Re: [R] legend problem [with ggplot[12]]
IF you provide us with the reproducible code that is requested in the posting guide, you may get some valuable help. I gues sthe is from ggplot2? Uwe Ligges Felipe Carrillo wrote: Hi: See the attached pdf graphic. The legend overlaps a little bit with the strip. Is there a way to move it over half a cm? Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'beside' option for boxplots
Example: boxplot(d1$b ~ d1$a, pars = list(boxwex = 0.15), at=c(1:3 - 0.1), xaxt=n) boxplot(d2$b ~ d2$a, add=T, pars = list(boxwex = 0.15), at=c(1:3 + 0.1), xaxt=n) axis(1, at=1:3) Uwe Malcolm Ryan wrote: Is there any way to get a boxplot of several data sets beside one another on the same graph, as there is for barplot? If I do: d1 - data.frame(a = c(rep(1:3, each = 3)), b = c(1:9)) d2 - data.frame(a = c(rep(1:3, each = 3)), b = c(9:1)) boxplot(d1$b ~ d1$a) boxplot(d2$b ~ d2$a, add=T) It will show the two datasets on the one graph, but the middle point will overlap. What I want is a layout more like: barplot(t(matrix(c(d1$b, d2$b), ncol=2)), beside=T) only showing the boxes from the plot above instead of bars. Can this be done? Malcolm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Understanding R Hist() Results...
Thank you again for all the R help folks who responded. I again appreciate all the help and insight and will investigate the options suggested. I guess I still doing a little head scratching at how the division occurred: It looks like the default hist(...) behavior is doing the following: HouseHist-hist(as.numeric(HouseYear_array)) HouseHist$counts [1] 2 1 4 4 8 8 That would equate to the following grouping of the years: [90, 91] (91, 92] (92, 93] (93, 94] (94, 95] (95, 96] However, the true division is something like the following: table(as.numeric(HouseYear_array)) 1990 1991 1992 1993 1994 1995 1996 1114488 Seems like hist behavior could have been: (89, 90] (90, 91] (91, 92] (92, 93] (93, 94] (94, 95] (95, 96] Of course, I haven't had any coffee yet... This goes with the following example: http://n2.nabble.com/What-is-going-on-with-Histogram-Plots-td3022645.htm --- On Thu, 6/4/09, ted.hard...@manchester.ac.uk ted.hard...@manchester.ac.uk wrote: From: ted.hard...@manchester.ac.uk ted.hard...@manchester.ac.uk Subject: RE: [R] Understanding R Hist() Results... To: R-help@r-project.org Cc: Jason Rupert jasonkrup...@yahoo.com Date: Thursday, June 4, 2009, 5:13 AM On 04-Jun-09 04:00:11, Jason Rupert wrote: Think I'm missing something to understand what is going on with hist(...) http://n2.nabble.com/What-is-going-on-with-Histogram-Plots-td3022645.htm l For my example I count 7 unique years, however, on the histogram there only 6. It looks like the bin to the left of the tic mark on the x-axis represents the number of entries for that year, i.e. Frequency. I guess it looks like the bin for 1990 is missing. Is there a better way or a different histogram R command to use in order to see all the age bins and them for them to be aligned directly over the year tic mark on the x-axis? Thanks again for any insights that can be provided. It's doing what it's supposed to -- which admitredly could be confusing when all your data lie on the exact boundaries between bins. From ?hist, by default include.lowest = TRUE, right = TRUE, and: If 'right = TRUE' (default), the histogram cells are intervals of the form '(a, b]', i.e., they include their right-hand endpoint, but not their left one, with the exception of the first cell when 'include.lowest' is 'TRUE'. In your data: sort(HouseYear_array) [1] 1990 1991 1992 1993 1993 1993 1993 1994 1994 [10] 1994 1994 1995 1995 1995 1995 1995 1995 1995 [20] 1995 1996 1996 1996 1996 1996 1996 1996 1996 and, with H-hist(as.numeric(HouseYear_array)) H$breaks # [1] 1990 1991 1992 1993 1994 1995 1996 so you get 2 (1990,1991) in the [1990-1] bin, 1 in the [1991-2] bin, 4 in [1992-3], and so on, exactly as observed. You can get what you're expecting to see by setting the 'breaks' parameter explicitly, and making sure the breakpoints do not coincide with data (which ensures that there is no confusion about what goes in which bin): hist(as.numeric(HouseYear_array),breaks=0.5+(1989:1996)) Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 04-Jun-09 Time: 11:13:22 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding a method to a generic in another package
The zoo package has lattice methods and does not use require and it puts lattice in Imports in the DESCRIPTION file. If a user wants to the use the zoo lattice functions then they must issue a library(lattice) call. As a result users do not have to load lattice but if they want to use it they must do it themselves. library(zoo) z - zoo(1:4) xyplot(z) # wrong library(lattice) xyplot(z) # yes! On Tue, Jun 2, 2009 at 5:14 PM, Jeffrey J. Hallman jhall...@frb.gov wrote: I am the maintainer of the 'tis' package. One of the functions in my package is 'nberShade'. A user wants to make nberShade generic, with the old version renamed as nberShade.default, all of which is fine with me. And he wants to add a new method, nberShade.ggplot, which works for objects of class ggplot. He also wants to add a method fortify.tis for the generic fortify defined in ggplot2. The nberShade.ggplot method uses a bunch of other functions from the ggplot2 package, and it's first line is require(ggplot2) From what he tells me, this function works. Where I'm having trouble is figuring out what I have to do to get the tis package to pass R CMD check. I really don't want to force users of the tis package to have to install ggplot2. What can I do? Is it enough to have Imports: ggplot2 in the DESCRIPTION file and import(ggplot2) in the NAMESPACE file? I've done that, but I still get this warning from R CMD check: * checking for unstated dependencies in R code ... WARNING 'library' or 'require' calls not declared from: ggplot2 See the information on DESCRIPTION files in the chapter 'Creating R packages' of the 'Writing R Extensions' manual. Well, I did read the manual, and it seems to say that what I'm doing is OK. So why am I getting the warning? -- Jeff __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using constrOptim() function
It looks to me like you need to add x to the arguments in your
call to constrOptim, something like the following:
constrOptim(beta_i, myFunction, NULL, ui, ci, mu = 1e-04, control = list(),
method = Nelder-Mead,outer.iterations = 100, outer.eps = 1e-05, x=x)
The first x in x=x tells constrOptim that this is a named
argument. Since it's NOT the name of an argument for constrOptim, it
must be part of the ... argument, which is offered to f, which in
this case is myFunction. The second x in x=x tells constrOptim
to pass your global variable x for this purpose.
If this does not work, please submit another post, providing a
simpler, self contained example. Define myFunction as something
simple to define that still generates your error. Then someone can copy
your code into R, run it and see your error. That will tend on average
to increase the probability of a response as well as its speed and
utility.
Hope this helps,
Spencer
Ali Mahani wrote:
I have a function myFunction(beta,x) where beta is a vector of coefficients
and x is a data frame (think of it as a matrix). I want to optimize the
function myFunction() by ONLY changing beta, i.e. x stays constant, with 4
constraints. I have the following code (with a separate source file for the
function):
rm(list=ls())
source('mySourceFile')
x=read.csv(myFile.csv,head=TRUE,sep=,)
beta_i=c(1,1,1,1,1,1,-1)
ui=rbind(c(1,0,0,0,0,0,0),c(0,1,0,0,0,0,0),c(0,0,1,0,0,0,0),c(0,0,0,0,0,0,-1))
ci=c(0,0,0,0)
constrOptim(beta_i, myFunction, NULL, ui, ci, mu = 1e-04, control = list(),
method = Nelder-Mead,outer.iterations = 100, outer.eps = 1e-05)
I am getting this error:
Error in f(theta, ...) : argument x is missing, with no default
If I replace myFunction with myFunction(beta,x) I get this error:
Error in beta * feature_1[i, ] : non-numeric argument to binary operator
If I try myFunction(beta_i,x) I get:
Error in constrOptim(beta_i, mirror_lf(beta_i, x), NULL, ui, ci, mu = 1e-04,
:
could not find function f
Clearly, I don;t have a good understanding of how to use constrOptim() or
optim() for that matter. Any guidance? Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Getting a column of values from a list - think I'm doing it thehard way
Or even:
library(chron)
ch - chron(HouseDates)
years(ch)
[1] 1990 1991 1992 1993 1994 1995 1996
Levels: 1990 1991 1992 1993 1994 1995 1996
# or
with(month.day.year(ch), year)
[1] 1990 1991 1992 1993 1994 1995 1996
On Thu, Jun 4, 2009 at 4:02 AM, ONKELINX, Thierry
thierry.onkel...@inbo.be wrote:
Dear Jason,
Have a look at years() from the chron package.
library(chron)
HouseDates - c(02/27/90, 02/27/91, 01/14/92, 02/28/93,
02/01/94, 02/01/95, 02/01/96)
HouseDatesFormatted-as.Date(HouseDates, %m/%d/%y)
years(HouseDates)
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Jason Rupert
Verzonden: donderdag 4 juni 2009 4:45
Aan: R-help@r-project.org
Onderwerp: [R] Getting a column of values from a list - think I'm doing
it thehard way
Example code it shown below.
I think I am doing this the hard way. I'm just trying to get the full
year value from an array of dates. An example array is shown below.
Right now, I'm using a for loop to pull the year out of a list where
the dates were split up into their individual components.
This seems to work, but just wondering if there is an easier way.
Thanks for any insights.
#*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
HouseDates - c(02/27/90, 02/27/91, 01/14/92, 02/28/93,
02/01/94, 02/01/95, 02/01/96)
# ?as.Date
HouseDatesFormatted-as.Date(HouseDates, %m/%d/%y)
HouseDatesFormatted
HouseDatesList-strsplit(as.character(HouseDatesFormatted), -,
fixed=TRUE)
HouseYear_array-NULL
length_array-length(HouseDatesList)
for(ii in 1:length_array)
{
HouseYear-HouseDatesList[[ii]][1]
HouseYear_array-c(HouseYear_array, HouseYear) }
as.character(HouseYear_array)
# Desired:
# [1] 1990 1991 1992 1993 1994 1995 1996
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd
is
door een geldig ondertekend document. The views expressed in this message
and any annex are purely those of the writer and may not be regarded as
stating
an official position of INBO, as long as the message is not confirmed by a
duly
signed document.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] IP-Address
edw...@web.de wrote:
Hi,
Unfortunately, they can't handle NA. Any suggestion? Some row for Ip
don't have ip address. This cause an error/ wrong result.
A quick fix could be to substitute ... or 0.0.0.0 for the NA
entries. (Use something like
ipch - as.character(df$ip)
ipch[is.na(df$ip)] - ...
connection - textConnection(ipch)
)
Eddie
library(gsubfn)
library(gtools)
library(rbenchmark)
n - 1
df - data.frame(
a = rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'), simplify=TRUE)
)
res - benchmark(columns=c('test', 'elapsed'), replications=10,
order=NULL,
peda = {
connection - textConnection(as.character(df$ip))
o - do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ]
},
peda2 = {
connection - textConnection(as.character(df$ip))
dfT - read.table(connection, sep='.', colClasses=rep(integer,
4), quote=, na.strings=NULL, blank.lines.skip=FALSE)
close(connection)
o - do.call(order, dfT)
df[o, ]
},
hb = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip)
dim(ip) - c(4, nrow(df))
ip - 256^3*ip[1,] + 256^2*ip[2,] + 256*ip[3,] + ip[4,]
o - order(ip)
df[o, ]
},
hb2 = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip);
dim(ip) - c(4, nrow(df))
o - sort.list(ip[4,], method=radix, na.last=TRUE)
for (kk in 3:1) {
o - o[sort.list(ip[kk,o], method=radix, na.last=TRUE)]
}
df[o, ]
}
)
print(res)
test elapsed
1 peda 4.12
2 peda2 4.08
3 hb 0.28
4 hb2 0.25
On Sun, May 31, 2009 at 12:42 AM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
edwin Sendjaja wrote:
Hi VQ,
Thank you. It works like charm. But I think Peter's code is faster.
What
is the difference?
i think peter's code is more r-elegant, though less generic. here's a
quick test, with not so surprising results. gsubfn is implemented in r,
not c, and it is painfully slow in this test. i also added gabor's
suggestion.
library(gsubfn)
library(gtools)
library(rbenchmark)
n = 1000
df = data.frame(
a=rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'),
simplify=TRUE))
benchmark(columns=c('test', 'elapsed'), replications=10, order=NULL,
peda={
connection = textConnection(as.character(df$ip))
o = do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ] },
waku=df[order(gsubfn(perl=TRUE,
'[0-9]+',
~ sprintf('%03d', as.integer(x)),
as.character(df$ip))), ],
gagr=df[mixedorder(df$ip), ] )
# peda 0.070
# waku 7.070
# gagr 4.710
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal,
self-contained, reproducible code.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Cochran’s Q statistic
Does anyone know which package include the computation of Cochrans Q statistic in R? jlfmssm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'beside' option for boxplots
Dear Malcom, Another option is to merge both dataset to one big dataset and then plot the big dataset. d1 - data.frame(a = c(rep(1:3, each = 3)), b = c(1:9), d = A) d2 - data.frame(a = c(rep(1:3, each = 3)), b = c(9:1), d = B) Dataset - rbind(d1, d2) library(ggplot2) Dataset$a - factor(Dataset$a) ggplot(Dataset, aes(x = a, y = b, colour = d)) + geom_boxplot() HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Uwe Ligges Verzonden: donderdag 4 juni 2009 13:45 Aan: Malcolm Ryan CC: r-help@r-project.org Onderwerp: Re: [R] 'beside' option for boxplots Example: boxplot(d1$b ~ d1$a, pars = list(boxwex = 0.15), at=c(1:3 - 0.1), xaxt=n) boxplot(d2$b ~ d2$a, add=T, pars = list(boxwex = 0.15), at=c(1:3 + 0.1), xaxt=n) axis(1, at=1:3) Uwe Malcolm Ryan wrote: Is there any way to get a boxplot of several data sets beside one another on the same graph, as there is for barplot? If I do: d1 - data.frame(a = c(rep(1:3, each = 3)), b = c(1:9)) d2 - data.frame(a = c(rep(1:3, each = 3)), b = c(9:1)) boxplot(d1$b ~ d1$a) boxplot(d2$b ~ d2$a, add=T) It will show the two datasets on the one graph, but the middle point will overlap. What I want is a layout more like: barplot(t(matrix(c(d1$b, d2$b), ncol=2)), beside=T) only showing the boxes from the plot above instead of bars. Can this be done? Malcolm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] IP-Address
Suggest you be more specific. The solution I posted does handle missing values
sorting them to the beginning:
Lines - id rank color status ip
+ 138 29746 yellow no 162.131.58.26
+ 138 29746 red yes 162.131.58.16
+ 138 29746 blue yes 162.131.58.10
+ 138 29746 red no 162.131.58.17
+ 138 29746 yellow no 162.131.58.14
+ 138 29746 red no 162.131.58.13
+ 138 29746 yellow no 162.132.58.15
+ 139 29746 green no 162.252.20.69
+ 140 29746 red yes 162.254.20.71
+ 141 29746 yellow no 163.253.7.153
+ 142 31804 green yes 163.253.20.114
+ 142 31804 green yes
+ 144 32360 black yes 161.138.45.226
DF - read.table(textConnection(Lines), header = TRUE, fill = TRUE)
library(gtools)
DF[mixedorder(DF$ip), ]
id rank color status ip
12 142 31804 greenyes
13 144 32360 blackyes 161.138.45.226
3 138 29746 blueyes 162.131.58.10
6 138 29746red no 162.131.58.13
5 138 29746 yellow no 162.131.58.14
2 138 29746redyes 162.131.58.16
4 138 29746red no 162.131.58.17
1 138 29746 yellow no 162.131.58.26
7 138 29746 yellow no 162.132.58.15
8 139 29746 green no 162.252.20.69
9 140 29746redyes 162.254.20.71
11 142 31804 greenyes 163.253.20.114
10 141 29746 yellow no 163.253.7.153
On Thu, Jun 4, 2009 at 6:39 AM, edw...@web.de wrote:
Hi,
Unfortunately, they can't handle NA. Any suggestion? Some row for Ip don't
have ip address. This cause an error/ wrong result.
