Seems you're missing the required header(s). Can't find the example in
the extensions manual but you probably need
#include
or
#include
HTH,
Peter
On Wed, Mar 14, 2012 at 11:03 PM, Erin Hodgess wrote:
> Dear R People:
>
> Here is something that I am sure is very simple. I'm just trying t
Do you really expect is to know how to figure out the error if you don't give
us the code that the error is pointing at (or the code preceding the error,
which may be at fault)? Please think before posting.
---
Jeff Newmiller
Hi Erin,
If you are just starting with including compiled code, I would highly
recommend the Rcpp + inline packages. I attached a small example of
how it works, but basically you get to inline C++ code in R files.
Cheers,
Josh
P.S. I wrote this snippet awhile ago when I did not know much about
G'day Erin,
On Thu, 15 Mar 2012 01:03:59 -0500
Erin Hodgess wrote:
> What is wrong, please?
Missing
#include
#include
In particular the latter is defining R_len_t. Guess that code snippet
in the manual is only showing the code of the function, but not the
complete file that needs to be com
Dear R People:
Here is something that I am sure is very simple. I'm just trying to
re-create the C convolution example in the Extensions manual. Here is
the subroutine:
void convolve(double *a, int *na, double *b, int *nb, double *ab)
{
R_len_t i, j, nab = *na + *nb - 1;
for(i = 0; i < nab; i++
Hi all,
Sorry if I seem a bit pissed because I am!
Basically, I have written some script and I wanted to make a "documentation"
for it using .Rd format.
I followed the example here in
http://stat.ethz.ch/R-manual/R-de
Dear R Users,
I am trying to plot a matrix (a Digital Elevation Model) using wireframe
[lattice] and color that matrix based on a separate/independent matrix of
the same resolution (both have the same number of rows and columns) as the
DEM. More specifically I would like to plot a DEM using wiref
Hi Camille,
I am surprised by your answer. If you do:
pred <- prediction(ROCR.simple$predictions*1000, ROCR.simple$labels)
x <- rev(unlist(pred@cutoffs))
You can see than x values are now between 0 and 1000. So, it should be probably
the same for your data.
Regards,
Pascal
--
View this me
Supply an 'str' of your dataframe so we can see what its structure is.
Do you have leading/trailing blanks in your ROUTE values. Print
them out and see what their number of characters (nchar) are. Are
they factors? You have not supplied enough information like a small
subset. I bet when you c
Hello,
>
> Thanks Rui.
>
> Sorry I forgot to mention what is the "var" in the previous message. It
> seems
> that when the correlation is small, it does not work.
> See below.
>
It works in my system: (your first example)
var <- matrix(c(1,0.05,0.05,1), nrow=2, ncol=2, byrow=T)
var
[,1] [
Hello.
I am beginning to analyze my work and have realized that a simple chi-square
analysis will not suffice for my research, with one notable reason is that data
are not discrete. Since my data fit the assumptions of a logistic regression,
I am moving forward with this analysis. With t
O yea sure. So for example:
grid <- read.table("table")
( i havent printed the output, as the table is 20,000 rows X 60 columns)
point_of_interest <- c("row1", "row2")
therefore all the other points in
plot(table)
are labelled green, but these two are labelled red.
But at the minute, becaus
This is the part of the function in my likelihood function
prob<-function(t1,t2,m1,m2,s_r,s_p,s_e,cor_m){
if ((t1==0) & (t2==0))
log_lik<-log(pmvnorm(lower=rep(-Inf, 2),
upper=c(m1/sqrt(s_r+s_p+s_e+1),m2/sqrt(s_r+s_p+s_e+1)),
mean=rep(0,2), corr=cor_m))
else if ((t1==0) & (t2==1))
log_lik<
Without a subset of data, it is hard to come up with a solution. Now
here is a way of determining what names are in common and then maybe
doing something:
B.names <- names(B.list)
A.names <- names(A.list)
common <- intersect(B.names, A.names)
for (i in common){
B.list[[i]] <- function(A.list[
I'll appreciate your help on this. I have values stored in a list as in
"mylist" below. I need to sum the values over all elements of the list
aggregated by the names of the matrices.
mylist <- list(matrix(c(0.2, 0.4), 1, 2, dimnames = list(NULL, c("1",
"2"))),
matrix(c(0.1, 0.5
I'm having a simple problem with subset. I'm choosing what I think is a valid
selection, but I either get everything or an empty dataframe. What am I doing
wrong?
