Hi
OTOH I wonder why cbind gives error as OP told us
x <- data.frame(x = 1:5)
y <- data.frame(y = 6:15)
merge(x,y)
cbind(x,y)
Gives different results but without any error.
Regards
Petr
>
> On Thu, Aug 2, 2012 at 4:52 PM, Ayyappa Chaturvedula
> wrote:
> > Michael,
> > Thank you for this ,
Hi Bert,
I won't post any more messages on this thread as problem has shifted from
Optimization in R to Graph Algorithms.
Rest fine
Khris.
On Aug 2, 2012, at 9:13 PM, Bert Gunter [via R] wrote:
> This discussion needs to be taken off (this) list, as it appears to
> have nothing to do with R.
Thanks for the response Petr
On Aug 2, 2012, at 11:11 PM, Petr Savicky [via R] wrote:
> On Thu, Aug 02, 2012 at 02:27:43AM -0700, khris wrote:
>
> >
> > On Aug 2, 2012, at 12:39 PM, Petr Savicky [via R] wrote:
> >
> > > On Wed, Aug 01, 2012 at 04:55:30AM -0700, khris wrote:
> > > > Hi Petr,
The only way then i believe is to reduce the width of the bars. Make it 0.5
in case if the numbers of variable allows you to do so.
--
View this message in context:
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Sent from the R help mailing list archive at N
I didn't do any research about this but I think it's the following:
If you run several t-tests to compare groups and then do a tukey-HSD you
won't get the same results either.
It's the same with the kruskal-wallis test. This happens because the
variance that is used for computing significant differ
No, My X label are so big so i cannot reduce margin and cex change won't
affect it much. Is there any way to write legend first and start plot
later.
Regards
--
View this message in context:
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Sent from the R h
Hi there,
I am doing multiple comparisons for data that is not normally distributed.
For this purpose I tried both functions kruskal{agricolae} and
kruskalmc{pgirmess}. It confuses me that these functions do not yield the
same results although they are doing the same thing, don't they? Can anyone
Hi,
I am working on bar plot and legend overlap plot. I attempted all position
like topright, bottomright, topleft and bottomleft still same pblm is there.
par(mar=c(5,22.5,2,2))
barplot(t(data[,2:3]) , beside=TRUE, col=c(rgb(.537, .769, .933),rgb(.059,
.412, .659)), width = 1, horiz=TRUE,cex.na
Hi,
For your first condition, this could be used:
gsub("(\\d)[.].*","\\1",49888.85)
[1] "49888"
Second condition:
formatC(4333.78889,width=8,format="f",digits=2,flag="0")
#[1] "04333.79"
formatC(884333.78889,width=8,format="f",digits=2,flag="0")
#[1] "884333.79"
Third condition:
sprintf
If you are looking for an introduction, and for Linux in particular, I
recommend the
last twelve chapters of "The Linux Command Line" (William Shotts) which
you can download in pdf from the following link.
http://linuxcommand.org/tlcl.php
Regards,
Jorge
On Thu, Aug 2, 2012 at 7:05 PM, Erin Hodge
Hi,
One more solution:
new<-"BABBBABB"
resAB<-count(unlist(strsplit(gsub("AB",1,new),""))==1)[2,]
resAB
# x freq
#2 TRUE 3
resBA<-count(unlist(strsplit(gsub("BA",2,new),""))==2)[2,]
resBA
# x freq
#2 TRUE 2
new1<-c("ABABAAABABBABABBA")
resAB<-count(unlist(strsplit(gsub
Hi,
You could also use these:
library(Biostrings)
gregexpr2("AB",new)
#[[1]]
#[1] 4 10 14
length(unlist(gregexpr2("AB",new)))
#[1] 3
length(unlist(gregexpr2("BA",new)))
#[1] 2
matchPattern("AB",new)
# Views on a 16-letter BString subject
#subject: BABBBABB
#views:
# start end wid
Hi everyone
I'm trying to estimate both the rho parameter and the degree of freedom of
t-ev copula from the data using the "fitCopula" function in the "copula"
package.
cop <- tevCopula(0.8)
est <- fitCopula(cop, x, method="ml");
However, there is always an error and it says:
Error: length(cop
I would suggest the following:
1. Change the margins of the third axis from 5.2 to say 2.5.
2. Reduce the cex in the legend from 1.5 to say 0.75
I think this should work for you.
