read.table() looks at the first five rows when determining how many columns
there are. If there are more columns in row 7 and you do not specify that in
the read.table() command directly, they will be wrapped to the next row.
This was discussed on the list within the last couple weeks.
Sarah
On
Daniel Kaschek daniel.kaschek at physik.uni-freiburg.de writes:
Hi Jia,
in order to test if the failing parameter estimation really has to do
with wrong initial values, you could first simulate data with your mixed
effects model. Just assume parameters, produce a data set with these
Yes, I did send an attachment - but I forgot that attachments might be
removed, no?
Here it is:
http://rapidshare.com/files/452815636/drmData.RData
Antje
On 16 March 2011 12:19, Mike Marchywka marchy...@hotmail.com wrote:
Date: Wed, 16 Mar
-Original Message-
From: Graham Williams [mailto:graham.willi...@togaware.com]
Sent: Saturday, March 12, 2011 7:19 PM
To: Xiaobo Gu
Cc: r-help@r-project.org
Subject: Re: No response after click the show Rules button on Tab
Associate.
On 10 March 2011 02:07, Xiaobo Gu
I have the following problem:
I have some string with numbers like k. I want to have a table like the
function table() gives. However I am not able to call the first row, the 1,
2, 3, 5, 6 or 9. I tried to do that by a data.frame, but that doesn't seem
to work either. The levels keep bothering
Dear reader,
I have the following problem:
I have some string with numbers like k. I want
to have a table like the function table() gives. However I am not able
to call the first row, the 1, 2, 3, 5, 6 or 9. I tried to do that by a
data.frame, but that doesn't seem to work either. The
Dear list,
I'm a quite new user of R-project, and I've a doubt on objects memory: I open a
new R session and the command memory.limits() gives me 1535 Mb of memory (the
PC has 2 Gb RAM and 32 bit), I create an integer vector object of 2e8 size, so
about 2e8*4 bytes (800Mb) of memory are
Hi,
I have a dataset with many observations some days while only one others. I
would like to calculate a mean value per day and then do regression analysis
on the means.
This is what I have:
YearDayTimeherring.density
200747 10.36 2.2
200747 11.50
Hi,
I must seek a favour regarding of R project,
can we make an application out of R. I mean a small application that
automatically runs and do the estimation automatically. Because the
things I do is that I copy codes from script to work book and then it
runs and gives the output. can it be
Hello Scott,
Thank you for the tips (I have posted the issue on the google group too), but
there's nothing on these or other websites that gives an answer on our question.
Following the traditional used rules of making a symbol plot, the used symbols
should have the same proportions, based on
Thanks, we're almost there. The 3rd statement needs to
satisfy fi_2[r,c]-fi_2[c,r] where rc.
On Tue, Mar 15, 2011 at 10:06 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
fi_2 - diag(1, i)
fi_2[lower.tri(fi_2)] - 1 - runif(sum(lower.tri(fi_2))) ^ .5
fi_2[upper.tri(fi_2)] -
On 3/15/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
Hi, all,
Does R have a const object concept like which is in C++ language? I
want to set some data frames as constant to avoid being modified
unintentionally. Thanks!
Although there is almost never a No in R,
I have the following problem:
I have some string with numbers like k. I want to have a table like the
function table() gives. However I am not able to call the first row, the 1,
2, 3, 5, 6 or 9. I tried to do that by a data.frame, but that doesn't seem
to work either. The levels keep bothering
I know this is a very elementary question... I could not find a solution
looking at old posts.
I am unable to access a variable outside the scope of a for loop, even when
the variable was defined before the loop:
haar - function() {
a = c(1.4560773, 2.3752412, 0.9798882, 3.0909252, 2.3986487,
This is my code:
mycols - rep(NULL, 430) ; mycols[c(1,3:5)] - rep(numeric, 4) ;
mycols[c(2)] - rep(character,1)
inp - read.table(myfile, skip=2, colClasses=mycols,fill=T)
head(inp)
Best,
Luis
On Wed, Mar 16, 2011 at 1:03 PM, David Winsemius dwinsem...@comcast.net wrote:
On Mar 16, 2011, at
Awesome,that worked! Thanks.
