Dear R People:
Hello again!
I found something unusual in the behavior of the "endpoints" function
from the xts package:
> x1 <- ts(1:24,start=2008,freq=12)
> dat <- seq(as.Date("2008/01/01"),length=24,by="months")
> library(zoo);library(xts)
> x2 <- zoo(1:24,order=dat)
> #Here is the surprise:
>
subset(data, grepl("[1-5]", section) & !grepl("0", section))
BTW
grepl("[1:5]", section)
does work. It checks for the characters 1, :, or 5.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kang Min
Sent: Thursday, 17 March 20
Eric -
What you mean to say is
t <- 3
z <- ifelse(t %in% c(1,2,3),1,0)
z
[1] 1
t <- 4
z <- ifelse(t %in% c(1,2,3),1,0)
[1] 0
Expressions don't recalculate themselves when you change
the value of a variable that they use. For that, you
would need a function:
makez = function(t)ifelse(
It doesn't work (in R) because it is not written in R. It's written in some
other language that looks a bit like R.
> t <- 3
> z <- t %in% 1:3
> z
[1] TRUE
> t <- 4
> z <- t %in% 1:3
> z
[1] FALSE
>
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-proje
Why doesn't this work and is there a better way ?
z <-ifelse(t==1 || 2 || 3, 1,0)
t <-3
z
[1] 1
t <-4
z
[1] 1
trying to say ...if t == 1 or if t== 2 or if t ==3 then true, otherwise
false
--
View this message in context:
http://r.789695.n4.nabble.com/Why-doesn-t-this-work-tp3383656p3383656.html
I have a new question, also regarding grepl.
I would like to subset rows with numbers from 1 to 5 in the section
column, so I used
subset(data, grepl("[1:5]", section))
but this gave me rows with 10, 1 and 5. (Why is this so?) So I tried
subset(data, grepl("[1,2,3,4,5]", section))
which worked.
On Mar 16, 2011, at 10:22 PM, Erin Hodgess wrote:
Dear R People:
I have a monthly time series which runs from January 1998 to
December 2010.
When I use tsp I get the following:
tsp(ibm$ts)
[1] 1998.000 2010.917 12.000
Is there an easy way to convert this to a seq.Date object, please
On Wed, Mar 16, 2011 at 10:22 PM, Erin Hodgess wrote:
> Dear R People:
>
> I have a monthly time series which runs from January 1998 to December 2010.
>
> When I use tsp I get the following:
>
>> tsp(ibm$ts)
> [1] 1998.000 2010.917 12.000
>
>
> Is there an easy way to convert this to a seq.Date
On Wed, Mar 16, 2011 at 4:00 PM, derek wrote:
> 1) In my very humble opinion R^2 can't be negative, at least for data for
> which it sound to use linear model.
!!! Your opinion, humble or not, counts for nothing. Thomas stated a
mathematical fact. I suggest you make an effort to understand what h
take a look at what the output of your cbind is; my guess is that it is a
character matrix. you probably want to do
data.frame(...,...,...)
without the cbind.
Sent from my iPad
On Mar 16, 2011, at 18:12, Martin Ralphs wrote:
> Dear R Help,
>
> I would be very grateful if somebody could exp
Hi Erin,
I am not sure what a "seq.Date object" is. My first thought is that
you are talking about the date method for seq(), but there are
hundreds of packages I do not know. In any case, here is what I think
you want.
Josh
## A small example is always nice
dat <- ts(1:12, frequency = 12,
s
Dear R People:
I have a monthly time series which runs from January 1998 to December 2010.
When I use tsp I get the following:
> tsp(ibm$ts)
[1] 1998.000 2010.917 12.000
Is there an easy way to convert this to a seq.Date object, please?
I would like to have something to the effect of
1998/0
On 11-03-16 07:55 PM, Peter Ehlers wrote:
> On 2011-03-16 15:02, Ben Bolker wrote:
>> ria.buffalo.edu> writes:
>>
>>>
>>> lm(y~x+0) yields the regression on x without the constant, i.e., y=bx+e,
>>> not y = a +e
>>>
>>> derek gmail.com>
>>> Sent by: r-help-bounces r-project.org
>>> 03/16/201
On Wed, Mar 16, 2011 at 8:20 PM, David Winsemius wrote:
>
> On Mar 16, 2011, at 7:58 PM, jctoll wrote:
>
>> Hi,
>>
>> I'm struggling to figure out the way to change the name of a column
>> from within a loop. The problem is I can't refer to the object by its
>> actual variable name, since that wi
On Mar 16, 2011, at 7:58 PM, jctoll wrote:
Hi,
I'm struggling to figure out the way to change the name of a column
from within a loop. The problem is I can't refer to the object by its
actual variable name, since that will change each time through the
loop. My xts object is A.
head(A)
> Using the 'CARET' package, is it possible to specify weights for features
> used in model prediction?
