Hi Russ,
One tool that might help could be ?methods and ?showMethods
For example:
## for S3
methods(class = "Date")
## for S4
showMethods(classes = "Date")
regarding getting the actual year, I would use (though there may be
better ways):
format.Date(as.Date("2010-01-01"), format = "%Y")
HTH,
Hi,
I'm still confused about how to find out what methods are defined for a
given class. For example, I know that
> today <- Sys.Date()
will produce an object of type Date. But I'm not sure what I can do with
Date objects or how I can find out.
> ?Date
refers me to the Date documentation pag
Hi Victor,
looking at the code for legend, it looks like the same 'cex' value is
used for the text in the legend as the title.
Here is a trick though: draw the legend twice, with different cex
values, and omitting title or text:
plot(1)
legend("topright",c("a","b"),title=" ")
legend("topright",
Hi Eric,
On Thu, Apr 28, 2011 at 6:46 PM, eric wrote:
> I have the following lines of code:
>
> ind <- rollapply(GSPC, 200, mean)
> signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
If you had looked at signal here, you would see that it is logical.
> signal[is.na(signal)] <- 0
but this is not doing
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Victor Gabillon
> Sent: April-28-11 8:22 PM
> To: r-help@r-project.org
> Subject: [R] Change the text size of the title in a legend of a R plot.
>
> Hello,
>
> Is it possible
On Thu, Apr 28, 2011 at 11:08:23PM -0400, Chee Chen wrote:
> Dear All,
> I would like to define a function: f(x,y,z) with three arguments x,y,z, such
> that: given values for x,y, f(x,y,z) is still a function of z and that I am
> still allowed to find the root in terms of z when x,y are given.
>
I just realized that I had misread what was wanted -- the code I wrote was
for mean=0, sd=1, not for mean=1. So for mean=m, and sd=s, lower limit L
and upper limit U, this approach will work:
n <- 1000
m <- 1
s <- 1
L <- .2
U <- .8
p_L <- pnorm(L, mean=m, sd=s)
p_U <- pnorm(U, mean=m, sd=s)
Actually, you are trying to *install* it
- on a Mac (without telling us)
- from source (why? people usually install binary packages there)
- asking on the wrong list (R-sig-mac is appropriate)
- with an incomplete R installation. Go to the CRAN Mac OS X pages
and read what they say about instal
Hi:
On Thu, Apr 28, 2011 at 7:28 PM, eric wrote:
> equire(quantmod)
> require(PerformanceAnalytics)
> rm(list=ls())
# Could you please not do this in the middle of a code chunk?
# Anyone who copies/pastes this into his/her session will lose everything
# (s)he had been doing.
> getSymbols("^GSPC
Here's one possibility:
funmaker = function(x,y,z)function(z)x + y + (x^2 - z)
uniroot(funmaker(1,3,z),c(0,10))$root
[1] 5
uniroot(funmaker(5,2,z),c(30,40))$root
[1] 32
(The third argument to the function doesn't really do anything.)
- Phil Spector
On Apr 28, 2011, at 11:07 PM, Lisa wrote:
I am trying to write a function to check unusual values in my
datasets and
correct them. As some R users suggested, I try to use readline() and
scan()
in my function. Suppose there are several unusual values in a
dataset. I
want to change the line
On Thu, 2011-04-28 at 23:08 -0400, Chee Chen wrote:
> Dear All,
> I would like to define a function: f(x,y,z) with three arguments x,y,z, such
> that: given values for x,y, f(x,y,z) is still a function of z and that I am
> still allowed to find the root in terms of z when x,y are given.
> For ex
On Apr 28, 2011, at 10:28 PM, eric wrote:
equire(quantmod)
require(PerformanceAnalytics)
rm(list=ls())
getSymbols("^GSPC", src="yahoo", from="1990-01-01", to=Sys.Date())
GSPC <-na.omit(Ad(GSPC))
ind <- rollapply(GSPC, 200, mean)
signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
signal[is.na(signal)]
chartSeries and its cousins plot each period as one observation. all
observations have equal time. It 'takes care of' irregular data by
assuming that the rate of information is equal to the number of
observations, which is a perfectly correct intuition and useful for many
things. If you want ins
On 29/04/11 09:40, Lisa wrote:
Dear all,
Does everyone know how to change the line numbers of scan()? For example,
scan(n = 3)
1: 4
2: 6
3: 9
Read 3 items
[1] 4 6 9
I just want to change the line numbers 1, 2, and 3 to, say a, b, and c that
look like this:
scan(n = 3)
a: 4
b: 6
c:
Hello,
Is it possible to change the text size of the title in a legend of a R plot?