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View this message in context:
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On Thu, Aug 2, 2012 at 8:57 PM, ivo welch wrote:
> I would like to insert a few modest size data frames directly into my
> R code. a short illustration example of what I want is
>
> d <- read.csv( _END_, row.names=1 )
> , "col1", "col2"
> "row1",1,2
> "row2",3,4
> __END__
>
> right now, the da
On 3 August 2012 at 03:15, William Dunlap wrote:
| > and you probably want closeAllConnections() immediately following
|
| No you do not want to close all connections. You should close each
| connection that you open, but not others (they may be used by other
| functions like sink() or capture.o
> and you probably want closeAllConnections() immediately following
No you do not want to close all connections. You should close each
connection that you open, but not others (they may be used by other
functions like sink() or capture.output()). Use something like:
readTableFromText <-funct
Colleagues
R 2.15.1
OS X
I have a lengthy script that generates a positive number that I display in a
graphic using text. The range of possible values is quite large and I am
looking for an efficient means to format.
1. If the number is large (e.g., > 10^7), I want to display only t
The test of moderator coefficients (QM) is chi-square distributed.You
can use the change in this value when you add a predictor to the model
as a chi-square test, with df equal to the change in df.
Jeremy
On 2 August 2012 05:54, Bexkens, Anika wrote:
> Dear Metafor users,
>
> I'd like to test a
Dear Diana,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Diana Marcela Martinez Ruiz
> Sent: August-02-12 5:00 PM
> To: j...@mcmaster.ca; tlum...@uw.edu; Ayuda en R
> Subject: Re: [R] summary(svyglm) Pr (> | t |) ?
>
>
> T
On Aug 2, 2012, at 5:57 PM, ivo welch wrote:
I would like to insert a few modest size data frames directly into my
R code. a short illustration example of what I want is
d <- read.csv( _END_, row.names=1 )
, "col1", "col2"
"row1",1,2
"row2",3,4
__END__
right now, the data sits in external
I'm not sure I entirely understand the question, but the closest thing
I can think of to a data frame literal, excepting dput(), would be
this:
d <- read.csv(textConnection("
a, b
1, cow
2, dog
3, cat"), header = TRUE)
and you probably want closeAllConnections() immediately following to
avoid a w
I would like to insert a few modest size data frames directly into my
R code. a short illustration example of what I want is
d <- read.csv( _END_, row.names=1 )
, "col1", "col2"
"row1",1,2
"row2",3,4
__END__
right now, the data sits in external files. I could put each column
into its own vec
Thank you very much for all your suggestions. I am very sorry my code is so
crude (it gives me a headache too!), I'm very new to R programing. I will
have to look at your suggestions/questions very carefully (I'm barely
holding on at this point) and get back to you (Dr. Winsemius) with some
answers
Hi Dan,
For question 1, yes you'll need geographic coordinates. I thinknit's
possible to get a shapefile of zip codes, but maybe someone else will know
the details.
For #2, you probably want maps instead of map, and you need to load a
package before you can use it:
install.packages("maps")
libra
Hi,
QUESTION TOPIC #1
I have some data I want to plot on a map. But what I have are home addresses:
street, City, State, complete postal code--i.e 95377-1234. Is there a way to
plot this data or do I need latitudinal and longitude coordinates? If so how do
I convert them? Is there a package tha
On Thu, Aug 2, 2012 at 6:05 PM, Erin Hodgess wrote:
> Dear R People:
>
> Sorry for the off topic question, but here it goes: could someone
> recommend a good bash shell programming book, please?
Don't know if it's the best, but it worked for me and it comes from a
semi-official source:
http://t
On Thu, Aug 2, 2012 at 5:53 PM, Jose Narillos de Santos
wrote:
> Many thanks Eik it works properly.
