How to determine restricted variable in SVAR and SVEC? There are some values
which set to be zero and others set to be NA.. How to determine values that set
to be 0? Thanks
Regards,
Arif
_
Hi Doug,
you can add the fitted curve using the following general paradigm:
## Plotting the data
plot(p~z)
## Defining grid of z values
## (100 values ensures a smooth curve in your case)
zValues - seq(min(z), max(z), length.out = 100)
## Adding predicted values corresponding to the grid
Hi all,
I have a problem with CPU usage while running the Rgui.exe
problem is
while I am running scripts on Rgui its taking 100% of CPU is there any
posibility to reduce the cpu consumption
are any package I can use to reduce CPU consumption
can any one help me out from this problem because
In article 5df5755d-a376-439a-b7f2-9901441db...@comcast.net,
dwinsem...@comcast.net says...
In insurance situation there is typically a cap on the covered losses
and there is also typically an amount below which it would not make
sense to offer a policy. So a minimum and a maximum are
Dear all,
Here I come with another stupid question. Suppose I want to use nls()
to fit a series of data (here modelled by generated points), then plot
the points and the fitting curve. I figured out some way of doing it:
x - runif(1:20, 0, 10)
y - 0.1*x^2 - rep(3, length(x)) + rnorm(length(x),
On 09/10/2009 12:42 AM, guangchuang yu wrote:
I used *lapply* and *for* to call the function *GetC* respectively. The
output is different!!! I can't figured out how this happen! It's so wierd.
x=lapply(ipi.go,GetC, ont = BP, org=cell.cycle)
unlist(x)
*IPI00011654 IPI00013683
cell cycle cell
On 10/09/2009 07:15 AM, David Winsemius wrote:
On Oct 8, 2009, at 3:28 PM, Danielle Dandreaux wrote:
...
The syntax line that appears to be causing problems is:
dataname=Igt2model.txt and the error message is: Error: unexpected
input in dataname=
Hi Danielle,
This sort of error often
On 10/09/2009 07:17 AM, Michael Yutzi wrote:
I have a heavy DATA saved in dbf format.
What I want is to bring that data to R with SQL statements. Like: I want
columns 1, 4, 5 and only when column 4 30.
Sorry asking it here instead of keep searching in manuals, but it seems that
there are too
Hi,
If you work on Windows you can reduce the priority of the Rgui.exe
process. You can do this in the task manager, right click Priority.
cheers,
Paul
venkata kirankumar wrote:
Hi all,
I have a problem with CPU usage while running the Rgui.exe
problem is
while I am running scripts on
Hi all,
In R, is there some functions or ways to create a Clustered-Stacked
Column Chart as the example in the following page
http://peltiertech.com/Excel/ChartsHowTo/ClusterStack.html?
I have browsed the R Graph Gallery (http://addictedtor.free.fr/graphiques/)
and searched the R site, and
Dear all,
I am using abline(lm ...) to insert a linear trendline through a portion of my
data (e.g. dataset[,36:45]). However, I am finding that whilst the trendline is
correctly displayed and representative of the data portion I've chosen, the
line continues to run beyond this data segment
I want to select a subset of an array, but I want to make a function so that
it can handle any number of dimensions.
This is probably best described with an example
x - 1:100
dim(x) - c(10,10)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]1 11 21 31 41 51 61
Hello David, seems there's sth wrong with my R.
I used to start R with Emacs+ESS, so i didn't find the error message till
today i ran R in an terminal :
si...@simon-t400:~$ R
R version 2.9.2 (2009-08-24)
Copyright (C) 2009 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
R is free
Hello joris, when i type the command:
?installed.packages
seems it works fine and shows me the man page of install.packages()
Then i followed ur instructions by first removing DAAG
remove.packages(DAAG)
Warning in remove.packages(DAAG) :
argument 'lib' is missing: using
Dear list,
I am trying to set up a propensity-weighted regression using the
survey package. Most of my population is sampled with a sampling
probability of one (that is, I have the full population). However, for
a subset of the data I have only a 50% sample of the full population.
