glsnow wrote
>>From your question it is not clear what your question/concerns really are,
> and from what we can see it could very well be that you do not understand
> the statistics that you are computing (not just the R implementation). We
> ask for a reproducible example because that helps us t
On Oct 17, 2013, at 9:11 PM, Steven LeBlanc wrote:
> Greets,
>
> I'm using nlminb() to estimate the parameters of a multivariate normal random
> sample with missing values and ran into an unexpected result from my call to
> dmvnorm()
There are at least 5 different version of dmvnorm. None of
I have a simple table, 2 columns and 1994 rows. First row, "Crime" is how
many crimes happen every month per 10 inhabitants and second row is
"Income" which contains the average income recorded in a city.
here's the head(dataset):
Crime Income
1 356.5152 4285.720
2 734.5625 4114.291
Greets,
I'm using nlminb() to estimate the parameters of a multivariate normal random
sample with missing values and ran into an unexpected result from my call to
dmvnorm() within the likelihood function. Particular details are provided
below. It appears that dmvnorm() makes a call to log(eigen
On Oct 17, 2013, at 2:56 PM, Ye Lin wrote:
> Hey R professionals,
>
> I have a large dataset and I want to run a loop on it basically creating a
> new column which gathers information from another reference table.
>
> When I run the code, R just freezes and even does not response after 30min
>
Hi all
I do not understand why I am getting the following error message. Can
anybody help me with this? Thanks in advance.
install.packages("cmprsk")
library(cmprsk)
result1 <-crr(ftime, fstatus, cov1, failcode=1, cencode=0 )
one.pout1 = predict(result1,cov1,X=cbind(1,one.z1,one.z2))
predict.cr
See ?pmax for getting the max for each year.
do.call('pmax', oil[-1])
Or equivalently:
pmax(oil$TX, oil$CA, oil$AL, oil$ND)
apply and which.max will give you the index:
i <- apply(oil[-1], 1, which.max)
which you can use to extract the state:
names(oil[-1])[i]
Jason
-Original Message--
On 17 October 2013 15:38, Timo Schmid wrote:
> I have some code in R with a lot of matrix multiplication and inverting. R
> can be very slow for larger matrices like 5000x5000.
> I have seen the new programming language Julia (www.julialang.org) which is
> quite fast in doing matrix algebra.
It
Hey R professionals,
I have a large dataset and I want to run a loop on it basically creating a
new column which gathers information from another reference table.
When I run the code, R just freezes and even does not response after 30min
which is really unusual. I tried sapply as well but does no
seq_along(x), integer(length(x)), is.na(x), or anything that produces an integer
(or numeric or logical) vector the length of x would work. I use integer() or
numeric()
to indicate I'm not using its value: it is just a vector in which to place the
return values of FUN().
Bill Dunlap
Spotfire, TI
Hi Bill,
#seq_along() worked in the cases you showed.
ave(seq_along(fac),fac,FUN=length)
#[1] 3 1 3 3
ave(seq_along(num), num, FUN=length)
#[1] 3 1 3 3
ave(seq_along(char), char, FUN=length)
#[1] 3 1 3 3
I thought, there might be some advantages in speed, but they were similar in
speed
Thanks, Bill.
But ?ave specifically says:
ave(x, ..., FUN = mean)
Arguments:
x
A numeric.
So that it should not be expected to work properly if the argument is
not (coercible to) numeric. Nevertheless, defensive programming is
always wise.
Cheers,
Bert
On Thu, Oct 17, 2013 at 1:34 PM, Willi
May I ask why:
count_by_class <- with(dat, ave(numeric(length(basel_asset_class)),
basel_asset_class, FUN=length))
should not be more simply done as:
count_by_class <- with(dat, ave(basel_asset_class, basel_asset_class,
FUN=length))
The way I did it would work if basel_asset_class we
May I ask why:
count_by_class <- with(dat, ave(numeric(length(basel_
asset_class)), basel_asset_class, FUN=length))
should not be more simply done as:
count_by_class <- with(dat, ave(basel_asset_class, basel_asset_class,
FUN=length))
?