>
> what I don´t know exactly is why you put
> c(xv,rev(xv)),c(efinal,rev(efinal2)),col="red" inside polygon. I see it
> works but for the moment I don´t really see the meaning of c() and why is
> n
Dear R People:
Sorry for the off topic question, but here it goes: could someone
recommend a good bash shell programming book, please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail
There's no need for 'strsplit' here.
new=c("BABBBABB")
A2B <- gregexpr("AB", new)
B2A <- gregexpr("BA", new)
length(A2B[[1]]) #3
length(B2A[[1]]) #2
Cheers,
Eloi
On 12-08-02 03:44 PM, David Winsemius wrote:
On Aug 2, 2012, at 2:27 PM, rnewbie565 wrote:
I am trying to count the numb
Many thanks Eik it works properly.
what I don´t know exactly is why you put
c(xv,rev(xv)),c(efinal,rev(efinal2)),col="red" inside polygon. I see it
works but for the moment I don´t really see the meaning of c() and why is
needed to put twice and so on.
If I had a vector of dates associated with e
You'll either need to do it in R or Oracle (I'm not sure you can do it
directly across the ODBC boundary) -- I'd actually think Oracle would
be more performant for this sort of thing.
But anyways, import your data from Oracle into R as a data.frame and
then look at the ?merge command.
Best,
Micha
Hello metafor users,
I'm using metafor to perform a single-effect summary estimate of the raw
proportion of patients experiencing a post-operative complication, and I'm
interested in seeing if this proportion differs between the three most
commonly used surgical techniques. The software is working
On Aug 2, 2012, at 2:27 PM, rnewbie565 wrote:
I am trying to count the number of times that the characters in a
string
change
For example-
new=c(BABBBABB)
Presumably you meant to quote that string.
I want to find the number of times that B changes to A and the
number of
times tha
- About the first question, I was not sure about what was the proper model (
a) or b) ) because I saw this at the end of the help for te ---> ?te :
n <- 500
v <- runif(n);w<-runif(n);u<-runif(n)
f <- test2(u,v,w)
y <- f + rnorm(n)*0.2
# tensor product of 2D thin plate regression spline and 1D c
HI David,
I realized it immediately after sending the message from email and corrected
it in Nabble as yahoo mail is down.
A.K.
__
If you reply to this email, your message will be added to the discussion below:
http://r.789695.n4.nabble.com/Subset-data-tp463
Thanks for the reply.
Maybe I am not describing this appropriately - but ultimately I am looking
for items in matrix B that co-vary with items in matrix A.
In the case below:
http://r.789695.n4.nabble.com/file/n4638966/Slide1.jpg
The values for Apples co-vary with Rice and Beans.
The aim is t
Thanks Elk,
That was the missing link.
#or
lapply(list6,function(x)max(x$target))
[[1]]
[1] 6
[[2]]
[1] 18
[[3]]
[1] 9
max(unlist(lapply(list6,function(x) max(x$target
A.K.
___
If you reply to this email, your message will be added to the disc
Hi,
I'm trying to read in some data from Cassandra in a column family called
DemoCF. I am able to connect to the cluster, describe and use the keyspace.
*> conn <- RC.connect(host='', port=)*
*> RC.describe.keyspace(conn2, 'DemoKS')*
$name
[1] "DemoKS"
$strategy_class
[1] "org.apache.cassandra.l
I am trying to count the number of times that the characters in a string
change
For example-
new=c(BABBBABB)
I want to find the number of times that B changes to A and the number of
times that A changes to B. I tried the grep command but I only figured out
the positions of when B changes to
Hi All,
I want to join a table (Dataset) that is created in R with a table that is
in oracle database. Can some one help me in accomplishing this task in R?
Example Code:
library(RODBC)
DB_CONNECT <- odbcConnect("DSN_NAME")
TABLE_JOIN <- sqlQuery(DB_CONNECT, "SELECT * FROM DB_TABLE WHERE COL_1
On Thu, Aug 2, 2012 at 4:52 PM, Ayyappa Chaturvedula
wrote:
> Michael,
> Thank you for this , it worked. I was thinking "by" is a required argument
> in merge function.
>
Well, it is "required" in a strict sense, but it has a default value
(defaulting to the shared names) so _you_ don't have to
On Thu, Aug 2, 2012 at 3:47 PM, danobolg321 wrote:
> This is a fairly entry level question - so any guidance on this would be
> appreciated...