In previous
Hello,
I am trying to run a for-loop in which I want to add glm output objects
sequentially into a list. I do not know how to give names to each object in
a list. I tried this :
List.glm - list()
for(i in 1:n)
list.glm - list(list.glm, glm(,,,))
but it is obviously unsuited…
Best
A couple of ways:
List.glm - list()
for(i in 1:n) list.glm[[i]] -glm(,,,)
OR
List.glm - lapply(1:n, function(x) glm(,,,))
On Fri, Oct 9, 2009 at 7:54 AM, P.Branco pjlbra...@yahoo.com wrote:
Hello,
I am trying to run a for-loop in which I want to add glm output objects
sequentially into
Dear All,
In R, is there a way (or a function) I can quickly check whether all the NA
values in one new created numerical variable happened are because of nbsp;
or something else in the original dataset? And how can we easily group these
NAs separately based on different reason (e.g. some NA
Dear list
I want to plot the QQ plot with some distributions like geometrical , lognormal
and truncated normal with confidence bands. does this options available.
Iam new to R. If you have any scripts and examples please kindly suggest me.
thank you
madan
[[alternative
-begin included message ---
With rpart we can get several terminals and draw it in the TREE plot.
Now I am trying to draw a plot like this: x-axis is each terminal's
value, and y-axis is those observe values. Does anyone has idea what
gramma should I use? Thanks in advance.
Try this:
x - array(1:1000, rep(10, 3))
vList - list(i = 4:6, j = 4:6, ... = 4:6)
do.call('[', c(list(x), vList))
On Fri, Oct 9, 2009 at 7:18 AM, Mark McDowall
markmcdowa...@googlemail.com wrote:
I want to select a subset of an array, but I want to make a function so that
it can handle any
Hi,
I am using line function to plot the line. And I would like to understand
Tukeyline algorithm. Since, the line function is calling the Tukeyline
algorithm(which is compiled code) using foreign function interface, I am not
able to look into the source code of this algorithm. Can somebody
On Oct 9, 2009, at 5:50 AM, Steve Murray wrote:
Dear all,
I am using abline(lm ...) to insert a linear trendline through a
portion of my data (e.g. dataset[,36:45]). However, I am finding
that whilst the trendline is correctly displayed and representative
of the data portion I've
Hello,
Today, Brian Ripley commited the revision 5 of R's svn repository.
I took this as an opportunity to do some data analysis of the log and
posted some code and graphics on my blog:
http://romainfrancois.blog.free.fr/index.php?post/2009/10/09/celebrating-R-commit-5
The plots of
If you want curve with substitute(...), you can try something about like this:
f - eval(parse(text = paste(substitute(, formula(yfit)[3], ,
as.list(coef(yfit))), sep = )))
curve(f, ...)
2009/10/9 Henrique Dallazuanna www...@gmail.com:
Try with predict:
plot(x, y)
lines(0:10, predict(yfit,
Dear Madan,
Please see the qq.plot() function in the car package.
I hope this helps,
John
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Madan Sigdel
Sent: October-09-09 8:54 AM
To: r-help@r-project.org
Subject: [R] QQ
Hi Steve,
However, I am finding that ... the trendline ... continues to run beyond
this data segment
and continues until it intersects the vertical axes at each side of the
plot.
Your best option is probably Prof. Fox's reg.line function in package car.
##
library(car)
?reg.line
reg.line
I don't think I've seen an R version, probably because the technique is not
very good for displaying data.
Have a look at http://chartsgraphs.wordpress.com/tag/r-and-excel/ for an
alternative method of displaying the data using lattice.
--- On Fri, 10/9/09, zhijie zhang rusers...@gmail.com
cb247343-29ec-495a-ba8e-d5a466bab...@comcast.net
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
Thanks for flagging up the 'segments' command. However=2C I'm having troubl=
e getting it to work - this is probably due to me
In further offline discussion with the poster it was explained that
the purpose of this is to display the icon object and not to return an
instantiatedProtoMethod which was what the code was attempting to do.
That is in proto, g$icon is not the icon object. It is the icon method
instantiated with
Hi Rob.
Without the file content_1.xml or any information
from the R call stack (e.g. options(error = recover)
and then run the command and dynamically explore the
state of affairs when the error occurs), there is
no way for us to know what the problem might be.
Somehow, the XML parser appears
The following code isn't working and we can't figure out why..
letters = c(A,B,C,D,E,F,G,H,I,J)
numbers = 1:3
for(i in 1:6){ #6 letters
for (j in 1:3) { #3 numbers
for (k in -1:1) { #answer -1,right or
Hi all,
I'm from Brazil.