>
> describe(OBDataSumm)
Description of OBDataSumm
Numeric
meanmedian varsd valid.n
SAMP
Try this.
x <- matrix(1:3, nrow = 1)
y <- matrix(1:6, nrow = 1)
z <- matrix(1:10, nrow = 1)
library(plyr)
do.call("rbind.fill.matrix", list(x,y,z))
Note that x,y,z need to be made into matrices before this works.
Michael
On Wed, Mar 14, 2012 at 6:31 PM, Zenonn87 wrote:
> Hello!
>
> How can I
Thanks Rui.
Sorry I forgot to mention what is the "var" in the previous message. It
seems
that when the correlation is small, it does not work.
See below.
> var <- matrix(c(1,0.05,0.05,1), nrow=2, ncol=2, byrow=T)
> var
[,1] [,2]
[1,] 1.00 0.05
[2,] 0.05 1.00
> qmvnorm(0.05, tail="upper", s
Errr...I'm still not sure how plot(table) gives you read and green
points, but it sounds like the easiest thing to do would be to use the
points() function to come back in and put new green points over the
red ones. E.g.
plot(runif(5000), runif(5000))
points(c(0.5, 0.25), c(0.5, 0.75), col = 2, pc
What distribution's log-likelihood are you using? If sigma is supposed
to be square-rooted, you may have to put a constraint or use of abs()
might suffice -- though, admittedly I'm not sure what that will do to
convergence behavior -- but it's hard to help without seeing the
function at hand.
Mich
Hello R people
I always wander what other people say about the R help. Finally after some
years of using, I decided that it is probably time to try to do something about
it, because the feeling of gritting teeth does not go away with years of usage.
:) Moreover, I think it is one of the few thi
Dear R Users
I am maximizing a user defined log likelihood function. It includes variance
parameter (sigma). I used R function optim with BFGS maximization method.
However, it stops before the solution saying “sqrt(sigma): NaNs produced”
Could anybody know a proper transformation for sigma which
Hello!
How can I top up with NA instead of repeat numbers when I am making
data.frame?
I have made vectors (6. peace) form a table with different length
(5,10,15,20,25,30 data). I want put these vectors into a data.frame but in
the smaller columns the numbers repeat continuously to fill up the em
Thanks Gabor. colClasses did the trick. Also, I'll be sure to minimize the
reproducible example the next time around using a text quote. Sorry about
the noob mistake.
--
View this message in context:
http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-
Hello,
li li-13 wrote
>
> Dear all,
>I need to use the "qmvnorm" function in mtvnorm package.
>
> Here is the error message I got
>> qmvnorm(0.05, tail="upper", sigma=var)$quantile
> Error in uniroot(pfct, interval = interval) :
> f() values at end points not of opposite sign
>
> There is
Thanks, this isn't actually homework though. I'm researching MCMC methods but
I've never really used R before so it's proved quite troublesome!
my code is now:
n=1
mu=0
sigma=1
lik<-function(theta) exp((-(theta-mu)^2)/2)
alpha<-function(theta,phi) min(lik(phi)/lik(theta),1)
theta1<-c(0,n)
the
Hi,
On Wed, Mar 14, 2012 at 2:21 PM, Dorothea Hill
wrote:
> Hi,
>
> I am trying to use a mantel test on two distance matrices. The code I have
> entered for each is:
>
> Gen_dists <- read.csv(file.choose(),
> stringsAsFactors = FALSE,
>
On Wed, Mar 14, 2012 at 6:07 PM, Aurélien PHILIPPOT
wrote:
> Dear R experts,
> I have a dataframe imported from a csv file (with read.csv).
>
> Here is an example:
>
> mm<- c("19860228", "19860331","19860430","19860531")
> id<-c("1","1","1","1")
> re<- c("C","0.25", "0.98", "1.
thanks for your suggestions, which work.
In addition, I updated Rstudio to the latest version and my old code works
again.