On Wed, Mar 16, 2011 at 6:46 AM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
mapply('/', l1, l2, SIMPLIFY = FALSE)
and
tapply(1:5, lapply(indxLi, as.numeric), sum)
On Tue, Mar 15, 2011 at 6:06 PM, Rohit Pandey rohitpandey...@gmail.com
wrote:
It isn't entirely clear to me what you want.
table() can function with many kinds of data, not just integers, so it
returns a vector with names.
For your case, with integer classes, you seem to possibly want:
x - table(k)
x - rbind(as.numeric(names(x)), as.numeric(x))
x
[,1] [,2] [,3]
No. First, please use path.expand(~) for this, and it does not
necessarily mean the home directory (and in principle it might not expand at
all). In practice I think it will always be *a* home directory, but on
Windows there may be more than one (and watch out for local/roaming profile
On Mar 16, 2011, at 8:13 AM, Sarah Goslee wrote:
read.table() looks at the first five rows when determining how many
columns
there are. If there are more columns in row 7 and you do not specify
that in
the read.table() command directly, they will be wrapped to the next
row.
This was
Hi,
Look at ?aggregate.
df - read.table(textConnection(Year Day Time herring.density
2007 47 10.36 2.2
2007 47 11.50 1.1
2007 47 14.24 1.4
2007 66 9.35 2.5), header=TRUE)
You could do this:
aggregate(herring.density~Year+Day, data=df, FUN=mean)
And for the second question:
No, defaults are evaluated in the evaluation frame of the function. That's
why you can use local variables in them, e.g. the way rgamma uses 1/rate as
a default for scale.
Oops, yes, I was getting confused with promises - non-missing
arguments are promises evaluated in the parent frame.
But
On Wed, Mar 16, 2011 at 9:07 AM, Luis Ridao luri...@gmail.com wrote:
This is my code:
mycols - rep(NULL, 430) ; mycols[c(1,3:5)] - rep(numeric, 4) ;
mycols[c(2)] - rep(character,1)
rep(NULL, 430) does not give you a vector of length 430; it gives you a NULL
vector, and at the end of this
What operating system are you working with? On windows, making it run by
double clicking on it from explorer is not going to work. You will
probably have to write a batch file that invokes Rterm or Rscript (see
documentation for which you want to use). So if your script file is
myscript.r, you
table() returns a named vector. You need the names and the values.
Or you could read my reply to one of the OTHER appearances of this
email in my inbox; I think I've seen three, though I only replied to one
and now this. It truly is unnecessary to send your query several times in
quick
Michael Friendly wrote:
This is just the flexibility of R. I've just discovered a new class of
geometries based on
pi - 2.3
?Constants
This is a must-fortune.
Dieter
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On Mon, Mar 14, 2011 at 9:41 AM, Strategische Analyse CSD Hasselt
csd...@fedpolhasselt.be wrote:
Hello,
we want to plot a proportional symbol map with ggplot. Symbols' area should
have the same proportions as the scaled variable.
Hereby an example we found on
?cat
cat prints text, and returns an invisible NULL. Also, it is general
practice to assign values using '-' even inside of functions, for reasons
detailed in ?-
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville
On Mar 16, 2011, at 9:07 AM, Jonathan P Daily wrote:
What operating system are you working with? On windows, making it
run by
double clicking on it from explorer is not going to work. You will
probably have to write a batch file that invokes Rterm or Rscript (see
documentation for which you
Hi,
Not sure what you are trying to do with the cat command, but cat
returns an invisible NULL (as described in the doc), so
a = cat(a,v)
just sets a to NULL
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Ravi Kulkarni
Dear all,
I have a dataframe which looks like this (dummy):
date-c(jan, feb, mar, apr, may, june, july,
aug,sep,oct,nov,dec)
col1-c(8.2,5.4,4.3,4.1,3.1,2.5,1.1,4.5,3.2,1.9,7.8,6.5)
col2-c(3.1,2.3,4.7,6.9,7.5,1.1,3.6,8.5,7.5,2.5,4.1,2.3)
dum-data.frame(cbind(date,col1,col2))
dum
date col1
require(reshape2)
melt(dum, id = 1)
On Wednesday, March 16, 2011 at 9:28 AM, pelt wrote:
Dear all,
I have a dataframe which looks like this (dummy):
date-c(jan, feb, mar, apr, may, june, july,
aug,sep,oct,nov,dec)
col1-c(8.2,5.4,4.3,4.1,3.1,2.5,1.1,4.5,3.2,1.9,7.8,6.5)
I think you meant:
a - c(a, v)
and not
a - cat(a, v)
On Wed, Mar 16, 2011 at 7:41 AM, Ravi Kulkarni ravi.k...@gmail.com wrote:
I know this is a very elementary question... I could not find a solution
looking at old posts.