-- Bert
On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap
On 10/17/2013 04:04 AM, Msugarman wrote:
Hi all,
I'm trying to graph the results of a weighted regression analysis. Is anyone
aware of a way to make my markers appear a different sizes to be consistent
with their respective weights?
Hi Mike,
Have a look at the "size_n_color" function in the pl
> What I need is to select only those records for which there are more than two
> default
> frequencies (defa_frequency),
Here is one way. There are many others:
> dat <- data.frame( # slightly less trivial example
basel_asset_class=c(4,8,8,8,74,3,74),
defa_frequency=(1:7)/8)
Hi,
You could also check ?data.table() as it could be faster.
#Speed comparison
set.seed(498)
oilT <-
data.frame(YEAR=rep(rep(1800:2012,50),100),state=rep(rep(state.abb,each=213),100),value=sample(2000:8,1065000,replace=TRUE),stringsAsFactors=FALSE)
system.time(res1 <-
oilT[as.logical(wi
On 10/17/2013 11:54 AM, Stock Beaver wrote:
> # Suppose I have a vector:
>
> myvec = c(1,0,3,0,77,9,0,1,2,0)
>
> # I want to randomly pick an element from myvec
> # where element == 0
> # and print the value of the corresponding index.
>
> # So, for example I might randomly pick the 3rd 0
> # and I
On Oct 16, 2013, at 10:04 AM, Msugarman wrote:
> Hi all,
>
> I'm trying to graph the results of a weighted regression analysis. Is anyone
> aware of a way to make my markers appear a different sizes to be consistent
> with their respective weights?
You have not produced any data or code. If usi
Typo fix below:
On Thu, Oct 17, 2013 at 3:05 PM, Sarah Goslee wrote:
> Not only does it not require a loop, this is a one-liner:
>
>> myvec <- c(1,0,3,0,77,9,0,1,2,0)
>> sample(which(myvec == 0), 1)
> [1] 4
>> sample(which(myvec == 0), 1)
> [1] 7
>> sample(which(myvec == 0), 1)
> [1] 2
>
> If the
Not only does it not require a loop, this is a one-liner:
> myvec <- c(1,0,3,0,77,9,0,1,2,0)
> sample(which(myvec == 0), 1)
[1] 4
> sample(which(myvec == 0), 1)
[1] 7
> sample(which(myvec == 0), 1)
[1] 2
If there's a possibility of not having zeros then you'll need to check
that separately, other
Hi folks,
Wondering if anyone might be able to help me on this one. I have just done
some geochemistry with X-ray Diffraction and Rietveld Refinement in order to
quantify the data. I have a observed spectra from my sample and a calculated
spectra from the Rietveld Refinement (in the attached image
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would print the corresponding index
# which is 7,
# My init
On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
I always get lost in simpleKey.
As this is my first use of it I take what's offered by those more
experienced than I.
The approach of directly modifying the trellis object usually works.
tmp <- xyplot(pct.quant ~ sampdate, data = ffg.st, gro
You may try:
mydat[with(mydat,ave(seq_along(basel_asset_class),basel_asset_class,FUN=length)>2),]
# basel_asset_class defa_frequency
#2 8 0.070
#3 8 0.030
#4 8 0.001
#or
library(plyr)
mydat[ddply(mydat,.(basel_asset_class
@Martin Morgan, Duncan Murdoch:
OK Thanks.
I did not understand the callNextMethod.
I will investigate this in detail.
This is great!
Thanks again,
Michael Meyer
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEAS
I always get lost in simpleKey. The approach of directly modifying
the trellis object usually works.