I tried to directly change the title.cex argument but it seems not to work.
Trying :
Horizo <- c(1,2,6,10,20)
legtext <- paste(Horizo,sep="")
legend("topleft", legend=legtext,col=col,text.col=col,lwd=lwd,
lty=l
On 29/04/11 08:56, Lisa wrote:
Hi, Rolf,
Thank you for your help. I am trying to use readline() and I have a question
about the maximum length of prompt string. The R help on readline() says:
“The prompt string will be truncated to a maximum allowed length, normally
256 chars (but can be change
I am trying to write a function to check unusual values in my datasets and
correct them. As some R users suggested, I try to use readline() and scan()
in my function. Suppose there are several unusual values in a dataset. I
want to change the line numbers in scan() to something like: unusual value
from the console ...
> table(signal)
signal
01
1286 3885
note there is no -1 value.
This is consistent with what I see if if plot(signal). When I issue that
statement from the console, I see signal vary between 0 and 1.0 but it never
goes to - 1
--
View this message in context:
http
equire(quantmod)
require(PerformanceAnalytics)
rm(list=ls())
getSymbols("^GSPC", src="yahoo", from="1990-01-01", to=Sys.Date())
GSPC <-na.omit(Ad(GSPC))
ind <- rollapply(GSPC, 200, mean)
signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
signal[is.na(signal)] <- 0
--
View this message in context:
http:/
Thanks for this clarification.
__
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 cod
what about something like:
dat <- data.frame(var1 = c("a","b"), var2 = c("c","d"))
dat$var3 <- sapply(dat, function(x) paste(dat$var1,dat$var2,sep=""))[,1]
Scott
On Thursday, April 28, 2011 at 10:22 AM, Abraham Mathew wrote:
I've kind of solved the issue.
>
> dat$Time <- paste(dat[,2], dat[,1]
Dear All,
I would like to define a function: f(x,y,z) with three arguments x,y,z, such
that: given values for x,y, f(x,y,z) is still a function of z and that I am
still allowed to find the root in terms of z when x,y are given.
For example: f(x,y,z) = x+y + (x^2-z), given x=1,y=3, f(1,3,z)= 1+
Hi:
It seems to work for me...here's a reproducible example.
set.seed(2053)
date <- seq(as.Date('1990-01-01'), by = 'days', length = 5000)
range(date)
# [1] "1990-01-01" "2003-09-09"
date <- sort(sample(date, 2000))
tdata <- data.frame(date = date, GSPC = rpois(2000, 1000))
library(zoo)
tdata2 <
Hi Eric,
tough to say. Please try to provide commented, minimal,
self-contained, reproducible code.
Cheers
Andrew
On Thu, Apr 28, 2011 at 06:46:16PM -0700, eric wrote:
> I have the following lines of code:
>
> ind <- rollapply(GSPC, 200, mean)
> signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
> s
On Thu, Apr 28, 2011 at 10:13 PM, Gabor Grothendieck
wrote:
> On Thu, Apr 28, 2011 at 4:49 PM, mathijsdevaan
> wrote:
>> Thanks, but it did not really improve the speed. Why is it that when I change
>> the layout of the matrix (which does not give the required results), the
>> speed increases tr
I have the following lines of code:
ind <- rollapply(GSPC, 200, mean)
signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
signal[is.na(signal)] <- 0
I never get a value of -1 for signal even though I know diff(ind , 5) is
less than zero frequently. It looks like when diff(ind , 5) is less than
zero, sign
On Thu, Apr 28, 2011 at 4:49 PM, mathijsdevaan wrote:
> Thanks, but it did not really improve the speed. Why is it that when I change
> the layout of the matrix (which does not give the required results), the
> speed increases tremendously? So:
>
> library(reshape2)
> library(zoo)
> z <- read.zoo(
Jennifer,
it looks like some of the columns that you are selecting don't exist.
What is the output of str(arc) before you run the code below?