I fit a Tobit model to FLUID MILK CONSUMPTION (DEPENDENT VARIABLE)
data using survreg (attached).
I am confused about the output interpretation and I would like yours
explanations.
Thanks,
Marcio Roberto Silva __
I have one column
x
97
94
91
90
NA
NA
NA
NA
I tried
book$r-ifelse(book$x!=NA,book$x+20,10)
I expect to get the result as follows
107
104
91
90
10
10
10
10
But what i was getting is empty column of variable r.How to solve this
--
View this message in context:
I tried this:
(c is the column vector with indices of those rows I want to replace)
table[c,]-replace(table[c,],c,newRows)
but it does not work and the error is:
new columns would leave holes after existing columns
Can anyone help please?
Thanks
--
View this message in context:
Dear colleagues,
I'm trying (and failing) to write the script required to generate a
chart that would help me assess the forecasting accuracy of a logistic
regression model by plotting the cumulative proportion of observed
events occurring in cases across the range of possible predicted
Hello R users,
I am writing a summary() for a custom class, and am to display the integers
right justified,
Say where x is the vector with integers, I am using the following:
cat(\t,format(x),\t...other columns)
this way I am trying to pass the format(x), to the cat function to display
it,
but
What you want is:
book$r-ifelse(is.na(book$r), 10,book$x+20)
On Fri, Oct 9, 2009 at 5:46 AM, premmad mtechp...@gmail.com wrote:
I have one column
x
97
94
91
90
NA
NA
NA
NA
I tried
book$r-ifelse(book$x!=NA,book$x+20,10)
I expect to get the result as follows
107
104
91
90
10
10
Because you missed your closing parens in the text function call. The closing
parens is closing the paste function, you need one more. I don't know what
you're trying to do here, but I am guessing there's a faster way?
-Original Message-
From: r-help-boun...@r-project.org
1) No need to post multiple times to the list
2) use the is.na function to test if a value is missing, not == or !=
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of premmad
Sent: Friday, October 09, 2009 5:05 AM
To:
try using 'sprintf'
On Fri, Oct 9, 2009 at 8:36 AM, sahil seth sahiils...@gmail.com wrote:
Hello R users,
I am writing a summary() for a custom class, and am to display the integers
right justified,
Say where x is the vector with integers, I am using the following:
A parenthesis is missing. You can figure that out easily watching the first
error.
text(0.5,0.5, text = paste(letters[i], +, numbers[j],=, letters [i+j+k]))
Xavier
- Mail Original -
De: Antje## ann3bu...@hotmail.com
À: r-help@r-project.org
Envoyé: Vendredi 9 Octobre 2009 17h36:34 GMT
The following code isn't working and we can't figure out why..
letters = c(A,B,C,D,E,F,G,H,I,J)
numbers = 1:3
for(i in 1:6){ #6 letters
for (j in 1:3) { #3 numbers
for (k in -1:1) { #answer -1,right
Hi. With your help, I've fixed the errors in my for loops.
But now, the for loop isn't working correctly. There should be a plot,
but there's no plot when I run the for loop..
This is the code:
letters = c(A,B,C,D,E,F,G,H,I,J)
numbers = 1:3
for(i in 1:6){ #6
Hello,
I need to put 2 or more different time series to one plot for comparison
each other. The problem is that the time series are irregular, moreover
the time lenghts and periods are not the same. Is there a way to manage
it in R?
Many thanks for any hint and advice in advance
Tomas
On Fri, Oct 9, 2009 at 4:54 PM, Anne Buunk ann3bu...@hotmail.com wrote:
text(0.5,0.5, text = paste(letters[i], +,
numbers[j],=, letters [i+j+k])
Missing ) on the end there. You have one ( for text( and one for
paste( but only one ).
Use an editor that matches
It plots on my system just fine. You might want to check what
directory (getwd()) that you are plotting in.
On Fri, Oct 9, 2009 at 12:17 PM, Anne Buunk ann3bu...@hotmail.com wrote:
Hi. With your help, I've fixed the errors in my for loops.
But now, the for loop isn't working correctly. There
Hi,
I have a strange behavior of plot function when trying to plot a
discriminant analysis obtained by fda.
