Dear List-Members,
since 3 weeks I have been heavily working on reproducing the results of an
economic paper. The method there uses the numerical solution of an integral
within nonlinear least squares. Within the integrand there is also some
parameter to estimate. Is that in the end possible to
Michael, it should be possible to get nls and integrate to work together.
Bt, there are several problems you need to consider first. The most
important is the definition of e.g. your f-function and how integrate() work.
The easiest way to show you the problem is like in the following.
The
Hello,
I wanted to create a histogram, but somehow I got stuck...
The interval limits are: x = 1, 2, 3, 3.5, 4.5, 5, 5.5
The interval widths are therefore: 1, 1, 0.5, 1, 0.5, 0.5
Nothing I tried worked... Can anyone help me please?
Thanks
Tobias
--
View this message in context:
Hi,
I created an ecdf and a boxplot. Now I would like to place the ecdf above
the boxplot.
But I only managed to align them horizontally. I used this code:
#---
par(mfrow=c(1,2), mar=c(5,3,3,1))
# ecdf
library(plotrix)
x =
Hi, All,
This is my program
ts1.sim - arima.sim(list(order = c(1,1,0), ar = c(0.7)), n = 200)
ts2.sim - arima.sim(list(order = c(1,1,0), ar = c(0.5)), n = 200)
tdata-ts(c(ts1.sim[-1],ts2.sim[-1]))
tre-c(rep(0,200),rep(1,200))
gender-rbinom(400,1,.5)
x-matrix(0,2,400)
x[1,]-tre
x[2,]-gender
fit
squall44 wrote:
Hi,
I created an ecdf and a boxplot. Now I would like to place the ecdf above
the boxplot.
But I only managed to align them horizontally. I used this code:
#---
par(mfrow=c(1,2), mar=c(5,3,3,1))
As in the message before: Please
squall44 wrote:
Hello,
I wanted to create a histogram, but somehow I got stuck...
The interval limits are: x = 1, 2, 3, 3.5, 4.5, 5, 5.5
The interval widths are therefore: 1, 1, 0.5, 1, 0.5, 0.5
Please read the help page more carefully! See ?hist and its argument
breaks.
Uwe Ligges
[EMAIL PROTECTED] wrote:
hej
i'm plotting time-series and label the x-axis as follows:
r - as.POSIXct(round(range(p1$time), month))
to define the time range for labeling the xaxis
plot(p1$time,p1$ p1, type=l, xaxt=n)
plots p1 against time
axis.POSIXct(1, at=seq(r[1], r[2],
Well, that was the first thing I tried. But the help only gives you the
commands and does not explain how to use it (I am a newbe). How do I use the
argument 'breaks'?
I tried:
#---
x = c(1, 2, 3, 3.5, 4.5, 5, 5.5)
breaks=c(1, 1, 0.5, 1, 0.5, 0.5)
hist(x,
breaks= breaks,
xlim=c(0,7),
Thanks for you answer Uwe,
I have the code par(mfrow=c(1,2), mar=c(5,3,3,1)) not from the help, but
from an example from the R Graph Gallery. I think the help is rather useless
for beginners (like me), because it does not explain anything and only gives
the commands.
Anyway, I always try first
squall44 wrote:
Well, that was the first thing I tried. But the help only gives you the
commands and does not explain how to use it (I am a newbe). How do I use the
argument 'breaks'?
I tried:
#---
x = c(1, 2, 3, 3.5, 4.5, 5, 5.5)
breaks=c(1, 1, 0.5, 1, 0.5, 0.5)
hist(x,
breaks=
This is described on the help page!
include.mean: Should the ARIMA model include a mean term? The default
is 'TRUE' for undifferenced series, 'FALSE' for differenced
ones (where a mean would not affect the fit nor predictions).
Further, if 'include.mean' is true, this
Hello,
Imagine a data frame like so:
Intensity0 Intensity1
1 767432.1 451743.4
2 3998988.0 4642145.0
3 818974.6 552315.8
and a vector like so:
[1] 1 2 1
How can I get R to produce a vector that contains the value in one column or
the other depending on the vector? The result
Please could anyone help me?
How can I fit a linear model where an intercept has no sense?
Thanks in advance..
Michael
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
First, I should admit that I didn't do a lot of searching beforehand, I'm
just cutting to the chase to ask the experts:
I'm currently running MySQL 5 queries with PHP 5.2.3 and returning results
to the end-user in web-page tables of ca. 50 rows by 5 columns. I'd like to
(more or less)
Hello --
Is there any R function that implements Wrapper Subset Selection method? I am
using 'leaps' package, and I do not know if 'leaps' works better than Wrapper.