Andrew
On Thu, Apr 28, 2011 at 04:49:41PM -0700, Jennifer Wessel wrote:
> This is part of my program. I am getting an error, that I cannot figure
> ou
Hi Derek,
I infer from the output that you're using a Mac. Generally including
that kind of information is useful.
Your computer lacks the requisite software to install the package.
Make is saying that it can't find 'gfortran'.
See this page for some more insight:
http://r.research.att.com/
On Apr 28, 2011, at 5:40 PM, Lisa wrote:
Dear all,
Does everyone know how to change the line numbers of scan()? For
example,
scan(n = 3)
1: 4
2: 6
3: 9
Read 3 items
[1] 4 6 9
I just want to change the line numbers 1, 2, and 3 to, say a, b, and
c that
look like this:
scan(n = 3)
On Apr 28, 2011, at 6:15 PM, derekverley wrote:
Hi all,
I'm trying to load the quantreg package but keep running into
problems no
matter which method I have tried. Does anybody know what this error
(below)
means in plain language and what I might do to get this installed.
I have
not h
I think that you probably need to provide the x values to nls. Try,
for example,
fit <- nls(species ~ a *(1 - exp(-b*samples)),start = list(a = 27, b = .15))
I hope that this helps,
Andrew
On Thu, Apr 28, 2011 at 01:56:43PM -0700, BornSurvivor wrote:
> I have the following data set, and I hav
Hi,
Try
> d <- data.frame(samples, species)
> fit = nls(species ~ a *(1 - exp(-b*samples)), start = list(a = 27, b =
.15), data = d)
> summary(fit)
Formula: species ~ a * (1 - exp(-b * samples))
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 35.824723.02073 11.860 6.10e-10 ***
b 0.0
This is part of my program. I am getting an error, that I cannot figure out,
any help would very much appreciated, thanks.
# subset variables
arc <- arc[,c("SNAP", "code", "ncode", "var", "n_total")]
Error in `[.data.frame`(arc, , c("SNAP", "code", "ncode", :
undefined col
This test SnowsPenultimateNormalityTest() is great :)
Best
--
View this message in context:
http://r.789695.n4.nabble.com/Kolmogorov-Smirnov-test-tp3479506p3482401.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org ma
Thank you all very kindly for your help.
-Rob
Robert Schutt III, MD, MCS
Resident - Department of Internal Medicine
University of Virginia, Charlottesville, Virginia
--
View this message in context:
http://r.789695.n4.nabble.com/fisher-exact-for-2x2-table-tp
Hi,
thanks for response.
>> The Kolmogorov-Smirnov test is designed for distributions on continuous
>> variable, not discrete like the >> poisson. That is why you are getting
>> some of your warnings.
I read in "Fitting distributions whith R" Vito Ricci page 19 that: "...
Kolmogorov-Smirnov t
There is both a technical and a theoretical element to my question...
Should I be able to use the outputs which arise from the fastbw function
as inputs to nomogram(). I seem to be failing at this, -- I obtain a
subscript out of range error.
That I can't do this may speak to technical failing
¿How to run several commands with one script?
Functions needed: SOURCE (script) + logical functions(if, while, for) +
scan(parameter A)
Im not going to put an example, but the thing is:
1) run the main script, which contains another SOURCE-funtion wich points to
an other script. source(sour
Dear all,
Does everyone know how to change the line numbers of scan()? For example,
> scan(n = 3)
1: 4
2: 6
3: 9
Read 3 items
[1] 4 6 9
I just want to change the line numbers 1, 2, and 3 to, say a, b, and c that
look like this:
> scan(n = 3)
a: 4
b: 6
c: 9
Read 3 items
[1] 4 6 9
Any he
Hi, Rolf,
Thank you for your help. I am trying to use readline() and I have a question
about the maximum length of prompt string. The R help on readline() says:
“The prompt string will be truncated to a maximum allowed length, normally
256 chars (but can be changed in the source code).”