Best,
Aurelien
2012/3/14 Petr Savicky
> On Wed, Mar 14, 2012 at 03:07:19PM -0700, Aurélien PHILIPPOT wrote:
> > Dear R experts,
> > I have a dataframe imported from a csv f
On Wed, Mar 14, 2012 at 03:07:19PM -0700, Aurélien PHILIPPOT wrote:
> Dear R experts,
> I have a dataframe imported from a csv file (with read.csv).
>
> Here is an example:
>
> mm<- c("19860228", "19860331","19860430","19860531")
> id<-c("1","1","1","1")
> re<- c("C","0.25", "
Dear Aurelien,
Thanks for the reproducible example. Here is one way:
mm<- c("19860228", "19860331","19860430","19860531")
id<-c("1","1","1","1")
re<- c("C","0.25", "0.98", "1.34")
mret <- data.frame(mm, id, re)
subset(mret, !is.na(as.numeric(as.character(re
HTH,
Jorg
Dear R experts,
I have a dataframe imported from a csv file (with read.csv).
Here is an example:
mm<- c("19860228", "19860331","19860430","19860531")
id<-c("1","1","1","1")
re<- c("C","0.25", "0.98", "1.34")
mret<-data.frame(mm, id, re)
mret<-as.numeric(as.character(mret
Heh, a one liner fix huh? This was going to be my alternative solution (got
this coded as soon as the character strings theory arrived, and yes I'll use
dput next time, sorry about that again and for all the miscommunication):
require(zoo); require(chron)
setwd("/home/knavero/Desktop")
rawData =
Hi Jean and Peter,
Thanks for the help. Both options are indeed faster than my initial
procedure.
Best,
Mathijs
--
View this message in context:
http://r.789695.n4.nabble.com/Merging-fully-overlapping-groups-tp4470999p4473013.html
Sent from the R help mailing list archive at Nabble.com.
_
"If you showed the output of dput(yourData) or even str(yourData)
others could see what types the columns are. Ordinary printed output can
make numeric, character, and factor data look the same. Comparisons
involving strings do not always return the same value as comparisons
involving the equival
Additionally, the dput(rawData) output:
dput(rawData)
structure(list(Date...Time = structure(1:1040, .Label = c("2/11/12 12:45
",
"2/11/12 13:00 ", "2/11/12 13:15 ", "2/11/12 13:30
",
"2/11/12 13:45 ", "2/11/12 14:00 ", "2/11/12 14:15
",
"
Hi guys,
Got a few days left and I need to model a random effect of species on the
body mass (logM) and temperature (K) slopes. This is what i've done so far
that works:
model1<-lme(logSSP~logM + K,random=~1|species,data=data1)
model2<-lme(logSSP~logM + K,random=~K|species,data=data1)
model3<-
Dear all,
I need to use the "qmvnorm" function in mtvnorm package.
Here is the error message I got
> qmvnorm(0.05, tail="upper", sigma=var)$quantile
Error in uniroot(pfct, interval = interval) :
f() values at end points not of opposite sign
There is no problem for 50th quantile.
> qmvnorm(0.
"I'm getting a 404 on the dropbox linklooking at your pastebin
link, my guess is that those numbers are in fact greater than zero,
but are very small so they appear as zeros in the print method.
If you want to send your data, it's easier to simply give us
dput(rawData) and copy and paste the r
On Mar 14, 2012, at 4:09 PM, John Smith wrote:
With most current version of R and RMS, the 4 curves are drew in
4 separate panels. Can anyone show me how can I get the figure
exactly like
the figure 7.8 in *Regression Modeling Strategies* (
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/RmS
This sounds like homework (in which case you should ask your instructor
for help), but among the problems I see just glancing at it:
* runif(-0.5,0.5) should be runif(1,-0.5,0.5)
* there are a host of problems involved with the fact that you have set
theta1 up as a two-column matrix, but repeat
On Mar 14, 2012, at 5:30 PM, Michael wrote:
How do I change the tick text on the x-axis of a plot and rotate
them 90
degree?
labs
[1] "[-0.185,-0.0997]" "(-0.0997,-0.0549]" "(-0.0549,-0.0293]"
[4] "(-0.0293,-0.00948]" "(-0.00948,0.00534]" "(0.00534,0.0178]"
[7] "(0.0178,0.035]" "(0.035,
How do I change the tick text on the x-axis of a plot and rotate them 90
degree?
