There is also clm() (for cumulative link models) from package ordinal
that has much the same interface that polr() has, but it does give you
p-values for the regression parameters. A simple example from
examples(clm):
library(ordinal)
data(wine)
fm1 <- clm(rating ~ contact + temp, data=wine)
(summ
On Mon, 29 Aug 2011, Dieter Menne wrote:
Gang Chen-4 wrote:
However, my problem is that the file generated
from a graph of fixed size is too large (in the order of 10MB) because of
many data points in multiple scatterplots. Any suggestions?
Generate pdf, open and save it in Adobe Acrobat w
Thank you, that's works just fine.
Luca
Il giorno 29/ago/2011, alle ore 23.48, H. T. Reynolds ha scritto:
> Hi,
>
> I use xtabs with the weight variable on the left hand side of the formula as
> in
>
> xtabs(weight ~ opinion + gender + ...)
__
R-h
Dan Abner wrote:
>
> I have been using the sink() function as follows:
>
> x<-rnorm(100)
> sink("C:\\Users\\dan\\Desktop\\Current Events\\myhw.txt",
> append=TRUE,split=TRUE)
> m1<-c(Mean=mean(x),SD=sd(x),Min=min(x),Max=max(x))
> m1
> sink()
>
> ===
>
> Is it possible to create tabular style
Gang Chen-4 wrote:
>
> However, my problem is that the file generated
> from a graph of fixed size is too large (in the order of 10MB) because of
> many data points in multiple scatterplots. Any suggestions?
>
Generate pdf, open and save it in Adobe Acrobat which does a compression
when the set
Marie Duarte wrote:
>
> Is anybody able to recommend packages to provide GUI like behaviour to
> interface with R that would facilitate, eg.
> * Rotatable surface plots (similar to LiveGraphics3D for mathematica,
> http://www.vis.uni-stuttgart.de/~kraus/LiveGraphics3D/examples/color_volumes.html
Hi all.
Using the following code:
plot(bsa, abs, col='blue', pch=16, xlim=c(0, 25), ylim=c(0.0, 1.0), xlab =
expression(paste(BSA, " (", mu, "g)", sep="")), ylab=expression(A[595]))
plot(unknown1, abs2, col='blue', pch=16, xlim=c(0, 25), ylim=c(0.0, 1.0),
xlab = expression(paste(Unknown_1, " (",
Why not
d <- ifelse( d1 > d2, d1, d2 )
or
d <- pmax( d1, d2 )
?
Apply operations may seem "beautiful", but I think the speed and simplicity of
vectorized operations are truly beautiful.
---
Jeff Newmiller The . .
Hallo
I am teaching a very elementary intro course about R, and want to show
the students how to find the distribution of the maximum of rolling
two balanced die. Is there perhaps a more elegant way to do this,
other than the way I am using below? (I would like to show them two
ways - the
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Heikki Kaskelma
> Sent: Monday, August 29, 2011 6:13 PM
> To: r-h...@stat.math.ethz.ch
> Subject: Re: [R] weird apply() behavior
>
> "array chip":
> >Hi, I had a weird results fr
Hello,
I'm fitting a repeated measures models, I don't know many things about it, I
need to help me with the model's graphic representation, and checking the
assumptions. My data set haves a cuantitative response variable and a
factor with 5 levels, this variable was measured three times each 20
Is something like this what you want?
x <- letters[1:4]
x
y <-do.call(paste,c( paste('"',x[1]), as.list(x[2:3]),
paste(x[4],'"'),sep="\n"))
y
cat(y,"\n")
-- Bert
On Mon, Aug 29, 2011 at 6:59 PM, Ben qant wrote:
> Unfortunately that didn't work. I just says the text is an invalid argument.
> I
Dear R users
I'd like to know if it is possible to create conditional plots in the
soiltexture package. I have data from the particle size distribution of
soils derived from two parent materials and I want to use the parent
material as conditioner.
Thanks a lot.
Alessandro Samuel-Rosa
Post-Gradu
"array chip":
>Hi, I had a weird results from using apply().
Consider:
y<-data.frame(list(a=c(1,NA), b=c('2k','0')))
do.call(args=y,
Vectorize(v=c("a","b"),
function(a,b) if(is.na(z<-b=="2k" & a==1)) 0
else z+0))
Heikki Kaskelma
Munkkiniemi
_
Hi,
The information on the following link concerning web interfaces to R appears
outdated:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#R-Web-Interfaces
Is anybody able to recommend packages to provide GUI like behaviour to
interface with R that would facilitate, eg.