I don’t
Thanks, but it did not really improve the speed. Why is it that when I change
the layout of the matrix (which does not give the required results), the
speed increases tremendously? So:
library(reshape2)
library(zoo)
z <- read.zoo(DF, split = 3, index = 2, FUN = identity) # Split on 3 and
index on
I have the following data set, and I have to find the line of best fit using
this equation,
y = a*(1 - exp(-b*x)).
samples = seq(1,20,by=1)
species = c(5,8,9,9,11,11,12,15,17,19,20,20,21,23,23,25,25,27,27,27)
plot(samples,species, main = "Accumulation Curve for Tree Species Richness",
xlab = "Sam
Hi all,
I'm trying to load the quantreg package but keep running into problems no
matter which method I have tried. Does anybody know what this error (below)
means in plain language and what I might do to get this installed. I have
not had problems downloading/installing/running packages in the
Good point. It would be absurdly inefficient if the upper and lower
limits on the interval of interest were, say, 0.2 and 0.201 instead of 0.2
and 0.8. Here's what I think is probably the best general approach:
Compute the CDF for the upper and lower limits of the interval and
generate unifo
Hi,
I realized that when I have irregular series to feed into lineChart,
the interval of each point in the chart does not seem to take care of
irregular time interval I specified in my input xts time series. But
rather, lineChart seems to take each point as equal spaced time
series. For example, I
Hi Arnaud,
the error is telling you that you don't have the "make" command. This
might be because you haven't installed the necessary software to
compile R packages. I suggest that you check the FAQ for Macintosh to
see how to do that.
Best wishes
Andrew
On Thu, Apr 28, 2011 at 01:30:58PM +02
Oh silly me--and I've been staring at that for a good hour. Thank you and
I'll keep your advice in mind.
On Thu, Apr 28, 2011 at 6:24 PM, Andrew Robinson <
a.robin...@ms.unimelb.edu.au> wrote:
> A couple of points here
>
> First, note that q doesn't increment in the code below. So, you're
>
Hi all,
I'm trying to fit models for data with three levels of nested random
effects: site/transect/plot. For example,
modelincrBS<-glmer(l.ru.ba.incr~shigo.av+pre.f.crwn.length+bark.thick.bh+Date+slope.pos.num+dens.T+dbh+leaf.area+can.pos.num+(1|site/transect/plot),
data=rws30.UL, family=gaussian
That method (creating lots of samples and throwing most of them away) is
usually frowned upon :-).
Try this: (I haven't, so it may well have syntax errors)
% n28<- dnorm(seq(.2,.8,by=.001),mean=1,sd=1)
% x <- sample(seq(.2,.8,by=.001), size=500,replace=TRUE, prob=n28)
And I guess in retrospe
A couple of points here
First, note that q doesn't increment in the code below. So, you're
getting the same variance each time.
Second, note that (t$Rec1==input3 & t$Rec2==input4) evaluates to F?T
or 0/1, and it's not clear from your code if that is what you intend.
Finally, it's much easi
I'm trying to find the variance of various outputs in a matrix:
for(l in 2:vl){
for(o in 1:(l-1)){
# Make sure the inputs are for the matrix "m"
input3=rownames(v)[o]
input4=colnames(v)[l]
r=t[(t$Rec1==input3 & t$Rec2==input4),output]
if(length(r)==0){
r=t[(t$Rec1==i
On Thu, 28 Apr 2011, viostorm wrote:
I have read the help page, or at least ?fisher.exact
I looked a bit on the Internet I guess it is applicable to > 2x2. I had
spoken to a biostatistician here who is quite excellent and was adamant
with me I could not do > 2x2.
I found this:
http://math
Kristian,
your example is rather complex and somewhat time consuming even on fast
computers, so it is not easy to track your problem down to its original
reason.
However, what is clear to me is that this is definitely *not* not a
problem of deSolve and most likely even not of optim, it is j
Hans,
You could parallelize it with the multicore package. The only other thing I
can think of is to use calls to .Internal(). But be vigilant, as this might
not be good advice. ?.Internal warns that only true R wizards should even
consider using the function. First, an example with .Intern
On Fri, 29 Apr 2011, Thomas Lumley wrote:
On Fri, Apr 29, 2011 at 8:01 AM, Mike Miller wrote:
On Thu, 28 Apr 2011, viostorm wrote:
I'm using fisher.exact on a 4x2 table and it seems to work.
Does anyone know exactly what is going on? I thought fisher.exact is
only for 2x2 tables.
You
Thanks Eik. As you said both packages give the same result, except labelling of
the x axis.
--- On Thu, 28/4/11, Eik Vettorazzi wrote:
From: Eik Vettorazzi
Subject: Re: [R] ROCR for combination of markers
To: "Rasanga Ruwanthi"
Cc: "R Help"
Date: Thursday, 28 April, 2011, 20:08
Hi Rasanga
Thanks, that worked.