>
> I have two sets of residuals, one of 100 values (A), and one of 10,000
> values (B). There are 50 paired sets of these values. The data is in the
> for
You can merge without a common variable:
x <- data.frame(x = 1:5)
y <- data.frame(y = 6:10)
merge(x,y)
works just fine (for one set of expectations)
If you need more help, please do make a reproducible example (as
requested of all R-help posts): the instructions here might help
http://stackove
See R FAQ 7.22 for the gory details -- It needs to go around the
qplot() statement.
In interactive use, most objects are "printed" after evaluation:
sometimes this doesn't happen because the returned value is marked
invisible. For grid graphics objects (of which ggplot objects are a
type), this "p
Hi Antonio,
It's hard to know for sure what the problem is, because we don't know
what your vef1 data.frame contains. My guess is that edad11 is not
numeric, so never equals 0 (although it might equal the character
string "0").
If you paste the results of str(vef1) in your reply we can see if my
Dear All,
I want to create a dataset for a NONMEM simulation. I have a dataframe with
individual PK parameters and want to create a dosing sceinario in a second
dataframe. I want to merge them both so that every individiual's PK
parameters are combined with the dosing scenario into one. I do not
This is a fairly entry level question - so any guidance on this would be
appreciated...
I have two sets of residuals, one of 100 values (A), and one of 10,000
values (B). There are 50 paired sets of these values. The data is in the
form of two matrices (100x50 and 10,000x50).
The goal is to corre
Thanks for the info, but with the anova () is obtained p_valor of Wald test
for the variable level of education, I want to see
the nievel p_valor in view of this variable as a dummy, so I wonder if in
summary (mod. logis) the Pr (> | t |)
lets me know if each level is significant in the mod
Dear Diana,
The Anova() function in the car package will produce a Wald test for each
term in a model fit by svyglm().
I hope this helps,
John
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Diana Marcela Martinez Ruiz
> Sen
Could you elaborate on this ? Where would the print /plot statement go ?
On Thu, Aug 2, 2012 at 4:38 PM, Michael Weylandt wrote:
> You need to explicitly print() the graphs to make them show -- or in
> recent ggplot2 versions I think you can also use plot() as an alias.
>
> Michael
>
> On Aug 2,
On Thu, Aug 2, 2012 at 1:23 PM, Abdul wrote:
> Hi everybody
> I need help to solve the following problem in finite element
We don't do homework on this list. Especially homework that has
essentially nothing to do with R.
-- Bert
> A field variable f(x,y)=xᵌ y is defined over a rectangle domain
On Aug 2, 2012, at 3:23 PM, Abdul wrote:
> Hi everybody
> I need help to solve the following problem in finite element
> A field variable f(x,y)=xᵌ y is defined over a rectangle domain
> Ω={K: 0≤x≥4 , 0≤y≥6” Given the expression
> g=∬_(0 0)^(6 4)▒〖X^3 Y dx dy〗
> And assume the following
On 02/08/2012 19:13, Jie wrote:
Dear All,
I am learning parallel in R and start with the package "snow". I did a test
about running time and the parallel version is much slower than the regulat
code. My laptop is X200s with dual core intel L9400 cpu.
The OS matters far more. Overheads on Wind
You need to explicitly print() the graphs to make them show -- or in recent
ggplot2 versions I think you can also use plot() as an alias.
Michael
On Aug 2, 2012, at 2:37 PM, "firdaus.janoos" wrote:
> Hello,
>
> I'm having some issues getting a ggplot figure to show up in the knitr
> output,
Abdul, this list is not for answering your homework questions.
On Thu, Aug 2, 2012 at 4:23 PM, Abdul wrote:
> Hi everybody
> I need help to solve the following problem in finite element
> A field variable f(x,y)=xᵌ y is defined over a rectangle domain
> Ω={K: 0≤x≥4 , 0≤y≥6” Given the expression
On 02/08/2012 18:06, K. Elo wrote:
Hi!
On Thu, 2 Aug 2012, Uwe Ligges wrote:
R.h should be part if your R installation, given the output above
probably in /usr/lib64/R/include ?
But there is no such file "R.h" in my system???