* Rotatable surface plots (si
Dear Rodrigo
The best thing to do is to label manually the points in the biplot. To do so
you have to use the following comand:
> text(location, "text to place")
*location* can be an x,y coordinate. Alternatively, the text can be placed
interactively via mouse by specifying location as *locator(
Perhaps you need to conform to the conventions mandated by RFC 2822 section 2.1
for record separators. Most libraries automatically handle this for you, but
perhaps your rJython library does not.
---
Jeff Newmiller The .
Unfortunately that didn't work. I just says the text is an invalid argument.
I also tried saving it in a variable name and passed that in, but that
didn't work. I get:
Error in ls(envir = envir, all.names = private) :
invalid 'envir' argument
...when I try to send the message.
Any other ideas?
Use the rms package to replace Design. Run anova(fit object from lrm) which
produces a matrix from which you can extract P-values. This also handles
the case of multiple degrees of freedom per predictor.
Frank
Debs Majumdar wrote:
>
> Hi,
>
> Are there any packages which prints out p-values
Thanks Bill and David!
John
- Original Message -
From: William Dunlap
To: array chip ; "r-help@r-project.org"
Cc:
Sent: Monday, August 29, 2011 5:21 PM
Subject: RE: [R] weird apply() behavior
apply() should come with a big warning that it was
written for matrices and can cause a lo
On Aug 29, 2011, at 8:17 PM, David Winsemius wrote:
On Aug 29, 2011, at 8:02 PM, array chip wrote:
Hi, I had a weird results from using apply(). Here is an simple
example:
y<-data.frame(list(a=c(1,NA),b=c('2k','0')))
y
a b
1 1 2k
2 NA 0
apply(y,1,function(x){x<-unlist(x);
apply() should come with a big warning that it was
written for matrices and can cause a lot of surprises
with data.frames. apply(X, ...) converts X to a
matrix and that is the kernel of your problem: if any
columns of X are not numeric then as.matrix(X) is
a character matrix whose columns are made
Hello everyone,
I have been using the sink() function as follows:
x<-rnorm(100)
sink("C:\\Users\\dan\\Desktop\\Current Events\\myhw.txt",
append=TRUE,split=TRUE)
m1<-c(Mean=mean(x),SD=sd(x),Min=min(x),Max=max(x))
m1
sink()
===
Is it possible to create tabular style output very much like the OD
On Aug 29, 2011, at 8:02 PM, array chip wrote:
Hi, I had a weird results from using apply(). Here is an simple
example:
y<-data.frame(list(a=c(1,NA),b=c('2k','0')))
y
a b
1 1 2k
2 NA 0
apply(y,1,function(x){x<-unlist(x);
That is quite unnecessary since apply coerces the r
Hi, I had a weird results from using apply(). Here is an simple example:
> y<-data.frame(list(a=c(1,NA),b=c('2k','0')))
> y
a b
1 1 2k
2 NA 0
> apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k' &
> !is.na(x[1]) & x[1]=='1') 1 else 0} )
This should print "1 0" as ou
Try:
paste(c("a", "b", "c"), collapse = "\n")
On Mon, Aug 29, 2011 at 8:56 PM, Ben qant wrote:
> Hello,
>
> Does anyone know how to convert this:
> > msg
> [1] "a"
> [2] "b"
> [3] "c"
>
>
> To:
>
> > msg
> "a
> b
> c"
>
> In other words, I need to convert a character vector to a single str
Hello,
Does anyone know how to convert this:
> msg
[1] "a"
[2] "b"
[3] "c"
To:
> msg
"a
b
c"
In other words, I need to convert a character vector to a single string with
carriage returns for each row.
Functionally, I'm attempting to send an email of a character vector in a way
that is re
Hi,
Are there any packages which prints out p-values for OLR's (like `ologit'
from Stata)? I want to run a bunch of OLRs and print the p-value for the first
coefficient from each of them.
I checked polr() under MASS and it doesn't.