*Ben Caldwell*
On Thu, Apr 28, 2011 at 2:02 PM, David Winsemius wrote:
>
> On Apr 28, 2011, at 3:53 PM, Benjamin Caldwell wrote:
>
> Hi folks,
>>
>> I'm dealing with a dataset with a lot of NAs, and want to use subset on
>> the
>> data without removing the NAs from the th
On Apr 28, 2011, at 3:53 PM, Benjamin Caldwell wrote:
Hi folks,
I'm dealing with a dataset with a lot of NAs, and want to use subset
on the
data without removing the NAs from the the dataframe; e.g.
rws50 <- subset(rw.fire.RW,shigo.av<50)
This removes instances with NA for the value in ad
On Apr 28, 2011, at 4:23 PM, viostorm wrote:
I have read the help page, or at least ?fisher.exact
Then it should have been clear that more than 2x2 tables can be used.
I looked a bit on the Internet I guess it is applicable to > 2x2. I
had
spoken to a biostatistician here who is quite
I have read the help page, or at least ?fisher.exact
I looked a bit on the Internet I guess it is applicable to > 2x2. I had
spoken to a biostatistician here who is quite excellent and was adamant with
me I could not do > 2x2.
I found this:
http://mathworld.wolfram.com/FishersExactTest.html
D
A couple of things to consider:
The Kolmogorov-Smirnov test is designed for distributions on continuous
variable, not discrete like the poisson. That is why you are getting some of
your warnings.
With a sample size over 10,000 you will have power to detect differences that
are not practically
Great! I got it. Thanks a bunch.
On Thu, Apr 28, 2011 at 4:02 PM, Duncan Murdoch wrote:
> On 28/04/2011 3:49 PM, Dat Mai wrote:
>
>> I currently have this code:
>>
>> for(j in 2:n){
>> for(i in 1:(j-1)){
>>
>> # Make sure the inputs are for the matrix "m"
>> input1=rownames(m)[i]
>>
On Apr 28, 2011, at 3:21 PM, Jannis wrote:
> On 04/28/2011 09:53 PM, Benjamin Caldwell wrote:
>> rws50<- subset(rw.fire.RW,shigo.av<50)
>
> quick and dirty would be to replace all NAs with -9 (or similar), use
> subset, and set all values ==-9 in the subset back to NA. There may be
> mo
You could always put the commands into a script or function, then instead of
retyping everything just run the script or function.
There is the zoomplot function in the TeachingDemos package which does what you
are suggesting (but my quick test did not work with your map). You could use
that as
On 04/28/2011 09:53 PM, Benjamin Caldwell wrote:
rws50<- subset(rw.fire.RW,shigo.av<50)
quick and dirty would be to replace all NAs with -9 (or similar),
use subset, and set all values ==-9 in the subset back to NA. There
may be more elegant solutions, though.
Jannis
_
On Fri, Apr 29, 2011 at 3:23 AM, ONKELINX, Thierry
wrote:
> Dear all,
>
> I'm working on a design with rotating panels. Each site is sampled every X
> year. Have a look at the code below the get an idea of the design. In reality
> the number of sites will be (much) higher.
>
> However I'm not su
On Fri, Apr 29, 2011 at 8:01 AM, Mike Miller wrote:
> On Thu, 28 Apr 2011, viostorm wrote:
>
>> I'm using fisher.exact on a 4x2 table and it seems to work.
>>
>> Does anyone know exactly what is going on? I thought fisher.exact is only
>> for 2x2 tables.
>
>
> You were wrong. I'm sure there's no
Hi
On 28/04/2011 3:00 p.m., Dario Strbenac wrote:
Hello,
I'm trying to follow the documentation of how to use gridBase, and
I've reached the minimal code example below as my best effort. Can
someone explain how to keep the column of boxplots on the same page
as the rectangles (even though I've
On 28/04/2011 3:49 PM, Dat Mai wrote:
I currently have this code:
for(j in 2:n){
for(i in 1:(j-1)){
# Make sure the inputs are for the matrix "m"
input1=rownames(m)[i]
input2=colnames(m)[j]
q=t[(t$Rec1==input1& t$Rec2==input2),output]
if(length(q)==0){
q=t[
On Apr 28, 2011, at 3:45 PM, viostorm wrote:
I'm using fisher.exact on a 4x2 table and it seems to work.