I am unable to access a variable outside the scope of a for loop,
b
[1] 2 1 1 1 1 1
pie(b)
Error in pie(b) : 'x' values must be positive.
Can someone help me?
And sorry i am an beginner
--
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Dear Gabor:
I did not have time to look at this issue these few days. Now I saw your
investigation. I am very grateful that you quickly identify the root cause
of this. It seems that a little caution needs to be exercised when applying
sqldf to text file with large number of blanks (I have no
Dear R-help ,
Here I try to explain the problem setting
problem setting:
we have four variables
y: outcome
x:exposure
u:confounder
z:Instrumental Variable
##we have instrumental variable as distance between centers and patients
place, we have many
##centers for exp:three centers c1,c1,c3 and
Try this:
reshape(dum, direction = 'long', idvar = 'date', varying =
list(c('col1', 'col2')))
On Wed, Mar 16, 2011 at 11:28 AM, pelt p...@knmi.nl wrote:
Dear all,
I have a dataframe which looks like this (dummy):
date-c(jan, feb, mar, apr, may, june, july,
aug,sep,oct,nov,dec)
It's ok, i solved it.
It wasn't an integer,
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__
R-help@r-project.org mailing list
'b' appears to be characters. try
pie(as.integer(b))
On Wed, Mar 16, 2011 at 9:28 AM, rens picca...@hotmail.com wrote:
b
[1] 2 1 1 1 1 1
pie(b)
Error in pie(b) : 'x' values must be positive.
Can someone help me?
And sorry i am an beginner
--
View this message in context:
On 16-Mar-11 13:28:00, rens wrote:
b
[1] 2 1 1 1 1 1
pie(b)
Error in pie(b) : 'x' values must be positive.
Can someone help me?
And sorry i am an beginner
The problem (as indicated by the marks around each value)
is that the variable 'b' is a vector of *characters*, not of
numbers.
Hi,
b is a character vector.
Try as.numeric(b).
You probably need to read some documentation about R.
Ivan
Le 3/16/2011 14:28, rens a écrit :
b
[1] 2 1 1 1 1 1
pie(b)
Error in pie(b) : 'x' values must be positive.
Can someone help me?
And sorry i am an beginner
--
View this message in
Thanks Sarah.
Best,
Luis
On Wed, Mar 16, 2011 at 1:19 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
On Wed, Mar 16, 2011 at 9:07 AM, Luis Ridao luri...@gmail.com wrote:
This is my code:
mycols - rep(NULL, 430) ; mycols[c(1,3:5)] - rep(numeric, 4) ;
mycols[c(2)] - rep(character,1)
On 16/03/11 13:04, Michael Friendly wrote:
On 3/15/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
Hi, all,
Does R have a const object concept like which is in C++ language? I
want to set some data frames as constant to avoid being modified
unintentionally. Thanks!
Dear all,
I've installed R-2.12.2 on a server with Suse 10 linux OS. When I try to create
images I got this error:
png(filename=hola.png)
Error in X11(paste(png::, filename, sep = ), width, height, pointsize, :
unable to start device PNG
In addition: Warning message:
In
Hi everybody.
I have list like this:
l-list(data.frame(q1=c(1,2,check),q2=c(3,check,5)),
data.frame(q1=c(check,1),q2=c(4,5)))
names(l)-c(A,B)
rownames(l[[1]])-c(aa,bb,cc)
rownames(l[[2]])-c(aa,bb)
Every object has the same number of columns but different number of rows.
Does anyone know if it
k-c(1,1,1,1,1,3,5,6,2,1,3,9,2,3,1,1,1)
table(k)
k
1 2 3 5 6 9
9 2 3 1 1 1
names(table(k))
[1] 1 2 3 5 6 9
as.numeric(names(table(k)))
[1] 1 2 3 5 6 9
x - table(k)
as.numeric(names(table(k)))[4]
[1] 5
You can always save the (numeric version of the) names as an object, too.
Does that help?