There's a lrm() function under Design which does print
Try this:
as.list(colSums(merge(m, n, all = TRUE), na.rm = TRUE))
On Mon, Aug 29, 2011 at 7:39 PM, zhenjiang xu wrote:
> Hi R users,
>
> Suppose I have two lists and the names of list 'm' are a subset of those of
> 'n', how can I sum the two lists with corresponding elements added together
> to
Hi R users,
Suppose I have two lists and the names of list 'm' are a subset of those of
'n', how can I sum the two lists with corresponding elements added together
to get list 'o'?
> n = list("a"=1,"b"=3,"c"=5)
> m = list('b'=4)
> o
$a
[1] 1
$b
[1] 7
$c
[1] 5
Thanks
--
Best,
Zhenjiang
On Aug 29, 2011, at 6:03 PM, Ayyappa Chaturvedula wrote:
Dear R group,
I am trying to plot uisng a for loop. Here is the script:
for (i in unique(cfn$ID)){
plot(cfn$TIME[cfn$ID==i],cfn$DV[cfn$ID==i],pch=16)}
I could access only the last plot of the series and cannot go back
to see
all the
Google on "R data table" please. Read the vignettes therein.
-- Bert
On Mon, Aug 29, 2011 at 1:59 PM, Abraham Mathew wrote:
> I didn't learn about data tables until recently. (They're never covered in
> any intro R books).
>
> In any case, I'm not sure what (if any) is the difference between a d
David -
I'm a little confused about why you would call that indexing error
"trifling". It seems like a pretty serious problem with the zoo
internals although it seems to have been fixed in the latest version of
zoo.
Gabor -
Thank you. After updating to the latest version of zoo *and* restart
On Aug 29, 2011, at 3:33 PM, R. Michael Weylandt wrote:
Hmm, I don't know what this means as trouble shooting, but I get the
following:
snipped
4
> x[as.Date('2001-01-05')] = 0
> x
Error in dimnames(x) <- dn :
length of 'dimnames' [1] not equal to array extent
> str(x)
‘zoo’ s
i have this data with me...i am only copying part of data here...
date time
"2011-05-23""04:31:17"
"2011-05-23""04:31:20"
"2011-05-23""04:31:22"
"2011-05-23""04:31:25"
"2011-05-23""04:31:27"
"2011-05-23""04:31:18"
"2011-05-23""04:31:20"
"2011-05
I didn't learn about data tables until recently. (They're never covered in
any intro R books).
In any case, I'm not sure what (if any) is the difference between a data
frame and a data table.
Can anyone provide a brief explanation?
Is one preferred over another or is it just dependent on the tas
Dear R group,
I am trying to plot uisng a for loop. Here is the script:
for (i in unique(cfn$ID)){
plot(cfn$TIME[cfn$ID==i],cfn$DV[cfn$ID==i],pch=16)}
I could access only the last plot of the series and cannot go back to see
all the plots. I appreciate your help in resolving this issue and also
So guys,
I am writing to ask for an opinion from you...
As I begin to write in the previous post, I developed a function
called myintegrate that is based in integrate but my function give
better results for the problem I am dealing. My development was
inspired in this post
http://r.789695.n4.nabb
I read somewhere that vector graphics such as eps or dpf are more favorable
than alternatives (jpeg, bmp or png) for publication because vector graphics
scale properly when enlarged. However, my problem is that the file generated
from a graph of fixed size is too large (in the order of 10MB) becaus
On Mon, Aug 29, 2011 at 4:59 PM, Gene Leynes wrote:
> David -
> I'm a little confused about why you would call that indexing error
> "trifling". It seems like a pretty serious problem with the zoo
> internals although it seems to have been fixed in the latest version of
> zoo.
At one time zo
On 29/08/2011 3:29 PM, Imbeaud (Inserm U674) wrote:
Hi,
I have to deal with a huge .txt table (~485.577 rows and 469 columns,> 1.5
Go file)
I used the read.table function
> tmp=read.table("data.txt", header=TRUE, sep="\t", fill=TRUE,
na.strings="NA", comment.char="", stringsAsFactors = FALSE)
Dear Group,
Has anyone conducted CR analysis using survey data i.e. with sampling
weights. I've already looked up CRAN task view in survival analysis. Any
lead is appreciated. Thanks!