Thanks,
Vu
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R-help@stat.math.ethz.ch mailing list
I'm having a Thursday morning mental block, any suggestions on the following
would be most appreciated...
I have (as an example)
surgery = c(d48, d67, dnc37, a75, d10, a78, d31,
d55, d1)
before each number part the possibilities are c(a, d, dnc), I'm trying
to split each element in
Thanks!
Joh
On Thursday 23 August 2007 12:01:50 you wrote:
x[cbind(1:nrow(x), the.vector)]
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
Johannes Graumann wrote:
Hello,
Imagine a data frame like
At 10:56 23/08/2007, Michal Kneifl wrote:
Please could anyone help me?
How can I fit a linear model where an intercept has no sense?
Well the line has to have an intercept somewhere I suppose.
If you use the site search facility and look for known intercept
you will get some clues.
Thanks in
A number of alternatives, such as:
lm(y ~ 0 + x)
lm(y ~ x -1)
See ?formula
On 8/23/07, Michal Kneifl [EMAIL PROTECTED] wrote:
Please could anyone help me?
How can I fit a linear model where an intercept has no sense?
Thanks in advance..
Michael
one way is the following:
data.frame(status = gsub([0-9], , surgery),
time = gsub([a-z], , surgery))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven,
?lm
Details
A formula has an implied intercept term. To remove
this use either y ~ x - 1 or y ~ 0 + x. See formula
for more details of allowed formulae.
Is this what you want?
--- Michal Kneifl [EMAIL PROTECTED] wrote:
Please could anyone help me?
How can I fit a linear model where
On 23-Aug-07 09:56:33, Michal Kneifl wrote:
Please could anyone help me?
How can I fit a linear model where an intercept has no sense?
Thanks in advance..
Michael
Presumably you mean where a non-zero intercept has no sense?
Suppose the model you want to fit is like
Response ~ B+C+X+Y
The FAQ Section 7 is a very useful place for new users
to find out any number of R idiosycracies. However
there is no numbering on the FAQ Table of Content or
on the Sections Tables of Contents.
An R-help list reply of Read FAQ 7.10 in response to
a question about converting a factor to numeric
This applies the indicated perl-style regular expression where the
first backreference (\\D+) is the non-digits and the second
backreference (\\d+) is the digits.
The two backreferences, but not the entire matched pattern itself,
are passed as arguments x and y to the function whose body is the
Note that googling
R FAQ 7.10
will get it on the first hit.
On 8/23/07, John Kane [EMAIL PROTECTED] wrote:
The FAQ Section 7 is a very useful place for new users
to find out any number of R idiosycracies. However
there is no numbering on the FAQ Table of Content or
on the Sections Tables
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Note that googling
R FAQ 7.10
will get it on the first hit.
True, however this is not exactly a self-contained
solution if one is actually trying to use the FAQ. I
may be old fashioned but when directed to the FAQ I
tend to go to the FAQ
try
5*which(tf)[cumsum(tf)]
Gladwin, Philip schrieb:
Hello,
What is the best way of solving this problem?
answer - ifelse(tf=TRUE, i * 5, previous answer)
where as an initial condition
tf[1] - TRUE
For example if,
tf - c(T,F,F,F,T,T,F)
over i = 1 to 7
then the output of the
Dear All,
I would like to know if it is possible to obtain the
optimal asset allocation with the fPortfolio library (or
others),
but setting at the beginning a desired level of Target Risk.
For example I can obtain the optimal asset allocation with
fPortfolio library or portfolio.optim()
DeaR
I am trying to use a dynamically create formula with nlsList and nlme, but I
cannot get the environment of the string-generated formal to work similarly
to the manually entered one.
Any idea?
Dieter
#-
library(nlme)
# Pinheiro/Bates p 280
fm1Indom.lis =
Hello,
I have two questions. I'd like to visualize data with a heatmap and I have the
following testcase:
x - rnorm(256)
nx - x + abs(min(x))
nnx - 255/max(nx) * nx
x - matrix(nnx, 16, 16)
rownames(x) -
c(A,B,C,D,E,F,G,I,H,J,K,L,M,N,O,P)
par(fin=c(8.0,8.0))
cp -
i'm trying to convert a zoo object of the following specifications:
str(z)
atomic [1:15642] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, index)='POSIXct', format: chr [1:15642] 2004-09-01 02:00:00
2004-09-01 03:00:00 2004-09-01 04:00:00 2004-09-01 05:00:00 ...
