejo C.S. wrote:
> Dear List,
>
> I can'f figure how to add point labels in the next plot (example from
> ?taxondive help page):
>
> library(vegan)
> data(dune)
> data(dune.taxon)
> taxdis <- taxa2dist(dune.taxon, varstep=TRUE)
> mod <- taxondive(dune, taxdis)
Dear R and statistics experts:
I have data of a behavioral experiment with the aim to investigate the effect
of a memory task on motor learning.
Question:
I would appreciate help in figuring out a possible formula to determine whether
motor learning across sessions differs between 2 groups.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Hasan Diwan
> Sent: Tuesday, November 29, 2011 10:16 AM
> To: Prof Brian Ripley
> Cc: R Project Help
> Subject: Re: [R] Weird Excel Time Format
>
> On 29 November 2011 09:32, Pro
look at just your data that is in that first id category and I bet you can
figure it out!
> myData[myData$id=='0m11',]
var1 var2 id
10 30.79 32.15 0m11
11 30.79 32.39 0m11
12 30.94NA 0m11
aggregate performs the na.rm step on the entire row thus, a mean of 30.79.
data.table and plyr pe
The semantics for na.rm are different for aggregate than for the other options.
The former removes any rows that contain an NA prior to performing the
computation, the latter methods work column-wise. You have to decide which is
correct for your purposes.
---
I am calculating the mean of each column grouped by the variable 'id'.
I do this using aggregate, data.table, and plyr. My aggregate results
do not match the other two, and I am trying to figure out what is
incorrect with my syntax. Any suggestions? Thanks.
Here is the data.
myData <- structure(l
By any chance is Kalem accessing the Internet through a proxy server which
requires a userid and password. There're are instructions in the Windows
FAQ on how to deal with this. He may need to obtain some information from
his IT department. Alternatively he could download the zip file and
install
without knowing much about your data or the base plotting...
I'd use the library ggplot2.
First, you'll need to format your dates to POSIXct
AggData$time <- as.POSIXct(AggData$time,format='%H:%M')
Then plotting is trivial.
ggplot(AggData,aes(x=time,y=value))+geom_points()
or +geom_line() if yo
You can see what the offending strings are with
> with(waterchem, levels(SC)[is.na(as.numeric(levels(SC)))])
[1] "-" "+"
Warning message:
In eval(expr, envir, enclos) : NAs introduced by coercion
but it may be easiest to use the colClasses argument to read.table
to force that column to be n
On Nov 29, 2011, at 1:18 PM, Rich Shepard wrote:
> I have a data frame with 1 factor, one date, and 37 numeric values:
> str(waterchem)
> 'data.frame': 3525 obs. of 39 variables:
> site : Factor w/ 64 levels "D-1","D-2","D-3",..: 1 1 1 1 1 ...
> $ sampdate : Date, format: "2007-12-12" "20
On Nov 29, 2011, at 2:18 PM, Rich Shepard wrote:
I have a data frame with 1 factor, one date, and 37 numeric values:
str(waterchem)
'data.frame': 3525 obs. of 39 variables:
site : Factor w/ 64 levels "D-1","D-2","D-3",..: 1 1 1 1 1 ...
$ sampdate : Date, format: "2007-12-12" "2008-03
Hi Rich,
Try looking at:
levels(waterchem$SC)
There must be something in that column that is triggering R to read it
as character. Potential examples include using "." to indicate
missing values or anything else that is not itself directly numeric.
You might also get some mileadge out of attemp
I have a model like this (Nelson and Siegel 1987)
http://r.789695.n4.nabble.com/file/n4120161/31f188c684764cd431619dbb59fed5ae.png";
border="0"/>
where tau and y are the maturities and yields, respectively, given to me in my
data file..
y<-c(4.863,5.662,6.41,6.864,7.153,7.352,7.409,7.474,7.503
Good afternoon everybody,
I'm quite new in functions optimization on R and, whereas I've read
lot's of function descriptions, I'm not sure of the correct settings for
function like "optimx" and "nlminb".
I'd like to minimize my parameters and the loglikelihood result of the
function.
My paramet
Hi All,
When we run the command : summary ( newmod<-gam(Dlq~ formula,family,,data) )
in R, the output would the effect of smoothness in R.
As of now to calculate the probability I am following the below approach:
1) Run the plot of the GAM , interpret the curves
2) Re Run the Regression a
Dear R useres, got the following problem. Given the AggData (listed below)
I need to plot AggData[,2] vs time (AggData[,1]) for chosen 'rows'. Ive done
already:
plot(AggData[rows,2], xaxt='n')
axis(1,at=seq(1,length(rows),1),sub("","", AggData[rows,1]))
which works, but I need to list only chose
Petr, thanks for pointing that out. Jim, you are exactly right! Thank you
for catching that. I did not realize in the other replies that they were
using log and not ln.