Kel
--
View this message in context:
http://r.789695.n4.nabble.com/Competing-Risk-with-survey-data-tp3777189
Greetings,
I am having trouble getting the function reformulate_ATSP_as_TSP to work for
me. I have provided a simple example of some of the code I've been using.
In particular, I'm not sure why I'm getting the error
"Error in dimnames(tsp) <- list(lab, lab) :
length of 'dimnames' [1] not e
Hi,
I have to deal with a huge .txt table (~485.577 rows and 469 columns, > 1.5
Go file)
I used the read.table function
> tmp=read.table("data.txt", header=TRUE, sep="\t", fill=TRUE,
na.strings="NA", comment.char="", stringsAsFactors = FALSE)
However, I encounter troubles in interpreting some "\t"
While R has library TSP to help solve traveling salesperson problems, does
anyone know if it has any libraries to help solve multiple traveling
salesperson problems? For instance, suppose one is planning school bus
routes and one has multiple buses. Thank you for your time.
--
View this message
Michael,
By the way, although I replied to David's email, I was responding to you as
well. Your results were exactly what I was expecting, but I didn't get your
results.
On Mon, Aug 29, 2011 at 1:51 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> How exactly do you mean it doesn
This seems like a very strange error.
In trying to troubleshoot this further I looked at the structure of x. The
new x has the length of the Index (2001-01-05 = 11327).
> library(zoo)
> x = zoo(1:5, as.Date('2001-01-01')+1:5)
> str(x)
zoo series from 2001-01-02 to 2001-01-06
Data: int [1:5]
I tried that, while I find the documentation a bit short, but the only
result I get from this is a probability distribution of my data (I'm
building a tree with 2 classes). How do I plot a tree where the counts
are show in each step/node?
BR,
Jay
On Aug 29, 9:40 pm, Weidong Gu wrote:
> ? predict
On Mon, Aug 29, 2011 at 2:45 PM, Gene Leynes wrote:
> Why doesn't this work?
>
> x = zoo(1:5, as.Date('2001-01-01')+1:5)
> x[as.Date('2001-01-05')]
> x[as.Date('2001-01-05')] = 0
> x
>
Make sure you have the most recent version of zoo which is this:
> packageVersion("zoo")
[1] ‘1.7.4’
--
Stat
On 29/08/2011 3:52 PM, Jie TANG wrote:
thank you , it works .
another problem is if can could define a variable to express the data.frame?
for example :
datanam<-c("newdata","newdata2")
plot(datanam[1][[newnam[1]]])
Use get():
plot(get(datanam[1])[[newnam[1]]]))
Duncan Murdoch
2011/8/30 J
thank you , it works .
another problem is if can could define a variable to express the data.frame?
for example :
datanam<-c("newdata","newdata2")
plot(datanam[1][[newnam[1]]])
2011/8/30 Justin Haynes
> try:
>
> newnam<-paste('newdatadat',dayno,sep='')
>
> plot(test[[newnam[1]]])
>
>
> On Mon,
well, if a pooled estimate of the residual standard error is not
desirable, then you just need to set argument 'pool' of lmList() to
FALSE, e.g.,
mlis <- lmList(yvar ~ . - clvar | clvar, data = df, pool = FALSE)
summary(mlis)
Best,
Dimitris
On 8/29/2011 9:20 PM, Dennis Murphy wrote:
Hi:
try:
newnam<-paste('newdatadat',dayno,sep='')
plot(test[[newnam[1]]])
On Mon, Aug 29, 2011 at 12:29 PM, Jie TANG wrote:
> hi, R-users
> I have a data.frame for example test$newdataday24 and test$newdataday48
> I can plot them by
> plot(test$newdataday24)
> but now i want to plot different
Hmm, I don't know what this means as trouble shooting, but I get the
following:
1) After library(zoo)
Attaching package: 'zoo'
The following object(s) are masked from 'package:base':
as.Date
and then for the first str(x)
zoo series from 2001-01-02 to 2001-01-06
Data: int [1:5] 1 2 3 4
hi, R-users
I have a data.frame for example test$newdataday24 and test$newdataday48
I can plot them by
plot(test$newdataday24)
but now i want to plot different data by define a variable to describe them
dayno<-c(24,48)
newnam<-paste("test$newdataday",dayno,sep="")
plot(newnam[1])
but i failed,
I would recommend using the new Bayesian package 'LaplacesDemon' available
on CRAN.