- attr(*, frequency)= num 0.000278
to
MASFERFC Team masferfc at gmail.com writes:
I'm currently running MySQL 5 queries with PHP 5.2.3 and returning results
to the end-user in web-page tables of ca. 50 rows by 5 columns. I'd like to
(more or less) simultaneously return to the browser a couple of canned
charts and graphs based on
On Thu, 23 Aug 2007 [EMAIL PROTECTED] wrote:
i'm trying to convert a zoo object of the following specifications:
str(z)
atomic [1:15642] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, index)='POSIXct', format: chr [1:15642] 2004-09-01 02:00:00
2004-09-01 03:00:00 2004-09-01 04:00:00 2004-09-01
for unweighted fits using `nls' I compute confidence intervals for the
fitted model function by using:
#---
se.fit - sqrt(apply(rr$m$gradient(), 1, function(x) sum(vcov(rr)*outer(x,x
luconf - yfit + outer(se.fit, qnorm(c(probex, 1 - probex)))
#---
where `rr'
Michael,
Assuming you want a model with an intercept of zero, I think we need to ask you
why you want an intercept of zero. When a normal regression indicates a
non-zero intercet, forcing the regression line to have a zero intercept changes
the meaning of the regression coefficients. If for
You might try rbind.fill in the reshape package.
Hadley
On 8/22/07, Kirsten Beyer [EMAIL PROTECTED] wrote:
Hello. I am looking for a function that will allow me to paste rows
together without regard for the numbers of columns in the datasets to
be joined. The only columns where it matters
Hi
I have used the hetcor function on dataset which consists of about 40
variables. Since the output is quite large entire output is not visible on
the R screen. Is there a way, I can retrieve the output that is not
visible?
Thank you
Regards
Upasna
--
On 8/22/07, Greg Tarpinian [EMAIL PROTECTED] wrote:
R2.3, WinXP
Dear all,
I am using the following functions:
f1 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(x))/exp(log(Phi4)))
f2 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(r)-log(x))/exp(log(Phi4)))
subject to the residual weighting
Can anyone help me solve this problem...thanks!
Consider a data frame, namely v, as such:
v
X1 X2 X3 X4 X5 X1 X2 X3 X4 X5
x1 1 2 -1 -1 -1 1 2 -1 -1 -1
y1 1 2 -1 -1 -1 1 2 3 -1 -1
What I would like to do is to create an array or data frame with only the
elements that appear in the
Thanks for the inputs, but it's not exactly what I want...
My problem is, that my timeseries (precipitation) contain lots of NA-values
and when I apply the code:
z- data.frame(a=p1$time.date, precip=p1$m1)
#gives a dataframe with precipitation and time (2004-01-01 02:12:00)
z-
On Thu, 23 Aug 2007, Marc Gilgen wrote:
Thanks for the inputs, but it's not exactly what I want...
You did not provide sufficient information. In particular, you still do
not provide a small reproducible example.
My problem is, that my timeseries (precipitation) contain lots of NA-values
On Thu, 23 Aug 2007, John Kane wrote:
The FAQ Section 7 is a very useful place for new users
to find out any number of R idiosycracies. However
there is no numbering on the FAQ Table of Content or
on the Sections Tables of Contents.
Hmm, doc/FAQ does have a numbered table of contents and
Try:
lapply(as.data.frame(t(DF)), function(x) unique(x[duplicated(x) x 0]))
On 8/23/07, dxc13 [EMAIL PROTECTED] wrote:
Can anyone help me solve this problem...thanks!
Consider a data frame, namely v, as such:
v
X1 X2 X3 X4 X5 X1 X2 X3 X4 X5
x1 1 2 -1 -1 -1 1 2 -1 -1 -1
y1 1
On 8/23/2007 11:28 AM, Prof Brian Ripley wrote:
On Thu, 23 Aug 2007, John Kane wrote:
The FAQ Section 7 is a very useful place for new users
to find out any number of R idiosycracies. However
there is no numbering on the FAQ Table of Content or
on the Sections Tables of Contents.
Hmm,
Deb,
Others have answered your question, I just want to give you something
else to think about.