~$ locate \/R.h
~$
Or with 'find':
# find / -name R.h
#
Any ide
Hi everybody
I need help to solve the following problem in finite element
A field variable f(x,y)=xᵌ y is defined over a rectangle domain
Ω={K: 0≤x≥4 , 0≤y≥6” Given the expression
g=∬_(0 0)^(6 4)▒〖X^3 Y dx dy〗
And assume the following bilinear interpolation shape functions are used to
discr
Hello,
I'm having some issues getting a ggplot figure to show up in the knitr
output, when placed in a loop.
Specifically, I have a loop inside a knitr chunk :
```{r fitting, warning=FALSE, fig.width=10, fig.height=10, fig.keep='high'}
for (t in 1:T)
{
# do a regression of tgt.vals ~ predi
Hello
I want to know if the summary of the logistic model with survey
Pr (> | t |) to test if the coefficient of the model is significant,
ie is the p_valor wald test for the model coefficients,
for I am interested to know if the three levels of the variable educational
level are significant
Dear All,
I have some questions regarding using the genpoisson{VGAM} .
genpoisson(llambda="elogit", ltheta="loge",
elambda=if(llambda=="elogit") list(min=-1,max=1) else list(),
etheta=list(), ilambda=NULL, itheta=NULL,
use.approx=TRUE, method.init=1, zero=1)
and
Hi Arun,
if you name the data.frame in the list, e.g.
mango <- list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0("a",1:6),target=data.frame(coconut=1:6))
banana <- list(target=data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0("a",1:5))
pineapple <- list(target=d
On Aug 2, 2012, at 12:00 PM, arun wrote:
Hello,
You need to use,
subset(vef1,vef1$edad1==0)
That should not be necessary (and quite defeats the usual value of
subset()'s nonstandard evaluation):
> vef1<-read.table(text="
+ age edad1
+ 12 1
+ 10 0
+ 9 0
+ 14 1
+ 15 1
Hi,
I have a question regarding whether it is possible to do post hoc tests on a
model fit with GAM {mgcv}. My response variable is abundance (no.
individuals per plot), and I have one continuous predictor (light) and one
factor (height) which includes 7 levels.
> mod2=gam(log_abundance~s(light
Dear All
I am trying to perform leave-one-out cross validation on a logistic
regression model using cv.glm from the boot package in R.
As I understand it, the standard cost function:
cost<-function(r,pi=0) mean(abs(r-pi)>0.5)
Uses a 50% risk threshold to classify cases as positive or negative a
Hello,
You need to use,
subset(vef1,vef1$edad1==0)
vef1<-read.table(text="
age edad1
12 1
10 0
9 0
14 1
15 1
",sep="",header=TRUE)
subset(vef1,vef1$edad1==0)
age edad1
2 10 0
3 9 0
A.K.
- Original Message -
From: "antonio.bso...@sapo.pt"
To: r-h
On Aug 1, 2012, at 8:09 PM, B787s wrote:
well, the article from this high impact factor journal mentioned
about each
step for building a nomogram including interpretation without going
too much
detail into statistical concept.
All I want to know is what the R code does to come up with the
Hmmm a trivial mockup suggests this should work:
x <- factor(sample(c(0, 1), 30, TRUE)
x == 0
so perhaps you could work up a reproducible example for us: see this
site for more details on how to do so:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Be
Dear R Users,
Recently I began to use R. I`m interested about comparing several
regression curves. REcently I found the package sm and the function
sm.ancova which I understand allows me to do this. I applied this function
to my data (seven regression curves) and I found that there are significant
I'm not surprised -- you're doing a fairly trivial calculation so the
overhead of setting up the parallelization is likely more than the
benefit from halving the computation time. Do it on a larger
calculation and I imagine the performance will be better -- I might
just be pulling this out of thin
I ran a negative binomial logit hurdle model and am now trying to plot the
effects of a continuous predictor variable (the only variable in my model)
on the count and zero component and the overall mean response. I'm confused
because for some values, the predicted overall mean is higher than the m
Hi Elk,
I tried to test with another case where coconut is in different position in
the sublist.
mango <- list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0("a",1:6),data.frame(coconut=1:6))
banana <- list(data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0("a",1:
Hello,
I was in a hurry at the time, when I replied.