R users,
I am trying to use a numeric vector to extract rows from a data frame. I want
to extract two rows
per unique WEEK (27 weeks) based on variable DOW_NUM by using pairs of vector
(s3) elements in the order
they appear. For example, vector initial elements 7 and 5 will be used to
Use dput:
dput(l, file = l_str.txt)
Then, to read again:
l - dget(file = 'l_str.txt')
On Wed, Mar 16, 2011 at 11:55 AM, andrija djurovic djandr...@gmail.com wrote:
Hi everybody.
I have list like this:
l-list(data.frame(q1=c(1,2,check),q2=c(3,check,5)),
On Wed, Mar 16, 2011 at 10:54 AM, Allan Engelhardt all...@cybaea.com wrote:
On 16/03/11 13:04, Michael Friendly wrote:
On 3/15/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
Hi, all,
Does R have a const object concept like which is in C++ language? I
want to
Hallo,
I modified a code given by Andrija, a contributor in the list to achieve two
objectives:
create 1000 samples from a list of 207 samples with each of the samples
cointaining 20 good and 20 bad. THis i have achievedcalcuate AUC each of the
1000 samples, this i get an error.
Please see
Soryy, I didn't explain well what I want. I would like to have a table in
csv on txt file like this:
$A
q1q2
aa 1 3
bb 2 check
cc check 5
$B
q1 q2
aa check 4
bb 1 5
The same as write.csv of any data frame.
On Wed, Mar 16, 2011 at 4:03 PM, Henrique
nqueralt at clinic.ub.es writes:
I've installed R-2.12.2 on a server with Suse 10 linux OS.
When I try to create images I got this error:
png(filename=hola.png)
Error in X11(paste(png::, filename, sep = ), width, height, pointsize, :
FAQ 7.19:
I want to fit some p-values to a beta distribution. But the problem is some
of the values have 0s and 1's. I am getting an error if I use the MASS
function to do this. Is there anyway to get around this?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
On 16/03/11 15:04, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 10:54 AM, Allan Engelhardtall...@cybaea.com wrote:
[...]
Yes, but you can still print(base::pi) and rm(pi) to get back to our flat
world, and you can't
assign(pi, 4, pos = package:base)
Error in assign(pi, 4, pos =
Dear R,
If I have remembered correctly, a square matrix is singular if and only if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a - matrix(c(1e20,1e2,1e3,1e3),2)
det(a)
[1] 1e+23
solve(a)
Error in solve.default(a) :
system is
Hi,
You could try this (others surely have better solutions):
out_file - file(file.csv, open=a) #creates a file in append mode
for (i in seq_along(l)){
write.table(names(l)[i], file=out_file, sep=,, dec=.,
quote=FALSE, col.names=FALSE, row.names=FALSE) #writes the name of the
list
Just as a heads-up: it is likely that unlocking the bindings in base
for pi, T, F, probably all BULTIN and SPECIAL functions, and possibly
more, will start signaling warnings in the near future. Doing this
may be useful at times for debugging but it can mess up assumptions
others make about how
On Wed, Mar 16, 2011 at 6:04 AM, Michael Friendly frien...@yorku.ca wrote:
On 3/15/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
Hi, all,
Does R have a const object concept like which is in C++ language? I
want to set some data frames as constant to avoid being
On Wed, Mar 16, 2011 at 11:49 AM, luke-tier...@uiowa.edu wrote:
Just as a heads-up: it is likely that unlocking the bindings in base
for pi, T, F, probably all BULTIN and SPECIAL functions, and possibly
more, will start signaling warnings in the near future. Doing this
may be useful at times
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 11:49 AM, luke-tier...@uiowa.edu wrote:
Just as a heads-up: it is likely that unlocking the bindings in base
for pi, T, F, probably all BULTIN and SPECIAL functions, and possibly
more, will start signaling warnings in the
If yoy write out the likelihood equations for an independent sample size n from
the beta(a,b) distribution:
L \propto \prod_i dbeta(y_i,a,b)
log(L) = constant + \sum_i dbeta(y_i,a,b,log=TRUE)
log(L)= constant + \sum_i (a-1) log(y_i) + (b-i) log(1-y_i)
you see that your problem comes from trying
Brilliant - that was really useful!
On Tue, Mar 15, 2011 at 3:46 PM, Ista Zahn iz...@psych.rochester.edu wrote:
Hi Claus,
On Tue, Mar 15, 2011 at 9:33 AM, Claus O'Rourke claus.orou...@gmail.com
wrote:
Hi,
I am trying to recursively apply a function to a selection of columns
in a
Jim Silverton jim.silverton at gmail.com writes:
I want to fit some p-values to a beta distribution. But the problem is some
of the values have 0s and 1's. I am getting an error if I use the MASS
function to do this. Is there anyway to get around this?