Eddie
library(gsubfn)
library(gtools)
library(rbenchmark)
n - 1
df - data.frame(
a = rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'), simplify=TRUE)
)
res - benchmark(columns=c('test', 'elapsed'), replications=10, order=NULL,
peda = {
connection - textConnection(as.character(df$ip))
o - do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ]
},
peda2 = {
connection - textConnection(as.character(df$ip))
dfT - read.table(connection, sep='.', colClasses=rep(integer,
4), quote=, na.strings=NULL, blank.lines.skip=FALSE)
close(connection)
o - do.call(order, dfT)
df[o, ]
},
hb = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip)
dim(ip) - c(4, nrow(df))
ip - 256^3*ip[1,] + 256^2*ip[2,] + 256*ip[3,] + ip[4,]
o - order(ip)
df[o, ]
},
hb2 = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip);
dim(ip) - c(4, nrow(df))
o - sort.list(ip[4,], method=radix, na.last=TRUE)
for (kk in 3:1) {
o - o[sort.list(ip[kk,o], method=radix, na.last=TRUE)]
}
df[o, ]
}
)
print(res)
test elapsed
1 peda 4.12
2 peda2 4.08
3 hb 0.28
4 hb2 0.25
On Sun, May 31, 2009 at 12:42 AM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
edwin Sendjaja wrote:
Hi VQ,
Thank you. It works like charm. But I think Peter's code is faster. What
is the difference?
i think peter's code is more r-elegant, though less generic. here's a
quick test, with not so surprising results. gsubfn is implemented in r,
not c, and it is painfully slow in this test. i also added gabor's
suggestion.
library(gsubfn)
library(gtools)
library(rbenchmark)
n = 1000
df = data.frame(
a=rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'),
simplify=TRUE))
benchmark(columns=c('test', 'elapsed'), replications=10, order=NULL,
peda={
connection = textConnection(as.character(df$ip))
o = do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ] },
waku=df[order(gsubfn(perl=TRUE,
'[0-9]+',
~ sprintf('%03d', as.integer(x)),
as.character(df$ip))), ],
gagr=df[mixedorder(df$ip), ] )
# peda 0.070
# waku 7.070
# gagr 4.710
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented, minimal,
self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
[R] visible code
Hello, Can anyone help me with the following: if one enters a function name in the R console then usually one sees the code of that function. But there are functions that one cannot see. For example I want to see the code of print.htest or t.test.default. These functions are non-visible. Is it possible to see the code anyway? Thanks in advance, Karin Groothuis-Oudshoorn [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a time interval from a single time variable
-- begin included message -- I am trying to set up a data set for a survival analysis with time-varying covariates. The data is already in a long format, but does not have a variable to signify the stopping point for the interval. The variable DaysEnrolled is the variable I would like to use to form this interval. This is what I have now: ... end inclusion I would have expected a dozen solutions from the list - data manipulation problems usually get a large following. It can be done in 4 lines, assuming that the parent data set is sorted by subject and time within subject. newdata$start - olddata$DaysEnrolled #start time = the current variable temp - olddata$DaysEnrolled[-1]# shift column up by one position temp[diff(olddata$id) !=0] - NA# NA for last line of each subject newdata$stop - c(temp, NA) # add the NA for the last subject I will leave it to others to compress this into a 1-line application of one of the apply functions. (Unreadable perhaps, but definitely more elegant :-) Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Hedges g statistic
which package have Hedges g statistic calculation in R? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cochran’s Q statistic
jlfmssm wrote: Does anyone know which package include the computation of Cochran’s Q statistic in R? install.packages(outliers) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cochran’s Q statistic
On Jun 4, 2009, at 7:48 AM, jlfmssm wrote: Does anyone know which package include the computation of Cochrans Q statistic in R? jlfmssm See this thread: https://stat.ethz.ch/pipermail/r-help/2006-September/113139.html HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] visible code
Yes. Uwe Ligges' R Help Desk article in this Rnews should answer your questions: www.r-project.org/doc/Rnews/Rnews_2006-4.pdf On Jun 4, 2009, at 9:08 AM, Karin Martijn wrote: Hello, Can anyone help me with the following: if one enters a function name in the R console then usually one sees the code of that function. But there are functions that one cannot see. For example I want to see the code of print.htest or t.test.default. These functions are non-visible. Is it possible to see the code anyway? Thanks in advance, Karin Groothuis-Oudshoorn David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Morlet wavelet analysis
Dear, I am using cwt function from Rwave package to perform Morlet wavelet analysis. d1-c(1.31673209591515, -0.171333455797346, -1.67618546079420, -0.931604651010362, -0.954614183224057, -1.19821794547758, 0.516096363353144, -0.0432811977638559, 0.737764943619919, 0.438046629673177, -0.208607167908743, -0.3091308821262, -1.42473112931594, 0.234125312118165, -0.307047554490597, 0.686201101422384, 1.40104418156371, -0.722985280679127, -1.47295298153946, -1.49070808616222, -0.294899345331949, -0.623882988251866, 0.95580872220543, 0.159226599747532, 0.258592921036826, -0.893605825950617, -1.48090318423471, -0.599454337019121, -0.315308442503905, 1.48221661623468, 1.07313149452230, 0.0330503877044541, 1.07332650281700, -1.37121080562242, -0.181568897416963, 1.55169357402125, -0.502404116831855, 2.19007041840759, 0.177374817139349, 1.37270541339308, 0.179612250662539, 1.33781757389541) d1-ts(d1deltat=1,start=c(1960,1)) library(Rwave) D1-cwt(d1,noctave=8,nvoice=1) After I did Morlet wavelet analysis by using function wavelet from dplR package. library(dplR) D2-wavelet(d1,Dt=1) But the wavelets from two tests are very different. What is the reason?? Thank you, Irina Irina Foss Environmental Research Institute North Highland College UHI Millennium Institute Castle Street Thurso, Caithness Scotland KW14 7JD United Kingdom Tel: +44 (0) 1847 889 587 Fax: +44 (0) 1847 890 014 http://www.eri.ac.uk/sub/staff.php?id=11 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cochran's Q statistic
jlfmssm schrieb: Does anyone know which package include the computation of Cochran?s Q statistic in R? jlfmssm [[alternative HTML version deleted]] I might be wrong but the packages rmeta and meta report a Q-statistic. For an example, see library(meta) meta1 metagen(meta1$TE, meta1$seTE, sm=RR) Bernd __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] order() with randomised order in ties?
On Thu, Jun 4, 2009 at 12:36 PM, Patrick Burns pbu...@pburns.seanet.com wrote: How about: order(x, runif(length(x))) Thanks - that is really elegant. Rainer Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Rainer M Krug wrote: Sorry for replying to my own post, but I found a solution. Still, a more elegant solution would be preferred. On Thu, Jun 4, 2009 at 12:02 PM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to use order() to get the order of a vector. But I would need a different behavior when ties occur: similar to the parameter ties.method = random in the rank() function, I would need to randomise the ties. Is this possible? The solution is to randomize the vector before submitting to order(): x - rep(1:10, 2) iS - sample( length(x) ) o - order( x[iS], na.last=NA, decreasing=TRUE) o [1] 8 16 12 17 2 9 7 15 10 11 4 14 3 5 13 20 1 6 18 19 x[iS][o] [1] 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 iS - sample( length(x) ) o - order( x[iS], na.last=NA, decreasing=TRUE) o [1] 14 19 13 20 2 18 3 10 1 15 4 9 11 12 6 7 8 16 5 17 x[iS][o] [1] 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 Thanks Rainer Example: x - rep(1:10, 2) order(x) [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 order(x) [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 ## I would need different order for the ties, as below in rank() example: rank(x, ties.method=random) [1] 1 4 6 7 10 12 13 15 18 19 2 3 5 8 9 11 14 16 17 20 rank(x, ties.method=random) [1] 2 4 5 7 9 12 14 15 18 19 1 3 6 8 10 11 13 16 17 20 Thanks Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dropping terms from regression w/ poly()
Hello r-help,
I'm fitting a model with lm() and using the orthogonal polynomials
from poly() as my basis:
dat - read.csv(ConsolidatedData.csv, header=TRUE)
attach(dat)
nrows - 1925
Rad - poly(Radius, 2)
ntheta - 14
Theta - poly(T.Angle..deg., ntheta)
nbeta - 4
Beta - poly(B.Beta..deg., nbeta)
model.1 - lm( Measurement ~ Block + Rad + Theta + Beta + Rad:Theta +
Rad:Beta + Theta:Beta)
Which works splendidly, and for my data set shows that the odd orders
in Theta and Beta are not significant (expected because it is a
symmetric measurement that I'm fitting). Now I want to simplify my
model by dropping all the odd-order terms. I couldn't figure out a
way to do that without defining new design matrices like this:
T2 - matrix(0,nrows,ntheta)
for(i in 1:ntheta){
T2[,i] - Theta[,i]
}
B2 - matrix(0,nrows,nbeta)
for(i in 1:nbeta){
B2[,i] - Beta[,i]
}
R2 -matrix(0,nrows,2)
R2[,1] - Rad[,1]
R2[,2] - Rad[,2]
And then fitting a model using the individual columns of T2, R2 and
B2. That lets me drop all the odd terms, but it is very cumbersome:
model.2 - lm( Measurement ~ Block + (R2[,1] + R2[,2]) + (T2[,1] +
T2[,2] + T2[,3] + T2[,4] + T2[,5] + T2[,6] + T2[,7] +
T2[,8] + T2[,9] + T2[,10] + T2[,11]+ T2[,12] + T2[,13] +
T2[,14]) + (B2[,1] + B2[,2] + B2[,3] + B2[,4]) + (R2[,1]
+ R2[,2]) : (T2[,1] + T2[,2] + T2[,3] + T2[,4] + T2[,5]
+ T2[,6] + T2[,7] + T2[,8] + T2[,9] + T2[,10] + T2[,11]+
T2[,12] + T2[,13] + T2[,14]) + (R2[,1] + R2[,2]) :
(B2[,1] + B2[,2] + B2[,3] + B2[,4]) + (T2[,1] + T2[,2] +
T2[,3] + T2[,4] + T2[,5] + T2[,6] + T2[,7] + T2[,8] +
T2[,9] + T2[,10] + T2[,11]+ T2[,12] + T2[,13] + T2[,14])
: (B2[,1] + B2[,2] + B2[,3] + B2[,4]))
because I have to call out each column explicitly. I'm new to R so I
have a suspicion that there is a much better way to accomplish this
(dropping odd terms from a model based on poly), but I haven't figured
it out. Any hints / suggestions? Thanks.
--
Joshua Stults
Website: http://j-stults.blogspot.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Plot and lm
I want to make a log-log plot with a regression line, but I can't figure out what I'm doing wrong. What I'm trying is: plot(mass,area, log=xy, pch=as.numeric(food)) abline(lm(mass~area)) or plot(mass,area, log=xy, pch=as.numeric(food)) islands$logmass - log(mass) islands$logarea - log(area) attach(islands) abline(lm(logmass~logarea)) But that does not show a line. Where am I going wrong? data: island, area,species,food,mass Ibiza , 577 , Anser n. sp., herb, 2.0 Ibiza , 577 ,Haliaeetus albicilla, carn, 4.8 Mauritius , 1874 , Raphus cucullatus, herb, 19 Mauritius , 1874 , Circus alphonsi, carn, 0.63 Mallorca , 3667 , Myotragus balearicus, herb, 40 Mallorca , 3667 , Aquila chrysaetos, carn, 4.2 Kreta , 8259 , Elephas creutzburgi, herb, 3200 ... /Par -- Par Leijonhufvud p...@hunter-gatherer.org I don't believe in reincarnation. I used to, but that was in another life. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Import ARIMA coefficients
Hello, I need to know how to import ARIMA coefficients. I already determined the coefficients of the model with other software, but now i need to do the forecast in R. For Example: I have a time series named x and i have fitted an ARIMA(1,0,1) (with other software) AR coef = -.172295 MA coef = .960043 (i know that this is not a good model, it's just an example) I try to import the coefficients doing this: fit - arima(x, order=c(1, 0, 1), fixed=c(-.172295, .960043, 0)) fit Call: arima(x = x, order = c(1, 0, 1), fixed = c(-0.172295, 0.960043, 0)) Coefficients: Error in se nrow(x$var.coef) : invalid 'y' type in 'x y' I will be very pleased if someone help me. _ Mais do que mensagens conheça todo o Windows Live. http://www.microsoft.com/windows/windowslive/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] RGtk2 help: Show list of column names from dataset and categorize as factor or numeric
Hi UseRs, I recently started working with the RGtk2 library. The documentation is comprehensive and I've tried learning from the examples in most Gtk tutorials (which have C code). This is a little problematic, but has helped quite a bit in getting me started. I would like to create a GUI for file selection, which then displays the column names from the selected file, and provides the user with checkboxes with each column name for the user to select. Two columns of check boxes (Factor Type, Numeric Type) and one column of names is what I would like to display. Moreover, I am planning to create a GUI tool that would have tabs in a notebook layout, each tab providing a certain functionality, beginning from basic charting of data, and going on to applying regression models and such on the data. This requires extensive knowledge in components that RGtk2 provides which could be implemented for the task outlined above. I have looked at the omegahat.org examples, but would like to see examples for such simple tasks as to how one could create a drop down list of column names to choose for x axis and another drop down allowing the choice of y axis, etc. Having made the choice to use RGtk2, I would appreciate if users could share their RGtk experience with me. Regards Harsh Singhal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RGtk2 help: Show list of column names from dataset and categorize as factor or numeric
I use gWidgetsRGtk2, which provides widgets such as notebook etc. You can start with gWidgetsRGtk2, and if you want to tailor the GUI according to your needs, you can modified them with RGtk2 package. Ronggui 2009/6/4 Harsh singhal...@gmail.com: Hi UseRs, I recently started working with the RGtk2 library. The documentation is comprehensive and I've tried learning from the examples in most Gtk tutorials (which have C code). This is a little problematic, but has helped quite a bit in getting me started. I would like to create a GUI for file selection, which then displays the column names from the selected file, and provides the user with checkboxes with each column name for the user to select. Two columns of check boxes (Factor Type, Numeric Type) and one column of names is what I would like to display. Moreover, I am planning to create a GUI tool that would have tabs in a notebook layout, each tab providing a certain functionality, beginning from basic charting of data, and going on to applying regression models and such on the data. This requires extensive knowledge in components that RGtk2 provides which could be implemented for the task outlined above. I have looked at the omegahat.org examples, but would like to see examples for such simple tasks as to how one could create a drop down list of column names to choose for x axis and another drop down allowing the choice of y axis, etc. Having made the choice to use RGtk2, I would appreciate if users could share their RGtk experience with me. Regards Harsh Singhal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Binning or grouping data
alamoboy wrote: Newbie here. Many apologies in advance for using the incorrect lingo. I'm new to statistics and VERY new to R. I'm attempting to group or bin data together in order to analyze them as a combined group rather than as discrete set. I'll provide a simple example of the data for illustrative purposes. Patient ID | Charges |Age | Race 1 | 100 |0 | Black 2 | 500 |3 | White 3 | 200 |5 | Hispanic 4 | 90 |7 | Asian 5 |400| 10 | Hispanic 6 |500| 16 | Black I'm trying to create three age categories--0 to 4, 5 to 11 and 12 to 17--and analyze their Charges by their Race. How do I go abouts to doing this? Thanks for any assistance! Sam Sam, In addition to functions mentioned by other respondents, you may wish to investigate findInterval(), which returns indices of bins. The resulting indices are very useful for subscripting as well as grouping. id [1] 1 2 3 4 5 6 age [1] 0 3 5 7 10 16 group - findInterval(age,breaks) group [1] 1 1 3 3 3 5 data.frame(id,age,group) id age group 1 1 0 1 2 2 3 1 3 3 5 3 4 4 7 3 5 5 10 3 6 6 16 5 Glen -- View this message in context: http://www.nabble.com/Binning-or-grouping-data-tp23864555p23872151.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] hist returning density larger than 1
The following code is giving me problems. I want to export densities
of a distribution to a csv file. At the bottom of the code I use the
hist function to generate the densities. But hist is returning values
greater than 1. I don't understand, why. Any help you can supply is
greatly appreciated.
