Dear r-help mailing list,
Some colleagues and I are working on a series of research studies related
to mailing list, and Stack Overflow. While I do understand that this
announcement would be technically off topic - it's about a study that
could involve R-Help users. So, please, If you have time an
Hi: You'd have to provide a dput of "model2" and "Country" for anyone to
give a definitive answer but my guess is that you have an orthogonal X
matrix which is causing you to fit the
model perfectly which causes the model residuals to be zero.
Also, you didn't explain what you're doing but modelli
On Oct 1, 2015, at 8:29 PM, Rolf Turner wrote:
> On 02/10/15 15:47, David Winsemius wrote:
>
>
>
>> On Oct 1, 2015, at 6:22 PM, Rolf Turner wrote:
>>>
>>> P.S. I have been unable to find a corresponding vector of the names
>>> of the days of the week, although I have a very vague recollection
On 02/10/15 15:47, David Winsemius wrote:
On Oct 1, 2015, at 6:22 PM, Rolf Turner wrote:
P.S. I have been unable to find a corresponding vector of the names
of the days of the week, although I have a very vague recollection
of the existence of such a vector. Does it exist, and if so what
is
On Oct 1, 2015, at 6:22 PM, Rolf Turner wrote:
> On 02/10/15 10:54, peter dalgaard wrote:
>
>>> On 01 Oct 2015, at 23:04 , Rolf Turner
>>> wrote:
>>>
>>> On 02/10/15 03:45, David L Carlson wrote:
>>>
>>>
>>>
If you want the month names:
> mnt <- c("Jan", "Feb", "Mar", "Apr",
I have a tkwidget table (say, tbl1) that may be reconfigured at various times
depending on user input. Is there an easy way to later extract table
properties? Something like...
nrow<-tkgetproperties(tbl1, rows)
Muchas thanks in advance.
-Dan
--
View this message in context:
http://r.789695.n
On 02/10/15 10:54, peter dalgaard wrote:
On 01 Oct 2015, at 23:04 , Rolf Turner
wrote:
On 02/10/15 03:45, David L Carlson wrote:
If you want the month names:
mnt <- c("Jan", "Feb", "Mar", "Apr", "May", "Jun",
+ "July", "Aug", "Sep", "Oct", "Nov", "Dec")
dimnames(tbl)$Month <- mnt
U
Dear All,
I am trying to use simulation to test mahalanobis distance matching method.
Somehow I do not think Matchit is giving me what I want
Below is the code:
n<-100
x1_contr<-runif(n,0,5)
x2_contr<-runif(n,0,5)
x_contr<-cbind(x1=x1_contr,x2=x2_contr)
x1_treat<-runif(n,1,6)
x2_treat<-runif(n,1,
Hi all,
I am trying to calculate the variance-covariance matrix for parameter Beta
under the null (Ho) using the "prop.odds" function in the timereg package.
In other words, I am looking for Var(Beta under the null).
For the Cox PH model, I used the "vcov" function and did the following:
I’m trying to estimate a model regressing the residuals on the country dummies
as follows;
model.resC <- lm(model2$res ~ as.factor(Country))
summary(model.resC)
As I call the model I get the following results regarding the residuals:
"ALL 90 residuals are 0: no residual degrees of freedom!"
Wha
> On 01 Oct 2015, at 23:04 , Rolf Turner wrote:
>
> On 02/10/15 03:45, David L Carlson wrote:
>
>
>
>> If you want the month names:
>>
>>> mnt <- c("Jan", "Feb", "Mar", "Apr", "May", "Jun",
>> + "July", "Aug", "Sep", "Oct", "Nov", "Dec")
>>> dimnames(tbl)$Month <- mnt
>
>
>
> Unnecessary
On 02/10/15 03:45, David L Carlson wrote:
If you want the month names:
mnt <- c("Jan", "Feb", "Mar", "Apr", "May", "Jun",
+ "July", "Aug", "Sep", "Oct", "Nov", "Dec")
dimnames(tbl)$Month <- mnt
Unnecessary typing; there is a built-in data set "month.abb" (in the
"base" package) that is
Replacing na.omit() with !is.na() appears to improve performance with time.
rm(list=ls())
test1 <- (rbind(c(0.1,0.2),0.3,0.1))
rownames(test1)=c('y1','y2','y3')
colnames(test1) = c('x1','x2');
test2 <- (rbind(c(0.8,0.9,0.5),c(0.5,0.1,0.6)))
rownames(test2) = c('y2','y5')
colnames(te
On Oct 1, 2015, at 2:26 AM, kirsada wrote:
> Please bear with me, I am very new to R.