The probability *density* of an
Numeric underflow. Try qr.solve(a)
Allan
On 16/03/11 15:28, Feng Li wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and only if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a-
On Wed, Mar 16, 2011 at 12:16 PM, luke-tier...@uiowa.edu wrote:
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 11:49 AM, luke-tier...@uiowa.edu wrote:
Just as a heads-up: it is likely that unlocking the bindings in base
for pi, T, F, probably all BULTIN and SPECIAL
On Wed, Mar 16, 2011 at 4:16 PM, luke-tier...@uiowa.edu wrote:
That would defeat the purpose. Unlocking things in base may be useful
for experimenting or debugging but it is not a good idea otherwise.
You are making experimenting in R more awkward? Then I'm throwing
That would defeat the
On Wed, 16 Mar 2011, Barry Rowlingson wrote:
On Wed, Mar 16, 2011 at 4:16 PM, luke-tier...@uiowa.edu wrote:
That would defeat the purpose. Unlocking things in base may be useful
for experimenting or debugging but it is not a good idea otherwise.
You are making experimenting in R more
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li m...@feng.li wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and only if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a - matrix(c(1e20,1e2,1e3,1e3),2)
det(a)
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. Please help!
Anna
__
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Dear R help,
Is there a way to extract a variable ( one column) from my summary
statistics to save it in a table form?
My post-factor analysis summary statistics gave me the output below. Is
there anyway to just save the mean of all the phi? The filename of this
summary output is postun2010sum.
Dear all,
result-summary(princomp(x,cor=TRUE))
result
Importance of components:
Comp.1Comp.2 Comp.3 Comp.4
Standard deviation 1.642136 1.0114376 0.45146892 0.27669119
Proportion of Variance 0.674153 0.2557515 0.05095605 0.01913950
Cumulative Proportion
Of course! Works fine now.
Thanks, Martyn and Jim...
Ravi
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Hi group,
I am trying to convert the organization of a data frame so I can do some
correlations between stocks,
I have something like this:
stock.returns -
data.frame(rbind(c(MSFT,20110301,0.05),c(MSFT,20110302,0.01),c(GOOG,20110301,-0.01),c(GOOG,20110302,0.04)))
colnames(stock.returns) -
When some libraries (e.g., VIM and Rcmdr) are loaded, they automatically
display a GUI window. Is there a way to load the library but suppress the
window?
greg
gregory carey
university of colorado
__
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Hi, there:
Is there any quick function to generate bivariate or multivariate gamma
distribution?
Thanks.
Yulei
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi
I have a vector m:
m
[1] ABC transporters
[2] 2
[3] Acetyl-CoA
[4] 1
[5] Energie
[6] 1
I know I can add line to graph with abline(), but I would like to print
R-squared, F-test value, Residuals and other statistics from lm() to a
graph. I don't know how to access the values from summary(), so that I can
use them in a following code or print them in a graph.
--
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On 16/03/11 09:20, fre wrote:
I have the following problem:
I have some string with numbers like k. I want to have a table like the
function table() gives. However I am not able to call the first row, the 1,
2, 3, 5, 6 or 9. I tried to do that by a data.frame, but that doesn't seem
The
How about?
sink(andrija.csv)
l
sink()
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Hi,
the boot function from the boot library seems to use an implausible
quantity of memory; is this to be expected, or am I doing something
wrong? Specifically, the vector I wish to bootstrap off is about 17000
entries long
length(fes)
[1] 16988
bm - function(x,indexes) mean(x[indexes])
then if
Hello Ivan,
Thank you very much for your comments, they were really useful and I’ll try
to memorize and use them in the future.
Getting back to my problem… well, I try to put it in a different way because
I’m afraid this is gonna be a little bit more difficult than I thought.
So, here is my
Dear community,
I have fitted a model using comands above, (rlm, lmrob or lmRob). I don't
have new data to validate de models obtained. I was wondering if exists
something similar to CVlm in robust regression. In case there isn't, any
suggestion for validation would be appreciated.
Thanks,
See ?seq for generating index vectors of different kinds.