# Set word path
dir-~/Research/MR Distribution Analysis/
setwd(dir)
getwd()
# Define Objects
Max.Shots=3
Max.Groups=5000
hist.brks - pretty(range(c(0,4)), 200)
for(sim in (2:Max.Shots)){
x.Data-matrix(nrow=Max.Groups,ncol=sim,byrow=TRUE)
y.Data-matrix(nrow=Max.Groups,ncol=sim,byrow=TRUE)
shot2mean.x-matrix(nrow=Max.Groups,ncol=sim,byrow=TRUE)
shot2mean.y-matrix(nrow=Max.Groups,ncol=sim,byrow=TRUE)
Euclid.Dist-matrix(nrow=Max.Groups,ncol=sim,byrow=TRUE)
mean.radius-vector(mode=numeric,length=Max.Groups)
# Generate the Simulation Data
###
# Number of Groups
for(Groups in (1:Max.Groups)) {
# Number of Shots per Group
for(Shots in (1:sim)) {
# Populate the Simulation Data
x.Data[Groups,Shots]-rnorm(n=1,mean=0,sd=1)
y.Data[Groups,Shots]-rnorm(n=1,mean=0,sd=1)
}
}
# Calculate the Component Distances from the Shots to the Group Mean
# Calculate the Euclidean Distance from the Shots to the Group Mean
# Calculate the Group Mean Radius
###
# Number of Groups
for(Groups in (1:Max.Groups)){
# Number of Shots per Group
for(Shots in (1:sim)){
shot2mean.x[Groups,Shots]-(x.Data[Groups,Shots]-
mean(x.Data[Groups,]))^2
shot2mean.y[Groups,Shots]-(y.Data[Groups,Shots]-
mean(y.Data[Groups,]))^2
Euclid.Dist[Groups,Shots]-sqrt(shot2mean.x[Groups,Shots]
+shot2mean.y[Groups,Shots])
}
mean.radius[Groups]-mean(Euclid.Dist[Groups,])
}
f-hist(mean.radius,breaks=hist.brks,plot=FALSE)
density.data-cbind(sim,f$mids,f$density)
colnames(density.data)-c(Simulation,MidPoint,Density)
if (sim==2) {Simulation.Data-density.data} else {Simulation.Data-
rbind(Simulation.Data,density.data)}
rm(x.Data)
rm(y.Data)
rm(shot2mean.x)
rm(shot2mean.y)
rm(Euclid.Dist)
rm(mean.radius)
}
write.table(
Simulation.Data,
file=paste(dir,/DATA/Simulation_Data.csv,sep=''),
sep=,,
quote=FALSE,
row.names=FALSE,
qmethod=double)
Steven Matthew Anderson
Anderson Research, LLC
Statistical Programming and Analysis
SAS (R) Certified Professional
adastr...@mac.com
Ad Astra per Aspera
שָׁלוֹם
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] error installing RCurl in SUSE SLES10-SP2
dear list, i'm trying to install the package RCurl into a linux system running SUSE Linux Enterprise Server 10 SP2 but i get compilation errors which i guess should be due to some missing additional software. i've been searching about the R-help archives without success so i hope somebody can point me out to this missing piece. in principle, i have the necessary curl software installed: $ rpm -qa | grep curl curl-7.15.1-19.11 curl-devel-7.15.1-19.11 curl-32bit-7.15.1-19.11 however as you'll see below, when i try to install it from R the compilation breaks. i've included at the end also the sessionInfo() output (it's R-2.9.0). i've also tried to install a newer version of RCurl (0.98) which i found in http://www.omegahat.org/RCurl but it gives the same errors. thanks! robert. install.packages(RCurl, repos=http://cran.r-project.org;) trying URL 'http://cran.r-project.org/src/contrib/RCurl_0.97-3.tar.gz' Content type 'application/x-gzip' length 1437829 bytes (1.4 Mb) opened URL == downloaded 1.4 Mb * Installing *source* package 'RCurl' ... checking for curl-config... /usr/bin/curl-config checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E Version has a libidn field Version has CURLOPT_URL No CURLOPT_NOPROXY enumeration value. No CURLINFO_CONDITION_UNMET enumeration value. No CURLINFO_REDIRECT_URL enumeration value. No CURLINFO_CERTINFO enumeration value. No CURLINFO_PRIMARY_IP enumeration value. No CURLINFO_APPCONNECT_TIME enumeration value. No CURLOPT_KEYPASSWD enumeration value. No CURLOPT_DIRLISTONLY enumeration value. No CURLOPT_APPEND enumeration value. No CURLOPT_KRBLEVEL enumeration value. No CURLOPT_USE_SSL enumeration value. No CURLOPT_TIMEOUT_MS enumeration value. No CURLOPT_CONNECTTIMEOUT_MS enumeration value. No CURLOPT_HTTP_TRANSFER_DECODING enumeration value. No CURLOPT_HTTP_CONTENT_DECODING enumeration value. No CURLOPT_NEW_FILE_PERMS enumeration value. No CURLOPT_NEW_DIRECTORY_PERMS enumeration value. No CURLOPT_POSTREDIR enumeration value. No CURLOPT_SSH_HOST_PUBLIC_KEY_MD enumeration value. No CURLOPT_OPENSOCKETFUNCTION enumeration value. No CURLOPT_OPENSOCKETDATA enumeration value. No CURLOPT_COPYPOSTFIELDS enumeration value. No CURLOPT_PROXY_TRANSFER_MODE enumeration value. No CURLOPT_SEEKFUNCTION enumeration value. No CURLOPT_SEEKDATA enumeration value. No CURLOPT_CRLFILE enumeration value. No CURLOPT_ISSUERCERT enumeration value. No CURLOPT_ADDRESS_SCOPE enumeration value. No CURLOPT_CERTINFO enumeration value. No CURLOPT_USERNAME enumeration value. No CURLOPT_PASSWORD enumeration value. No CURLOPT_PROXYUSERNAME enumeration value. No CURLOPT_PROXYPASSWORD enumeration value. No CURLOPT_SSH_HOST_PUBLIC_KEY_MD5 enumeration value. No CURLOPT_NOPROXY enumeration value. No CURLOPT_TFTP_BLKSIZE enumeration value. No CURLOPT_SOCKS5_GSSAPI_SERVICE enumeration value. No CURLOPT_SOCKS5_GSSAPI_NEC enumeration value. Version has CURLOPT_PROTOCOLS Version has CURLOPT_REDIR_PROTOCOLS configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/cursos/MBI/soft/R/R-2.9.0/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include-fpic -g -O2 -c base64.c -o base64.o gcc -std=gnu99 -I/cursos/MBI/soft/R/R-2.9.0/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include-fpic -g -O2 -c curl.c -o curl.o curl.c:553: warning: 'struct curl_certinfo' declared inside parameter list curl.c:553: warning: its scope is only this definition or declaration, which is probably not what you want gcc -std=gnu99 -I/cursos/MBI/soft/R/R-2.9.0/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include-fpic -g -O2 -c curlInit.c -o curlInit.o gcc -std=gnu99 -I/cursos/MBI/soft/R/R-2.9.0/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include-fpic -g -O2 -c curl_base64.c -o curl_base64.o gcc -std=gnu99 -I/cursos/MBI/soft/R/R-2.9.0/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include-fpic -g -O2 -c enums.c -o enums.o In file included from CurlOptEnums.h:10, from enums.c:79: CURLOptTable.h:136: error: 'CURLOPT_LOCALPORT' undeclared here (not in a function) CURLOptTable.h:137: error:
Re: [R] order() with randomised order in ties?
On Thu, Jun 4, 2009 at 12:45 PM, Rainer M Krug r.m.k...@gmail.com wrote: On Thu, Jun 4, 2009 at 12:36 PM, Patrick Burns pbu...@pburns.seanet.com wrote: How about: order(x, runif(length(x))) Thanks - that is really elegant. One thing: it is saver to use sample(length(x)) instead of runif(length(x)), as runif() might also produce ties. It is quite unlikely that those result in ties for both vectors, but not impossible. That problem is avoided with sample(length(x)) Rainer Rainer Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Rainer M Krug wrote: Sorry for replying to my own post, but I found a solution. Still, a more elegant solution would be preferred. On Thu, Jun 4, 2009 at 12:02 PM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to use order() to get the order of a vector. But I would need a different behavior when ties occur: similar to the parameter ties.method = random in the rank() function, I would need to randomise the ties. Is this possible? The solution is to randomize the vector before submitting to order(): x - rep(1:10, 2) iS - sample( length(x) ) o - order( x[iS], na.last=NA, decreasing=TRUE) o [1] 8 16 12 17 2 9 7 15 10 11 4 14 3 5 13 20 1 6 18 19 x[iS][o] [1] 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 iS - sample( length(x) ) o - order( x[iS], na.last=NA, decreasing=TRUE) o [1] 14 19 13 20 2 18 3 10 1 15 4 9 11 12 6 7 8 16 5 17 x[iS][o] [1] 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 Thanks Rainer Example: x - rep(1:10, 2) order(x) [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 order(x) [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 ## I would need different order for the ties, as below in rank() example: rank(x, ties.method=random) [1] 1 4 6 7 10 12 13 15 18 19 2 3 5 8 9 11 14 16 17 20 rank(x, ties.method=random) [1] 2 4 5 7 9 12 14 15 18 19 1 3 6 8 10 11 13 16 17 20 Thanks Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ordered Twoing criterion in classification trees
Dear R users, I'm using the rpart package to build classification trees. I'm interested in implementing the ordered Twoing as a splitting criterion. Does anyone have experience with this task ? Thank you for your help Paolo Paolo Radaelli Dipartimento di Metodi Quantitativi per le Scienze Economiche ed Aziendali Facoltà di Economia Università degli Studi di Milano-Bicocca Via Bicocca degli Arcimboldi, 8 20126 Milano Italy e-mail paolo.radae...@unimib.it Tel +39 02 6448 3163 Fax +39 02 6448 3105 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] hist returning density larger than 1
Hi, If I understand your problem correctly, you didn't need to send us your entire code. A simple reproducible example that showed just your problem would have been sufficient. Take a look at this, and reread the help for hist(). testdata - runif(1000) testdata.hist - hist(testdata) testdata.hist$density [1] 1.19 1.01 0.89 0.79 0.92 1.04 1.13 0.93 [9] 1.02 1.08 # See the help for the actual definition of density (admittedly not as clear as # one might want). sum(testdata.hist$density * diff(testdata.hist$breaks)) [1] 1 # What you appear to actually want: testdata.hist$counts/1000 [1] 0.119 0.101 0.089 0.079 0.092 0.104 0.113 0.093 0.102 0.108 Sarah On Thu, Jun 4, 2009 at 11:23 AM, Steven Matthew Anderson adastr...@mac.com wrote: The following code is giving me problems. I want to export densities of a distribution to a csv file. At the bottom of the code I use the hist function to generate the densities. But hist is returning values greater than 1. I don't understand, why. Any help you can supply is greatly appreciated. -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Small mystery : passing a subset= argument to lme|lm through ...
Dear list,
I have problems involving passing a subset= argument through
I'm trying to augment the set of defined analyses for mice (homonymous
package) with a call to lme. This package create multiple imputations of
missing data in a mids object, each completed data set may be obtained
through the complete(data, set) function.
sessionInfo()
R version 2.9.0 (2009-04-17)
x86_64-pc-linux-gnu
locale:
LC_CTYPE=fr_FR.UTF-8;LC_NUMERIC=C;LC_TIME=fr_FR.UTF-8;LC_COLLATE=fr_FR.UTF-8;LC_MONETARY=C;LC_MESSAGES=fr_FR.UTF-8;LC_PAPER=fr_FR.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=fr_FR.UTF-8;LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices datasets utils methods
base
other attached packages:
[1] odfWeave_0.7.10 XML_2.3-0 lattice_0.17-25
loaded via a namespace (and not attached):
[1] grid_2.9.0 MASS_7.2-47 nlme_3.1-92 rpart_3.1-44
# First attempt :
foo-function(formula, data, ...) {
library(nlme)
call - match.call()
if (!is.mids(data))
stop(The data must have class mids)
analyses-lapply(1:data$m, function(i) lme(formula,
data=complete(data,i),
...))
object - list(call = call, call1 = data$call, nmis = data$nmis,
analyses = analyses)
oldClass(object) - c(mira, oldClass(object))
return(object)
}
bar1-foo(ttotal~nbpradio, data=Data1.Imp, random=~1|ctr)
class(bar1$analyses[[1]])
[1] lme # Fine : the random= argument *HAS BEEN* passed to lme()
# through ... ; therefore it's (usually) possible.
# Further tests (not shown) show that results are reasonable (i. e.
# compliant with expectations...).
bar2-foo(log(ttotal)~log(nbpradio), data=Data1.Imp, random=~1|ctr,
subset=nbpradio0)
Erreur dans eval(expr, envir, enclos) :
..2 utilisé dans un mauvais contexte, aucun ... où chercher
Further exploration show that a subset= argument *DOES NOT GET PASSED*
to lme(), which complaints furthermore not to find a ... argument.
Since lme is a method private to the nlme package, I can't trace the
source of error.
[Later : Same observation with lm() instead of lme()...]
Now, ISTR that an analogous problem *has already been discussed* on this
list, but even a prolonged use of RSiteSearch() did not retrieve it.
http://finzi.psych.upenn.edu/R/Rhelp02/archive/10139.html hints that
Modelling and graphics functions with formulas all have slightly
different nonstandard evaluation rules, and it turns out that lme()
doesn't evaluate extra arguments like subset in the parent environment,
but this does not tell me why a ... argument is *not* found. Different
frame of evaluation ?
May some kind soul point me to the right direction ? (He may even
further my shame by telling me how he retrieved the relevant
information...).