For what model?
> And for the 'knn' implementation, is there a way
> to choose a distance metric (i.e. Mahalanobis distance)?
>
No, sorry.
Max
__
Seconded
On 03/16/2011 05:37 PM, Bert Gunter wrote:
Ha! -- A fortunes candidate?
-- Bert
If this is really a time series, then you will have serious validity
problems due to auto-correlation among non-independent units. (But if you
are just searching for a way to pull the wool over the eyes of
THANK U VERY MUCH!!
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-read-Cumulative-proportion-in-using-princomp-tp3381881p3383396.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing l
Hey everyone,
I just saw a demonstration of a model that is built entirely in R that
allows for transportation greenhouse gas related scenario testing and it
used an awesome GUI utilizing iplots, rJava and gWidgets libraries. Very
cool stuff. I am a somewhat seasoned R user but cannot seem to
Thank you very much! I will check out help page for extraction operator.
Many thanks,
Haillie
--
View this message in context:
http://r.789695.n4.nabble.com/Saving-summary-outputs-in-a-table-form-tp3381959p3383316.html
Sent from the R help mailing list archive at Nabble.com.
__
Dear R Help,
I would be very grateful if somebody could explain why this is happening. I
am trying to plot a lattice barchart of a vector of numbers with age
bandings that I have brought together into a data frame. When I plot the
variables in their raw vector form, it works. When I try to plot
Using the 'CARET' package, is it possible to specify weights for features
used in model prediction? And for the 'knn' implementation, is there a way
to choose a distance metric (i.e. Mahalanobis distance)?
Thanks,
~Kendric
[[alternative HTML version deleted]]
___
1) In my very humble opinion R^2 can't be negative, at least for data for
which it sound to use linear model.
Or the data would have to be utterly wrong to fit them with linear model.
2) I don't want to fit data with linear model of zero intercept.
3) I dont know if I understand correctly. Im 100%
I have been reading about autocorrelation in linear models over the last
couple of days, and I have to say the more I read, the more confused I
get. Beyond confusion lies enlightenment, so I'm tempted to ask R-Help for
guidance.
Most authors are mainly worried about autocorrelation in the resi
The values appear, indeed, however the grid has now shifted so that it is
not aligned with the ticks on the axes. I'm using
grid(nx=NULL, ny=NULL)
that should do the trick to follow the axes' ticks, but it does not.
I would appreciate any sugestions how to solve it.
Regards,
Sara
pln <- read.t
Hi,
I'm struggling to figure out the way to change the name of a column
from within a loop. The problem is I can't refer to the object by its
actual variable name, since that will change each time through the
loop. My xts object is A.
> head(A)
A.Open A.High A.Low A.Close A.Volume A.
On 2011-03-16 15:02, Ben Bolker wrote:
ria.buffalo.edu> writes:
lm(y~x+0) yields the regression on x without the constant, i.e., y=bx+e,
not y = a +e
derek gmail.com>
Sent by: r-help-bounces r-project.org
03/16/2011 03:49 PM
Would someone like to (please!) write this up and submi
How about using the prediction accuracy or ROC curve if you have validation
data set?
On Wed, Mar 16, 2011 at 11:42 AM, Alexx Hardt wrote:
> Am 16.03.2011 19:34, schrieb Anna Gretschel:
>
>> Am 16.03.2011 19:21, schrieb Alexx Hardt:
>>
>> And to be on-topic: Anna, as far as I know anova's are
Firstly, the way you have constructed your data frame in the example will
convert everything to factors. What you need to do is actually a bit simpler:
###
dum <- data.frame(date, col1, col2)
###
One way to turn this into the kind of data frame you want is to convert the
main part of i
Hi all,
I'm using the R GUI on a Mac. It is easy to produce a
number of plots and put them into a Quartz window, and then
use Command-left or Command-right to flip between them.