> labs
[1] "[-0.185,-0.0997]" "(-0.0997,-0.0549]" "(-0.0549,-0.0293]"
[4] "(-0.0293,-0.00948]" "(-0.00948,0.00534]" "(0.00534,0.0178]"
[7] "(0.0178,0.035]" "(0.035,0.0566]" "(0.0566,0.0932]"
[10] "(0.0932,0.183]"
Thanks for the idea. I think this example works fast mainly due to the limited
number of matches. For each fdf1$chr there are only 2 potential matches in
fdf2. In reality there are only 24 possible values for chr (1-22 and X, Y).
When I replace the chr seq with more realistic values, I run o
On Wed, Mar 14, 2012 at 12:43 PM, knavero wrote:
> attached http://r.789695.n4.nabble.com/file/n4472408/dataout_2471_843.csv
> dataout_2471_843.csv
>
Here is how the problem can be presented in a self contained, minimal,
reproducible fashion (as per last two lines on every message to
r-help):
l
On Tue, Mar 13, 2012 at 08:56:33PM -0700, mdvaan wrote:
> Hi,
>
> I have data on individuals (B) who participated in events (A). If ALL
> participants in an event are a subset of the participants in another event I
> would like to remove the smaller event and if the participants in one event
> are
Craig Lyon rogers.com> writes:
>
> Hi,
>
> I am trying to run a generalized linear regression using a negative binomial
> error distribution. However, I want to use an overdispersion parameter that
> varies (dependent on the length of a stretch of road) so glm.nb will not do.
>
> >From what I'
Load package fBasics
library(fBasics)
colors()
should give you what you want. Also helpful are
colorTable()
colorLocator()
Which I found by searching for colors from the R console help menu.
> Date: Wed, 14 Mar 2012 22:55:39 +0800
> From: totang...@gmail.com
> To: r-help@r-project.org
Marianna,
You can use merge for that (or match). Using merge:
MyData <- data.frame(
V1=c("red-j", "red-j", "red-j", "red-j", "red-j", "red-j"),
V4=c(10.5032, 9.3749, 10.2167, 10.8200, 9.2831, 8.2838),
redNew=c("appearance blood-n", "appearance ground-n", "appearance
sea-n", "appeara
Since you found readOGR, you might want to look at writeOGR.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 2/22/12 3:43 PM, "gztourek" wrote:
>Hi,
>
>I am new to R and am a very basic user. I'm importing SHP files,
You could try doing it without a loop (.C or other):
(rgnsnp <- merge(region,snps))
(rgnsnp[with(rgnsnp,STOP>=POS & POS >= START),])
Here is my test for merge+search on 100k/200k:
fdf1 <- data.frame(chr=1:10,p=runif(10),d=sample(10))
fdf2 <- data.frame(chr=rep(1:10,2),s=runif(2
Just took a closer look at this, Bill Dunlap's guess was correct (mine
wasn't): your data is stored as a character (or at least, mine is
after the commands below) -- character sorting is highly platform
dependent so that's probably why I got a different answer than you.
rawData <- zoo(as.numeric(c
Here is an example of the start of a self-contained way to reproduce your
problem:
> library(chron)
> library(zoo)
> rawData <-
> read.zoo("http://r.789695.n4.nabble.com/file/n4472408/dataout_2471_843.csv";,
> header = TRUE,
+ FUN = as.chron, format = "%m/%d/%Y %H:%M", index.column = 2
Oops, this uses one of my own functions %like%
"%like%" <- function(x,y) grepl(y,x,ignore=TRUE)
Here's an improved version of word.tif ...
word.tif <- function(filename="Word_Figure_%03d.tif", zoom=5, width=17,
height=10, pointsize=10, ...) {
if (!grepl("[.]ti[f]+$", filename,ignore.case=TRUE)
That sure doesn't resemble the output of dput().