Sincerely,
Emmanuel Charpentier
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Binning or grouping data
alamoboy wrote: Newbie here. Many apologies in advance for using the incorrect lingo. I'm new to statistics and VERY new to R. I'm attempting to group or bin data together in order to analyze them as a combined group rather than as discrete set. I'll provide a simple example of the data for illustrative purposes. Patient ID | Charges |Age | Race 1 | 100 |0 | Black 2 | 500 |3 | White 3 | 200 |5 | Hispanic 4 | 90 |7 | Asian 5 |400| 10 | Hispanic 6 |500| 16 | Black I'm trying to create three age categories--0 to 4, 5 to 11 and 12 to 17--and analyze their Charges by their Race. How do I go abouts to doing this? Thanks for any assistance! Sam Oops! My use of bins other than you described was not part of some obscure strategy. Should have been as shown below: id [1] 1 2 3 4 5 6 age [1] 0 3 5 7 10 16 breaks [1] 0 5 12 17 group - findInterval(age,breaks) data.frame(id,age,group) id age group 1 1 0 1 2 2 3 1 3 3 5 2 4 4 7 2 5 5 10 2 6 6 16 3 -- View this message in context: http://www.nabble.com/Binning-or-grouping-data-tp23864555p23872705.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help understanding output from aov and from anova
Hi Steve, Thanks for your response and also for the useful information about the latest version warning. On the earlier version that I was using (2.6.2 Win), I was expecting at least an error or warning in response to submitting an obviously corner case condition that should result in NaN, but it does not yield any. The t-test does so. Actually, when the groups have two samples each (and all values are identical), the P-value results in an NaN which is what one would expect. It is great to know that it has been corrected at least through a warning in the latest version. R version 2.6.2 (2008-02-08) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN 3-900051-07-0 snip #No warning/error provided when there are 2 samples in 1 group, and 1 in the other vtot=c(200,200,200) fac=as.factor(c(1,1,2)) anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum SqMean Sq F value Pr(F) fac1 1.4722e-27 1.4722e-27 0. 0.6667 Residuals 1 4.4166e-27 4.4166e-27 T-test works as expected with an error t.test(vtot~fac,var.equal=TRUE) Error in t.test.default(x = c(200, 200), y = 200, var.equal = TRUE) : data are essentially constant # P-value is NaN when we have at least two samples in each group vtot=c(200,200,200,200) fac=as.factor(c(1,1,2,2)) anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum Sq Mean Sq F value Pr(F) fac1 0 0 Residuals 2 0 0 anova(lm(vtot~fac))[1,5] [1] NaN Best, Suman. On Wed, Jun 3, 2009 at 9:24 PM, Steven McKinney smckin...@bccrc.ca wrote: Hi Suman, What version of R are you running? In R 2.9.0 running your first example yields a warning Warning message: In anova.lm(lm(vtot ~ fac)) : ANOVA F-tests on an essentially perfect fit are unreliable so some adept R developer has taken the time to figure out how to warn you about such a problem. Perhaps someone will add this to aov() at some point as well. The only variability in this problem is that introduced by machine precision rounding errors. The exercise of submitting data with no variability to a program designed to assess variability cannot be expected to produce meaningful output, so there's nothing to understand except the issue of machine precision. Machine roundoff error is an important topic, so I'd recommend learning about that issue, which will do most to help understand these examples. Best SteveM R version 2.9.0 (2009-04-17) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. vtot=c(7.29917, 7.29917, 7.29917) #identical values fac=as.factor(c(1,1,2)) #group 1 has first two elements, group 2 has anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum Sq Mean Sq F value Pr(F) fac 1 1.6818e-30 1.6818e-30 0. 0.6667 Residuals 1 5.0455e-30 5.0455e-30 Warning message: In anova.lm(lm(vtot ~ fac)) : ANOVA F-tests on an essentially perfect fit are unreliable summary(aov(vtot~fac)) Df Sum Sq Mean Sq F value Pr(F) fac 1 1.6818e-30 1.6818e-30 0. 0.6667 Residuals 1 5.0455e-30 5.0455e-30 fac=as.factor(c(1,2,2)) anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum Sq Mean Sq F value Pr(F) fac 1 6.7274e-30 6.7274e-30 1.3340e+32 2.2e-16 *** Residuals 1 5.043e-62 5.043e-62 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Warning message: In anova.lm(lm(vtot ~ fac)) : ANOVA F-tests on an essentially perfect fit are unreliable Steven McKinney, Ph.D. Statistician Molecular Oncology and Breast Cancer Program British Columbia Cancer Research Centre email: smckin...@bccrc.ca tel: 604-675-8000 x7561 BCCRC Molecular Oncology 675 West 10th Ave, Floor 4 Vancouver B.C. V5Z 1L3 Canada -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Suman Sundaresh Sent: Wednesday, June 03, 2009 3:55 PM To: r-help@r-project.org Subject: [R] Need help understanding output from aov and from anova Hi all, I noticed something strange when I ran aov and anova. vtot=c(7.29917, 7.29917, 7.29917) #identical values fac=as.factor(c(1,1,2)) #group 1 has first two elements, group 2 has the 3rd element When I run: anova(lm(vtot~fac)) Analysis of Variance Table
Re: [R] Getting a column of values from a list - think I'm doing it the hard way
If you want to play R golf then:
sapply(HouseDatesList, '[[', 1)
[1] 1990 1991 1992 1993 1994 1995 1996
Does the same thing in fewer keystrokes, but Ronggui's solution is more
readable.
--
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-boun...@r-
project.org] On Behalf Of Ronggui Huang
Sent: Wednesday, June 03, 2009 8:57 PM
To: Jason Rupert
Cc: R-help@r-project.org
Subject: Re: [R] Getting a column of values from a list - think I'm
doing it the hard way
2009/6/4 Jason Rupert jasonkrup...@yahoo.com:
Example code it shown below.
I think I am doing this the hard way. I'm just trying to get the
full year value from an array of dates. An example array is shown
below. Right now, I'm using a for loop to pull the year out of a
list where the dates were split up into their individual components.
This seems to work, but just wondering if there is an easier way.
Thanks for any insights.
#*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
HouseDates - c(02/27/90, 02/27/91, 01/14/92, 02/28/93,
02/01/94, 02/01/95, 02/01/96)
# ?as.Date
HouseDatesFormatted-as.Date(HouseDates, %m/%d/%y)
HouseDatesFormatted
HouseDatesList-strsplit(as.character(HouseDatesFormatted), -,
fixed=TRUE)
sapply(HouseDatesList,function(x) x[[1]])
[1] 1990 1991 1992 1993 1994 1995 1996
HouseYear_array-NULL
length_array-length(HouseDatesList)
for(ii in 1:length_array)
{
HouseYear-HouseDatesList[[ii]][1]
HouseYear_array-c(HouseYear_array, HouseYear)
}
as.character(HouseYear_array)
# Desired:
# [1] 1990 1991 1992 1993 1994 1995 1996
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
HUANG Ronggui, Wincent
PhD Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Least Squares Method
Dear Helpers, I need to fit a gamma function on a distribution. I want to use the Method of the Least Squares for minimizing the sum of squared residuals (SSE). I don't know how to do this. I guess I need to calculate the best fit parameter values and then somehow comparing my empirical distribution with the theorical one (in this case gamma). This is what I've done so far: rate is the name of my distribution (12000 data point) mean- mean(rate) # mean of the empirical distribution var -var(rate)# variance of the empirical distribution l.est - mean/var# lambda estimate (scale param.) a.est - (mean^2)/var # alfa estimate (shape param.) should I create a random gamma distribution and then comparing it with my data? But how? And then Is there a way to visualize in a graph the fitting with gamma? Thanks a lot for your help!! Alessandra -- View this message in context: http://www.nabble.com/Least-Squares-Method-tp23872037p23872037.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] loglilelihood
Hi, I tried fitting loglinear model using the glm(catspec). The data used is FHtab. . An independence model was fitted. Here summary() and fitmacro( ) give different values for AIC. I understand that fitmacro( ) takes the likelilhood ratio L2(deviance) to calculate AIC and uses the formula AIC= L2- d.f(deviance)*2 and this AIC is used in nested models. (Am I right?) The value given by loglik( ) is used by summary to calculate AIC using the formula AIC= -2 logL + K* e.d.f. Can anybody tell me how this log L is calculated? It is not logL=Sum(ni* log (pi)), where pi= fitted frequency / total no. of observations. The help available at extract AIC( ) also is not giving the answer. Please help regards Subha [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] wrong labels and colors of points in graph/plot
Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(!is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Import ARIMA coefficients
On Thu, 4 Jun 2009 15:21:52 +0100 Daniel Mail d20...@live.com.pt wrote: DM I need to know how to import ARIMA coefficients. I already DM determined the coefficients of the model with other software, but DM now i need to do the forecast in R. So why then don't you fit the model then in R Would be much more convenient... Stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help needed with ridge regression and choice of lambda with lm.ridge!!!
Hi, I'm a beginner in the field, I have to perform the ridge regression with lm.ridge for many datasets, and I wanted to do it in an automatic way. In which way I can automatically choose lambda ? As said, right now I'm using lm.ridge MASS function, which I found quite simple and fast, and I've seen that among the returned values there are HKB estimate of the ridge constant and L-W estimate of the ridge constant, together with GCV values. I found on the web other studies where people simply choose among one of these quantities. It will be perfect to me to do the same, but how? Which are the decisional criteria, if there are criteria? HKB, L-W or none of these ? Another (for me) important question: Aren't the lambda in general supposed to increase with the increasing of the number of predictors ? Isn't the ridge regression supposed to work fine even with number of predictors number of observations? At least I was said so... But if I have a dataset of 16 observations and 34 predictors I get: fmr-lm.ridge(y~0+ .+., data=x, lambda = seq(0,10,0.01)) select(fmr) modified HKB estimator is -1.850770e-28 modified L-W estimator is -2.012264e-28 smallest value of GCV at 0.01 and similar values if I reduce the number of predictor in the dataset, all numbars between 17and 34. but if I build a dataset with only 16 predictors (euqal to the number of rows) I get: select(fmr) modified HKB estimator is 0.1511719 modified L-W estimator is 3.322775 smallest value of GCV at 0.51 And at the same way, other accettable values for any smaller dataset... Please, could anybody help me? Thanks in advance Giulio _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 5-D density? was Re: R help
(copy of earlier reply which was not addressed to r-help. Also added informative Subject:) yes .. is not responsive to the question of how you propose to display or examine such a mathematical object. My signature was perhaps a lame (certainly an ineffective) effort at getting you to acknowledge that there might, just might, be some conceptual difficulties in viewing a 5+1 = 6 dimensional object. Can you point to any examples of such a procedure being applied in a manner that you find helpful? It is certainly feasible to calculate distances in 5-space, but the next step, displaying the concentration of those distances as a function of the coordinates, would be the tough nut. (1000 data points is not a problem. That amount of data should easily fit in any device that can run R.) -- David On Jun 4, 2009, at 12:06 AM, arijit kumar debnath wrote: yes.. I want a density estimator for 5 dimensional data.I want it to be efficient since I don't have access to a fast computer.My data set is quite large(about 1000 data points) and I have only a pentium 4 1.7 Ghz (512 Mb RAM)computer. On Thu, Jun 4, 2009 at 7:25 AM, David Winsemius dwinsem...@comcast.net wrote: We can get you a kde2d but you really want a kde5d? What sort of display are you proposing? -- David Earthling Winsemius On Jun 3, 2009, at 9:26 PM, arijit kumar debnath wrote: I want to fit a kernel density to a data-set of size 1000 and each data point being of dimension 5. The default density function does'nt work for me. Is there any other function available which can do the job efficiently.? I don't have access to a very fast computer. David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wrong labels and colors of points in graph/plot
No attachment came through the mailserver. Did you follow the directions in the posting guide regarding acceptable types of attachments? I also do not see any code that would let us reproduce an input process. One method that might work is to offer the results of dput(coefficients) as text within your message. -- David On Jun 4, 2009, at 12:22 PM, Katharina May wrote: Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(! is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Return variable assignments from a function
I still think that fortune(181) applies here. Someday you (or another user
that you give your function to) will run this, then realize that you/they had
an A, B, or c variable that has just been overwritten that you/they wanted to
keep. (also 'c' is one of the variable names recommended against, there is a
very often used function 'c').
You are also still breaking the association between A, B, and c. Keeping
things grouped into a single object is one of the powerful concepts in R/S.
If you know that one function (funcA) is only going to be called by another
function (funcB) and you want funcB to have access to things computed in funcA,
then the easiest/most straight forward is to return the values from funcA to
funcB and use the ret$A type syntax. Using '-' to have funcA assign values
in funcB is safer (and often simpler) than using 'assign', but you still need
to be careful.
If you want to use the return values at the command line and just want to avoid
all the typing of object$... then use the 'with' and 'within' functions (or
other functions that take a list of objects as the data to operate on).
If you tell us more about what you are trying to accomplish, maybe we can show
you a way to use the power of the R way of thinking to make your task easier
rather than trying to work around it.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: hydeb...@gmail.com [mailto:hydeb...@gmail.com] On Behalf Of Scott
Hyde
Sent: Wednesday, June 03, 2009 9:30 PM
To: r-help@r-project.org
Subject: Re: Return variable assignments from a function
As a followup to my question yesterday, what if I were to return the
argument as a list, and then unwrap the list with the function I've
written called objects. Is there any problems with doing it? It
works to use it inside other functions. For example:
=
objects - function(alist) {
for (vars in names(alist))
assign(vars,alist[[vars]],pos=sys.frame(-1))
}
simple - function(m,n) {
A=matrix(c(3,3,2,3),2,2)
B=m
c=1:n
list(A=A,B=B,c=c)
}
rm(A,B,c) #just in case they exist
stuff=simple(2,3)
objects(stuff)
A
[,1] [,2]
[1,]32
[2,]33
B
[1] 2
c
[1] 1 2 3
=
-Scott
*
Scott K. Hyde
Assistant Professor of Statistics and Mathematics
College of Math and Sciences
Brigham Young University -- Hawaii
Laie, HI 96762
On Tue, Jun 2, 2009 at 5:30 PM, Scott Hyde hy...@byuh.edu wrote:
I'd like to perform return variable assignments like matlab. For
example, the following function would return A, B, and c to the script
that called it.
=
function [A,B,c] = simple(m,n)
A=[ 3 2; 3 3]
B=m
c=1:n
=
I'd like to do similar assignments in R, but I seem to be able to
only return one variable. I tried to use a list to return all the
arguments, but then each has to be referred to using the list. For
example:
=
simple - function(m,n) {
A=matrix(c(3,3,2,3),2,2)
B=m
c=1:n
list(A=A,B=B,c=c)
}
stuff=simple(2,3)
stuff
$A
[,1] [,2]
[1,] 3 2
[2,] 3 3
$B
[1] 2
$c
[1] 1 2 3
=
Then I could assign each variable like this (which is what I'd like
to avoid):
=
A=stuff$A
B=stuff$B
c=stuff$c
rm(stuff) #stuff isn't needed anymore.
=
I've even toyed with the superassignment operator, which also works,
but I think it doesn't work for functions of functions. The following
example works.
=
simple2 - function(m,n) {
A - matrix(c(3,3,2,3),2,2)
B - m
c - 1:n
}
stuff2=simple2(2,3)
stuff2
[1] 1 2 3
A
[,1] [,2]
[1,] 3 2
[2,] 3 3
B
[1] 2
c
[1] 1 2 3
=
In the example below, I call the function ten inside the function
nine. I'm expecting that the variable b should change only in the
function nine (and not in the global environment). In other words, I
think the line (nine) b= 9 should be (nine) b= 10.
Can someone help me know how to do this correctly?
-Scott
=
nine = function(a) {
b - 9
ten(a)
print(paste((nine) b=,b))
}
ten = function(d) {
b - 10
print(paste((ten) b=,b))
print(paste((ten) d=,d))
d
}
nine(5)
[1] (ten) b= 10
[1] (ten) d= 5
[1] (nine) b= 9
b
[1] 10
=
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Re: [R] Using WinBUGS from R: A Multi-Way Array Problem
Your output example below looks the same as the input. But I think the 'aperm' function may be what you are looking for, read its help page and run the example to see if that will work for you. -- 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-boun...@r- project.org] On Behalf Of Khurram Nadeem Sent: Wednesday, June 03, 2009 11:37 PM To: r-help@r-project.org Subject: [R] Using WinBUGS from R: A Multi-Way Array Problem Please suggest a way out to the following problem. I have a T by n data matrix (say Y) where coulmns are time series of length T. To do some analysis in WinBUGS I need to construct my data as follows. yy-rep(Y,k) ## this will be a vector Yk-array(yy,dim=c(T,n,k)) ## data array Here the definition of dim indices is first index: T rows second index: n columns third index: for kth T by n array EXAMPLE T=3 n=2 k=2 Y-matrix(c(1,2,3,4,5,6), ncol=n) ## my data matrix yy-rep(Y,k) Yk-array(yy,dim=c(T,n,k)) Yk # this produces the following R output. , , 1 [,1] [,2] [1,]14 [2,]25 [3,]36 , , 2 [,1] [,2] [1,]14 [2,]25 [3,]36 That is, I have copied the orignal data k=2 times. WinBUGS will be supplied the following data d-list(Yk=Yk, T=T,n=n,k=k). Now in WinBUGS I have to define a multivariate stochastic node as as follows e[i,1:n,kk]~dmnorm( , ) ## i= 1,2,...,T ; kk = 1,2,...,k But Winbugs accepts only something like e[ , , 1:n]~dmnorm( , ). That is 1:n has to be given at the leftmost position. This means that I need to change the definition of dim indices in R as follows. first index: for kth T by n array second index: T rows third index: n columns. Specifically, I want the above output as 1, , [,1] [,2] [1,]14 [2,]25 [3,]36 2, , [,1] [,2] [1,]14 [2,]25 [3,]36 I would appriciate any help in this regard. Khurram Nadeem PhD Student Department of Math. Stat. Sciences University of Alberta [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wrong labels and colors of points in graph/plot
sorry, the attachment can be found here for download: http://www.wzw.tum.de/waldinventur/fileadmin/coefficient2.RData katharina wrote: Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(!is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/wrong-labels-and-colors-of-points-in-graph-plot-tp23873337p23873849.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Excel Export in a beauty way
Also see this post at
https://stat.ethz.ch/pipermail/r-help/2008-July/169149.html
The same idea is discussed in a SAS proceedings paper (but it is NOT
specific to SAS)
www.lexjansen.com/wuss/2005/data_presentation/dp_using_*sas*_with_xml.pdfhttp://www.lexjansen.com/wuss/2005/data_presentation/dp_using_sas_with_xml.pdf
As Marc indicated, the autofit column widths is a bit confusing. It would
be better called set the column width to a number large enough to show all
current contents.
