Spencer:
Hmmm
Well, I don't get what's going on here -- as.character.default is
internal -- but could you method-ize a simple synonym:
asChar<- function(e,...)UseMethod("asChar")
asChar.call <- function(e,...)deparse(e,...)
asChar.default <- function(e,...)as.character(e,...)
> asChar(xDy)
[
I just started to use the Glmnet function for doing analysis of multinomial
data. See the attached code for the analysis of the Fisher iris data as an
example. My question is. When I want to determine the coefficients for the best
lambda value, it returns more than one set of coefficents.. Which
Hallo,
I have a table in which I would like to insert the min and max values of
another column (date and time in as.POSIXct Format).
This is my row table, where "su" and "sa" are still the same as "Vollzeit":
head(treat)
Vollzeit Datum Zugnacht su
Hallo,
I have a table in which I would like to insert the min and max values of
another colum (date and time in as.POSIXct Format).
This is my row table, where "su" and "sa" are still the same as "Vollzeit":
head(treat)
Vollzeit Datum Zugnacht su
Hi y'all,
I'm using quantmod's getSymbols function, which retrieves data in XTS
format.
I'm trying to pass "IBM" into the "ticker" variable, then write the table
referencing "ticker." However, when I run the write.table command, it
writes "IBM", not the data inside IBM.
Do you have any thoughts
On Thu, 8 May 2014 09:01:06 PM Jason Rupert wrote:
> > split(df, Status)
>
> Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) :
> object 'Status' not found
>
> > names(PCR_duplicatedCheck_df)
>
> [1] "Key" "MinCreated" "MaxUpdated"
> [4] "Status"
>
> I am tot
Luigi
I suspect that it is in accessing the correct groups within the panel function
I suggest that printing the values before plotting within the panel function
may give you some idea.
Without specific data I cannot suggest anything else
Duncan
-Original Message-
From: Luigi Marongiu
> split(df, Status)
Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) :
object 'Status' not found
> names(PCR_duplicatedCheck_df)
[1] "Key" "MinCreated" "MaxUpdated"
[4] "Status"
I am totally confused...what do I need to try next?
___
On 5/8/2014 8:05 PM, Bert Gunter wrote:
[1] "x$y"
Spencer:
Does
deparse(substitute(x$y))
[1] "x$y"
do what you want?
No: The problem is methods dispatch. class(quote(x$y)) =
'call', but as.character(quote(x$y)) does NOT go to "as.character.call".
deparse(quote(x$y)) retur
[1] "x$y"
Spencer:
Does
deparse(substitute(x$y))
[1] "x$y"
do what you want?
Cheers,
Bert
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Thu,
"as.character.call" seems not to work as an alias for deparse.
Consider the following:
xDy <- quote(x$y)
class(xDy)
"call"
as.character.call <- function(x, ...)deparse(x, ...)
as.character(xDy)
[1] "$" "x" "y"
# fails
str(xDy)
# language x$y
as.character.language <- function(x, ...)"langua
Jenushka tulane.edu> writes:
>
> I'm a beginning R user.
>
> The data: Volume of nectar in flowers under 4 different treatments,
> nested
> for individual (measures were taken mutliple times from different
> flowers of
> the same individual- never the same flower).
This is really more of
I cannot run your code because not all the variables are defined, but
try not doing the nested replacements, rftab$masskg[...] <- newValue,
in the loop. Instead, pull out masskg as a new stand-along object
before the start of the loop and put it back into rftab at the end of
the loop. E.g., someth
Hello everybody,
I have written a nested for-loop, but as length(uc) > 170,000, this would
take VERY long. I have tried to use sapply or something but I cannot get it
to work, I would be happy if someone could point out to write this more
efficiently. Thank you all,
Ludwig
ergsens <- data.frame(
On May 8, 2014, at 9:49 AM, Abhinaba Roy wrote:
> Hi R helpers,
>
> I have a dataframe like
>
> ID Yr_Mnth AMT_PAID AMT_DUEpaidToDue
> CS0026A201301 320.48 19040.168319328
> CS0026A2013024881.31157080.310753119
> CS0026A2
Hi R helpers,
I have a dataframe like
ID Yr_Mnth AMT_PAID AMT_DUEpaidToDue
CS0026A201301 320.48 19040.168319328
CS0026A2013024881.31157080.310753119
CS0026A2013037609.04255850.297402384
CS0026A201304
Hi,
May be:
indx <- !duplicated(as.character(interaction(df1[,-3])))
merge(df1[indx,],df2)
A.K.