Stacked bar plots may not be the best tool to use for this situation,
they are ok for comparing the bottom and top groups, but all the other
groups don't line up in a stacked bar plot making those
Hi R-Help:
I want to produce a map of the US with different colors for selected
states.
I installed the map package, and then used:
library(maps)
I can see that by using
map( 'state' )
you get the state boundaries, also.
How do I fill in different colors for the different states? I see
Does the following code do something like what you want (but with the
built in data):
plot(as.integer(state.region), state.x77[,'Murder'],
+ xaxt='n', xlim=c(0,5), col=as.integer(state.region))
axis(1, at=1:4, labels=levels(state.region))
oo - order(order( as.integer(state.region),
Hi,
I am using this sm.survival() to plot smoothed survival curve against a
single continuous covariate. I did a logarithmic transformation on the
covariate. Now in the plot, I want the tick mark value to be transformed
back to original values. For instance, my current x axis tick mark values
John == John Kane [EMAIL PROTECTED] writes:
Apologies for the poor quality of the screen capture.
I think the first one is a screen cap of
http://cran.r-project.org/doc/FAQ/R-FAQ.html. Is that correct?
The faq that is part of the r-doc-html package from Debian also
has the same bulleted
Thank you Duncan. That is exactly what I meant. The
next level (for example, after clicking on R
Miscellanea) should also have numbers.
Whether or not it's possible in html as implimented on
the R website is another matter of course.
--- Duncan Murdoch [EMAIL PROTECTED] wrote:
On 8/23/2007
Hello I need a tiny peace of help. (PPC Mac Os X 10.4.10; R 2.5.1)
I have a data.frame with numeric and factor variables.
I would like to convert same of the factors to ordered factors.
I tried with:
attach (table)
as.ordered (V3)
but this gives me only the V3 Vector as ordred back but in the
Yes that is it. Thanks
--- Michael A. Miller [EMAIL PROTECTED] wrote:
John == John Kane [EMAIL PROTECTED]
writes:
Apologies for the poor quality of the screen
capture.
I think the first one is a screen cap of
http://cran.r-project.org/doc/FAQ/R-FAQ.html. Is
that correct?
The
Hello everyone,
I will like to know if there is an R package to compute exact confidence
intervals for the ratio of two binomial proportions.
Tony.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
Thanks, John. That works... The following is a trivial but complete
application that uses glayout to
position the widgets, then lets you edit a parameter and execute a
function that uses the parameter:
##
### glayout example
wtesta = gwindow(Sample
Hi,
I am not sure if this is a bug and I apologize if it is something I
didn't read carefully in the R extension manual. My initial search on the
R help and R devel list archive didn't find useful information.
I am using .Call (as written in the R extension manual) for the C code
and have
Dear Rs:
Hi, I am trying to read a big text file (nrows=243440, ncols=144). It
seems the computational time of all the read methods
(scan,readtable,read.delim) is not linear to the number of rows I
want to read in: things became really slow once I tried to read in
10 lines compare to
Hi Folks,
Does anyone know if (apparently not on CRAN) there is
any archive of older versions of R packages?
Or is it the case that not only are old versions on
CRAN humanesly destoyed, but all older versions out
in the wild are hunted down and shot?
(I can find older versions of the R core,
As shown below, the build process fails with only vague messages,
leaving me clueless as to how to resolve.
Thanks, in advance, for any help that you may offer.
Mike
--
# ./configure --prefix=/SOURCES/R-2.5.1 --with-iconv=no
...
...
...
R is now configured
The attach function only attachs a copy of the data (changes to the data don't
show up in the attached copy). Also you need to tell R what to do with the
result of as.ordered (where to save it).
Try something like:
table$V3 - as.ordered(table$V3)
Or
table - transform(table,
#Hi R-users,
#I have an example DF like this:
y1 - rnorm(10) + 6.8
y2 - rnorm(10) + (1:10*1.7 + 1)
y3 - rnorm(10) + (1:10*6.7 + 3.7)
y - c(y1,y2,y3)
x - rep(1:3,10)
f - gl(2,15, labels=paste(lev, 1:2, sep=))
g - seq(as.Date(2000/1/1), by=day, length=30)
DF - data.frame(x=x,y=y, f=f, g=g)
DF$wdays
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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.
On Thu, 23 Aug 2007, Yupu Liang wrote:
Dear Rs:
Hi, I am trying to read a big text file (nrows=243440, ncols=144). It
seems the computational time of all the read methods
(scan,readtable,read.delim) is not linear to the number of rows I
want to read in: things became really slow once I
Another option is to read it into a database and from there into R.