Kevin
On Thu, Jun 4, 2009 at 12:37 AM, Patrick Connolly
p_conno...@slingshot.co.nz wrote:
On Wed, 03-Jun-2009 at 08:54AM -0500, Marc Schwartz wrote:
[...]
For example, using the Perl package that I do for the WriteXLS
package, it is not possible to use the AutoFit capability to
easily set all column widths wide enough to visually allow for the
data contained within each. From the available Perl package
documentation, this appears to be a run time only feature, which
means that it would require Excel to be available and running and
the user either going through the column formatting menu, or
creating a macro to automate the process.
I've used that Perl script for some time and find it very useful
(thanks Marc). I can't see a lot of point in trying to get the column
widths done by it. In Excel, it's very easy to highlight every column
and then double click on the line separating any two column names and
that will set ALL the widths to their optimum in one go.
(Just in case anyone didn't know.)
If that's not sufficient, one of those other methods might suit.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] documentation / An Intro to R / list manipulation
Hello R users http://cran.r-project.org/doc/manuals/R-intro.html An Introduction to R 6.2 Constructing and modifying lists After a short but successful struggle with nested associative arrays (using named lists), I think the following documentation change might be beneficial. The existing description and example (see below) implies that the given statement would transfer the component name as well (not so): Lst[5] - list(matrix=Mat) - new text for section 6.2 - Lists, like any subscripted object, can be extended by specifying additional components. For example Lst[5] - list(Mat) This new component can also be named, using either a character string (as shown) or suitable string variable names(Lst)[5] - matrix - test code - Mat - matrix() Lst - list() Lst[1] - list(matrix = Mat) str(Lst) # List of 1 # $ : logi [1, 1] NA names(Lst)[1] - matrix str(Lst) # List of 1 # $ matrix: logi [1, 1] NA many thanks to the R team in general Robbie ps: I'm using R for visualizing data, including directed graphs (via adjacency lists) and timeseries, and for summarizing relatively diverse but consistent datasets --- Robbie Morrison PhD student -- policy-oriented energy system simulation Technical University of Berlin (TU-Berlin), Germany University email (redirected) : morri...@iet.tu-berlin.de Webmail (preferred) : rob...@actrix.co.nz [from IMAP client] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot and lm
Could you provide a reproducible example even with fake data would be fine or dput() yours. On Thu, Jun 4, 2009 at 10:11 AM, Par Leijonhufvud p...@hunter-gatherer.org wrote: I want to make a log-log plot with a regression line, but I can't figure out what I'm doing wrong. What I'm trying is: plot(mass,area, log=xy, pch=as.numeric(food)) abline(lm(mass~area)) or plot(mass,area, log=xy, pch=as.numeric(food)) islands$logmass - log(mass) islands$logarea - log(area) attach(islands) abline(lm(logmass~logarea)) But that does not show a line. Where am I going wrong? data: island, area,species,food,mass Ibiza , 577 , Anser n. sp., herb, 2.0 Ibiza , 577 , Haliaeetus albicilla, carn, 4.8 Mauritius , 1874 , Raphus cucullatus, herb, 19 Mauritius , 1874 , Circus alphonsi, carn, 0.63 Mallorca , 3667 , Myotragus balearicus, herb, 40 Mallorca , 3667 , Aquila chrysaetos, carn, 4.2 Kreta , 8259 , Elephas creutzburgi, herb, 3200 ... /Par -- Par Leijonhufvud p...@hunter-gatherer.org I don't believe in reincarnation. I used to, but that was in another life. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fast way of finding top-n values of a long vector
Try adding a version that uses sort with the partial argument, that should be
faster than regular sort (for long enough test vectors) and possibly faster
than the max solutions when finding more than just the largest 2.
Also, for your max solutions, what happens when the 2 largest values are
identical?
--
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-boun...@r-
project.org] On Behalf Of Allan Engelhardt
Sent: Thursday, June 04, 2009 2:18 AM
To: r-help@r-project.org
Subject: [R] Fast way of finding top-n values of a long vector
If x is a (long) vector and n length(x), what is a fast way of
finding the top-n values of x?
Some suggestions (calculating the ratio of the two top values):
library(rbenchmark)
set.seed(1); x - runif(1e6, max=1e7); x[1] - NA;
benchmark(
replications=20,
columns=c(test,elapsed),
order=elapsed
, sort = {a-sort(x, decreasing=TRUE, na.last=NA)[1:2]; a[1]/a[2];}
, max = {m-max(x, na.rm=TRUE); w-which(x==m)[1]; m/max(x[-w],
na.rm=TRUE);}
, max2 = {w-which.max(x); max(x, na.rm=TRUE)/max(x[-w], na.rm=TRUE);}
)
# test elapsed
# 3 max2 0.772
# 2 max 1.732
# 1 sort 4.958
I want to apply this code to a few tens of thousands of vectors so
speed
does matter. In C or similar I would of course calculate the result
with a single pass through x, and not with three passes as in 'max2'.
Allan.
PS: I know na.last=NA is the default for sort, but there is no harm in
being explicit in how you want NA's handled.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] wrong labels and colors of points in graph/plot
Was not able to find NCStats on CRAN or BioConductor repositories. You are wasting our (or mine at any rate) time by giving us dead ends. You have been bitten by the factor snake in the grass. Color should be text rather than a factor if you want the desired results. See if this is illuminating and eventually helpful: str(coefficient2) 'data.frame': 26 obs. of 11 variables: $ site_no : num 29 10 17 18 19 25 9 33 37 41 ... $ intercept: num 0.603 2.767 3.6 1.86 2.753 ... $ slope: num 0.912 0.867 0.841 0.99 0.911 ... $ average_height : num 90.4 2096.2 990 1950 2184 ... $ average_crown_length : num NA 1066 592 645 795 ... $ average_dbh : num 11.8 271.2 105.1 169.2 220.9 ... $ average_age : num NA 80.3 19.3 50.9 68.2 ... $ average_BM_branches : num NA 66649 12679 11246 38049 ... $ average_BM_leaves_needles: num 2.28 31051.41 7727.07 9322.6 27112.2 ... $ average_BM_total : num 36.7 538194.3 62332.9 163137.2 339177.5 ... $ color: Factor w/ 4 levels violet,yellow,..: 1 2 2 2 2 2 1 3 1 1 ... as.character(coefficient2$color) [1] violet yellow yellow yellow yellow yellow violet black violet violet black [12] yellow violet yellow violet violet violet yellow violet violet black green [23] green green green green coefficient2$color - as.character(coefficient2$color) Now do your plots -- David On Jun 4, 2009, at 12:54 PM, katharina wrote: sorry, the attachment can be found here for download: http://www.wzw.tum.de/waldinventur/fileadmin/coefficient2.RData katharina wrote: Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(! is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/wrong-labels-and-colors-of-points-in-graph-plot-tp23873337p23873849.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
Re: [R] Plot and lm
stephen sefick ssef...@gmail.com [2009.06.04] wrote: Could you provide a reproducible example even with fake data would be fine or dput() yours. Sorry, I don't understand what you mean. Im my post was the first 8 lines of my data, imported into R with islands - read.table(islands.csv, sep=,, h=T) and the code was a cut-and-past from my .Rhistory. When I run it I get a nice graph, but no line from abline (unless it is vertical or horizontal and superimposed uppon one of the axes)... Which turns out ot be the case: Just running lm I get lm(mass~area) Call: lm(formula = mass ~ area) Coefficients: (Intercept) area 3.615e+025.967e-05 lm(logmass~logarea) Call: lm(formula = logmass ~ logarea) Coefficients: (Intercept) logarea -1.3480 0.4747 Forcing the graph with ylim=c(0,001,1) I see a line from the latter, but (no surprise) none from the former. Now I just need to fix my assumptions such that I produce a line that is an actual regession line... Thanks for making me think it through! /Par -- Par Leijonhufvud p...@hunter-gatherer.org The best comment I heard about Starship Troopers was Based on the back cover of a book by RAH. -- Paul Tomblin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wrong labels and colors of points in graph/plot
Sorry for the problem with NCStats, I used it from this page not realizing that this is not 'official': http://www.rforge.net/NCStats/files/ Thank you for helping me with the color issue, I finally know now what has happened with the colors of the points my plots. David Winsemius wrote: Was not able to find NCStats on CRAN or BioConductor repositories. You are wasting our (or mine at any rate) time by giving us dead ends. You have been bitten by the factor snake in the grass. Color should be text rather than a factor if you want the desired results. See if this is illuminating and eventually helpful: str(coefficient2) 'data.frame': 26 obs. of 11 variables: $ site_no : num 29 10 17 18 19 25 9 33 37 41 ... $ intercept: num 0.603 2.767 3.6 1.86 2.753 ... $ slope: num 0.912 0.867 0.841 0.99 0.911 ... $ average_height : num 90.4 2096.2 990 1950 2184 ... $ average_crown_length : num NA 1066 592 645 795 ... $ average_dbh : num 11.8 271.2 105.1 169.2 220.9 ... $ average_age : num NA 80.3 19.3 50.9 68.2 ... $ average_BM_branches : num NA 66649 12679 11246 38049 ... $ average_BM_leaves_needles: num 2.28 31051.41 7727.07 9322.6 27112.2 ... $ average_BM_total : num 36.7 538194.3 62332.9 163137.2 339177.5 ... $ color: Factor w/ 4 levels violet,yellow,..: 1 2 2 2 2 2 1 3 1 1 ... as.character(coefficient2$color) [1] violet yellow yellow yellow yellow yellow violet black violet violet black [12] yellow violet yellow violet violet violet yellow violet violet black green [23] green green green green coefficient2$color - as.character(coefficient2$color) Now do your plots -- David On Jun 4, 2009, at 12:54 PM, katharina wrote: sorry, the attachment can be found here for download: http://www.wzw.tum.de/waldinventur/fileadmin/coefficient2.RData katharina wrote: Hi there, I trying to solve this problem for the whole day not going anywhere, so I really hope maybe somebody can help me in this community... I've got an object coefficient2 which I want to plot in differerent ways, with colors and labels added to the points, but somehow there seems to be a problem if a value is NA within the independent variable, resulting in false labels and false colors for the points. plot(coefficient2$intercept ~ coefficient2$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coefficient2$color) highlight(coefficient2$intercept ~ coefficient2$average_height, lbls=coefficient2$site_no,col=Red, cex = .6) plot(coefficient2$intercept ~ coefficient2$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coefficient2$color ) highlight(coefficient2$intercept ~ coefficient2$average_dbh, lbls=coefficient2$site_no,col=Red, cex = .6 If I create a temporary object for each plot excluding any NA values for the x axis variable, somehow all points are displayed and the labels are correct except for the color (I have e.g. no clue why some points are red which I do not define at all and no ones are yellow which I use several times). #create temp container for all coefficients with average heights for plotting coef_avheight - coefficient2[which(! is.na(coefficient2$average_height)),] plot(coef_avheight$intercept ~ coef_avheight$average_height, main=intercepts ::: height, ylab=intercepts, xlab=average height per site [cm], xlim=c(20,3020), ylim=c(-2,5), col=coef_avheight$color) highlight(coef_avheight$intercept ~ coef_avheight$average_height, lbls=coef_avheight$site_no,col=Red, cex = .6) #create temp container for all coefficients with average dbh for plotting coef_avdbh- coefficient2[which(!is.na(coefficient2$average_dbh)),] plot(coef_avdbh$intercept ~ coef_avdbh$average_dbh, main=intercepts ::: dbh, ylab=intercepts, xlab=average dbh per site [mm], xlim=c(-10,360), ylim=c(-2,5),col=coef_avdbh$color ) highlight(coef_avdbh$intercept ~ coef_avdbh$average_dbh, lbls=coef_avdbh$site_no,col=Red, cex = .6) Maybe someone can explain me the color issue and the problem with the NA values which results in a wrong labeling and to few points being displayed? I'm new to R as you can guess and my code isn't really elegant but I really cannot get faults within it... Attached you can find the referred R object (coefficient2). highlight requires library(NCStats)... Thank you very, very much, Katharina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in
[R] Error Catching?
Hi, Is there an easy way to catch errors, in order to arrange for r-scripts to exit gracefully? I'm thinking of something along the lines of using is.na with an if/else statement, but for errors. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to tell if as.numeric succeeds?
Suppose I have a vector of strings. I'd like to convert this to a vector of numbers if possible. How can I tell if it is possible? as.numeric() will issue a warning if it fails. Do I need to trap this warning? If so, how? In other words, my end goal is a function that takes a vector of strings and returns either a numeric vector or the original vector. Assuming this doesn't already exist, then to write it I'd need a function that returns true if its input can be converted to numeric and false otherwise. Any suggestions will be appreciated. -- View this message in context: http://www.nabble.com/how-to-tell-if-as.numeric-succeeds--tp23874936p23874936.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error Catching?
On Thu, Jun 4, 2009 at 6:56 PM, Brigid Mooney bkmoo...@gmail.com wrote: Hi, Is there an easy way to catch errors, in order to arrange for r-scripts to exit gracefully? I'm thinking of something along the lines of using is.na with an if/else statement, but for errors. Try 'try'. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error Catching?
On Jun 4, 2009, at 12:56 PM, Brigid Mooney wrote: Hi, Is there an easy way to catch errors, in order to arrange for r-scripts to exit gracefully? I'm thinking of something along the lines of using is.na with an if/else statement, but for errors. Thanks! See ?try and ?tryCatch HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot and lm
I can't copy and paste your example right out of you email and into an R session. ?dput and then see if you can copy and paste it into an R session and make it work. That way it is easier for everyone and you have a better chance of getting helpful responses. HTH Stephen Sefick On Thu, Jun 4, 2009 at 1:47 PM, Par Leijonhufvud p...@hunter-gatherer.org wrote: stephen sefick ssef...@gmail.com [2009.06.04] wrote: Could you provide a reproducible example even with fake data would be fine or dput() yours. Sorry, I don't understand what you mean. Im my post was the first 8 lines of my data, imported into R with islands - read.table(islands.csv, sep=,, h=T) and the code was a cut-and-past from my .Rhistory. When I run it I get a nice graph, but no line from abline (unless it is vertical or horizontal and superimposed uppon one of the axes)... Which turns out ot be the case: Just running lm I get lm(mass~area) Call: lm(formula = mass ~ area) Coefficients: (Intercept) area 3.615e+02 5.967e-05 lm(logmass~logarea) Call: lm(formula = logmass ~ logarea) Coefficients: (Intercept) logarea -1.3480 0.4747 Forcing the graph with ylim=c(0,001,1) I see a line from the latter, but (no surprise) none from the former. Now I just need to fix my assumptions such that I produce a line that is an actual regession line... Thanks for making me think it through! /Par -- Par Leijonhufvud p...@hunter-gatherer.org The best comment I heard about Starship Troopers was Based on the back cover of a book by RAH. -- Paul Tomblin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to tell if as.numeric succeeds?
On Jun 4, 2009, at 1:01 PM, Steve Jaffe wrote:
Suppose I have a vector of strings. I'd like to convert this to a
vector of
numbers if possible. How can I tell if it is possible?
as.numeric() will issue a warning if it fails. Do I need to trap this
warning? If so, how?
In other words, my end goal is a function that takes a vector of
strings and
returns either a numeric vector or the original vector. Assuming this
doesn't already exist, then to write it I'd need a function that
returns
true if its input can be converted to numeric and false otherwise.
Any suggestions will be appreciated.
In the case of as.numeric(), it will return NA for any elements in the
vector that cannot be coerced to numeric:
as.numeric(c(a, 1, c, 2))
[1] NA 1 NA 2
Warning message:
NAs introduced by coercion
So there are at least three scenarios:
1. All of the elements can be coerced
2. Some of the elements can be coerced
3. None of the elements can be coerced
You could feasibly do something like this, which will suppress the
warning, so there is no textual output in the case of a warning:
suppressWarnings(as.numeric(c(a, 1, c, 2)))
[1] NA 1 NA 2
Then you just need to check to see if any() of the elements are not
NAs, which means that at least one value was successfully coerced:
NumCheck - function(x)
{
Result - suppressWarnings(as.numeric(x))
if (any(!is.na(Result))) TRUE else FALSE
}
NumCheck(c(a, 1, c, 2))
[1] TRUE
NumCheck(letters[1:4])
[1] FALSE
NumCheck(c(1, 2, 3, 4))
[1] TRUE
Alter the internal logic in the function relative to NA checking
depending up what you may need.