Still, I ran the following:
rawData <- read.zoo("dataout_2471_843.csv", header = TRUE, FUN =
as.chron, format = "%m/%d/%Y %H:%M", index.column = 2, sep = ",",
aggregate = function(x) tail(x,1))
rawData$Meter.ID = NULL
rawData[rawData$KW.ch..1..
Thank you Michael,
Yes it appeared odd to me too... I'll email them back to verify.
Thanks!
Aurelie
On 2012-03-14, at 4:56 PM, R. Michael Weylandt wrote:
> Hmmm, that seems an odd request: how could one make a plot in Excel
> natively (excepting ASCII art)?
>
> Perhaps write and ask them if the
With most current version of R and RMS, the 4 curves are drew in
4 separate panels. Can anyone show me how can I get the figure exactly like
the figure 7.8 in *Regression Modeling Strategies* (
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/RmS/course2.pdf)
Thanks
On Tue, May 17, 2011 at 4:04
On Tue, Mar 13, 2012 at 08:56:33PM -0700, mdvaan wrote:
> Hi,
>
> I have data on individuals (B) who participated in events (A). If ALL
> participants in an event are a subset of the participants in another event I
> would like to remove the smaller event and if the participants in one event
> are
Perhaps using the win.metafile() graphics device (instead of pdf()) would work.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of R. Michael Weylandt
> Sent: Wednesday,
Hi,
I am trying to run a generalized linear regression using a negative binomial
error distribution. However, I want to use an overdispersion parameter that
varies (dependent on the length of a stretch of road) so glm.nb will not do.
>From what I've read I should be able to do this using GLM by s
I have found the TIF format (with lossless compression) most suitable for
inclusion in Word documents.
Here's my function for producing a tif plot...
word.tif <- function(filename="Word_Figure.tif", zoom=5, res=96*zoom, width=17,
height=10, pointsize=10, bg='white') {
if (!filename %like% "
Hi,
I am trying to use a mantel test on two distance matrices. The code I have
entered for each is:
Gen_dists <- read.csv(file.choose(),
stringsAsFactors = FALSE,
na.strings = c(" "),
Dear R-ers,
still the newbie. With a question about coordinates of a vector appearing or
not in a data.frame.
I have a data.frame (MyData) with 3 columns which looks like this:
V1V4 redNew
red-j 10.5032 appearance blood-n
Hi all,
I'm trying to write a MH algorithm in R for a standard normal distribution,
I've been trying for a good week or so now with multiple attempts and have
finally given up trying to do it on my own as I'm beginning to run out of
time for this, would somebody please tell me what is wrong with my
attached http://r.789695.n4.nabble.com/file/n4472408/dataout_2471_843.csv
dataout_2471_843.csv
--
View this message in context:
http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p4472408.html
Sent from the R help mailing list archive at Nab
Hmmm, that seems an odd request: how could one make a plot in Excel
natively (excepting ASCII art)?
Perhaps write and ask them if they can take any image formats? R
exports to most of them quite nicely: png, eps, jpeg, bitmap, svg,
etc.
Michael
On Wed, Mar 14, 2012 at 3:49 PM, Aurelie Cosandey G
Hi all,
I have created forest plots using the package "meta" that I submitted as pdf
for publication. I just received an email from the editor asking me if I could
send these files in an Excel or MS Word format such that they can treat them as
tables with box plots.
I am not sure if I can do t
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Davis, Brian
> Sent: Wednesday, March 14, 2012 2:28 PM
> To: r-help@R-project.org
> Subject: [R] Needing a better solution to a lookup problem.
>
> I have a solution (actually a f
On Mar 14, 2012, at 3:27 PM, Davis, Brian wrote:
I have a solution (actually a few) to this problem, but none are
computationally efficient enough to be useful. I'm hoping someone
can enlighten me to a better solution.
I have data frame of chromosome/position pairs (along with other
dat
I have a solution (actually a few) to this problem, but none are
computationally efficient enough to be useful. I'm hoping someone can
enlighten me to a better solution.
I have data frame of chromosome/position pairs (along with other data for the
location). For each pair I need to determine
If you showed the output of dput(yourData) or even str(yourData)
others could see what types the columns are. Ordinary printed output can
make numeric, character, and factor data look the same. Comparisons
involving strings do not always return the same value as comparisons
involving the equivale
Thank you. The binomial()$linkinv() is good to know.