> tmp <- xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_feed_grp,
> type =
+ 'p', pch = 19, key = simpleKey(text = levels(ffg.st$func_feed_grp), space =
+ 'right', points = T, lines =
Correction. (2nd para first three lines)
Pl read following line
What I need is to select only those records for which there are more than two
default frequencies (defa_frequency), Thus, there is only one default frequency
= 0.150 w.r.t basel_asset_class = 4 whereas there are default frequenci
I am sorry perhaps was not able to put the question properly. I am not
looking for the subset of the data.frame where the basel_asset_class is > 2. I
do agree that would have been a basic requirement. Let me try to put the
question again.
I have a data frame as
mydat = data.frame(basel_ass
If all your data is numeric then you can use an array instead of a data
frame and arrays can easily be 3, 4, or higher dimensional. Or you can use
a data frame with a column each for x, y, z, and time; with possible other
columns representing groups or other attributes, essentially a 3
dimensional
On 17-10-2013, at 18:48, Tim Umbach wrote:
> Hi there,
>
> another beginners question, I'm afraid. Basically i want to selct the
> maximum of values, that correspond to different variables. I have a table
> of oil production that looks somewhat like this:
>
> oil <- data.frame( YEAR = c(2011,
The simplest approach is to specify the cex parameter in the call to plot.
plot(1:3, 1:3, cex=3:1) for example will plot the 1st point 3 times as
big, the 2nd 2 times as big, and the 3rd at the standard size.
You can get more control by using the symbols function instead of the plot
function and
Hi,
You may try:
unlist(lapply(seq_len(nrow(oil)),function(i) oil[i,-1][which.max(oil[i,-1])]))
# CA ND
#4 6
#or
library(reshape2)
datM <- melt(oil,id.var="YEAR")
datM[as.logical(with(datM,ave(value,list(YEAR),FUN= function(x) x%in%
max(x,]
# YEAR variable value
#3 2011
I received the following error message with the multicore package:
install.packages("multicore")
Warning in install.packages :
package multicore is not available (for R version 3.0.2)
Warning in install.packages :
package multicore is not available (for R version 3.0.2)
Warning message:
pa
On 10/17/2013 08:54 AM, Michael Meyer wrote:
Suppose you have a base class "Base" which implements a function "Base::F"
which works in most contexts but not in the context of "ComplicatedDerived"
class
where some preparation has to happen before this very same function can be
called.
You woul
I work very mutch with the packages RWeka and multicore. If you try to run
J48 or any tree of RWeka with multicore we hava some errors.
Example I:
library(RWeka);
library(multicore);
mclapply(1:100, function(i) {
J48(Species ~., iris);
});
Output: "Error in .jcall(o, \"Ljava/lang/Class;\"
>From your question it is not clear what your question/concerns really are,
and from what we can see it could very well be that you do not understand
the statistics that you are computing (not just the R implementation). We
ask for a reproducible example because that helps us to help you, just a
c
On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
That should have worked.
That's what I thought when I first tried it.
I think something else is interfering. Did you redefine either T or F?
Not intentionally.
Please send the output from dput(head(ffg.st)) so we can experiment in
your
Hi there,
another beginners question, I'm afraid. Basically i want to selct the
maximum of values, that correspond to different variables. I have a table
of oil production that looks somewhat like this:
oil <- data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
You want the offset function in the formula:
lm( A ~ B + I(B^2) + offset(C), data=Dataset)
This will force the coefficient on C to be 1, if you wanted a coefficient
of another value then just do the multiplication yourself, e.g. offset( 2 *
C ) for a slope of 2.
Also you can use poly(B,2) to fit
That should have worked. I think something else is interfering.
Did you redefine either T or F?
Please send the output from dput(head(ffg.st))
so we can experiment in your setting.
Rich
On Thu, Oct 17, 2013 at 12:12 PM, Rich Shepard wrote:
> On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
>
>
On 17 Oct 2013, at 13:44 , Hermann Norpois wrote:
> Hello,
>
>
> my dots of 0 and 2 are quite close to the marging. So I would like to move
> the 0 and the 2 both towards the 1. I wish to be my dots more centered.