HTH,
Marc Schwartz
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newton method again
Roslina Zakaria wrote:
Hi Ravi,
I did ask you some question regarding newton method sometime ago.. Now I
have fixed the problem and I also wrote 2 looping code (ff1 and ff2) to
evaluate the modified Bessel function of the first kind and call them in
the newton code. But I dont't understand why it gives the error message
but still give the result but it is incorrect(too big and too small).
ff1 - function(bb,eta,z){
r - length(z)
for (i in 1:r) {
sm - sum(besselI(z[i]*bb,eta)/besselI(z[i]*bb,eta+1))*z[i]}
sm
}
ff1(bb,eta,z)
ff2 - function(bb,eta,z,k){
r - length(z)
for (i in 1:r) {
sm1 -
sum((z[i]*bb/2)*(psigamma((0:k)+eta+1,deriv=0)/(factorial(0:k)*gamma((0:k)+eta+1
sm2 - sum((besselI(z[i]*bb,eta)*log(z[i]*bb/2) -
sm1)/besselI(z[i]*bb,eta))}
sm2
}
ff2(bb,eta,z,10)
newton.input3 - function(pars)
{ ## parameters to be approximated , note: eta - alpha3-0.5
eta - pars[1]
bt3 - pars[2]
bt4 - pars[3]
rho - pars[4]
b1 - (pars[2]-pars[3])^2+4*pars[2]*pars[3]*pars[4]
b2 - sqrt(b1)
bb - b2/(2*pars[2]*pars[3]*(1-pars[4]))
bf2 -
(pars[3]+2*pars[2]*pars[4]-pars[2])/(2*pars[2]^2*(pars[4]-1)*b2)
bf3 -
(pars[2]+2*pars[3]*pars[4]-pars[3])/(2*pars[3]^2*(pars[4]-1)*b2)
bf4 -
(2*pars[2]*pars[3]*pars[4]+pars[2]^2+pars[3]^2)/(2*pars[2]*pars[3]*(pars[4]-1)^2*b2)
zsm - sum(z)
psigm - psigamma(pars[1]+0.5,deriv=0)
pdz - log(prod(z))
erh - (1+2*pars[1])*(pars[4]-1)
brh1 - 2*pars[2]*pars[3]*pars[4]+pars[2]^2+pars[3]^2
brh2 - 2*pars[2]*pars[3]*(pars[4]-1)^2
k - 1000
## function
fn1a - pdz -r*(2*psigm + log(b1))/2
fn2a - (pars[2]*r*erh+zsm)/(2*pars[2]^2*(1-pars[4]))
fn3a - (pars[3]*r*erh+zsm)/(2*pars[3]^2*(1-pars[4]))
fn4a -
(pars[2]*pars[3]*r*erh+(pars[2]+pars[3])*zsm)/(-2*pars[2]*pars[3]*(pars[4]-1)^2)
## function that involve modified Bessel function of 1st kind
fn1b - ff2(bb,eta,z,k)
fn2b - bf2*ff1(bb,eta,z)
fn3b - bf3*ff1(bb,eta,z)
fn4b - bf4*ff1(bb,eta,z)
## final function
fn1 - fn1a + fn1b
fn2 - fn2a + fn2b
fn3 - fn3a + fn3b
fn4 - fn4a + fn4b
fval - c(fn1,fn2,fn3,fn4)
## output
list(fval=fval)
}
library(BB)
start - c(0.7,0.8,0.6,0.4)
dfsane(pars=start,fn=newton.input3)
newton.input3(start)
library(BB)
start - c(0.7,0.8,0.6,0.4)
dfsane(pars=start,fn=newton.input3)
Error in dfsane(pars = start, fn = newton.input3) : element 1 is empty;
the part of the args list of 'length' being evaluated was:
(par)
newton.input3(start)
$fval
[1] 103.0642 452.5835 823.6637 -1484.3209
There were 50 or more warnings (use warnings() to see the first 50)
Here is my data:
z
[1] 4.2 11.2 0.8 20.4 16.6 3.8 1.2 4.0 10.8 10.2 6.6 25.6 18.2 4.6
15.0 1.2 12.0 25.4 6.4 1.6 4.8 10.0 3.0
[24] 7.0 1.8 15.0 8.6 11.2 5.4 1.8 23.2 10.8 25.4 6.0 6.0 5.0 1.4
11.0 8.4 7.4 6.4 2.6 8.6 15.8
You could also try package nleqslv which implements Newton and Broyden
methods for solving systems of equations.
I have tried to run your problem but you are not providing all the
information required.
Moreover your example contains errors: for example where are the arguments
defined in the call of ff1 on the line ff1(bb,eta,z) right after the
definition of ff1?
Where is the variable r used in the lines calculating fn1a, fn2a etc. in
function newton.input3?
Is it the same as in ff1 and ff2? length(z)?
When I insert r-length(z) in newton.input3() I get the results shown in
your post for $fval.
The warnings are being given by factorial(0:k): In factorial(0:k) : value
out of range in 'gammafn'
Why are you assigning pars[1], pars[2] etc to scalars and then afterwards
not or hardly using them?
You code is inefficient since you are calling ff1 in newton.input3 three
times with exactly the same input.
I have tried to run your code in nleqslv but it appears to run very slowly
so I can't help you any further at this point in time.
What is the purpose of the loop in function ff1
for (i in 1:r) {
sm - sum(besselI(z[i]*bb,eta)/besselI(z[i]*bb,eta+1))*z[i]}
(on returning from the function sm will contain the value obtained for i=r)
?
Given the presentation of your problem, I cannot make head or tail of what
you are trying to do so I can't help you any further.
Berend Hasselman
--
View this message in context:
http://www.nabble.com/newton-method-tp22653758p23875467.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fast way of finding top-n values of a long vector
Greg Snow wrote:
Try adding a version that uses sort with the partial argument, that should be
faster than regular sort (for long enough test vectors) and possibly faster
than the max solutions when finding more than just the largest 2.
I find the documentation for the partial argument in sort very difficult
to understand, but it seems to be something like this:
library(rbenchmark)
set.seed(1); x - runif(1e6, max=1e7); x[1] - NA;
benchmark(
replications=20,
columns=c(test,elapsed),
order=elapsed
, sort = {a-sort(x, decreasing=TRUE, na.last=NA)[1:2]; a[1]/a[2];}
, qsrt = {a-sort(x, decreasing=TRUE, na.last=NA, method=quick)[1:2];
a[1]/a[2];}
, part = {a-sort.int(-x, partial=1:2, na.last=NA)[1:2]; a[1]/a[2];}
, max1 = {m-max(x, na.rm=TRUE); w-which(x==m)[1];
m/max(x[-w],na.rm=TRUE);}
, max2 = {w-which.max(x); max(x, na.rm=TRUE)/max(x[-w], na.rm=TRUE);}
)
# test elapsed
# 5 max2 0.846
# 4 max1 1.957
# 3 part 2.752
# 2 qsrt 4.561
# 1 sort 5.577
Completely agree on your point about partial sort being faster for
larger values of n and certainly giving more scalable code which was
also what I was looking for so thanks for that tip!
Also, for your max solutions, what happens when the 2 largest values are
identical?
It returns those two values, just like the sort solution:
x - c(999,NA,1:10,NA,999)
m-max(x, na.rm=TRUE); w-which(x==m)[1]; c(m, max(x[-w],na.rm=TRUE))
# [1] 999 999
sort(x, decreasing=TRUE, na.last=NA)[1:2]
# [1] 999 999
Allan.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot and lm
I think the problem is that plot's log axes are to the base 10 so the lm() call needs to use log10, not log. E.g., x-101:200 y-sqrt(x)+1+runif(100) plot(x,y,log=xy) abline(col=red, lm(log(y)~log(x))) # nothing plotted abline(col=blue, lm(log10(y)~log10(x))) # passes through points Expand the y axis a bit and you can see where the base-e line went: plot(x,y,log=xy, ylim=c(1,100)) abline(col=red, lm(log(y)~log(x))) # plotted well above the points abline(col=blue, lm(log10(y)~log10(x))) # passes through points Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of stephen sefick Sent: Thursday, June 04, 2009 10:04 AM To: Par Leijonhufvud Cc: R-help Subject: Re: [R] Plot and lm Could you provide a reproducible example even with fake data would be fine or dput() yours. On Thu, Jun 4, 2009 at 10:11 AM, Par Leijonhufvud p...@hunter-gatherer.org wrote: I want to make a log-log plot with a regression line, but I can't figure out what I'm doing wrong. What I'm trying is: plot(mass,area, log=xy, pch=as.numeric(food)) abline(lm(mass~area)) or plot(mass,area, log=xy, pch=as.numeric(food)) islands$logmass - log(mass) islands$logarea - log(area) attach(islands) abline(lm(logmass~logarea)) But that does not show a line. Where am I going wrong? data: island, area,species,food,mass Ibiza , 577 , Anser n. sp., herb, 2.0 Ibiza , 577 , Haliaeetus albicilla, carn, 4.8 Mauritius , 1874 , Raphus cucullatus, herb, 19 Mauritius , 1874 , Circus alphonsi, carn, 0.63 Mallorca , 3667 , Myotragus balearicus, herb, 40 Mallorca , 3667 , Aquila chrysaetos, carn, 4.2 Kreta , 8259 , Elephas creutzburgi, herb, 3200 ... /Par -- Par Leijonhufvud p...@hunter-gatherer.org I don't believe in reincarnation. I used to, but that was in another life. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] logical indexing multidimensional arrays
Suppose I have an n-dimensional array and a logical vector as long as the first dimension. I can extract an n-dimensional subarray with a[ i, , , , .. ,] where there are n-1 commas (ie empty indices) Is there an alternative notation that would better lend itself to more generic use, e.g. to write a function that takes 'a' and 'i' and returns a[i, , , .. ,]? -- View this message in context: http://www.nabble.com/logical-indexing-multidimensional-arrays-tp23875985p23875985.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] type = 'b' with Grid
Dear all,
I feel like I've been reinventing the wheel with this code (implementing
type = 'b' for Grid graphics),
http://econum.umh.ac.be/rwiki/doku.php?id=tips:graphics-grid:linesandpointsgrob
Has anyone here attempted this with success before? I found suggestions
of overlapping large white points to mask the lines but it's not ideal.
I welcome any comments on the code. (e.g., should it perhaps use
polylines rather than segments?, is there an obvious optimization to
make?, ...)
Regards,
baptiste
# code reproduced below,
barbedGrob - function( x = 1:10/12,
y = sin(1:10)/3+0.5,
size=1, shape=21, space=1,
colour=red, fill=blue,
linetype=1, linewidth=1){
n - length(x)
dx - diff(x)
dy - diff(y)
# duplicate the points to make split segments
new.x - rep(x, each=2)[-c(1, 2*length(x))]
new.y - rep(y, each=2)[-c(1, 2*length(y))]
new.size - rep(size, each=2, length=2*n)[-c(1, 2*n)]
length - sqrt(dx^2 + dy^2) # length of initial segments
exclusion - 0.5*space*convertX(unit(new.size, char), npc, TRUE)
scaling - exclusion / rep(length, each=2) # exclusion factor around
each point
start - seq(1, by=2, length(new.x)) # starting points
end - seq(2, by=2, length(new.x)) # end points
x.start - scaling[start] * dx[(start+1)/2] + new.x[start] # shift the
points
y.start - scaling[start] * dy[(start+1)/2] + new.y[start] # keeping the
direction of the initial segments
x.end - new.x[end] - scaling[end] * dx[end/2]
y.end - new.y[end] - scaling[end] * dy[end/2]
grob.lines - segmentsGrob(
x0 = x.start, y0 = y.start,
x1 = x.end, y1=y.end,
default.units=native,
gp = gpar(
col = colour,
lex = linewidth, lty = linetype, lineend = butt
)
)
grob.points - pointsGrob(x, y, pch=shape, size=unit(size, char),
gp = gpar(
col = colour,
fill = fill,
lex = linewidth, linejoin = mitre
)
)
gTree(children = gList(grob.lines,grob.points))
}
# example of use
g -
barbedGrob(size=sample(1:3, 10, repl=T),
fill=alpha(white, 0.3),
col=alpha(cadetblue4, 0.8),
linewidth=5, space=1.2)
pushViewport(vp=viewport(width=1, height=1))
grid.rect(gp=gpar(fill=thistle2))
grid.grill(gp=gpar(col=lavenderblush1, lwd=3, lty=3))
grid.draw(g)
--
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] For Social Network Analysis-Graph Analysis - How to convert 2 mode data to 1 mode data?
All, There is a simple solution to this problem using R's matrix algebra commands. I describe in detail at http://www.stanford.edu/~messing/Affiliation%20Data.html http://www.stanford.edu/~messing/Affiliation%20Data.html . Best wishes, -Solomon -- View this message in context: http://www.nabble.com/For-Social-Network-Analysis-Graph-Analysis---How-to-convert-2-mode-data-to-1-mode-data--tp17157380p23875403.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] IP-Address
Thank you Peter,
This solved the problem.
edw...@web.de wrote:
Hi,
Unfortunately, they can't handle NA. Any suggestion? Some row for Ip
don't have ip address. This cause an error/ wrong result.
A quick fix could be to substitute ... or 0.0.0.0 for the NA
entries. (Use something like
ipch - as.character(df$ip)
ipch[is.na(df$ip)] - ...
connection - textConnection(ipch)
)
Eddie
library(gsubfn)
library(gtools)
library(rbenchmark)
n - 1
df - data.frame(
a = rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'), simplify=TRUE)
)
res - benchmark(columns=c('test', 'elapsed'), replications=10,
order=NULL,
peda = {
connection - textConnection(as.character(df$ip))
o - do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ]
},
peda2 = {
connection - textConnection(as.character(df$ip))
dfT - read.table(connection, sep='.', colClasses=rep(integer,
4), quote=, na.strings=NULL, blank.lines.skip=FALSE)
close(connection)
o - do.call(order, dfT)
df[o, ]
},
hb = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip)
dim(ip) - c(4, nrow(df))
ip - 256^3*ip[1,] + 256^2*ip[2,] + 256*ip[3,] + ip[4,]
o - order(ip)
df[o, ]
},
hb2 = {
ip - strsplit(as.character(df$ip), split=., fixed=TRUE)
ip - unlist(ip, use.names=FALSE)
ip - as.integer(ip);
dim(ip) - c(4, nrow(df))
o - sort.list(ip[4,], method=radix, na.last=TRUE)
for (kk in 3:1) {
o - o[sort.list(ip[kk,o], method=radix, na.last=TRUE)]
}
df[o, ]
}
)
print(res)
test elapsed
1 peda 4.12
2 peda2 4.08
3 hb 0.28
4 hb2 0.25
On Sun, May 31, 2009 at 12:42 AM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
edwin Sendjaja wrote:
Hi VQ,
Thank you. It works like charm. But I think Peter's code is faster.