Ben Bolker
Sent by: r-help-boun...@r-project.org
08/29/2011 02:50 PM
To
cc
Subject
Re: [R] Bayesian functions for mle2 object
Billy.Requena gmail.com> writes:
>
> Hi everybody,
>
> I'm interested
Hi:
Dimitris' solution is appropriate, but it needs to be mentioned that
the approach I offered earlier in this thread differs from the
lmList() approach. lmList() uses a pooled measure of error MSE (which
you can see at the bottom of the output from summary(mlis) ), whereas
the plyr approach subd
On Aug 29, 2011, at 3:02 PM, Gene Leynes wrote:
Although I'm not sure what you're talking about with pop-up windows...
I got (as expected) assignment, so I assumed you were not expecting
assignment.
Weird, this is what I'm getting in either R 2.13.0 or R 2.12.0:
> library(zoo)
Warning
Although I'm not sure what you're talking about with pop-up windows...
Weird, this is what I'm getting in either R 2.13.0 or R 2.12.0:
> library(zoo)
Warning: package 'zoo' was built under R version 2.13.1
> x = zoo(1:5, as.Date('2001-01-01')+1:5)
> x[as.Date('2001-01-05')]
2001-01-05
4
You can do this using function lmList() from package nlme, without
having to split the data frames, e.g.,
library(nlme)
mlis <- lmList(yvar ~ . - clvar | clvar, data = df)
mlis
summary(mlis)
I hope it helps.
Best,
Dimitris
On 8/29/2011 5:37 PM, Nilaya Sharma wrote:
Dear All
Sorry for th
Michael Parent ufl.edu> writes:
>
> Thanks!
>
> "This problem isn't uniquely defined. Are you
> willing to generate more samples than you need and then throw
> away extreme values? Or do you want to 'censor'
> extreme values (i.e. set values <= 1 to 1 and values >=7 to 7)?"
>
> I'd like the
Hi:
This is straightforward to do with the plyr package:
# install.packages('plyr')
library('plyr')
set.seed(1234)
df <- data.frame(clvar = rep(1:4, each = 10), yvar = rnorm(40, 10, 6),
var1 = rnorm(40, 10, 4), var2 = rnorm(40, 10, 4),
var3 = rnorm(40, 5, 2), var
Hello everyone,
I working in a public health project and we have created a Decision Tree for
categorical variables usign the package rpart. Our goal is to develop a model
(Using the ROC tool) in order to predict presence/ausent of diabetes and get a
better understanding of what are the import
On Aug 29, 2011, at 2:45 PM, Gene Leynes wrote:
Why doesn't this work?
x = zoo(1:5, as.Date('2001-01-01')+1:5)
x[as.Date('2001-01-05')]
x[as.Date('2001-01-05')] = 0
x
I think this is especially bad because it doesn't cause an error.
It lets
you do something to x, but then you can't see x
How exactly do you mean it doesn't work? Copied from my GUI:
> x = zoo(1:5, as.Date('2001-01-01')+1:5)
> x[as.Date('2001-01-05')]
2001-01-05
4
> x[as.Date('2001-01-05')] = 0
> x
2001-01-02 2001-01-03 2001-01-04 2001-01-05 2001-01-06
1 2 3 0 5
Billy.Requena gmail.com> writes:
>
> Hi everybody,
>
> I'm interested in evaluating the effect of a continuous variable on the mean
> and/or the variance of my response variable. I have built functions
> expliciting these and used the 'mle2' function to estimate the coefficients,
> as follows:
Why doesn't this work?
x = zoo(1:5, as.Date('2001-01-01')+1:5)
x[as.Date('2001-01-05')]
x[as.Date('2001-01-05')] = 0
x
I think this is especially bad because it doesn't cause an error. It lets
you do something to x, but then you can't see x again to see what it did.
[[alternative HTML
Dear all,
I'm forecasting health services utilization using Lee-Carter method.