You could do one thing: extract sublist using:
list4 <- list(mango, banana, pineapple)
b1<-list()
for(i in 1:3){
b1[[i]]<-list()
b1[[i]]<-list4[[i]][1]
}
b1
b<-list()
for(i in 1:3){
b[[i]]<-list()
b[[i]]<-lapply(b1[[i]][[1]],FUN=functio
Hi Elk,
Thanks for the input. Somehow, I got entangled with which.max. I should
have used max.
lapply(list4,function(x) max(x[[1]]$coconut))
[[1]]
[1] 6
[[2]]
[1] 18
[[3]]
[1] 9
max(unlist(lapply(list4,function(x) max(x[[1]]$coconut
#[1] 18
A.K.
- About the visualization, my question is more about interpretation. In the
case of :
model_name <- gam ( bm ~ t + te (t_year, temp_W, temp_sept, k = 5, bs = c(
“cc”,”cr”,”cr”)), data = data)
* a)* vis.gam (model_name , view= c(“t_year”, “temp_W”))
*b)* vis.gam (model_name , view= c(“t_year
Can you tell me what does the option "zero" in genpoisson{VGAM} mean?
zero An integer vector, containing the value 1 or 2. If so, *ë* or
*theta*respectively are modelled as an intercept only. If set to
NULL then both linear/additive predictors are modelled as functions of the
explanatory variable
Hi all,
I have made a lot of attempts with ts package and arima model for
forecasting.
Now I am looking on zoo package.
I follow Gabor's steps from another post on leap years
https://stat.ethz.ch/pipermail/r-help/2011-February/269069.html
and I am confused in that step:
# put dates (without Feb
Hi
Need a little help. This is easy question, but i´m new to R.
I want to subset a data frame called vef1 using a variable that is a factor
called edad11, 0 if under 11 years, 1 otherwise.
I tried:
vef1.sub<-subset(vef1, edad11==0)
The code runs correctly but i get a empty dataframe.
If i use thi
Dear All,
I am learning parallel in R and start with the package "snow". I did a test
about running time and the parallel version is much slower than the regulat
code. My laptop is X200s with dual core intel L9400 cpu.
Should I make more clusters than 2? Or how to improve the performance?
# insta
On Thu, Aug 02, 2012 at 02:27:43AM -0700, khris wrote:
>
> On Aug 2, 2012, at 12:39 PM, Petr Savicky [via R] wrote:
>
> > On Wed, Aug 01, 2012 at 04:55:30AM -0700, khris wrote:
> > > Hi Petr,
> > >
> > > It been sometime since I wrote. But here are the latest developments.
> > >
> > > When I
On 02/08/12 18:02, Bert Gunter wrote:
Well, geez! Without trying to upstage SImon, why not google it yourself?!
http://en.wikipedia.org/wiki/Akaike_information_criterion
-- one or two slightly surprising statements in there though :-)
I quite like the coverage in Davison (2003) "Statistical Mod
On Thu, Aug 2, 2012 at 5:58 AM, loyolite270 wrote:
> oh sorry ..
>
> The detailed description of the problem is given below
>
> dataFrame is matrix of dim 100X100 with some values
Odd name for something that's not a dataFrame (i.e., data.frame != matrix)
> a is a vector of length 1
> x is a
I still try to figure out, what you are finally asking for, but maybe
res<-sapply(list4,function(x)max(x[[1]]$coconut))
names(res)<-c("mango", "banana", "pineapple")
res
max(res)
is worth a try?
Especially i did not get the point of doing something like
x[which.max(x)] because this is in any cas
On Thu, Aug 2, 2012 at 8:16 AM, Cloneberry wrote:
> Hi,
> just during these vacation days, I'm trying to approach with multicore
> package
> and I have some troubles with foreach.
> What I'm trying to do is to extract a data in coordinate (ii,jj) from a
> matrix2,
> only if the data in the same co
Bert, thanks for the very useful link. I am duly chastised for not
having searched thoroughly enough.
Will
On 8/2/2012 10:02 AM, Bert Gunter wrote:
Well, geez! Without trying to upstage SImon, why not google it yourself?!
http://en.wikipedia.org/wiki/Akaike_information_criterion
(seemed pret
Hey all,
I was wondering about the order of execution of functions and found that, in
fact, I should look at the topic of buffering output to the console.
Which I did - below are some links of related topics, in case anyone wants to
read up on it.
I decided to change the concerning lines in t
Hi!