What
is the difference?
i think peter's code is more r-elegant, though less generic. here's a
quick test, with not so surprising results. gsubfn is implemented in
r, not c, and it is painfully slow in this test. i also added gabor's
suggestion.
library(gsubfn)
library(gtools)
library(rbenchmark)
n = 1000
df = data.frame(
a=rnorm(n),
b = rnorm(n),
c = rnorm(n),
ip = replicate(n, paste(sample(255, 4), collapse='.'),
simplify=TRUE))
benchmark(columns=c('test', 'elapsed'), replications=10,
order=NULL, peda={
connection = textConnection(as.character(df$ip))
o = do.call(order, read.table(connection, sep='.'))
close(connection)
df[o, ] },
waku=df[order(gsubfn(perl=TRUE,
'[0-9]+',
~ sprintf('%03d', as.integer(x)),
as.character(df$ip))), ],
gagr=df[mixedorder(df$ip), ] )
# peda 0.070
# waku 7.070
# gagr 4.710
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal,
self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 5-D density? was Re: R help
I don't want to plot or visualize the plot. I just want to clculate the density at certain prespecified points.For that,knowin the funcional form is enough.I just want a function that returns me back the estimate of density at some point On Thu, Jun 4, 2009 at 10:04 PM, David Winsemius dwinsem...@comcast.netwrote: (copy of earlier reply which was not addressed to r-help. Also added informative Subject:) yes .. is not responsive to the question of how you propose to display or examine such a mathematical object. My signature was perhaps a lame (certainly an ineffective) effort at getting you to acknowledge that there might, just might, be some conceptual difficulties in viewing a 5+1 = 6 dimensional object. Can you point to any examples of such a procedure being applied in a manner that you find helpful? It is certainly feasible to calculate distances in 5-space, but the next step, displaying the concentration of those distances as a function of the coordinates, would be the tough nut. (1000 data points is not a problem. That amount of data should easily fit in any device that can run R.) -- David On Jun 4, 2009, at 12:06 AM, arijit kumar debnath wrote: yes.. I want a density estimator for 5 dimensional data.I want it to be efficient since I don't have access to a fast computer.My data set is quite large(about 1000 data points) and I have only a pentium 4 1.7 Ghz (512 Mb RAM)computer. On Thu, Jun 4, 2009 at 7:25 AM, David Winsemius dwinsem...@comcast.netwrote: We can get you a kde2d but you really want a kde5d? What sort of display are you proposing? -- David Earthling Winsemius On Jun 3, 2009, at 9:26 PM, arijit kumar debnath wrote: I want to fit a kernel density to a data-set of size 1000 and each data point being of dimension 5. The default density function does'nt work for me. Is there any other function available which can do the job efficiently.? I don't have access to a very fast computer. David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 5-D density? was Re: R help
To address the creation of a 5-D density object: http://finzi.psych.upenn.edu/R/library/ks/html/kda.kde.html http://finzi.psych.upenn.edu/R/library/ks/html/kde.html http://finzi.psych.upenn.edu/R/library/locfit/html/locfit.raw.html -- David Winsemius On Jun 4, 2009, at 12:34 PM, David Winsemius wrote: (copy of earlier reply which was not addressed to r-help. Also added informative Subject:) yes .. is not responsive to the question of how you propose to display or examine such a mathematical object. My signature was perhaps a lame (certainly an ineffective) effort at getting you to acknowledge that there might, just might, be some conceptual difficulties in viewing a 5+1 = 6 dimensional object. Can you point to any examples of such a procedure being applied in a manner that you find helpful? It is certainly feasible to calculate distances in 5-space, but the next step, displaying the concentration of those distances as a function of the coordinates, would be the tough nut. (1000 data points is not a problem. That amount of data should easily fit in any device that can run R.) -- David On Jun 4, 2009, at 12:06 AM, arijit kumar debnath wrote: yes.. I want a density estimator for 5 dimensional data.I want it to be efficient since I don't have access to a fast computer.My data set is quite large(about 1000 data points) and I have only a pentium 4 1.7 Ghz (512 Mb RAM)computer. On Thu, Jun 4, 2009 at 7:25 AM, David Winsemius dwinsem...@comcast.net wrote: We can get you a kde2d but you really want a kde5d? What sort of display are you proposing? -- David Earthling Winsemius On Jun 3, 2009, at 9:26 PM, arijit kumar debnath wrote: I want to fit a kernel density to a data-set of size 1000 and each data point being of dimension 5. The default density function does'nt work for me. Is there any other function available which can do the job efficiently.? I don't have access to a very fast computer. David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Minor tick marks for date/time ggplot2 (this is better, but not exactly what I want)
On Mon, Jun 1, 2009 at 2:18 PM, stephen sefick ssef...@gmail.com wrote: library(ggplot2) melt.updn - (structure(list(date = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149, 11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149), class = Date), site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(unrestored, restored), class = factor), value = c(1.10799962473684, 0.732152347032395, 0.410438861475827, 0.458941230025228, 0.429883166858706, 0.831083728521569, 0.601942073736539, 0.81855597155132, 1.12612228239269, 0.246006569972335, 0.940239233910111, 0.98645360143702, 0.291191536260016, 0.346271105079473, 1.36216149279675, 0.878585508942967, 0.525184260519839, 0.803247305232454, 1.08086182748669, 1.24915815325761, 0.971046497346528, 0.936835411801682, 1.26957337598606, 0.337691543740682, 0.90931142298893, 0.950891472223867, 0.290354002109368, 0.426509990013021)), .Names = c(date, site, value), row.names = c(NA, -28L), class = data.frame)) #I would also like to add tick marks to this graph is possible with no label for the months in between the years qplot(date, value, data=melt.updn, shape=site, ylab=Distance ,main=Euclidean Distances Time Series, xlim=c(as.Date(2002-1-1), as.Date(2006-3-1)))+geom_line()+theme_bw()+geom_vline(x=as.numeric(as.Date(2002-11-01))) + opts(panel.grid.major = theme_line(colour=grey, size=0.75), panel.grid.minor=theme_line(colour=grey, size=0.25)) Unfortunately that's currently not possible - tick marks are always associated with major grid lines, and more importantly currently scale_date only lets you specify the time between ticks, not their labels. Are the minor monthly grid lines not good enough? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 5-D density? was Re: R help
thank you. This is just what I was looking for. On Fri, Jun 5, 2009 at 3:16 AM, David Winsemius dwinsem...@comcast.netwrote: To address the creation of a 5-D density object: http://finzi.psych.upenn.edu/R/library/ks/html/kda.kde.html http://finzi.psych.upenn.edu/R/library/ks/html/kde.html http://finzi.psych.upenn.edu/R/library/locfit/html/locfit.raw.html -- David Winsemius On Jun 4, 2009, at 12:34 PM, David Winsemius wrote: (copy of earlier reply which was not addressed to r-help. Also added informative Subject:) yes .. is not responsive to the question of how you propose to display or examine such a mathematical object. My signature was perhaps a lame (certainly an ineffective) effort at getting you to acknowledge that there might, just might, be some conceptual difficulties in viewing a 5+1 = 6 dimensional object. Can you point to any examples of such a procedure being applied in a manner that you find helpful? It is certainly feasible to calculate distances in 5-space, but the next step, displaying the concentration of those distances as a function of the coordinates, would be the tough nut. (1000 data points is not a problem. That amount of data should easily fit in any device that can run R.) -- David On Jun 4, 2009, at 12:06 AM, arijit kumar debnath wrote: yes.. I want a density estimator for 5 dimensional data.I want it to be efficient since I don't have access to a fast computer.My data set is quite large(about 1000 data points) and I have only a pentium 4 1.7 Ghz (512 Mb RAM)computer. On Thu, Jun 4, 2009 at 7:25 AM, David Winsemius dwinsem...@comcast.net wrote: We can get you a kde2d but you really want a kde5d? What sort of display are you proposing? -- David Earthling Winsemius On Jun 3, 2009, at 9:26 PM, arijit kumar debnath wrote: I want to fit a kernel density to a data-set of size 1000 and each data point being of dimension 5. The default density function does'nt work for me. Is there any other function available which can do the job efficiently.? I don't have access to a very fast computer. David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] RPostgreSQL segfault with LEFT JOIN
Hi, I recently upgraded to R 2.9.0 on linux x86. After doing so, I switched to the RPostgreSQL package for interfacing with a postgresql database. I am using postgresql 8.3.7. A query that works from the postgresql terminal is causing a segfault when executed from R. My sessionInfo, the error message, and the R code used to generate the error are listed below. I have noticed that a trivial query (SELECT 1 as value) or other queries seem to work fine. It is only when I enable the LEFT JOIN (see below) that I get a segfault. Could this be related to the treatment of null values? Any ideas? Thanks! Dylan Here is the code that caused the error # libs library(RPostgreSQL) ## query DB q - SELECT deb_lab_data.* -- matrix_wet_color_hue as hue, matrix_wet_color_value as value, matrix_wet_color_chroma as chroma FROM deb_lab_data -- LEFT JOIN horizon USING (pedon_id, hz_number) WHERE deb_lab_data.pedon_id ~~ '%SJER%' ORDER BY deb_lab_data.pedon_id, deb_lab_data.top ASC # create an PostgreSQL instance and create one connection. drv - dbDriver(PostgreSQL) conn - dbConnect(drv, host=localhost, dbname=XXX, user=XXX) query - dbSendQuery(conn, q) x - fetch(query, n = -1) # extract all rows Here is the error message in R: row number 0 is out of range 0..-1 *** caught segfault *** address (nil), cause 'memory not mapped' Traceback: 1: .Call(RS_PostgreSQL_exec, conId, statement, PACKAGE = .PostgreSQLPkgName) 2: postgresqlExecStatement(conn, statement, ...) 3: is(object, Cl) 4: is(object, Cl) 5: .valueClassTest(standardGeneric(dbSendQuery), DBIResult, dbSendQuery) 6: dbSendQuery(conn, q) Here are the details on my R install: R version 2.9.0 (2009-04-17) i686-pc-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] RPostgreSQL_0.1-4 DBI_0.2-4 -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RPostgreSQL segfault with LEFT JOIN
On 4 June 2009 at 16:17, Dylan Beaudette wrote: | Hi, | | I recently upgraded to R 2.9.0 on linux x86. After doing so, I switched to the | RPostgreSQL package for interfacing with a postgresql database. I am using | postgresql 8.3.7. | | A query that works from the postgresql terminal is causing a segfault when | executed from R. | | My sessionInfo, the error message, and the R code used to generate the error | are listed below. | | I have noticed that a trivial query (SELECT 1 as value) or other queries seem | to work fine. It is only when I enable the LEFT JOIN (see below) that I get a | segfault. Could this be related to the treatment of null values? As per some recent messages on the r-sig-db list, I think that the error is due to a bug in the handling of 'schema.table' queries. If you just use 'select ... from table' you're fine. Not sure if this helps you -- someone has to go in and fix the bug. Dirk | | Any ideas? | Thanks! | Dylan | | Here is the code that caused the error | | # libs | library(RPostgreSQL) | | ## query DB | q - | SELECT deb_lab_data.* | -- matrix_wet_color_hue as hue, matrix_wet_color_value as value, | matrix_wet_color_chroma as chroma | FROM deb_lab_data | -- LEFT JOIN horizon USING (pedon_id, hz_number) | WHERE deb_lab_data.pedon_id ~~ '%SJER%' | ORDER BY deb_lab_data.pedon_id, deb_lab_data.top ASC | | # create an PostgreSQL instance and create one connection. | drv - dbDriver(PostgreSQL) | conn - dbConnect(drv, host=localhost, dbname=XXX, user=XXX) | query - dbSendQuery(conn, q) | x - fetch(query, n = -1) # extract all rows | | | Here is the error message in R: | | row number 0 is out of range 0..-1 | | *** caught segfault *** | address (nil), cause 'memory not mapped' | | Traceback: | 1: .Call(RS_PostgreSQL_exec, conId, statement, PACKAGE | = .PostgreSQLPkgName) | 2: postgresqlExecStatement(conn, statement, ...) | 3: is(object, Cl) | 4: is(object, Cl) | | 5: .valueClassTest(standardGeneric(dbSendQuery), DBIResult, dbSendQuery) | 6: dbSendQuery(conn, q) | | | | | Here are the details on my R install: | | R version 2.9.0 (2009-04-17) | i686-pc-linux-gnu | | locale: | LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C | | attached base packages: | [1] stats graphics grDevices utils datasets methods base | | other attached packages: | [1] RPostgreSQL_0.1-4 DBI_0.2-4 | | | | | -- | Dylan Beaudette | Soil Resource Laboratory | http://casoilresource.lawr.ucdavis.edu/ | University of California at Davis | 530.754.7341 | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide http://www.R-project.org/posting-guide.html | and provide commented, minimal, self-contained, reproducible code. -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RPostgreSQL segfault with LEFT JOIN
On Thursday 04 June 2009, Dirk Eddelbuettel wrote: On 4 June 2009 at 16:17, Dylan Beaudette wrote: | Hi, | | I recently upgraded to R 2.9.0 on linux x86. After doing so, I switched | to the RPostgreSQL package for interfacing with a postgresql database. I | am using postgresql 8.3.7. | | A query that works from the postgresql terminal is causing a segfault | when executed from R. | | My sessionInfo, the error message, and the R code used to generate the | error are listed below. | | I have noticed that a trivial query (SELECT 1 as value) or other queries | seem to work fine. It is only when I enable the LEFT JOIN (see below) | that I get a segfault. Could this be related to the treatment of null | values? As per some recent messages on the r-sig-db list, I think that the error is due to a bug in the handling of 'schema.table' queries. If you just use 'select ... from table' you're fine. Not sure if this helps you -- someone has to go in and fix the bug. Dirk Thanks Dirk, After some further investigation, I see that the query works fine if I *do not use column aliases* : # segfaults: q - SELECT deb_lab_data.* , matrix_wet_color_hue as hue, matrix_wet_color_value as value, matrix_wet_color_chroma as chroma FROM deb_lab_data LEFT JOIN horizon USING (pedon_id, hz_number) WHERE deb_lab_data.pedon_id ~~ '%SJER%' ORDER BY deb_lab_data.pedon_id, deb_lab_data.top ASC # works fine: q - SELECT deb_lab_data.* , matrix_wet_color_hue, matrix_wet_color_value, matrix_wet_color_chroma FROM deb_lab_data LEFT JOIN horizon USING (pedon_id, hz_number) WHERE deb_lab_data.pedon_id ~~ '%SJER%' ORDER BY deb_lab_data.pedon_id, deb_lab_data.top ASC Very strange... Dylan | Any ideas? | Thanks! | Dylan | | Here is the code that caused the error | - |--- # libs | library(RPostgreSQL) | | ## query DB | q - | SELECT deb_lab_data.* | -- matrix_wet_color_hue as hue, matrix_wet_color_value as value, | matrix_wet_color_chroma as chroma | FROM deb_lab_data | -- LEFT JOIN horizon USING (pedon_id, hz_number) | WHERE deb_lab_data.pedon_id ~~ '%SJER%' | ORDER BY deb_lab_data.pedon_id, deb_lab_data.top ASC | | # create an PostgreSQL instance and create one connection. | drv - dbDriver(PostgreSQL) | conn - dbConnect(drv, host=localhost, dbname=XXX, user=XXX) | query - dbSendQuery(conn, q) | x - fetch(query, n = -1) # extract all rows | - |--- | | Here is the error message in R: | - |--- row number 0 is out of range 0..-1 | | *** caught segfault *** | address (nil), cause 'memory not mapped' | | Traceback: | 1: .Call(RS_PostgreSQL_exec, conId, statement, PACKAGE | = .PostgreSQLPkgName) | 2: postgresqlExecStatement(conn, statement, ...) | 3: is(object, Cl) | 4: is(object, Cl) | | 5: .valueClassTest(standardGeneric(dbSendQuery), DBIResult, | dbSendQuery) 6: dbSendQuery(conn, q) | - |--- | | | | Here are the details on my R install: | - |--- R version 2.9.0 (2009-04-17) | i686-pc-linux-gnu | | locale: | LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UT |F-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;L |C_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C | | attached base packages: | [1] stats graphics grDevices utils datasets methods base | | other attached packages: | [1] RPostgreSQL_0.1-4 DBI_0.2-4 | - |--- | | | | -- | Dylan Beaudette | Soil Resource Laboratory | http://casoilresource.lawr.ucdavis.edu/ | University of California at Davis | 530.754.7341 | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide | http://www.R-project.org/posting-guide.html and provide commented, | minimal, self-contained, reproducible code. -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help Needed
HI, I am Angshuman a postdoc in Buck Institute, Novato, CA. I am using random forest in R. I have a problem. I have a training file and a test file. I need to generate model file to classify a set of data of the test file. I need to know the command for that. Please let me know. Thank you. Angshuman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ROracle: cannot insert several columns
Hi all, I've been playing with ROracle (0.5-9) for a few days and I can't wrap my mind around this one. Here's a sample of my R (2.4.0) session. my.df-data.frame(prd_id=c(123,456),vol_factor=c(.123,.456)) my.df prd_id vol_factor 1123 0.123 2456 0.456 library(ROracle) Loading required package: DBI conn-dbConnect(Oracle,***/*...@***) dbGetQuery(conn,create table mytable (prd_id number, vol_factor number)) dbGetQuery(conn,insert into mytable (prd_id,vol_factor) values(123,.123)) dbGetQuery(conn,insert into mytable (prd_id,vol_factor) values(456,.456)) dbGetQuery(conn,select * from mytable) PRD_ID VOL_FACTOR 0123 0.123 1456 0.456 the above works as expected. Now let's try to insert new rows into mytable using a prepared statement and bind variables: ps-dbPrepareStatement(conn,insert into mytable (prd_id,vol_factor) values (:1,:2),bind=c(rep(numeric,2))) res-dbExecStatement(ps,my.df) dbCommit(conn) [1] TRUE dbGetQuery(conn,select * from jdomenge_test) PRD_ID VOL_FACTOR 0123 0.123 1456 0.456 2123 NA 3456 NA so the 2 new rows were appended, except the values in the second column were seemingly not read... the same happens with dbWriteTable: dbWriteTable(conn,mytable,df,append=T,row.names=F) [1] TRUE dbGetQuery(conn,select * from mytable) PRD_ID VOL_FACTOR 0123 0.123 1456 0.456 2123 NA 3456 NA 4123 NA 5456 NA I'm clueless at this point, could find no answer in the help files or on the web... for a while now. Any help would be *greatly* appreciated. Thanks in advance, Jean-Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Wiring or Arduino package/scripts
Thanks... This will be very usefull. Warren Young wrote: Jorge Cornejo wrote: Hi, I am looking any way to communicant with Arduino (http://www.arduino.cc/) Wiring (http://www.wiring.org.co) boards and read data generate with sensor on these. On Linux you can simply open the file /dev/ttyS0 in read/write mode to talk on the first serial port on the PC. (ttyS1 for the second port, etc.) If your PC doesn't have serial ports, you can get a cheap USB to serial adapter, most of which are based on common chips like the PL2303, which will also work, though with a different /dev node, ttyUSB0 or something like that. Type dmesg | tail after plugging the adapter in to see what /dev/ node it was assigned. Avoid the expensive adapters. They often use special chips, requiring drivers that aren't built into the OS. The cheaper ones are actually better, because the chances are better that they're using some generic chip which your OS already knows how to talk to. All of that also applies to Mac OS X and other Unixy type systems. If you're on Windows, you may be screwed. But, you can always install Linux on another partition, or in a virtual machine system. :) It doesn't appear that R has a built-in way to control the bit rate and such. You can do that from the command line with setserial on Linux, but it's probably easier to just stick with the defaults -- 115200 bps, 8N1 -- and make your Arduino use that. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Wiring-or-Arduino-package-scripts-tp23846389p23879858.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antialiasing plots and text on different devices
I have a question about antialiasing when R generates bitmaps. (This follows
a thread on the ggplot2 mailing list.)