/Henrik
On Wed, Mar 16, 2011 at 7:43 AM, rens picca...@hotmail.com wrote:
Hi
I have a vector m:
m
[1] ABC transporters
[2] 2
[3] Acetyl-CoA
[4] 1
[5] Energie
[6] 1
[7] FAD Biosynthese
[8] 1
[9] Glyoxylate and
hi, list
R is undoudtedly my favorite statistic tool, however, the data
inputnpart has long been a pain. most data I have to deal with are
irregular and contains special character.
Recently I get a tab delimited data, read.table(filename,sep=\t)
constantly return erors for certain rows does not
Hi,
The values are calculated on the fly in the summary function
stats:::print.summary.princomp
using
vars - result$sdev^2
vars - vars/sum(vars)
cumsum(vars)
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of ian_zhangty
Peter Langfelder wrote:
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li lt;m...@feng.ligt; wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and only
if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a -
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 12:16 PM, luke-tier...@uiowa.edu wrote:
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 11:49 AM, luke-tier...@uiowa.edu wrote:
Just as a heads-up: it is likely that unlocking the bindings in
On Mar 16, 2011, at 12:53 PM, Anna Gretschel wrote:
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. Please help!
You should do more searching. I can
On 2011-03-16 10:31, Martyn Byng wrote:
Hi,
The values are calculated on the fly in the summary function
stats:::print.summary.princomp
using
vars- result$sdev^2
vars- vars/sum(vars)
cumsum(vars)
Or you could use the prcomp summary function (maybe you
should use prcomp() for you model?):
Dear Anna,
What is your goal in obtaining a value for R^2 ? I believe it is not
provided for a non-linear model, because it does not make much sense.
It certainly will not have the same interpretation as in a linear
model, and all the ways it *could* be defined come with their own
sets of
Am 16.03.2011 18:15, schrieb David Winsemius:
On Mar 16, 2011, at 12:53 PM, Anna Gretschel wrote:
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. Please
You can't. R^2 has no (consistent, sensible) meaning for nonlinear
models. If you don't understand why not, get local help or do some
reading.
Cheers,
Bert
On Wed, Mar 16, 2011 at 9:53 AM, Anna Gretschel ana-...@web.de wrote:
Dear List,
how can I obtain the value of r suqared for a
Am 16.03.2011 18:19, schrieb Joshua Wiley:
Dear Anna,
What is your goal in obtaining a value for R^2 ? I believe it is not
provided for a non-linear model, because it does not make much sense.
It certainly will not have the same interpretation as in a linear
model, and all the ways it *could*
Its useful for being able to set defaults for arguments that do not
have defaults. That cannot break existing programs.
Until the next program decides do co change those defaults and either
can't or does and you end up with incompatible assumptions. It also
make the code with the added
R users,
I found a solution to my own question. I just needed to insert (j+(j-1)):(2*j)
as the indices for vector s3. No need to respond.
Thank you
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Hi,
?summary.lm
describes what summary statistics get calculated and returned, so
ll - lm(y ~ x)
ss - summary(ll)
ss$fstatistic
for example would give the F statistic
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of
Is there any way that this could be made into a fortune -- perhaps by
omitting the poster's identity?
yes there are threads concidering this topic but they are all about the
theory not about how to get the value of r^2 for a non-linear model in R.
:=)
Cheers,
Bert
Anna: I say this because you
Here's one way:
ans = reshape(stock.returns,idvar='Date',
+varying=list(names(stock.returns)[-1]),
+direction='long',
+times=names(stock.returns)[-1],
+v.names='Return',timevar='Ticker')
rownames(ans) = NULL
ans
Date
Hi,
You should carefully and thoroughly read the help page for the
extraction operators. See ?[
Let's say your output below is in a data.frame called df. You can do
something like this:
df[, Mean]
To select only some specific phi rows, you probably need to set some
regular expressions,
Dear Bert,
so what can I do to obtain a goodness of fit for a non-linear model if
r² does not work?
And here comes my next question: is it apropriate to comopare a linear
and a non-linear model with anova()?
Thank you so much for answering,
Anna
Am 16.03.2011 18:54, schrieb Bert Gunter:
On 2011-03-16 08:47, derek wrote:
I know I can add line to graph with abline(), but I would like to print
R-squared, F-test value, Residuals and other statistics from lm() to a
graph. I don't know how to access the values from summary(), so that I can
use them in a following code or print them
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