Ben
On Wed, Mar 14, 2012 at 12:23 PM, Patrick Breheny
wrote:
> Actually, I responded a bit too quickly last time, without really reading
> through your example carefully. You're fitting a logistic regression model
> and plotting the results o
Also, the term "error" means something specific so try not to misuse
it (it will misdirect people as to the nature of your problem). An
error looks like what you get if you do this:
sqrt(list() + `if` / data)
Michael
On Wed, Mar 14, 2012 at 2:38 PM, R. Michael Weylandt
wrote:
> I'm getting a 40
I'm getting a 404 on the dropbox linklooking at your pastebin
link, my guess is that those numbers are in fact greater than zero,
but are very small so they appear as zeros in the print method.
If you want to send your data, it's easier to simply give us
dput(rawData) and copy and paste the re
My apologies: I was thinking of "rolling" more in the sense of
"sliding" than "creeping". I'm not aware of anything suited for that,
but it's not hard to "roll your own" with something like
dats <- data.frame(x = 1:100, y = rnorm(100))
for(i in 80:100){
lm(y ~ x, data = dats[1:i, ])
}
Michae
This code performs the same operation in about 1/10th the time on my
machine.
Give it a try.
look <- function(i) {
# look for subsets
dif <- m[, i] - m
apply(dif, 2, min) > -0.5
}
nosubsets <- function(df) {
# eliminate events that are subsets of other even
What sort of plot are you using? I'm not really clear on what your
data as a whole look like: if you use dput() you can create a
representation and we can work from there.
In addition to the resources Josh recommended, the following sites can
direct you to all sorts of graphical goodies, all of wh
Actually, I responded a bit too quickly last time, without really
reading through your example carefully. You're fitting a logistic
regression model and plotting the results on the probability scale. The
better way to do what you propose is to obtain the confidence interval
on the scale of th
That was embarrassingly easy. Thanks again Patrick! Just correcting a
little typo to his reply. this is probably what he meant:
pred = predict(fit,data.frame(x=xx),type="response",se.fit=TRUE)
upper = pred$fit + 1.96 * pred$se.fit
lower = pred$fit - 1.96 * pred$se.fit
# For people who are interes
Hi all,Â
I was trying to use glht() from multcomp package to construct a contrast on
interaction term
in a linear model to do some comparisons. I am little uncertain on how to
construct contrasts on a 3-way interaction containing a continuous variable,
and hope someone can confirm what I did i
How to stack these subplots horizontally and vertically together in a nice
way?
Thank you!
On Wed, Mar 14, 2012 at 11:04 AM, Clint Bowman wrote:
> ?quantile on the individual bins, make your deciles, then plot the ten
> series as usual with your x values at the midpoint of the bins.
>
> Clint B
Thank you!
How large is large sample?
What about 50 data points in D1 and another 50 data points in D2?
Thanks a lot!
On Wed, Mar 14, 2012 at 11:49 AM, peter dalgaard wrote:
>
> On Mar 14, 2012, at 16:21 , Michael wrote:
>
> > How to test the statistical significance of the difference of two
1. Is this homework? -- we don't do homework.
2. This is not an R question -- posting to stats.stackexchange.com or
other statistics websites is usually more appropriate for such non-R
statistical questions
3. General approach: Combine all data; model it with both a simpler
(fewer parameter, sing
On Mar 14, 2012, at 16:21 , Michael wrote:
> How to test the statistical significance of the difference of two
> univariate Linear Regression betas?
>
> Hi all,
>
> There are two samples of data: D1 and D2.
>
> On data D1 we do a univariate Linear Regression and get the coefficient
> beta1.
>
As you can see:
time is from 1 to 1460, 1 means they the measure Tem for the globe at the
first 6 hours , 2=after 12 hours .; and so on 1460 *6 = 8760 hours which
equals 1 year
first of all I want to convert all data from kelvin to degree
second I want to convert from 6 hourly to daily a
I have data that I would like to analyze in R with the following format:
*,M1,,M2,
*,S1,S2,S1,S2,
p1,m1s1v1,m1s2v1,m2s1v1,m2s2v1
p2,m1s1v2,m1s2v2,m2s1v2,m2s2v2
..,...,...,...,
so I have a method M1 and M2 and each of these have attributes that I would
like to analyze for the different problem
Here's the exact error I'm receiving:
http://pastebin.com/mNsPauwk
Tracked each output along the way. Starting to think there's a bug in the
source code.