I'd attached the data for reproduction. I made the same analysis (linear
discriminant analysis) but using lda and fda (default args, using
polyreg). The resulting coefficients and trace
Google is your friend! (search on Tukey median line fit); e.g.
http://www.amstat.org/publications/jse/v14n2/morrell.html
Bert Gunter
Genentech Noclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of shanmukha patel
Hi all,
Thank you for your help. Now I am able to select every 5th row of the data
from the main data set (x)
using
sub1- x[seq(1, nrow(x), by=5), ]
So sub1 contains one fith of the data set X. I want also create another
data set that will contain the remaining data set from X (ie., four
Thanks David,
for answering this question.
On Oct 8, 2009, at 4:33 AM, msig...@yahoo.com wrote:
Dear all
when I try to install distr, the following error appears, I am using
R in windows. can u suggest me?
I'm not a Windows user, but the obvious questions would be How?. What
commands
It probably means that your data is not in the right format. PLEASE do
read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
If you show the data, it might help. It sounds like something is
empty, but it is hard to
sub3 - x[-seq(1, nrow(x), by=5), ]
Notice the '-' in front of the seq() command. This will select
everything but what is in the sequence.
From: Ashta [mailto:sewa...@gmail.com]
Sent: Friday, October 09, 2009 12:42 PM
To: Nutter, Benjamin
Cc:
You are emitting weird hex-stuff from your mail client. And from what
I can tell that is not a reproducible example even if hex-ation
problem gets fixed. There is a worked example in segments.
On Oct 9, 2009, at 10:45 AM, Steve Murray wrote:
On Oct 9, 2009, at 12:41 PM, Ashta wrote:
Hi all,
Thank you for your help. Now I am able to select every 5th row of
the data
from the main data set (x)
using
sub1- x[seq(1, nrow(x), by=5), ]
So sub1 contains one fith of the data set X. I want also create
another
data set that will
plot.zoo and xyplot.zoo in the zoo package can both do that:
library(zoo)
z - zoo(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40),
as.Date(c(1992-01-10, 1992-01-17, 1992-01-24, 1992-01-31,
1992-02-07, 1992-02-14, 1992-02-21, 1992-02-28, 1992-03-06,
1992-03-13)))
# z and lag(z) have
b2a99330-1d1d-49cc-b287-81f37c16b...@comcast.net
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
Thanks Mark=2C the reg.line trick seemed to work really well.
David - hopefully the hex-text will have gone now - if not=2C please
On Oct 9, 2009, at 6:42 AM, Zhen Lin wrote:
I tried this:
(c is the column vector with indices of those rows I want to replace)
table[c,]-replace(table[c,],c,newRows)
but it does not work and the error is:
new columns would leave holes after existing columns
replace() is supposed to work
Thanks to Dirk for pointing it out in my blog : thomas and tlumley
are the same person. Also martyn and plummer, and paul and
murrell ... updated number of commits per author below
sort( table( simple$author ), decreasing=T )
ripleymaechler hornik pd murdoch
Uwe Ligges wrote:
Uwe Ligges wrote:
Kim Vanselow wrote:
Dear R-Users and Developers,
I want to calculate something like k-means clustering, but with
ordinal data (Braun-Blanquet) to combine this classification
technique with a NMDS-Ordination.
I found an algorithm especially developed
Dear List,
Is there a color palette avaliable similar to what is used in satellite ocean
color imagery? I.e. a gradient with blue on one end and red on the other, with
yellow in the middle? I have tried topo.colors(n) but that comes out more
yellow on the end. I am looking for something
See ?colorRampPallete
On Fri, Oct 9, 2009 at 3:51 PM, Tim Clark mudiver1...@yahoo.com wrote:
Dear List,
Is there a color palette avaliable similar to what is used in satellite ocean
color imagery? I.e. a gradient with blue on one end and red on the other,
with yellow in the middle? I
On Fri, Oct 9, 2009 at 7:51 PM, Tim Clark mudiver1...@yahoo.com wrote:
Dear List,
Is there a color palette avaliable similar to what is used in satellite ocean
color imagery? I.e. a gradient with blue on one end and red on the other,
with yellow in the middle? I have tried topo.colors(n)
On Fri, 2009-10-09 at 11:51 -0700, Tim Clark wrote:
Dear List,
Is there a color palette avaliable similar to what is used in satellite ocean
color imagery? I.e. a gradient with blue on one end and red on the other,
with yellow in the middle? I have tried topo.colors(n) but that comes out
Dear list,
I would like to start some R workshops at King's College London, and
to do so, I would like to use the Use R! logo at
http://www.agrocampus-ouest.fr/math/useR-2009//useR%21%202008_fichiers/useR-middle.png
Since it seems to be difficult to get a shell account at KCL, I also went
ahead
Thanks! The colorRampPalette() did just what I need.