I mostly use R on Linux, although I sometimes use it in Mac and Windows as
well. On Linux, I've found that plotting shapes 15-18 via cairo results in
bad-looking output. The points are not antialiased, and they are jagged and
misshapen. Plots generated in Windows also aren't antialiased, but at least
the vector shapes seem to be aligned to pixel boundaries so that the raster
images look consistent.
You can see this in the scatterplots I've posted here:
http://stdout.org/~winston/X/r-antialias/pch.html
Also on that page are pch symbol charts from a modified version of pchShow()
from the pch help page. I rendered PNG's of the chart on Mac, Linux, and
Windows, using png() and setting type to Xlib, cairo, quartz, if available
on the given platform. In Windows, I did it without the type argument (in
Windows, the flag wasn't available). Finally, I also installed the Cairo
package and used CairoPNG. Please note the distinction between cairo,
which is built-in, and Cairo, which is an installed package.
(Side note: there's no Mac-Xlib image, but only because I had some X server
issues on that computer.)
Here are some observations and questions that hopefully someone can answer:
- With cairo, all shapes are antialiased except 15-18. Why do cairo and
Cairo give different results for shapes 15-18?
- With CairoPNG, all shapes are antialiased. Title text looks different
between the platforms, though. On Mac, it's normal text; on Linux, it's
bold; and on Windows, it's italic. I believe that font.main was set to 2, so
it should be bold. Why does Cairo render text so differently on different
platforms?
It would be nice not to have to tailor scripts to fit the quirks of whatever
platform I happen to be at. For example, to have the title render properly
in bold and have all shapes be anti-aliased, here's what I would need to do:
- Mac: png(type=quartz)
- Linux: CairoPNG()
- Windows: not possible
Ideally, I would like to use the same command on all platforms to generate
good antialiased graphs with similar-looking fonts. Is such a thing
possible?
-Winston
This is the code that I used to generate all the images for each platform:
pchShow -
function(extras = c(*,., o,O,0,+,-,|,%,#),
cex = 3, ## good for both .Device==postscript and x11
col = red3, bg = gold, coltext = brown, cextext = 1.2,
main = paste(plot symbols : points (... pch = *, cex =,
cex,)))
{
nex - length(extras)
np - 26 + nex
ipch - 0:(np-1)
k - floor(sqrt(np))
dd - c(-1,1)/2
rx - dd + range(ix - ipch %/% k)
ry - dd + range(iy - 3 + (k-1)- ipch %% k)
pch - as.list(ipch) # list with integers strings
if(nex 0) pch[26+ 1:nex] - as.list(extras)
plot(rx, ry, type=n, axes = FALSE,
xlab = x-axis label, ylab = y-axis label,
main = main)
abline(v = ix, h = iy, col = lightgray, lty = dotted)
for(i in 1:np) {
pc - pch[[i]]
## 'col' symbols with a 'bg'-colored interior (where available) :
points(ix[i], iy[i], pch = pc, col = col, bg = bg, cex = cex)
if(cextext 0)
text(ix[i] - 0.3, iy[i], pc, col = coltext, cex = cextext)
}
}
sysname - paste(version$platform, _, version$major, ., version$minor,
sep=)
# Make the scatterplot
set.seed(123)
png(paste(sysname, -default_png_scatterplot.png, sep=), width=300,
height=300)
plot(rnorm(20),rnorm(20), pch=16, main=paste(sysname,pch=16))
dev.off()
# Get the possible types on this computer (cairo, Xlib, etc)
types - NULL
if (capabilities()[cairo])types - c(types, cairo)
if (capabilities()[X11]) types - c(types, Xlib)
if (capabilities()[aqua]) types - c(types, quartz)
# Generate the images
if (is.null(types)) {
# If no types available
png(paste(sysname, -default_png.png, sep=), width=300,
height=300)
pchShow(cex=.9, cextext=.8, main=paste(sysname, default PNG
output))
dev.off()
} else {
for (i in 1:length(types)) {
# If types are available, use them
png(paste(sysname, -, types[i], .png, sep=), width=300,
height=300,
type=types[i])
pchShow(cex=.9, cextext=.8, main=paste(sysname,types[i]))
dev.off()
}
}
# Make the CairoPNG version
library(Cairo)
CairoPNG(paste(sysname, -, CairoPNG, .png, sep=), width=300,
height=300)
pchShow(cex=.8, cextext=.8, main=paste(sysname, CairoPNG))
dev.off()
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] RGtk2 help: Show list of column names from dataset and categorize as factor or numeric
On Thu, Jun 4, 2009 at 7:53 AM, Harsh singhal...@gmail.com wrote: Hi UseRs, I recently started working with the RGtk2 library. The documentation is comprehensive and I've tried learning from the examples in most Gtk tutorials (which have C code). This is a little problematic, but has helped quite a bit in getting me started. I would like to create a GUI for file selection, which then displays the column names from the selected file, and provides the user with checkboxes with each column name for the user to select. Two columns of check boxes (Factor Type, Numeric Type) and one column of names is what I would like to display. This would use what is known as a GtkTreeView widget. There is RGtkDataFrame object that will tie an R data frame directly to a GtkTreeView as its GtkTreeModel. See help(RGtkDataFrame). Using a GtkTreeView is pretty complicated, but powerful. Basically, you create your RGtkDataFrame from an R data.frame and pass it as the model to the gtkTreeView constructor. You then need to add columns to the tree view to map the data to the view. Your best bet is to check out the demos included with RGtk2, like editableCells and treeStore. Moreover, I am planning to create a GUI tool that would have tabs in a notebook layout, each tab providing a certain functionality, beginning from basic charting of data, and going on to applying regression models and such on the data. help(GtkNotebook) This requires extensive knowledge in components that RGtk2 provides which could be implemented for the task outlined above. I have looked at the omegahat.org examples, but would like to see examples for such simple tasks as to how one could create a drop down list of column names to choose for x axis and another drop down allowing the choice of y axis, etc. See help(GtkComboBox) and gtkComboBoxNewText(). Having made the choice to use RGtk2, I would appreciate if users could share their RGtk experience with me. I wrote a paper on RGtk2, but for some reason it has never been published. Probably time to write a book. Regards Harsh Singhal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R FAQ - web interfaces section very out of date.
I just checked every project in the Web interfaces section of the FAQ (http://cran.r-project.org/doc/FAQ/R-FAQ.html#R-Web-Interfaces ). There are many dead links in this section. Of the links that work, many projects themselves contain links that don't work, appear to have not been maintained in many years, and/or use deprecated versions of R. Perhaps this section could be broken into subsections of Active or Working Projects and Historical Interest? The R FAQ serves at the public face of R to some extent, and it seems that improving this section would be *very* helpful to the R community. -christian graduate student University of New Mexico Biology -- Far better an approximate answer to the right question, which is often vague, than the exact answer to the wrong question, which can always be made precise -- j.w. tukey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Install RCurl in Linux
Hello all, I had both curl and curl-devel (both 7.15) installed on my x86_64/CentOS machine. However, I still got problem when I do R CMD INSTALL RCurl  Error message below, since library and/or include path is missing. I tried R CMD INSTALL RCurl --configure-args='--libdir=/usr/lib64/ --includedir=/usr/include/' Same result. Any hint/suggestion would be appreciated. ... No CURLOPT_NOPROXY enumeration value. No CURLINFO_CONDITION_UNMET enumeration value. No CURLINFO_REDIRECT_URL enumeration value. No CURLINFO_CERTINFO enumeration value. No CURLINFO_PRIMARY_IP enumeration value. No CURLINFO_APPCONNECT_TIME enumeration value. No CURLOPT_KEYPASSWD enumeration value. No CURLOPT_DIRLISTONLY enumeration value. No CURLOPT_APPEND enumeration value. No CURLOPT_KRBLEVEL enumeration value. ... Version has CURLOPT_PROTOCOLS Version has CURLOPT_REDIR_PROTOCOLS configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/local/lib64/R/include -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include   -fpic -g -O2 -c enums.c -o enums.o In file included from CurlOptEnums.h:10,                 from enums.c:79: CURLOptTable.h:145: error: âCURLOPT_SOCKOPTFUNCTIONâ undeclared here (not in a function) CURLOptTable.h:146: error: âCURLOPT_SOCKOPTDATAâ undeclared here (not in a function) CURLOptTable.h:147: error: âCURLOPT_SSL_SESSIONID_CACHEâ undeclared here (not in a function) CURLOptTable.h:148: error: âCURLOPT_SSH_AUTH_TYPESâ undeclared here (not in a function) CURLOptTable.h:149: error: âCURLOPT_SSH_PUBLIC_KEYFILEâ undeclared here (not in a function) CURLOptTable.h:150: error: âCURLOPT_SSH_PRIVATE_KEYFILEâ undeclared here (not in a function) CURLOptTable.h:151: error: âCURLOPT_FTP_SSL_CCCâ undeclared here (not in a function) make: *** [enums.o] Error 1 ERROR: compilation failed for package âRCurlâ * Removing â/usr/local/lib64/R/library/RCurlâ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] OT: Inference for R - Interview
Dear All, Slightly off -non technical topic ( but hey it is Friday) Following last week's interview with REvolution Computing which makes enterprise versions of R, here is another interview with the rapidly growing company Blue Reference CEOPaul van Eikeren at http://www.decisionstats.com/2009/06/04/interview-inference-for-r/ http://www.decisionstats.com/2009/06/04/interview-inference-for-r/ Paul talks on his product, Inference for R- a add on plugin which makes a R GUI within Office Excel available for 199$ a year ( and *separate academic*program as well) for enhanced analytics as well as graphical capabilities. Best Regards, Ajay Ohri www.decisionstats.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with using subset from two different tables
Hi all, I am new to R-project my problem is I tried to get subset from two different tables its giving error but if i m tring for geting results from one table its working actually i have to take values from two tables with applying different conditions on two tables like kk- is an object of one table and fk- is an object of another table here i have to get values from these tables like subset(kk fk,kk$Rmaxtgavcg 1.256 fk$rmaxtgavcg 3.25,select=c(uniqueid)) my doubt is, that if any thing like this expression is there in R-project or i have to go for two different subsets and then adding those two one after another with checking the common uniqueid's any help is more precious to me Thanks in advance kiran. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: Inference for R - Interview
Is it really necessary to further advertise this company which already spams R-help subscribers? Hadley On Thu, Jun 4, 2009 at 10:41 PM, Ajay ohri ohri2...@gmail.com wrote: Dear All, Slightly off -non technical topic ( but hey it is Friday) Following last week's interview with REvolution Computing which makes enterprise versions of R, here is another interview with the rapidly growing company Blue Reference CEOPaul van Eikeren at http://www.decisionstats.com/2009/06/04/interview-inference-for-r/ http://www.decisionstats.com/2009/06/04/interview-inference-for-r/ Paul talks on his product, Inference for R- a add on plugin which makes a R GUI within Office Excel available for 199$ a year ( and *separate academic*program as well) for enhanced analytics as well as graphical capabilities. Best Regards, Ajay Ohri www.decisionstats.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Install RCurl in Linux
It seems you need a later curl -- 7.15 is rather old (and I think incomplete as a version number) and 7.19.5 is current. It is helpful to tell us what version of packages (here RCurl) you are trying to use, as well as your precise OS. On Thu, 4 Jun 2009, heyi xiao wrote: Hello all, I had both curl and curl-devel (both 7.15) installed on my x86_64/CentOS machine. However, I still got problem when I do R CMD INSTALL RCurl ?? Error message below, since library and/or include path is missing. I tried R CMD INSTALL RCurl --configure-args='--libdir=/usr/lib64/?? --includedir=/usr/include/' That should be part of the default paths. Same result. Any hint/suggestion would be appreciated. ... No CURLOPT_NOPROXY enumeration value. No CURLINFO_CONDITION_UNMET enumeration value. No CURLINFO_REDIRECT_URL enumeration value. No CURLINFO_CERTINFO enumeration value. No CURLINFO_PRIMARY_IP enumeration value. No CURLINFO_APPCONNECT_TIME enumeration value. No CURLOPT_KEYPASSWD enumeration value. No CURLOPT_DIRLISTONLY enumeration value. No CURLOPT_APPEND enumeration value. No CURLOPT_KRBLEVEL enumeration value. ... Version has CURLOPT_PROTOCOLS Version has CURLOPT_REDIR_PROTOCOLS configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/local/lib64/R/include?? -DHAVE_LIBIDN_FIELD=1 -DHAVE_CURLOPT_URL=1 -DHAVE_CURLOPT_PROTOCOLS=1 -DHAVE_CURLOPT_REDIR_PROTOCOLS=1 -I/usr/local/include?? -fpic?? -g -O2 -c enums.c -o enums.o In file included from CurlOptEnums.h:10, from enums.c:79: CURLOptTable.h:145: error: ??CURLOPT_SOCKOPTFUNCTION?? undeclared here (not in a function) CURLOptTable.h:146: error: ??CURLOPT_SOCKOPTDATA?? undeclared here (not in a function) CURLOptTable.h:147: error: ??CURLOPT_SSL_SESSIONID_CACHE?? undeclared here (not in a function) CURLOptTable.h:148: error: ??CURLOPT_SSH_AUTH_TYPES?? undeclared here (not in a function) CURLOptTable.h:149: error: ??CURLOPT_SSH_PUBLIC_KEYFILE?? undeclared here (not in a function) CURLOptTable.h:150: error: ??CURLOPT_SSH_PRIVATE_KEYFILE?? undeclared here (not in a function) CURLOptTable.h:151: error: ??CURLOPT_FTP_SSL_CCC?? undeclared here (not in a function) make: *** [enums.o] Error 1 ERROR: compilation failed for package ??RCurl?? * Removing ??/usr/local/lib64/R/library/RCurl?? [[alternative HTML version deleted]] -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