Does anyone know exactly what is going on? I thought fisher.exact
is only
for 2x2 tables.
Have you read the help page?
--
David Winsemius, MD
West Hartford, CT
___
On Thu, 28 Apr 2011, viostorm wrote:
I'm using fisher.exact on a 4x2 table and it seems to work.
Does anyone know exactly what is going on? I thought fisher.exact is
only for 2x2 tables.
You were wrong. I'm sure there's nothing wrong with the program. You
will find that with bigger tabl
Hi folks,
I'm dealing with a dataset with a lot of NAs, and want to use subset on the
data without removing the NAs from the the dataframe; e.g.
rws50 <- subset(rw.fire.RW,shigo.av<50)
This removes instances with NA for the value in addition to anything <50.
Any suggestions much appreciated.
*B
I currently have this code:
for(j in 2:n){
for(i in 1:(j-1)){
# Make sure the inputs are for the matrix "m"
input1=rownames(m)[i]
input2=colnames(m)[j]
q=t[(t$Rec1==input1 & t$Rec2==input2),output]
if(length(q)==0){
q=t[(t$Rec1==input2 & t$Rec2==input1),output]
}
I'm using fisher.exact on a 4x2 table and it seems to work.
Does anyone know exactly what is going on? I thought fisher.exact is only
for 2x2 tables.
Note: I can't use chi-squared because I have a couple of cells with 0 and <
5 observations.
--
View this message in context:
http://r.789695
On Apr 28, 2011, at 3:38 PM, David Winsemius wrote:
On Apr 28, 2011, at 3:13 PM, Abraham Mathew wrote:
I'm using the subset() function in R.
dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13))
subset(dat, Number >= 10)
However, I want to find the number of all rows who meet the
On Apr 28, 2011, at 3:13 PM, Abraham Mathew wrote:
I'm using the subset() function in R.
dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13))
subset(dat, Number >= 10)
However, I want to find the number of all rows who meet the Number>=10
condition.
I've done this in the past with s
Try this:
sum(dat$Number >= 10)
On Thu, Apr 28, 2011 at 3:13 PM, Abraham Mathew wrote:
> I'm using the subset() function in R.
>
> dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13))
>
> subset(dat, Number >= 10)
>
> However, I want to find the number of all rows who meet the Number>=10
On Wed, Apr 27, 2011 at 4:46 AM, Stat Consult wrote:
> Dear ALL
>
> I want to load "HTSanalyzeR", It 's necessary to load "igraph" package.
> This time I see this error:
> library(igraph)
>
> library(HTSanalyzeR)
> Loading required package: GSEABase
> Loading required package: Biobase
> Error: pa
I tried logspline function using a lower bound 0 for my data, it works
like a charm. When the I changed the xlim only positive part, the
vertical line was also gone. That's exactly what I expected.
Thanks.
-JJ
Greg Snow wrote:
You might want to use the logspline package instead of the density
I'm using the subset() function in R.
dat <- data.frame(one=c(6,7,8,9,10), Number=c(5,15,13,1,13))
subset(dat, Number >= 10)
However, I want to find the number of all rows who meet the Number>=10
condition.
I've done this in the past with something like colSums or rowSums or another
similar fun
Yup, thats the culprit. Thx.
//M
On 28. apr. 2011, at 18.36, Achim Zeileis wrote:
> On Thu, 28 Apr 2011, moleps wrote:
>
>> sessionInfo yields the following:
>
> OK, the "Design" package causes the problem here. When you load the Design
> package, it provides a new Surv() and related methods
Hello R users
When I want to zoom in on a map or any plot in R, I usually use the locator
function to reset the
limit of the plot and do something like:
library(maptools)
library(mapdata)
map('worldHires',ylim=c(40,55),xlim=c(-90,-60))
points(-71,47,col="red",pch=16)
text(-73,49,"Hello")
Thanks a lot for all your replies.
This may be a bad question. But, for me I was improved by asking this
question.
Thanks,
Jian-Feng,
2011/4/28 David Winsemius
>
> On Apr 28, 2011, at 12:09 PM, Ravi Varadhan wrote:
>
> Surely you must be joking, Mr. Jianfeng.
>>
>>
> Perhaps not joking and
Hi Rasanga,
I think there is nothing wrong with that plot (you may compare both
curves obtained by ROC from Epi and plot.roc from pROC).