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Fri, 10/9/09, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote:
From: Barry Rowlingson b.rowling...@lancaster.ac.uk
Subject: Re: [R] Satellite ocean color palette?
To:
I am attempting to graph 12 months of temperatures, delineate the months with a
vline and place the names of the months at the top of the graph.
So far I have gotten everything to work except the names, despite getting a
similar graph to work yesterday the day before yesterday with Baptise A's
Chicagoland R Users:
We are pleased to announce a Fall meetup for Chicagoland R users. This
is open to anyone with an interest in R: practioners, researchers,
casual users and other interested parties.
WHEN: October 29, 2009 @5:30
WHERE: Jak's Tap www.jakstap.com
A short series of so-called
Hi
You may find that you need to use the compilation tools. I gave up on
using R from Mandriva repositories as the R base is usually out of
date and the selection of packages available depends on the interests
of those who produced them. If I knew how rpm spec files worked I
might do something
Date: Fri, 09 Oct 2009 18:18:05 +0200
From: Tomas Lanczos lanc...@fns.uniba.sk
Sender: r-help-boun...@r-project.org
Precedence: list
Hello,
I need to put 2 or more different time series to one plot for comparison
each other. The problem is that the time series are irregular, moreover
Hi,
file.info is producing data.frame with ctime variable. Help file says
that on Unix this is 'last status change' and on Windows 'creation
time'. Is there a way to get 'last status change' on Windows using
some R function?
Thanks,
Johannes
__
Hi,
How to plot the same types of graphics on the same R graphic device? Suppose
that we want to plot a vector y against x (using plot for instance). How is it
possible to plot y against x for different values of these two vectors on the
same device so that the plots could be compared?
Cheers,
Hi All
I am running a linear regression using the lm object.
In the event that my independent variable is the same across all
observations the regression slope is returned as an NA.
For example, if I have the following
y=c(10,12,17)
x=c(5,5,5)
lm = lm(y~x)
produces the following
That comes out as an NA because X'X is not invertible because it is not full
rank (one row/column is a linear combination of the other(s)). And that
means there is no unique solution to the system.
y=c(10,12,17)
x=c(5,5,5)
X=cbind(1,x)
X
t(X)%*%X
solve(t(X)%*%X)
Therefore, nope, there is now
On 09-Oct-09 21:12:18, Brecknock, Peter wrote:
Hi All
I am running a linear regression using the lm object.
In the event that my independent variable is the same across all
observations the regression slope is returned as an NA.
For example, if I have the following
y=c(10,12,17)
Daniel
Thanks very much for the reply.
If the data fails the underlying assumptions of regression wouldn't it make
sense to suppress all the output and not just the slope coefficient?
Incidently, if I run this simple example in Excel it returns the slope as 0.
Intuitively, this makes sense
On 09-Oct-09 21:45:04, Brecknock, Peter wrote:
Daniel
Thanks very much for the reply.
If the data fails the underlying assumptions of regression wouldn't it
make sense to suppress all the output and not just the slope
coefficient?
Incidently, if I run this simple example in Excel it
?lines
?points
Something like :
x - 1:10
y1 - rnorm(10, 4, 3)
y2 - rnorm(5,5,2)
ymax - max(c(y1,y2))
ymin - min(c(y1,y2))
plot(x, y1, col='red',ylim = c(ymin,ymax))
lines(y2, col='green')
--- On Fri, 10/9/09, carol white wht_...@yahoo.com wrote:
From: carol white wht_...@yahoo.com
On Oct 9, 2009, at 5:45 PM, Brecknock, Peter wrote:
Daniel
Thanks very much for the reply.