> And: I dont need so much space between 0,1 and 2.
>
> How does it work?
> I tr
"Kindly guide" ...
This is a very basic question, so the kindest guide I can give is to read
an Introduction to R (ships with R) or a R web tutorial of your choice so
that you can learn how R works instead of posting to this list.
Cheers,
Bert
On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin
Hello,
my dots of 0 and 2 are quite close to the marging. So I would like to move
the 0 and the 2 both towards the 1. I wish to be my dots more centered.
And: I dont need so much space between 0,1 and 2.
How does it work?
I tried:
plot (data, axes=FALSE, main=i, ylab= expression (z^2))
Hi,
Try:
dat <- read.table(text="
ID Name
1 Andy
2 John
3 Amy",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2 <- read.table(text="
ID Date Value
1 2013-10-01 10
1 2013-10-02 15
2 2013-10-01 7
2 2013-10-03 10
2 2013-10-04 15
3 2013-10-01
On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
par.settings = list(
superpose.points = list(col = rainbow(7), pch = 19),
superpose.lines = list(col = rainbow(7))
)
I had tried that, too. Legend symbols stubbornly remain unfilled.
Thanks, Richard,
Rich
___
par.settings = list(
superpose.points = list(col = rainbow(7), pch = 19),
superpose.lines = list(col = rainbow(7))
)
On Thu, Oct 17, 2013 at 11:48 AM, Rich Shepard wrote:
> On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
>
>> put the pch into the par.settings
>
>
> Richard,
>
> Tried thi
Hi,
I think based on your title, the output you provided is not clear. If it
depends on Date, there should be four columns.
library(reshape2)
res1 <- dcast(merge(dat,dat2,by="ID"),ID+Name~Date,value.var="Value")
colnames(res1)[3:6] <- c("First", "Second", "Third", "Fourth")
rownames(res1) <- 1
Hi,
I have some code in R with a lot of matrix multiplication and inverting. R can
be very slow for larger matrices like 5000x5000.
I have seen the new programming language Julia (www.julialang.org) which is
quite fast in doing matrix algebra. So my idea is to set up the simulations in
R and st
Dear all,
I have a problem where I must represent points with XYZ coordinates
changing over time. I will do a number of operations on this data such as
calculating the YZ-projection distance of the points to the origin over
time, the frequency spectrum of the X-T data etc. I am trying to find a
go
Hi,
I have a set of matrix data named invest consists of 450 observations (75
countries, 6 years) with 7 variables (set as I, pop, inv, gov, c, life, d;
which each is numeric[450]). The procedure is modify from code provided
by B.E. Hansen at http://www.ssc.wisc.edu/~bhansen/progs/ecnmt_00.ht
Greetings,
I have an S4 class "B" (Base) which defines a function f=f(this="B",...)
Dervided from B we have a derived class D which also defines a function
f=f(this="D",...)
In the definition of D::f we want to call the version B::f and could do this by
simply calling
f(baseClassObject(this),
Hi all,
I'm trying to graph the results of a weighted regression analysis. Is anyone
aware of a way to make my markers appear a different sizes to be consistent
with their respective weights?
Thanks,
-Mike Sugarman
Wayne State University
--
View this message in context:
http://r.789695.n4.nab
Hi David,
That worked brilliantly! Many thanks. I also had trouble getting subplot()
to work with either TeachingDemos or Hmisc.
Best,
Mark
--
View this message in context:
http://r.789695.n4.nabble.com/map-with-inset-tp4678341p4678426.html
Sent from the R help mailing list archive at Nabble.
Katherine,
There are multiple ways to do this and I highly recommend you look into a
basic R manual or search the forums. One quick example would be:
mysub <- subset(mydat, basel_asset_class > 2)
Cheers,
Charles
On Thu, Oct 17, 2013 at 1:55 AM, Katherine Gobin
wrote:
> Dear Forum,
>
> I have
On Thu, 17 Oct 2013, Richard M. Heiberger wrote:
put the pch into the par.settings
Richard,
Tried this again, but I'm not finding the proper location within
par.settings.
par.settings = list(superpose.points = list(col = rainbow(7)),
superpose.lines = list(col = rainbow(7)), pch = 19)
If
>So why not start with some statistical textbook? There are plenty of them
available in CRAN.