I have a routine to run LC method in R package, and I understood all steps
to model and forecasting the rates by this method, except two things:
1) how to adjust the estimated admission rates by the total number of
adm
On Aug 29, 2011, at 2:40 PM, Andra Isan wrote:
Hello All,
I have a data frame consisting of 4 columns (id1, id2, y, pred)
where pred is the predicted value based on the glm function and my
data frame is called "all". "data" is another data frame that has
all data but I want to put together
? predict.rpart
Weidong Gu
On Mon, Aug 29, 2011 at 12:49 PM, Jay wrote:
> Hi,
>
> when I have made a decision tree with rpart, is it possible to "apply"
> this tree to a new set of data in order to find out the distribution
> of observations? Ideally I would like to plot my original tree, with
>
Hello All,
I have a data frame consisting of 4 columns (id1, id2, y, pred)
where pred is the predicted value based on the glm function and my data frame
is called "all". "data" is another data frame that has all data but I want to
put together some important columns from my original data frame
?tryCatch
HTH,
Dennis
On Mon, Aug 29, 2011 at 9:04 AM, s1oliver wrote:
> Hi, beginner to R and was having some problems scraping data from tables in
> html using the XML package. I have included some code below.
>
> I am trying to loop through a series of html pages, each of which contains a
> s
Thanks!
"This problem isn't uniquely defined. Are you willing to generate more samples
than you need and then throw away extreme values? Or do you want to 'censor'
extreme values (i.e. set values <= 1 to 1 and values >=7 to 7)?"
I'd like the retain a normal distribution so I wouldn't want to
Dear All
Sorry for this simple question, I could not solve it by spending days.
My data looks like this:
# data
set.seed(1234)
clvar <- c( rep(1, 10), rep(2, 10), rep(3, 10), rep(4, 10)) # I have 100
level for this factor var;
yvar <- rnorm(40, 10,6);
var1 <- rnorm(40, 10,4); var2 <- rnorm(40,
Hi, beginner to R and was having some problems scraping data from tables in
html using the XML package. I have included some code below.
I am trying to loop through a series of html pages, each of which contains a
single table from which I want to scrape data. However, some of the pages
are blank
Hi,
when I have made a decision tree with rpart, is it possible to "apply"
this tree to a new set of data in order to find out the distribution
of observations? Ideally I would like to plot my original tree, with
the counts (at each node) of the new data.
Reagards,
Jay
_
Hi:
integrate() is not a vectorized function. This appears to work:
sapply(1:2, function(x) func(x, 0.1, 0.1, sad = Exp))
[1] 0.250 0.125
In this case, sapply() is a disguised for loop.
HTH,
Dennis
On Mon, Aug 29, 2011 at 9:45 AM, . . wrote:
> Ooops,
>
> sorry!
>
> The problem occurs when
>
>
If your main goal is to look at a data frame and you are ok with scrolling,
then look at the View function (note capitalization) as an alternative to just
printing the data frame.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
On Aug 29, 2011, at 15:39 , Sebastian Bauer wrote:
>>
>> > rr <- data.frame(a = c(1,1,1,1,2), b=c(1,2,2,3,1))
>>
>> > ave(order(rr$a, rr$b), rr$a, rr$b )
>> [1] 1.0 2.5 2.5 4.0 5.0
>
> Actually, this may be a solution I was looking for! Note that it assumes that
> rr to be sorted already (hen
The grconvertX and grconvertY functions may be helpful in finding the endpoints
to use.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
Hi:
The bars *are* ordered in the same way, but when you use coord_flip(),
the left category goes on top and the right category goes on the
bottom. Is this what you want?
ggplot(df, aes(x = name, y = value, fill = type)) +
geom_bar(position = position_dodge()) +
coord_flip() +
scale_fill_ma
Do you mean things like treatment of categorical variables in regression
procedures (which have different defaults in different procedures in SAS),
and different default as to the reference category in logistic regression?
Jeremy
On 29 August 2011 04:46, n wrote:
> Hello all,
>
> I am looking
Hi David,
Unfortunately I need to use the "should have been" frequencies if the sample
corresponded perfectly in terms of some "reference" variables to the
population.
That is, if in my sample I observe V1_R1=10%, V1_R2=50%, V3_R3=40% while the
same known population distribution is V1_R1=20%,
On Aug 27, 2011, at 3:37 PM, Simon Zehnder wrote:
Dear R users,
I am using R right now for a simulation of a model that needs a lot of
memory. Therefore I use the *bigmemory* package and - to make it
faster -
the *doMC* package. See my code posted on http://pastebin.com/dFRGdNrG
Now, if I
You are somewhere in Circles 3 and 4 of
'The R Inferno'.