On Thursday, May 8, 2014 12:34 PM, Massimo Bressan
wrote:
yes, thank you for all your replies, they worked out correctly indeed...
...but because of my fault, by then working on my real data I ful
yes, thank you for all your replies, they worked out correctly indeed...
...but because of my fault, by then working on my real data I fully
realised that I should have mentioned something that is changing (quite
a lot, in fact) the terms of the problem...
please would you consider the follow
Thanks. I guess I could have searched for that. Apologies.
I'll have to try it on my tablet.
Kevin
On 05/08/2014 11:57 AM, Jeremy Miles wrote:
It exists:
https://play.google.com/store/apps/details?id=com.appsopensource.R
No graphics.
Jeremy
On 8 May 2014 05:44, Kevin E. Thorpe mailto:ke
It exists:
https://play.google.com/store/apps/details?id=com.appsopensource.R
No graphics.
Jeremy
On 8 May 2014 05:44, Kevin E. Thorpe wrote:
> This is a question asked purely out of idle curiosity (and may also be in
> wrong list). Are there plans for porting R to Android devices or
> chro
Hi Mark,
The problem here is that the constructor expects there to be at least
one observation per location. The nb.l list has neighbourhood
information for 166 locations, while the 'obs' data contains
observations for only 99 of them (unique(obs$xy.idx)).
The solution probably requires more
Hi,
May be this helps:
merge(unique(df1),df2)
A.K.
On Thursday, May 8, 2014 5:46 AM, Massimo Bressan
wrote:
given this "bare bone" example:
df1 <- data.frame(id=rep(1:3,each=2), item=c(rep("A",2), rep("B",2),
rep("C",2)))
df2 <- data.frame(id=c(1,2,3), who=c("tizio","caio","sempronio"))
Hi Roger and Simon,
Thanks for the replies. Simon's suggestion of an isolated or missing
neighbourhood doesn't hold either.
I've attached the code below - its my attempt to solve the FELSPLINE
sausage using mrf rather than a soap smoother. Its a bit convoluted, but
should run ok. I thought this w
Hi Mark,
I'm not sure what is happening here - there is no chance that nb.l
contains a neighbourhood not in the levels of obs$xy.idx, I suppose?
i.e. is
all(names(nb.l)%in%levels(obs$xy.idx))
also TRUE? Here is some code illustrating what nb should look like (and
in response to Roger Bivand
Thank you very much. This was exactly what I was looking for.
--
View this message in context:
http://r.789695.n4.nabble.com/Outlier-Detection-with-k-Means-tp4690098p4690186.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-pr
This is a question asked purely out of idle curiosity (and may also be
in wrong list). Are there plans for porting R to Android devices or
chromebooks? Maybe it's as simple as compiling the source, but I don't
know what tools are available.
One of the current advantages of R is it runs on all
Hello,
There are some alternatives without using sqldf or another package.
1.
tmp2 <- aggregate(item ~ id, data = df1, FUN = unique)
Then merge() like you've done.
2.
tmp3 <- merge(df1, df2)
tmp3[!duplicated(tmp3), ]
Hope this helps,
Rui Barradas
Em 08-05-2014 10:44, Massimo Bressan escr
Mark Payne gmail.com> writes:
>
> Hi,
>
> Does anyone have an example of a Markov Random Field smoother (MRF) in MGCV
> where they have specified the neighbourhood directly, rather than supplying
> polygons? Does anyone understand how the rules should be? Based on the
> columb example, I have s
given this "bare bone" example:
df1 <- data.frame(id=rep(1:3,each=2), item=c(rep("A",2), rep("B",2),
rep("C",2)))
df2 <- data.frame(id=c(1,2,3), who=c("tizio","caio","sempronio"))
I need to group the first dataframe "df1" by "id" and then merge with
the second dataframe "df2" (again by "id")
dear all members
is there anyone explain to me the code below and how can i transfer this
code to winbugs program.
q[i,1]=qnorm(runif(1,min=.5,max=1),0,1)
thanks in advance
thanoon
[[alternative HTML version deleted]]
__
R-help@r-project.org
thanks Michael.
That fits in with the response from JMP and with the experience of colleagues.
Guess I'll just stick with R :)
cheers Bob
-Original Message-
From: Meyners, Michael [mailto:meyner...@pg.com]
Sent: 08 May 2014 07:20
To: Samuel J Gardner; Robert Douglas
Hi Johannes,
Below code gives good results for me; note that trying multiple
starting is often important in fitting mixture models, even in simple
cases like this.
Note also that the sigma and nu parameters in gamlssMX are fitted on a
log scale, hence the possible occurrence of negative results.
ht
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