However, Quartz seems to hold a maximum of about 15 plots,
and then discards anything plotted before that.
This
Hello R users,
I have this regex see [1] for apache log lines. I tried using R to parse
some data (only because I wanted to stay in R).
A sample line is [2]
(a) I saved the line in [1] into "~/tmp/a.txt" and [2] into "/tmp/a.txt"
pat <- readLines("~/tmp/a.txt")
test <- readLines("/tmp/a.txt")
te
Thanks all of you for the very interesting discussion. I think I get it
now.
Feng
On 03/16/2011 09:25 PM, rex.dw...@syngenta.com wrote:
Feng,
Your matrix is *not* (practically) singular; its inverse is.
The message said that the *system* was singular, not the matrix.
Remember Cramer's Rule:
ria.buffalo.edu> writes:
>
> lm(y~x+0) yields the regression on x without the constant, i.e., y=bx+e,
> not y = a +e
>
> derek gmail.com>
> Sent by: r-help-bounces r-project.org
> 03/16/2011 03:49 PM
>
Would someone like to (please!) write this up and submit it to
Kurt Hornik for inclu
On Thu, Mar 17, 2011 at 10:01 AM, derek wrote:
> It states summary.lm:
>
> r.squared R^2, the ‘fraction of variance explained by the model’,
>
> R^2 = 1 - Sum(R[i]^2) / Sum((y[i]- y*)^2),
>
> where y* is the mean of y[i] if there is an intercept and zero otherwise.
>
> Why to use different f
On Wed, Mar 16, 2011 at 1:12 PM, wrote:
> On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
>
>> On Wed, Mar 16, 2011 at 12:16 PM, wrote:
>>>
>>> On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
>>>
On Wed, Mar 16, 2011 at 11:49 AM, wrote:
>
> Just as a heads-up: it is likely that unl
Looks like the problem may be that R is automatically passing this flag to
the g++ compiler: "-arch x86_64" which appears to be causing trouble for
opencv. Does anyone know how to suppress this flag?
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-make-sure-R-s-g-compiler
Ha! -- A fortunes candidate?
-- Bert
>
> If this is really a time series, then you will have serious validity
> problems due to auto-correlation among non-independent units. (But if you
> are just searching for a way to pull the wool over the eyes of the
> statistically uninformed, then I guess th
I am new to the R language. I am trying to plot multiple figures on one page
through a loop, but the code just produce one graph on one page. Can someone
show some light on what's wrong?
Here is my code:
library("quantreg")
tcdata<-read.table("mydata.txt",header=TRUE)
postscript("myfigure.ps")
ba
It states summary.lm:
r.squared R^2, the ‘fraction of variance explained by the model’,
R^2 = 1 - Sum(R[i]^2) / Sum((y[i]- y*)^2),
where y* is the mean of y[i] if there is an intercept and zero otherwise.
Why to use different formula when intercept is set to zero?
I tried to compute R^2
On Mar 16, 2011, at 3:19 PM, Justin Haynes wrote:
I have a very large dataset with columns of id number, actual value,
predicted value. This used to be a time series but I have dropped the
time component. So I now have a data.frame where the id number is
repeated but each value in the actual
On 16-03-2011, at 21:11, David Winsemius wrote:
>
> On Mar 16, 2011, at 1:32 PM, Berend Hasselman wrote:
>> .
>> svd(a) indicates the problem.
>>
>> largest singular value / smallest singular value=1e17 (condition number)
>> --> reciprocal condition number=1e-17
>> and the standard solve c
You can get fairly bad results from solve() and
other linear algebra routines without warning.
E.g., the following function makes a 2 by 2 matrix which
would have determinate 1.0 if we had infinite precision
arithmetic and actually will produce one of determinant
1 on a computer if you use the "ri
Try this:
a <- m[c(TRUE, FALSE)]
On Wed, Mar 16, 2011 at 11:43 AM, rens 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 dicarboxylate metabolism"
> [10
lm(y~x+0) yields the regression on x without the constant, i.e., y=bx+e,
not y = a +e
derek
Sent by: r-help-boun...@r-project.org
03/16/2011 03:49 PM
To
r-help@r-project.org
cc
Subject
[R] Strange R squared, possible error
k=lm(y~x)
summary(k)
returns R^2=0.9994
lm(y~x) is supposed
On 3/16/2011 8:04 AM, taby gathoni wrote:
data<-data.frame(id=1:(165+42),main_samp$SCORE,
x=rep(c("BAD","GOOD"),c(42,165)))
> f<-function(x) {
+ str.sample<-list()
+ for (i in 1:length(levels(x$x)))
+ {
+ str.sample[[i]]<-x[x$x==levels(x$x)[i]
,][sample(tapply(x$x,x$x,length)[i],20,rep=T),]
+
?summary.lm
The R^2 section explains that R^2 is computed differently depending
on whether or not an intercept is in the model.