David, thank you for the lessons. I will improve my question asking skills.
Thanks to all,
Xu
On Tue, Nov 29, 2011 at 9:31 AM,
I have a data frame with 1 factor, one date, and 37 numeric values:
str(waterchem)
'data.frame': 3525 obs. of 39 variables:
site : Factor w/ 64 levels "D-1","D-2","D-3",..: 1 1 1 1 1 ...
$ sampdate : Date, format: "2007-12-12" "2008-03-15" ...
$ CO3 : num 1 1 6.7 1 1 1 1 1 1
On 11/29/2011 01:39 PM, Hasan Diwan wrote:
On 29 November 2011 10:26, Jeff Newmiller wrote:
For my part, I highly recommend exporting to CSV before importing to R.
Will look into doing so...
I don't think this point can be stressed enough. I have had too many
bad experiences with dates i
On 29 November 2011 10:26, Jeff Newmiller wrote:
> You haven't indicated how you are accessing the Excel file, or whether it is
> an XLS or XLSX file. It sounds like you might be using rcom or a dependent
> package, in which case you may need to read the Excel COM interface
> documentation more
Omitting the leading zero is dangerous, since the decimal point can disappear
in a poor hardcopy leading to later misinterpretation.
---
Jeff NewmillerThe . . Go Live...
DCN:B
You may also want to deal with a possible
leading negative sign:
> lambda <- c(0, 0.005, 0.01, 0.02, 0.04, 0.08,
-0.005, -0.01, -0.02, -0.04, -0.08, 1000)
> gsub("^0(\\..*[^0])0*$","\\1", lambda)
[1] "0" ".005" ".01"".02"".04"".08"
[7] "-0.005" "-0.01
You haven't indicated how you are accessing the Excel file, or whether it is an
XLS or XLSX file. It sounds like you might be using rcom or a dependent
package, in which case you may need to read the Excel COM interface
documentation more carefully. In any event, you can't expect as.Date to
und
On 29 November 2011 09:32, Prof Brian Ripley wrote:
> If all else fails, read the help page. There are examples on ?as.Date of
> reading Excel dates.
I did, it seems there is either (a) a problem with my code, or (b) a
problem with the documentation. See below:
> rawtimeColumn
[1] 39406.577662037
kerry1912 wrote on 11/29/2011 09:51:34 AM:
>
> I'm using a histogram and want to overlay this onto a barplot. I need
the
> histogram lines to be thicker in order for it to stand out more on top
of
> the barplot. Is there a command in order to do this?
>
> Thanks in advance.
This topic has be
.. and if you want to simultaneously handle possible multiple trailing
zeros (not sure whether this could even happen)
(somewhat but not completely tested)
> lambda <- c(0, 0.005, 0.01, 0.02, 0.04, 0.08)
> gsub("^0(\\..*[^0])0*$","\\1",lambda)
[1] "0"".005" ".01" ".02" ".04" ".08"
Note th
If all else fails, read the help page. There are examples on
?as.Date of reading Excel dates.
But I don't believe the time you give. (0.577662037 is just before
13:51:50).
On Tue, 29 Nov 2011, Hasan Diwan wrote:
I have a 10-column XLS file, with 2 date fields. As far as I can tell,
they we
Here's one way to get rid of leading zeros before the
decimal point:
> gsub("^0\\.", "\\.", as.character(lambda))
[1] "0"".005" ".01" ".02" ".04" ".08"
Sarah
On Tue, Nov 29, 2011 at 12:04 PM, Michael Friendly wrote:
> A simple question, but I can't find something to do what I want:
>
> G
A simple question, but I can't find something to do what I want:
Given: a vector of numbers, like
lambda <- c(0, 0.005, 0.01, 0.02, 0.04, 0.08)
Desired: format them in minimal space for use as plot labels, ie,
without leading or tailing 0s. For this example:
lambdaf <- c("0", .005", ".01", "
I have a 10-column XLS file, with 2 date fields. As far as I can tell,
they were configured identically in Excel 2010. One of these fields
resembles "39406.577662037", whilst in Excel, it is shown as
"2007-11-20 13:42:20". Applying as.Date() with the default format
doesn't do it. Any ideas as to wh
Hi,
what's mean "/" in command:
betareg(inf~Grupo/Sexo, data=dados)
it's a effect nested?