>
> My question is regarding the use of the improveProb function in the Hmisc
> package. I have two logistic models, the only difference being that the
> second model contains my novel marker of interest. I am
Hello,
I studied the effect of a hurricane in Cozumel on understory birds. I
have bird abundances (i.e. counts) registered always on the SAME six
sites (i.e. blocks). I have data for: before the hurricane, first year
after the hurricane, second year after the hurricane. I each of these
time periods
I have limited statistical experience from my coursework in undergrad running
simple linear regressions and performing chi-square tests. I have some data,
~5000 survey results on individuals, each with a score from a scale of 1-12 on
how security conscious they are (determined by their answers t
I reworked Frank Schwidom's solution to make it shorter than its original
version.
test1 <- (rbind(c(0.1,0.2),0.3,0.1))
rownames(test1)=c('y1','y2','y3')
colnames(test1) = c('x1','x2');
test2 <- (rbind(c(0.8,0.9,0.5),c(0.5,0.1,0.6)))
rownames(test2) = c('y2','y5')
colnames(test2) = c(
On Thu, 1 Oct 2015, Marc Schwartz wrote:
On behalf of The R Foundation for Statistical Computing, this is an
announcement that, effective October 15, 2015, the Nabble online forums
will no longer be a supported vehicle for posting new threads and/or
replying to existing threads on R-Help.
+2
You should always reply to the list since other posters may have other
suggestions. Assuming your data frame is called rain:
> str(rain)
'data.frame': 2192 obs. of 4 variables:
$ Year : int 1960 1960 1960 1960 1960 1960 1960 1960 1960 1960 ...
$ Month : int 1 1 1 1 1 1 1 1 1 1 ...
$ Day
Hi
Assuming you data.frame is dat
then you have to restrict the data going to a function to be counted so that it
counts values > 0
with(dat, tapply(Amount, list(Year, Month), function(x) length(x[x > 0])) )
gives
1 2 3 4 5 6 7 8 9 10 11 12
1960 24 15 2 12 19 22 18 24 22 20 30 29
I would recommend that you use auglag() rather than constrOptim.nl() in the
package "alabama." It is a better algorithm, and it does not require feasible
starting values.
Best,
Ravi
-Original Message-
From: Rainer M Krug [mailto:rai...@krugs.de]
Sent: Thursday, October 01, 2015 3:37
Envoyé de mon iPhone
> Le 1 oct. 2015 à 15:17, Ravi Varadhan a écrit :
>
> I would recommend that you use auglag() rather than constrOptim.nl() in the
> package "alabama." It is a better algorithm, and it does not require
> feasible starting values.
Thanks - that was one question I wanted
library(dplyr)
df <- data.frame(z = rep(c("A", "B")), x = 1:6, y = 7:12) %>%
arrange(z)
temp <- reshape(df, v.names = c("x", "y"), idvar = c("x", "y"), timevar = "z",
direction = "wide")
lA <- na.omit(temp[,c("x.A", "y.A")])
lB <- na.omit(temp[,c("x.B", "y.B")])
df.long <- as.data.frame(cbind(lA,
Dear R-help mailing list (and Emmanuel, the ape package maintainer),
I would like to change a non binary phylo object to hclust, but this does
not seem to work smoothly.
Here is a small R code to demonstrate the problem:
# an hclust tree with 3 branches from the root
hc <- hclust(dist(c(1:2, 4,5
Dear Michael,
You'll need a combination of dplyr and tidyr
library(dplyr)
library(tidyr)
data.frame(id = rep(1:3, 2), z = rep(c("A", "B")), x = 1:6, y = 7:12) %>%
arrange(z) %>%
gather(variable, value, -z, -id) %>%
mutate(newcol = paste(z, variable, sep = ".")) %>%
select(-z, -variable)
Thanks a lot Giorgio for your Help
Regards,
Maram
On 1 October 2015 at 14:27, Giorgio Garziano
wrote:
> If you are running a 32-bit Windows, there are following upper limits:
>
>
>
>
> https://cran.r-project.org/bin/windows/base/rw-FAQ.html#There-seems-to-be-a-limit-on-the-memory-it-uses_0021
>
I have a data frame with a structure similar to the following. The variable
z is a grouping variable; x and y are measurement variables.
library(dplyr)
df <- data.frame(z = rep(c("A", "B")), x = 1:6, y = 7:12) %>%
arrange(z)
z x y
1 A 1 7
2 A 3 9
3 A 5 11
4 B 2 8
5 B 4 10
6 B 6 12
I nee
I am new to GBM and I am trying to run it on "train.1" dataset(dim(train.1)
3994743) and checking predictions on "test"(dim(test) 2 192) using the
codes:
gbmFit1 <- train(as.factor(train.1$Labels)~., data = train.1[,-43], method =
"gbm", trControl = fitControl,verbose = FALSE)
gbm_dev
If you are running a 32-bit Windows, there are following upper limits:
https://cran.r-project.org/bin/windows/base/rw-FAQ.html#There-seems-to-be-a-limit-on-the-memory-it-uses_0021
starts by:
memory.limit(size=1920)
and try increasing value of size as a parameter for memory.limit().