-- Bert
On Wed, Mar 16, 2011 at 12:49 PM, derek wrote:
> k=lm(y~x)
> summary(k)
> returns R^2=0.9994
>
> lm(y~x) is supposed to find coef. a anb b in y=a*x+b
>
> l=lm
Hi Derek,
R^2 doesn't mean the same thing when you omit the intercept, as has
been discussed on this list before. See
http://r.789695.n4.nabble.com/lm-without-intercept-td3312429.html
Best,
Ista
On Wed, Mar 16, 2011 at 3:49 PM, derek wrote:
> k=lm(y~x)
> summary(k)
> returns R^2=0.9994
>
> lm(y~
Feng,
Your matrix is *not* (practically) singular; its inverse is.
The message said that the *system* was singular, not the matrix.
Remember Cramer's Rule: xi = |Ai| / |A|
The really, really large determinant of your matrix is going to appear in the
denominator of your solutions, so, essentially,
k=lm(y~x)
summary(k)
returns R^2=0.9994
lm(y~x) is supposed to find coef. a anb b in y=a*x+b
l=lm(y~x+0)
summary(l)
returns R^2=0.9998
lm(y~x+0) is supposed to find coef. a in y=a*x+b while setting b=0
The question is why do I get better R^2, when it should be otherwise?
Im sorry to use the wor
Thanks for showing me the link to the code / your response / your work in
general.
It seems that the real magic is happening in the call to the function
attributes, via the line
attr(x, "srcref")
I'm guessing that attributes must be defined somewhere deep inside the R
machinery (since I didn't fin
I have a very large dataset with columns of id number, actual value,
predicted value. This used to be a time series but I have dropped the
time component. So I now have a data.frame where the id number is
repeated but each value in the actual and predicted columns are
unique.
I assume I need to
Thanks a lot for your answer.
Martin Patenaude-Monette
MSc. Candidate
Département de biologie
Université du Québec à Montréal
2011/3/15 Terry Therneau
> --- included text --
> I have done model selection between candidate Cox models, using AICc
> calculated with penalized log likelihoods. The
I am trying to read a table from MySQL, I have loaded the file in "ts"
database, in table name ACC. but i am unable to read it in R through
getSymbol function.
mysql> show databases;
++
| Database |
++
| information
Hi Stefan,
thats really interesting - I never though of trying to benchmark Linux-64
against OSX (a friend who works on large databases, says OSX performs better
than Linux in his work!). Thanks for posting your comparison, and your hints
:)
i) I guess you have a very fast CPU (Core i7 or so, I g
Thank you that is very helpful.
--
View this message in context:
http://r.789695.n4.nabble.com/adding-linear-regression-data-to-plot-tp3357946p3382615.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
ht
require(reshape2)
dcast(stock.returns, Date ~ Ticker)
The numbers were changed from their original values, but if you originally
created the values Return as.numeric they should stay the same
On Wednesday, March 16, 2011 at 9:37 AM, chris99 wrote:
Hi group,
>
> I am trying to convert the organi
On Mar 16, 2011, at 1:32 PM, Berend Hasselman wrote:
Peter Langfelder wrote:
On Wed, Mar 16, 2011 at 8:28 AM, 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 co
On Mar 16, 2011, at 1:21 PM, Henrik Bengtsson wrote:
See ?seq for generating index vectors of different kinds.
/Henrik
Most people will probably understand seq(...) better than they
understand modulo arithmetic, but logical indexing could also be used
for the task:
((1:12) %/% 2) ==
Thanks everyone for different solutions.
Every solution works very well. For my purpose, this function sink does what
I was looking for.
Andrija
On Wed, Mar 16, 2011 at 4:23 PM, Roman Luštrik wrote:
> How about?
>
> sink("andrija.csv")
> l
> sink()
>
> --
> View this message in context:
> http:
On Mar 16, 2011, at 17:37 , Yong Wang wrote:
> 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(filenam
If you do ?Startup then you get the help page that describes all that R does as
it starts up and there are a few places in there that it describes where you
can put things to be run automatically.