--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
__
R-help@r-project
I think everyone is worried about your data, not what a correlation
matrix is. Now I think you may be even more turned around: you want a
correlation between the values of the correlation matrix?
Just dput() your data object and copy it into your email and we'll see
if it's possible to calculate a
Hi Michael,
thank you so much for your fast reply.
On the image below there is an example of what I mean.
I need the correlation between the values on the fields with "?".
http://r.789695.n4.nabble.com/file/n4119734/corr_ex.png
But my source data is not in a matrix. It looks like the table in
I'm using a histogram and want to overlay this onto a barplot. I need the
histogram lines to be thicker in order for it to stand out more on top of
the barplot. Is there a command in order to do this?
Thanks in advance.
--
View this message in context:
http://r.789695.n4.nabble.com/Making-the-li
Hi:
This looks like the one:
library('lattice')
data(Oats, package = 'MEMSS')
print(xyplot(yield ~ nitro | Block, Oats, groups = Variety, type = c("g", "b"),
auto.key = list(lines = TRUE, space = 'top', columns = 3),
xlab = "Nitrogen concentration (cwt/acre)",
Do you have multiple data points for Car/Train etc?
And do you want to see if there are differences between in mean/medians these
modes of transport?
If so explore anova, kruskal-wallis
.
> Date: Tue, 29 Nov 2011 16:57:57 +0100
> From: lig...@statistik.tu-dortmund.de
> To: falk.hilli...@tw
Hi,
Unfortunately I do not have an answer to this question yet, but it is
something I'm currently examining. We're hoping to construct a template for
health economic evaluation using decision tree (eventually working up to
more complex modelling methodologies). I'll keep you posted on our progress
> -Original Message-
> What I like to know is if I apply an ANOVA to this data and
> choose the control group as the reference group (using the
> relevel function) what groups exactly are compared? Are only
> all treated groups 1, 2, 3 tested against the control group
> or are all possi
On 29.11.2011 16:41, Geophagus wrote:
hi @ all,
I have problem with creating a matrix for a cor() function.
I try to use the cor() function on a matrix to test the correlation between
each value in a column.
Maybe like corr(x, method = xyz).
My x has two columns maybe like this:
MEDIA VALUE
C
I'm not sure how you mean to calculate correlation if you have a
single observation of each mediumcan you provide your data (or a
subset thereof) so we can see what you are actually working with and
if correlation makes sense.
Michael
On Tue, Nov 29, 2011 at 10:41 AM, Geophagus
wrote:
> hi @
Looks like a typical plot produced using the 'lattice' package. There
is plenty of documentation on the use of the package. Run some of the
examples.
On Tue, Nov 29, 2011 at 10:43 AM, syrvn wrote:
> Hello,
>
>
> can anybody tell me how to produce a plot like the one in
>
> http://cran.r-project
hi @ all,
I have problem with creating a matrix for a cor() function.
I try to use the cor() function on a matrix to test the correlation between
each value in a column.
Maybe like corr(x, method = xyz).
My x has two columns maybe like this:
MEDIA VALUE
Car 23
Train26
Plane 25
Cab
Great, many thanks.
On 11/29/2011 3:09 PM, Duncan Murdoch wrote:
On 29/11/2011 8:36 AM, Mathew Brown wrote:
Hi there,
I'm running R on windows 7 with Rstudio. Everyday I receive a zip file
where a bunch of half-hourly files are zipped together.
I then use
xx=unzip(ind)
to get xx, which consi
Hi:
Here's one approach. I assume that your first 1000 matrices have a
single 1 in each matrix, the next set of 1000 have two 1's, ..., and
the last one has 99 1's. (No point in doing all 1's since they're all
constant.) If that's the case, then try the following.
# Each row represents a differen
I normally use the raster or clim.pact pckages to read netcdf (.nc) files.
This has always worked out for me until this weekend every time i try to
read a .nc file i get the following error
Program: C:\Program Files\RStudio\bin\x64\rsession.exe File: posixio.c,
Line 417 Expression: offset >= 0 Thi
Hello,
can anybody tell me how to produce a plot like the one in
http://cran.r-project.org/web/packages/lme4/vignettes/Implementation.pdf
on page 13, Figure 6?
The data is stored in:
library(nlme)
data(Oats)
Cheers
--
View this message in context:
http://r.789695.n4.nabble.com/Help-neede
On Nov 29, 2011, at 7:32 AM, Grant McDonald wrote:
Dear all, I am finding difficulty in the following, I would like to
create an empty matrix e.g. 10x10 of 0s and sequentially fill this
matrix with randomly placed a 1s until it is saturated. Producing 100
matrices of sequentially increasing de
Folks:
On Tue, Nov 29, 2011 at 6:24 AM, Sarah Goslee wrote:
> I have to admit I'm not entirely sure what your question is. How to
> put a 1 in a random position in a matrix?