I wasn't implying, that I haven't read any textbook, or didn't do any
research. I read some textbooks/Papers/etc. during the research about what
to do and came across the wilcox test. I meant to imply th
Dear Forum,
I have a data frame as
mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15,
0.07, 0.03, 0.001))
> mydat
basel_asset_class defa_frequency
1 2 0.150
2 8 0.070
3 8 0.030
4
Quote
By the way, your use of the syntax D::f and B::f suggests that you're
thinking from a C++ point of view. That's very likely to lead to
frustration: the S4 object system is very different from C++. Methods
don't belong to classes, they belong to generics. There is no such thing
as D::f
This is the world-famous "fizzbuzz" problem. You should be able to find
lots of implementations by Googling that word. Here's a pointless
collection I wrote once:
# a really dumb fizzbuzz alg competition
#fbfun1 is 2.5x faster than fbfun2
# fbfun3 is 10x faster than fbfun1
# fbfun1 is 2x faster
Milan is correct.
The prefix is used when saving the XML content that is represented in
a different format in R.
To get the prefix
on the XML content that you save, use a document object
doc = newXMLDoc()
root = newXMLNode("foo", doc = doc)
saveXML(doc)
Sorry for the confusion.
D
On 10
put the pch into the par.settings
On Thu, Oct 17, 2013 at 11:17 AM, Rich Shepard wrote:
> When I specify pch = 19 for a scatter plot the points are filled circles.
> Deapite reading ?points and trial-and-error experimentation I have not found
> how to have the legend symbols (now open circles)
When I specify pch = 19 for a scatter plot the points are filled circles.
Deapite reading ?points and trial-and-error experimentation I have not found
how to have the legend symbols (now open circles) filled.
An example command is:
xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_fe
Thanks to all who replied.
Here are two versions of a function (sans sanity checks) that do what I
want:
foo1 <- multifoo(1:2, "A")
foo2 <- multifoo(1:2, "B")
mfoo <- list(A=foo1, B=foo2)
class(mfoo) <- c("foolist", "list")
#' flatten a list of lists
# from Duncan Murdoch
flatten <- functio
Does this get you the rest of the way?
> mfoo2 <- unlist(mfoo, recursive = FALSE)
> names(mfoo2) <- unlist(lapply(mfoo, names))
> class(mfoo2) <- "foolist"
> str(mfoo2)
List of 4
$ A1:List of 2
..$ x: int 3
..$ y: int 10
..- attr(*, "class")= chr "foo"
$ A2:List of 2
..$ x: int [1:2] 6 4
On 17/10/2013 9:15 AM, Michael Friendly wrote:
I have functions that generate lists objects of class "foo" and lists of
lists of these, of class
"foolist", similar to what is shown below.
You can use c() to join lists. So in the example below,
c(mfoo$A, mfoo$B)
will give you a list with the
unlist(mfoo, recursive = FALSE) gets you pretty close.
Best,
Ista
On Thu, Oct 17, 2013 at 9:15 AM, Michael Friendly wrote:
> I have functions that generate lists objects of class "foo" and lists of
> lists of these, of class
> "foolist", similar to what is shown below.
>
> How can I flatten some
On 17/10/2013 9:01 AM, Michael Meyer wrote:
Sorry,
if the previous message seems without context.
Indeed, the first message was bounced by filtering rules (triggered by subject
heading than which nothing could be more benign or less liable to suspician).
It was:
Greetings,
I have an S4 class
I have functions that generate lists objects of class "foo" and lists of
lists of these, of class
"foolist", similar to what is shown below.