I use Int
The “4096” was just an example.
Try:
memory.limit(size=3968)
Furthermore, to overcome memory size limits vs. in memory R data management
beyond your 4Gb,
you may explore package “ff”.
--
Cheers,
GG
From: Maram SAlem [mailto:marammagdysa...@gmail.com]
Sent: giovedì 1 ottobre 2015 14:12
To: G
Thanks a lot Giorgio, I used
memory.limit(size=4096)
but got
don't be silly!: your machine has a 4Gb address limit
I'm working on my Ph.D. thesis and I have a huge code of which this is just
a very small part, so does this error mean that I need a new computer with
extended capabilites to be a
Check your memory size by:
memory.limit()
try to increase it by:
memory.limit(size=4096)
From: Maram SAlem [mailto:marammagdysa...@gmail.com]
Sent: giovedì 1 ottobre 2015 13:22
To: Giorgio Garziano
Cc: r-help@r-project.org
Subject: Re: [R] (subscript) logical subscript too long
Thanks Giorgi
I got it sorted thank you. As you point out, the original code works fine
(there was a type mismatch with it).
On 1 Oct 2015, at 12:00, PIKAL Petr wrote:
Hi
Are agents and agent different objects? Or it is a typo? Better would be to
provide some real or fake data by dput or at least result o
Thanks Giorgio, I got it.
I managed to reach the matrix s whose rows represent all the possible
combinations. Here is the code:
> n=12
> m=7
> D<-matrix(0,nrow=n-m+1,ncol=m-1)
> for (i in 1:m-1)
+ {
+ D[,i]<-seq(0,n-m,1)
+ }
> ED <- do.call(`expand.grid`,as.data.frame(D))
> ED<-as.matrix(ED)
+1
On Thu, Oct 1, 2015 at 5:55 AM, Marc Schwartz wrote:
> Greetings all,
>
> On behalf of The R Foundation for Statistical Computing, this is an
> announcement that, effective October 15, 2015, the Nabble online forums
> will no longer be a supported vehicle for posting new threads and/or
> repl
Hi Sarah
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Sarah
> Goslee
> Sent: Thursday, October 01, 2015 1:01 PM
> To: Thomas Chesney
> Cc: r-help@r-project.org
> Subject: Re: [R] Increment certain values in a vector
>
> Hi,
>
> On Thursday, Octobe
Hi,
On Thursday, October 1, 2015, Thomas Chesney <
thomas.ches...@nottingham.ac.uk> wrote:
> How can I manipulate values of only certain entries in a vector, based on
> what the existing values are?
>
> So for instance if I want to add one to each of the following values, or
> multiply them by a
Hi
Are agents and agent different objects? Or it is a typo? Better would be to
provide some real or fake data by dput or at least result of str(agent) and
str(agents)
Actually your code shall work in case agent and agents have same length and
expected order.
agents[which(agent$membership == 1
Greetings all,
On behalf of The R Foundation for Statistical Computing, this is an
announcement that, effective October 15, 2015, the Nabble online forums will no
longer be a supported vehicle for posting new threads and/or replying to
existing threads on R-Help.
This decision was not made lig
Please bear with me, I am very new to R.
My question is regarding the use of the improveProb function in the Hmisc
package. I have two logistic models, the only difference being that the
second model contains my novel marker of interest. I am trying to calculate
NRI and IDI to compare models.
I h
How can I manipulate values of only certain entries in a vector, based on what
the existing values are?
So for instance if I want to add one to each of the following values, or
multiply them by a specific factor:
agents[which(agent$membership == 1)]
how would I do this please?
If there was ju
Ravi Varadhan writes:
> Hi Rainer,
> It is very simple to specify the constraints (linear or nonlinear) in
> "alabama" . They are specified in a function called `hin', where the
> constraints are written such that they are positive.
OK - I somehow missed the part that, when the values x are val
Hi
keep the conversation on list, others can help you too. Your dput sending was
OK but HTML post tends to twist the message content unexpectedly so it negated
your effort.
I still insist that
aggregate(mydat$amount>0.001, list(mydat$month, mydat$year), sum)
gives you the answer you want. Di
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