--
View this message in context:
http://r.789695.n4.nabble.com/logical-test-not-functioning-correctly-on-zoo-series-what-the-hell-tp4471654p
Il 3/14/2012 4:43 PM, William Dunlap ha scritto:
I see you have a solution, but why do you want the
results of split() in matrix form? E.g., does it make
a nicer printout, is it needed to interface with other
R functions, is it needed to interface with other
I need matrix to export the data in t
"You're missing a comma between 0 and "]" in the last line if your goal
is to retrieve the rows that satisfy that condition (and if the
condition makes any sense). Haven't tested the rest of your code,
though.
?Extract
> The outputs that I'm getting, however, are
> printing out 0's down th
The predict() function has an option 'se.fit' that returns what you are
asking for. If you set this equal to TRUE in your code:
pred <- predict(fit,data.frame(x=xx),type="response",se.fit=TRUE)
will return a list with two elements, 'fit' and 'se.fit'. The pointwise
confidence intervals will
Hi: I'm jot sure if it's exactly what you want but check out Hotelling's
paper from 1940. It
should be in the archives because I answered this question before ( not
from you ).
If you can't find it, I'll find the title actually, here's the title:
Hotelling, The Selection of Variates For Use i
Thanks David for the details and pointer to bitops functions. Buried a
bit deep that was.
I like to think the memory constraints of win xp keeps my code lean and
efficient. RAM is like a suburban garage, the bigger it is, the more
useless junk people stuff in there.
Michael Folkes
-Origina
On Mar 14, 2012, at 11:47 AM, knavero wrote:
"> Here's the raw data I'm working with (will be available
temporarily):
http://dl.dropbox.com/u/41922443/dataout_2471_843.csv
Nothing appears. "
^ Clicking on the link should prompt you to download a csv file and
save it
somewhere in your H
On Mar 14, 2012, at 11:53 AM, ilai wrote:
On Wed, Mar 14, 2012 at 8:56 AM, Thomas Hoffmann
wrote:
Hi all,
I still fail to plot an axis title with the following expression:
plot(0,xlab=expression('(SOC [' * kgm^{-2} * '])' * ^{-2}))
the xlab should look like: (SOC [kgm^2])^0.25
with an o
Hi John,
Thanks again. That looks like an easy and convenient approach. Regards,
Chris
--
View this message in context:
http://r.789695.n4.nabble.com/MANOVA-and-Extra-Sums-of-Squares-Tests-tp4470077p4472265.html
Sent from the R help mailing list archive at Nabble.com.
___
"> Here's the raw data I'm working with (will be available temporarily):
>
> http://dl.dropbox.com/u/41922443/dataout_2471_843.csv
Nothing appears. "
^ Clicking on the link should prompt you to download a csv file and save it
somewhere in your HDD.
--
View this message in context:
http://r.7896
The error seems clear enough to me... you are trying to mix vectors of
different lengths where one length is not a multiple of the other length.
Your discussion is pretty confused though, referring to possible data frames x,
x.sub, x.res and Dataset. You might benefit from reviewing ?str and ?dp
?quantile on the individual bins, make your deciles, then plot the ten
series as usual with your x values at the midpoint of the bins.
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology
"> The outputs that I'm getting, however, are
> printing out 0's down the columns. I've tried various methods assuming
> various theories, read the R manual via "?" for different possible
> solutions, Googled stuff, tried the ifelse function which produces
> the same
> error, tried creating logic
Yes.
please look at RExcel. You can download it from rcom.univie.ac.at
The wiki page there has many papers discussing similar projects (follow the
Literature link in the left column). Followup should be on the rcom
mailing list.
Rich
On Wed, Mar 14, 2012 at 7:12 AM, burcy wrote:
> Hi All
>
1 - 100 of 173 matches
Mail list logo