RSQLite has the capability of reading certain text files directly into
an SQLite database without going through R and from there one
can read it into R. You can use RSQLite to do that. Alternately this
post describes how the
On Thu, 23 Aug 2007, [EMAIL PROTECTED] wrote:
Hi Folks,
Does anyone know if (apparently not on CRAN) there is
any archive of older versions of R packages?
Yes. On CRAN. At the bottom of the page listing all the packages there is
a section
-
Related Directories
Archive
On Thu, 23 Aug 2007, [EMAIL PROTECTED] wrote:
Hello everyone,
I will like to know if there is an R package to compute exact confidence
intervals for the ratio of two binomial proportions.
Probably not. This is a tricky business as the references below make
clear.
To wit, small sample
Hi,
Maybe this is more a programming questions than a specific R-project question,
but maybe there is someone who can point me in the right direction.
I have a picture of cirkels which I took with a digital camera.
Now I want to use the diameter of the cirkels on the picture for analysis in R.
map('state', col=terrain.colors(50), fill=T)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 23/08/07, Smith, Phil (CDC/CCID/NCIRD) [EMAIL PROTECTED] wrote:
Hi R-Help:
I want to produce a map of the US with different colors for selected
states.
I installed the
On 23 August 2007 at 19:37, (Ted Harding) wrote:
| Does anyone know if (apparently not on CRAN) there is
| any archive of older versions of R packages?
See
$CRAN/src/contrib/Archive/
which has further subdirectories A to Z each of which contain all known
versions of all packages
On 8/23/07, Bart Joosen [EMAIL PROTECTED] wrote:
Maybe this is more a programming questions than a specific R-project
question, but maybe there is someone who can point me in the right direction.
I have a picture of cirkels which I took with a digital camera.
Now I want to use the diameter
Hi,
Maybe this is more a programming questions than a specific R-project question,
but maybe there is someone who can point me in the right direction.
I have a picture of cirkels which I took with a digital camera.
Now I want to use the diameter of the cirkels on the picture for analysis in R.
Dear R community,
I am trying to code a fairly complex equation for optim(). My current
approach is too slow for optim().
I have a function that takes a double integral (hopefully correctly) across
two terms, e.g.,
doubleint - function(c1,c2) {integrate(function(y) {
sapply(y, function(y)
On 8/23/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 8/23/2007 11:28 AM, Prof Brian Ripley wrote:
On Thu, 23 Aug 2007, John Kane wrote:
The FAQ Section 7 is a very useful place for new users
to find out any number of R idiosycracies. However
there is no numbering on the FAQ Table of
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Thu, 23 Aug 2007, John Kane wrote:
The FAQ Section 7 is a very useful place for new
users
to find out any number of R idiosycracies.
However
there is no numbering on the FAQ Table of Content
or
on the Sections Tables of Contents.
Hi Bart,
If you only have 36 circles, the fastest way would be to use some image
processing software and measure the circles by hand. One option is to
use ImageJ, which you can download here
http://rsb.info.nih.gov/ij/
Julian
Bart Joosen wrote:
Hi,
Maybe this is more a programming
On Thu, 23 Aug 2007, Scott Stark wrote:
Dear R community,
I am trying to code a fairly complex equation for optim(). My current
approach is too slow for optim().
I have a function that takes a double integral (hopefully correctly) across
two terms, e.g.,
doubleint - function(c1,c2)
Pardon my ignorance, but is there a difference in cor.test between
exact=FALSE and exact=NULL when method=spearman?
Take for example:
x-c(1,2,2,3,4,5)
y-c(1,2,2,10,11,12)
cor.test(x,y, method=spearman, exact=NULL)
This gives an error message,
Warning message: Cannot compute exact p-values with
Please note clarifications in below. My apologies for any confusion.
Thanks again,
Scott
-- Forwarded message --
From: Scott Stark [EMAIL PROTECTED]
Date: Aug 23, 2007 1:03 PM
Subject: Expedite scalar f(x) evaluation over vectors
To: [EMAIL PROTECTED]
Dear R community,
I am
Hi Bart,
I have never used image processing software in R (I
was doing this with Matlab), but here is what I would
have done algorithmically:
1) convert the picture to gray-scale
2) find a threshold value which separates the circles
from the background and convert your image to black
and white
3)
This won't work since it produces a matrix (try this).