As usual, both functions plot (1-specificity) at the x-coordinate
against sensitivity on y. Perhaps a bit confusing is the labelling of
the latter function, it ha
On 28/04/2011 11:00 AM, cpoirier wrote:
Dear Ranjan,
I'm afraid the command "options(help_type = 'text')" will not work if R is
closed and opened again as the 'help_type' argument is set to 'html' by
default in Windows (same for me).
That's a default, but you can change it when you install R i
Actually, it is plotting the points in the decreasing order. See:
plot(xx)
Are you looking to reverse the x axis? Perhaps you will find an answer
in either ?par or ?axis.
On Thu, Apr 28, 2011 at 2:09 PM, Bogaso Christofer
wrote:
> Hi all, please consider this plot:
>
>
>
> xx <- seq(4, 0.01, by
Try this:
plot(xx, yy, type="l", xlim=rev(range(xx)))
___
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
B
Hi all, please consider this plot:
xx <- seq(4, 0.01, by = -0.04)
yy <- rnorm(xx)
plot(xx, yy, type="l")
Here you see my original 'xx' was in decreasing order, however R puts it in
the increasing order. I understand that in any plot x and y axis grow is
increasing order, however I am wond
Hallo everybody,
I'm wondering whether it might be possible to speed up the following code:
Error<-rnorm(1000, mean=0, sd=0.05)
estimate<-(log(1.1)-Error)
DCF_korrigiert<-(1/(exp(1/(exp(0.5*(-estimate)^2/(0.05^2))*sqrt(2*pi/(0.05^2))*(1-pnorm(0,((-estimate)/(0.05^2)),sqrt(1/(0.05^2))-1)
On Apr 28, 2011, at 12:09 PM, Ravi Varadhan wrote:
Surely you must be joking, Mr. Jianfeng.
Perhaps not joking and perhaps not with correct statistical
specification.
A truncated Normal could be simulated with:
set.seed(567)
x <- rnorm(n=5, m=1, sd=1)
xtrunc <- x[x>=0.2 & x <=0.8]
r
I've kind of solved the issue.
dat$Time <- paste(dat[,2], dat[,1], sep=" ")
head(dat)
dat[,c(1,2)] <- NA
head(dat)
> head(dat)
Year Month Number Time
1 NA NA 0 Jan 2002
2 NA NA 0 Feb 2002
3 NA NA 0 March 2002
4 NA NA 1 April 2002
5 NA NA 0 May 2002
6 NA NA 0 June 2002
unfortunately, can I delete
Dear Ranjan,
I'm afraid the command "options(help_type = 'text')" will not work if R is
closed and opened again as the 'help_type' argument is set to 'html' by
default in Windows (same for me).
You should better modify the 'C:/Program Files/R/R-X.XX.X/etc/Rprofile.site'
file with a text editor, a
Hello,
I am trying to create a matrix that generates an output from 2 inputs. The
problem I have is that I'm assigning a large set of data to various
categories (that are used for the matrix). This technically means that I
will have multiple output values for each unique set of inputs--I want to
a
Hi folks, I have a simple question that I just can't solve.
I'm trying to merge two columns in my data frame.
> sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: i686-pc-linux-gnu (32-bit)
> head(dat)
Year Month Number
2002 Jan 0
2002 Feb 0
2002 March0
2002 April
On Thu, 28 Apr 2011, moleps wrote:
sessionInfo yields the following:
OK, the "Design" package causes the problem here. When you load the Design
package, it provides a new Surv() and related methods. This clashes with
the computations of ctree() based on Surv(). So it's better not to load
bo
On Apr 28, 2011, at 12:30 PM, петрович wrote:
Hello again.
Following a great idea from Jim and Dennis, I began to use the
function
ave() in my big data.
I proved to use this command in my data with FUN=max(x,na.rm=T),
Try instead:
look at the 3rd example of help(ave)
and I
received a
sessionInfo yields the following:
> sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats4tcltk splines grid stats graphics grDevices
utils datasets metho
Hello again.
Following a great idea from Jim and Dennis, I began to use the function
ave() in my big data.
I proved to use this command in my data with FUN=max(x,na.rm=T), and I
received a Warning that translated looks like
Warning in max (x, na.rm = T):
no arguments to max; returning -Inf
Is th
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