On Thu, 2 Aug 2012, Uwe Ligges wrote:
R.h should be part if your R installation, given the output above
probably in /usr/lib64/R/include ?
But there is no such file "R.h" in my system???
~$ locate \/R.h
~$
Or with 'find':
# find / -name R.h
#
Any ideas?
Kind regards,
Kimmo
__
Well, geez! Without trying to upstage SImon, why not google it yourself?!
http://en.wikipedia.org/wiki/Akaike_information_criterion
(seemed pretty clear to me...)
-- Bert
On Thu, Aug 2, 2012 at 9:54 AM, Will Shadish wrote:
> Simon, could you clarify this paragraph. We are submitting to a psych
On Aug 2, 2012, at 16:01 , Sara S wrote:
> Dear all
>
> I'm trying to caculate the MLEs for parameters of Inverse Gaussian
> distribution (in a k-sample problem with common mean) by using
> EM-Algorithm. I found some package for EM-Algorithm that are useful
> for missing or incomplete data and
Simon, could you clarify this paragraph. We are submitting to a
psychology journal and I am certain they will be asking us about why it
is still ok to use AIC for non-nested models. Thanks. Will Shadish
On 8/2/2012 9:49 AM, Simon Wood wrote:
For AIC model comparison, the usual advice applies th
Ricardo,
Your second construction (b) is the correct one (in a you are asking for
one marginal to be a 2 dimensional cubic regression spline, which
doesn't exist in mgcv).
For visualization, would the example at the end of ?te be the thing to
do? In 3d I find looking at a series of 2d slices
You can use a table to compare your predicted values and the response.
Here is an example: http://www.cyclismo.org/tutorial/R/tables.html
On Thu, Aug 2, 2012 at 5:52 AM, Abraham Mathew wrote:
> I'm developing a naive bayes in R. I have the following data and am trying
> to predict on returned (cl
Sorry. I've used:
> library(rms)
I realise I still have a lot to learn to ask questions well - it took me a
long time to compile this one, but I've obviously missed important things.
Please see below for the session info.
> sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: i386-pc-mingw32/i
Hi Michael,
Thank you so much for the help Michael! When I put my package 'PKG2' (on which
PKG1 is depending on) in the regular place, perhaps a place R CMD check won't
look at, then when I try to build PKG1, they will say PKG2 required but not
found, so I have to put it under a directory R look
Hi Josh,
Thank you so much for your detailed explanation. It's very clear, and now I can
understand the difference between depends and imports.
Thank you so much for your help!! I really really appreciate it!
Yours,
Xuan
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com
Dear all
I'm trying to caculate the MLEs for parameters of Inverse Gaussian
distribution (in a k-sample problem with common mean) by using
EM-Algorithm. I found some package for EM-Algorithm that are useful
for missing or incomplete data and are not helpful for solving my
problem.
(Exactly, the p
Hey guys,
I am using the "urca" and "vars" package to estimate a VECM.
I already estimated the VECM with the ca.jo, put restrictions on the
cointegration and loading matrix via "alrtest", "blrtest" and "ablrtest" and
finally reesimated the restricted VECM with cajorls.
Is it possible to transfor
Hi,
Try this:
Here, in the example dataset, you have 3 banks. Suppose, I want to delete 2
banks randomly out of 3, (you can name your specific banks to delete),
dat1<-read.table(text="
Year Name totalliabilties assets
1990 a 90 10
1991 a 89 48
1
This discussion needs to be taken off (this) list, as it appears to
have nothing to do with R.
-- Bert
On Thu, Aug 2, 2012 at 2:27 AM, khris wrote:
>
> On Aug 2, 2012, at 12:39 PM, Petr Savicky [via R] wrote:
>
>> On Wed, Aug 01, 2012 at 04:55:30AM -0700, khris wrote:
>> > Hi Petr,
>> >
>> > It
Hello,
If I understand it now, can't you do this:
Individual cases:
list4<-list(mango[[1]],banana[[1]],pineapple[[1]])
b<-list()
for(i in 1:3){
b[[i]]<-list()
b[[i]]<-lapply(list4[[i]],FUN=function(x)x[which.max(x)])
}
b1<-data.frame(do.call(rbind,b))
row.names(b1)<-c("mango","banana","pineap
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