I have done this for a doctor before who wanted to show the demonstration I
showed him to others,
Thanks a lot!! It helped!
2011/3/15 Sara Szeremeta
> Hi
>
> I am trying to plot two simple graphs with a grid in background. The axis
> and grid appears in correct position, but the actual data are not there
> Can somebody provide me a hint what is missing?
>
> The code is:
>
> pln <- rea
Try this:
xtabs(Return ~ Date + Ticker, stock.returns)
On Wed, Mar 16, 2011 at 11:37 AM, chris99 wrote:
> 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(
Am 16.03.2011 19:34, schrieb Anna Gretschel:
Am 16.03.2011 19:21, schrieb Alexx Hardt:
And to be on-topic: Anna, as far as I know anova's are only useful to
compare a submodel (e.g. with one less regressor) to another model.
thanks! i don't get it either what they mean by fortune...
It's an
The optimal way of doing it depends on how you want to use the result. An
easy way has been recommended - if you have
boo <- list(first=data.frame(a=1:5, b=2:6), second=data.frame(a=6:10,
b=7:11))
.. then
sink("boo.txt")
boo # or: print(boo)
sink()
... will put it all in the same file, the sa
Am 16.03.2011 19:29, schrieb Heiman, Thomas J.:
Hi Anna,
AIC and BIC are good criteria for determining degree of model fit..
Sincerely,
tom
Thomas Heiman, PhD
Info Systems Eng, Sr
The MITRE Corporation | Center for Enterprise Modernization
Office: 703-983-2951 | thei...@mitre.org
-Origin
Am 16.03.2011 19:11, schrieb Joshua Wiley:
(Are fortunes determined by voting? There is precedence for seconding
desired fortunes at least)
Wait, what do you mean by fortune? Is there a
statistics-quote-package-something for the unix shell program 'fortune'
? If so, I want want want!
And
On Wed, Mar 16, 2011 at 10:54 AM, Bert Gunter wrote:
> 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-linea
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 i
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:
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 expression
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 Ticke
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
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 d
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 po
>> 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 ad
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* b
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 wrote:
> Dear List,
>
> how can I obtain the value of r suqared for a non-linear model? Fo
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 help
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 probl
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$sdev^2
vars<- vars/sum(vars)
cumsum(vars)
Or you could use the prcomp summary function (maybe you
should use prcomp() for you model?):
On Mar 16, 2011, at 12:53 PM, Anna Gretschel wrote:
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. Please help!
You should do more searching. I can remembe
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 12:16 PM, wrote:
On Wed, 16 Mar 2011, Gabor Grothendieck wrote:
On Wed, Mar 16, 2011 at 11:49 AM, wrote:
Just as a heads-up: it is likely that unlocking the bindings in base
for pi, T, F, probably all BULTIN and SPEC
Peter Langfelder wrote:
>
> On Wed, Mar 16, 2011 at 8:28 AM, Feng Li wrote:
>> Dear R,
>>
>> If I have remembered correctly, a square matrix is singular if and only
>> if
>> its determinant is zero. I am a bit confused by the following code error.
>> Can someone give me a hint?
>>
Hi,
The values are calculated on the fly in the summary function
stats:::print.summary.princomp
using
vars <- result$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
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
See ?seq for generating index vectors of different kinds.
/Henrik
On Wed, Mar 16, 2011 at 7:43 AM, rens 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
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, u...
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 refr
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
How about?
sink("andrija.csv")
l
sink()
--
View this message in context:
http://r.789695.n4.nabble.com/export-list-to-csv-tp3381992p3382062.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat
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 first
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.
--
View this message in
Hi
I have a vector m:
m
[1] "ABC transporters"
[2] "2"
[3] "Acetyl-CoA"
[4] "1"
[5] "Energie"
[6] "1"
Hi, there:
Is there any quick function to generate bivariate or multivariate gamma
distribution?
Thanks.
Yulei
__
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
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
__
R-help@r-project.org mailing list
https://stat.ethz
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.
Of course! Works fine now.
Thanks, Martyn and Jim...
Ravi
--
View this message in context:
http://r.789695.n4.nabble.com/Scope-of-variable-tp3381485p3381932.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
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
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. Al
1 - 100 of 185 matches
Mail list logo