If you have a function to apply over more
than one argument, then 'mapply' will do
that.
But you don't need to do that -- you can do
the operation you want efficiently:
*) create your resulting matrix with all zeros,
no reason for this to
Ooops,
sorry!
The problem occurs when
func(1:2,0.1,0.1,sad=Exp)
On Mon, Aug 29, 2011 at 12:27 PM, R. Michael Weylandt
wrote:
> Can't help, code runs fine on my machine once you change "valu" to "value."
> Are you sure it fails in a vanilla run of R and isn't caused by any other
> choices you h
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Bert Gunter
> Sent: Monday, August 29, 2011 7:07 AM
> To: Jim Lemon
> Cc: r-help@r-project.org
> Subject: Re: [R] Asking Favor For "Remove element with Particular Value In
> Vect
Hi,
comments below.
On Mon, Aug 29, 2011 at 8:12 AM, Ben qant wrote:
> Correction. My solution didn't work either Didn't return the correct
> values. Can you post an example that takes three arguments? I'm working on
> how to do this now.
> thanks...sorry. I"m new to R and R.oo.
>
> Ben
>
>
Simon,
Though we're please to see another use of bigmemory, it really isn't
clear that it is gaining you
anything in your example; anything like as.big.matrix(matrix(...))
still consumes full RAM for both
the inner matrix() and the new big.matrix -- is the filebacking really
necessary. It also do
It doesn't help to post this twice, but it may help to know why this is of
interest.
Frank
n wrote:
>
> Hello all,
>
> I am looking for theories and statistical analyses where the defaults
> employed in R and SAS are different. As a result, the outputs under
> the defaults should (at least sligh
On 29/08/2011 9:23 AM, behave wrote:
Hi there
I'm trying to configure R to get access to the internet.
Using the Internet Explorer a proxy .pac script is used.
Reading some older threads I found that I can use the --internet2 option.
When choosing a mirror I get the error: "407 Proxy Authentica
On Aug 29, 2011, at 9:15 AM, Campbell, Desmond wrote:
Petr, Jorge, Daniel,
Yes you could also use outer() instead of expand.grid().
This is quite useful to know.
Also I didn't know you could turn a matrix into a vector by setting
its dimensions to NULL like that. I always used as.vector( m
Hello R users
I'm using MuMIn but for some reason I'm not getting the adjusted confidence
interval and uncoditional SE whe I use model.avg().
I took into consideration the steps provided by Grueber et al (2011)
Multimodel inference in ecology and evolution: challenges and solutions in
JEB.
I cre
Thank you very much,friend.
--
View this message in context:
http://r.789695.n4.nabble.com/Asking-Favor-For-Remove-element-with-Particular-Value-In-Vector-tp3772779p3776435.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-proje
Hi there
I'm trying to configure R to get access to the internet.
Using the Internet Explorer a proxy .pac script is used.
Reading some older threads I found that I can use the --internet2 option.
When choosing a mirror I get the error: "407 Proxy Authentication Required".
This seems reasonable
Thank you very much,friend.
--
View this message in context:
http://r.789695.n4.nabble.com/Asking-Favor-For-Remove-element-with-Particular-Value-In-Vector-tp3772779p3776430.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-proje
In fact by fiddling with the "at" and "colorkey" options, I was able to get
the result I expected. Now the colors are assigned correctly, as well as the
colorkey. Here's my code:
# see data in the original
data$level[is.na(data$level)] <- 10 # assign a value above the scale to NA
values
levelplo
Thank you very much,friend.
--
View this message in context:
http://r.789695.n4.nabble.com/Asking-Favor-For-Remove-element-with-Particular-Value-In-Vector-tp3772779p3776427.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-proje
Petr, Jorge, Daniel,
Yes you could also use outer() instead of expand.grid().
This is quite useful to know.
Also I didn't know you could turn a matrix into a vector by setting its
dimensions to NULL like that. I always used as.vector( m ).
And (as I've just discovered) you can use it to reconfi
Hi all,
I am trying to do a barplot in ggplot2 and want to make sure that the legend
order is consistent with the bar order, that is the legend order is orig and
match; and the bars are ordered in the same way. It seems to me that I can only
control one of them. Any idea?
library(ggplot2)
df <
1 - 100 of 140 matches
Mail list logo