What should work is
x[(1:nrow(x)) + nrow(x)*(v-1)]
--- Johannes Graumann [EMAIL PROTECTED]
wrote:
Thanks!
Joh
On Thursday 23 August 2007 12:01:50 you wrote:
x[cbind(1:nrow(x), the.vector)]
Patrick Burns
[EMAIL PROTECTED]
On Thu, Aug 23, 2007 at 03:44:36PM -0700, Scott Stark wrote:
Please note clarifications in below. My apologies for any confusion.
Thanks again,
Scott
-- Forwarded message --
From: Scott Stark [EMAIL PROTECTED]
Date: Aug 23, 2007 1:03 PM
Subject: Expedite scalar f(x)
Correct, I didn't notice that the coma was inside the
cbind(). Sorry...
--- Rolf Turner [EMAIL PROTECTED] wrote:
On 24/08/2007, at 12:51 PM, Moshe Olshansky wrote:
This won't work since it produces a matrix (try
this).
On the contrary, Patrick's solution is correct. I
tried
On 24/08/2007, at 12:51 PM, Moshe Olshansky wrote:
This won't work since it produces a matrix (try this).
On the contrary, Patrick's solution is correct. I tried it. It
works just fine.
cheers,
Rolf Turner
What should work
Hi Bart,
One more comment:
You do not really need the morphological closing to
close the holes inside the circles. Another
possibility is to reverse the black-and-withe picture,
i.e. make the holes and background be 1 and the
circles 0, label the connected components and then
only the component
Hi,
I am getting stuck with kalman filter a little bit. If anyone could provide
some help, I would really appreciated. Basically, I have some of the
observations that are irregularly spaced on the time axis. Says, each data of
mine is supposed to be of 1 millisec interval. However, in some
Hello,
I am sure I am not the only person with this problem.
I have a list with n elements, each consisting of a single column matrix
with different row lengths. Each row has a name ranging from A to E. Here
is an example:
alph[[1]]
A 1
B 2
C 3
D 4
alph[[2]]
A 1
C 3
D 4
alph[[3]]
A 1
D 4
E 5
Here are two solutions. The first repeatedly uses merge and the
second creates a zoo object from each alph component whose time
index consists of the row labels and uses zoo's multiway merge to
merge them.
# test data
m - matrix(1:5, 5, dimnames = list(LETTERS[1:5], NULL))
alph -
Hi Gabor,
Thank you. The native solution works just fine, though there is an
interesting side effect, namely, that with very large lists the rows of
the output become scrambled though the corresponding columns are correctly
sorted. The zoo package solution does not work on large lists: there is
I'm wondering if R has functions to return pvalues with given x-squares and
df. I googled for it but couldn't seem to find one. Appreciate any helps
here.
An example: df=4, x- c(33.69, 32.96, 30.90) which are the statistic for
chi-square, I'd like to get the corresponding pvalues for each values
What is 'R 2-5.2.1'? AFAIK there is no such version.
I can tell you the most likely issue: is your Perl is pre 5.6.1 (very old
indeed)?
The current R-patched (2.5.1 patched) requires Perl 5.6.1, and we do
suggest that you install that rather than 2.5.1.
(Interestingly, all versions of R
On 8/24/07, Christopher Marcum [EMAIL PROTECTED] wrote:
Hi Gabor,
Thank you. The native solution works just fine, though there is an
interesting side effect, namely, that with very large lists the rows of
the output become scrambled though the corresponding columns are correctly
sorted. The
At 12:41 AM 8/24/2007, Karen wrote:
I'm wondering if R has functions to return pvalues with given x-squares and
df. I googled for it but couldn't seem to find one. Appreciate any helps
here.
An example: df=4, x- c(33.69, 32.96, 30.90) which are the statistic for
chi-square, I'd like to get the
Hi Gabor,
My apologies. Both solutions work just fine on large lists (n=1000,
n[[i]]=500). A memory problem on my machine caused the error and
fail-to-sort. Thank you!
PS - The zoo method is slightly faster.
Best,
Chris
Gabor Grothendieck wrote:
On 8/24/07, Christopher Marcum [EMAIL
OK. One other thought. The R merge command has a sort= argument
that you can try out. See ?merge
On 8/24/07, Christopher Marcum [EMAIL PROTECTED] wrote:
Hi Gabor,
My apologies. Both solutions work just fine on large lists (n=1000,
n[[i]]=500). A memory problem on my machine caused the
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