If the data fails the underlying assumptions of regression wouldn't
it make sense to suppress all the output and not just the slope
coefficient?
Incidently, if I run this simple example in Excel
Fair point.
Thanks.
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Friday, October 09, 2009 5:02 PM
To: Brecknock, Peter
Cc: Daniel Malter; r-help@r-project.org
Subject: Re: [R] lm output
On Oct 9, 2009, at 5:45 PM, Brecknock, Peter wrote:
Daniel
Hi Carol,
It isn't at all clear exactly what you are trying to do, but you might want to
read the help for either points() and lines() [to put more than one data pair
on a single plot], or for par, specifically mfrow and mfcol, or for layout [to
put more than one plot on a single device window].
On 09/10/2009 5:05 PM, johannes rara wrote:
Hi,
file.info is producing data.frame with ctime variable. Help file says
that on Unix this is 'last status change' and on Windows 'creation
time'. Is there a way to get 'last status change' on Windows using
some R function?
What does last status
I'm just learning R (I don't know any other programming languages),
and I have a question. I am trying to figure out how to ask for user
input (say, a set of 3 numbers) then put those numbers into an array.
I've looked around, but I haven't been able to find any answers that I
Dear R Helpers,
I have a pretty large dataframe (150,000 variables, 10,000 entries for
each) and have to run a regression on each of the variables. Recorded are
the pvals.
I wrote a function and use sapply. The function looks something like this:
calcpval-function(x){
modela -
On Oct 9, 2009, at 6:36 PM, Sharon Beckett wrote:
I'm just learning R (I don't know any other programming languages),
and I have a question. I am trying to figure out how to ask for user
input (say, a set of 3 numbers) then put those numbers into an
array.
I've looked around, but I
Just so we complete (partially) the web-record...
Some further hunting determined that the apos was an apostrophe -
obvious only in retrospect! Removal of this character has resolved the
Entity errors, but not the xmlParseStartTag: invalid element name errors.
I had assumed that the two
tim.colors() in library fields
On Fri, Oct 9, 2009 at 2:51 PM, Tim Clark mudiver1...@yahoo.com wrote:
Dear List,
Is there a color palette avaliable similar to what is used in satellite ocean
color imagery? I.e. a gradient with blue on one end and red on the other,
with yellow in the
Thanks a lot. Maybe someone else has the method to solve that.
2009/10/9 John Kane jrkrid...@yahoo.ca
I don't think I've seen an R version, probably because the technique is not
very good for displaying data.
Have a look at http://chartsgraphs.wordpress.com/tag/r-and-excel/ for an
-- Forwarded message --
From: Khanh Nguyen kngu...@cs.umb.edu
Date: Fri, Oct 9, 2009 at 10:10 PM
Subject: Re: [R] Creating a Clustered-Stacked Column Chart
To: zhijie zhang rusers...@gmail.com
May be you can try to look into ggplot2
http://had.co.nz/ggplot2/position_fill.html
library(lattice)
barchart(Titanic, scales = list(x = free),
auto.key = list(title = Survived))
Or if you prefer vertical:
barchart(Titanic, scales = list(x = free),
auto.key = list(title = Survived), horizontal=FALSE)
There are adjustments available to the space between
On 10/09/2009 11:36 PM, sahil seth wrote:
Hello R users,
I am writing a summary() for a custom class, and am to display the integers
right justified,
Say where x is the vector with integers, I am using the following:
cat(\t,format(x),\t...other columns)
this way I am trying to pass the
No, actually it does better to not suppress all output, because it tells you
where the trouble comes from by just showing the NA for the slope. The
intercept the regression gives you is the mean of y in this case. As for the
slope, Ted's graphic is illustrative as to why no slope can be estimated.
It appears several that of my scripts are beginning to reaching maturity, so I
am curious if it is possible to add an external GUI to run the scripts from
this simplified GUI interface.
The scripts are fairly rudimentary so the GUI only needs a few radial buttons
and a could of numeric
On Fri, 9 Oct 2009, Duncan Murdoch wrote:
On 09/10/2009 5:05 PM, johannes rara wrote:
Hi,
file.info is producing data.frame with ctime variable. Help file says
that on Unix this is 'last status change' and on Windows 'creation
time'.
No, that is not what it says. It actually says
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