How can I flatten something like this to remove the top-level list
structure, i.e.,
return a single-level list of "foo" objects, of class "foolist"?
foo
Sorry,
if the previous message seems without context.
Indeed, the first message was bounced by filtering rules (triggered by subject
heading than which nothing could be more benign or less liable to suspician).
It was:
Greetings,
I have an S4 class "B" (Base) which defines a function f=f(this=
Greetings,
Meanwhile I have figured out how do do it only to find out that I have more
serious problems.
Generally calling Base::f on the base class object is not what you want,
instead you want to call
Base::f on the full object for the following reasons:
If the base class is virtual, then Bas
Consider the function f(x) = x on the open interval (0,1). It does not have a
maximum.
That is what your likelihood function will look like. The MLE does not exist.
Chris
(Although if everything is continuous and you are okay with limits there is an
extension that gets you to Terry's original
Or faster (both computational speed and amount of code):
colSums(is.na(rbind(c1, c2, c3)))
On Thu, Oct 17, 2013 at 4:34 AM, Carl Witthoft wrote:
> mattbju2013 wrote
> > Hi guys this is my first post, i need help summing the number of NA's in
> a
> > few vectors
> >
> > for example..
> >
> > c1
hello togehter,
i have a little problem, maybe you can help me.
I have a data.frame like this one:
IDName
1 Andy
2 John
3 Amy
and a data.frame like this:
ID DateValue
12013-10-0110
12013-10-0215
22013-10-017
22013-10-0310
22013-
mattbju2013 wrote
> Hi guys this is my first post, i need help summing the number of NA's in a
> few vectors
>
> for example..
>
> c1<-c(1,2,NA,3,4)
> c2<-c(NA,1,2,3,4)
> c3<-c(NA,1,2,3,4)
>
> how would i get a result that only sums the number of NA's in the vector?
> the.result.i.want<-c(2,0,1,
On 10/17/2013 06:17 PM, catalin roibu wrote:
Dear R users,
I want to extract column's from different data frame with different row
length.
How can I do this in R?
Hi catalin,
If I understand your question, which I think is:
I want to extract columns from different data frames with differing
For the record. I have found a possible sollution:
nn <- nodeapply(z)
n.names= names(unlist(nn[[1]]))
ext <- unlist(sapply(n.names, function(x) grep("split.varid.", x, value=T)))
ext <- gsub("kids.split.varid.", "", ext)
ext <- gsub("split.varid.", "", ext)
dep.var <- as.character(terms(z)[1][[2]]
> -Original Message-
> I am doing a polynomial linear regression with 2 independent variables
> such as :
>
> lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset))
>
> R return me a coefficient per independent variable, and I would need
> the coefficient of the C parameter to equal 1.
Leavi
Dear all,
I have been trying to find a simple solution to my problem without success,
though i have a feeling a simple syntaxe detail coul make the job.
I am doing a polynomial linear regression with 2 independent variables such as :
lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset))
R return me
Le mercredi 16 octobre 2013 à 23:45 -0400, Earl Brown a écrit :
> I'm using the "XML" package and specifically the saveXML() function but I
> can't get the "prefix" argument of saveXML() to work:
>
> library("XML")
> concepts <- c("one", "two", "three")
> info <- c("info one", "info two", "info t
Wow!! Thank you so much for your suggestions! For now, A.K's suggestion #1
is perfect for me!
Thank you very much!
Best,
Charles
On Thu, Oct 17, 2013 at 2:34 AM, William Dunlap wrote:
> You could bump up the day each time an hour was less than the previous
> one. E.g.,
> testtime <-
> c("
Dear R users,
I want to extract column's from different data frame with different row
length.
How can I do this in R?
Thank you very much!
best regards!
CR
--
---
Catalin-Constantin ROIBU
Lecturer PhD, Forestry engineer
Forestry Faculty of Suceava
Str. Universitatii no. 13, Suceava, 720229, R
82 matches
Mail list logo