>
> mat <- matrix(0, 10, 10)
> mat[sample(1:nrow(mat), 1), sample(1:ncol(mat), 1)] <- 1
This is unnecessary. In R: matrice
Hi again,
Working with my real data and not with the little example i sent to the list i
discovered that segm_distance function from package "pracma" does not converge
to 0 in all cases, even if i increase the number of iteration to 10,000 for
example. It seems that it depends on the "initiali
On 29.11.2011 13:09, narendarreddy kalam wrote:
i have R 2.14
2.14.0 I guess.
version.and i have downloaded bayesQR package from following
link
http:// http://cran.r-project.org/web/packages/bayesQR/index.ht ml
my OS is Windows7.i have downloaded Windows binary:bayesQR_1.3.zip file
On Nov 29, 2011, at 2:30 AM, Xu Wang wrote:
David,
Did my reply get orphaned
All replies are "orphaned". You are asked to include context if your
question relies on code that has previously been posted.
or are you trying to help me realize that asking
why something does not work is not
example(heatmap.2) works for me, hence it is either your data, your
version of some package or your version of R that causes the problems,
we do not have information of any of these although the posting guide
asks you to provide this for each posting. So we cannot help.
Uwe Ligges
On 29.11.2
Let's see... you could:
- read the directions at http://cran.r-project.org or the longer ones
here http://cran.r-project.org/doc/manuals/R-admin.pdf
- look at the help for install.packages()
- actually look at the menus for R, as there's an install from local
file option there somewhere in the Wind
I have to admit I'm not entirely sure what your question is. How to
put a 1 in a random position in a matrix?
mat <- matrix(0, 10, 10)
mat[sample(1:nrow(mat), 1), sample(1:ncol(mat), 1)] <- 1
will do so, but if you need to fill a random position that is *currently zero*
then you'll need to wrap i
On 29/11/2011 8:36 AM, Mathew Brown wrote:
Hi there,
I'm running R on windows 7 with Rstudio. Everyday I receive a zip file
where a bunch of half-hourly files are zipped together.
I then use
xx=unzip(ind)
to get xx, which consists of :
[1] "./2011/A20112961503.flx" "./2011/A20112961503.log" ".
On 29.11.2011 07:06, Indrajit Sengupta wrote:
What have you tried so far - can you explain? "fitdistrplus" package is the
default package for fitting distributions.
It is a contributed packages, and perhaps it is a good one (I do not
know), but calling it the *default* ... who defined that?
On 29.11.2011 13:15, R. Michael Weylandt wrote:
I think you are looking for the system() command.
... and you certainly do not want to use savePlot but rather use the
pdf() device directly.
Uwe Ligges
Michael
On Tue, Nov 29, 2011 at 7:11 AM, Christof Kluß wrote:
Hi Jim, Hi Ken
Am 27
use pattern matching (regular expressions): e.g.,
myFileNames[grepl("slt$", myFileNames)]
On Tue, Nov 29, 2011 at 8:36 AM, Mathew Brown
wrote:
>
>
> Hi there,
> I'm running R on windows 7 with Rstudio. Everyday I receive a zip file
> where a bunch of half-hourly files are zipped together.
> I
Hi
I used to make heatmaps using following commands but now some thing
has changed as I get
library(gplots)
heatmap.
2(qtl.map,Rowv=F,dendrogram="column",col=colorRampPalette(c("blue","lightblue","black","black","yellow","red")),breaks=seq(-4.01,4.01,length.out=51),
density.info="none", )
follow
Hi there,
I'm running R on windows 7 with Rstudio. Everyday I receive a zip file
where a bunch of half-hourly files are zipped together.
I then use
xx=unzip(ind)
to get xx, which consists of :
[1] "./2011/A20112961503.flx" "./2011/A20112961503.log"
"./2011/A20113211730.slt" "./2011/A201132118
Hi List,
Being new to R, I am trying to apply boot.stepAIC() for Model selection by
bootstrapping the stepAIC() procedure. I had gone through the discussion in
various thread on the variable selection methods. Understood the pros and
cons of various method, also going through the regression modelli
Dear all, I am finding difficulty in the following, I would like to
create an empty matrix e.g. 10x10 of 0s and sequentially fill this
matrix with randomly placed a 1s until it is saturated. Producing 100
matrices of sequentially increasing density., This process needs to be
randomized 1000 ti
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