Dear all,
Lattice provides automatic coloring for subgroups on each panel by the
simple use of a groups statement. For an application I want to change
these colors to a predifined set. This works well using a panel function
in stead of the default as long as there are only points in the graphs.
Hello Ali,
It looks to me like you have two questions: 1) the error you get from
scatterplot3d 2) how to calculate the monthly mean.
If you could provide some sample data that would make it a lot easier
to help you understand why the error occurred and what to do about it.
For now, maybe this wi
Hi:
Just to add to the discussion, see the following article by Russell Lenth on
the subject:
http://www.stat.uiowa.edu/techrep/tr378.pdf
Dennis
On Thu, Sep 2, 2010 at 3:59 PM, C Peng wrote:
>
> Agree with Greg's point. In fact it does not make logical sense in many
> cases. Similar to the us
Hi,
I'd like to be able to run a .exe in the background whenever library(package-x)
is called. Is this possible?
~Aks
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PLE
On 2010-09-02 22:32, Gabor Grothendieck wrote:
On Thu, Sep 2, 2010 at 7:39 PM, Marlin Keith Cox wrote:
This data are kilojoules of energy that are consumed in starving fish over a
time period (Days). The KJ reach a lower asymptote and level off and I
would like to use a non-linear plot to show
Hi Dejian,
Thanks for the reply!
I finally found the problem. It is actually in the "panel.cor" function.
Adding "..." in the function and "text" call fixed everything.
Best,
...Tao
- Original Message
> From: Dejian Zhao
> To: r-help@r-project.org
> Sent: Thu, September 2, 2
On 2010-09-02 22:16, Jocelyn Paine wrote:
Greg, thanks for the suggestion. That's useful to know for future work.
It's not so good in this case, because I'm making the plots for a
colleague who doesn't know R, and it would be a bother for me to have to
send him several files and him reassemble th
On Thu, Sep 2, 2010 at 7:39 PM, Marlin Keith Cox wrote:
> This data are kilojoules of energy that are consumed in starving fish over a
> time period (Days). The KJ reach a lower asymptote and level off and I
> would like to use a non-linear plot to show this leveling off. The data are
> noisy an
try to use difftime() instead of as.difftime().
On Thu, Sep 2, 2010 at 10:32 PM, Dunia Scheid wrote:
> Hello all,
>
> I've 2 strings that representing the start and end values of a date and
> time.
> For example,
> time1 <- c("21/04/2005","23/05/2005","11/04/2005")
> time2 <- c("15/07/2009", "03/
Hi,
I'm doing a generalized linear mixed model, and I currently use an R function
called "glmm". However, in this function they use a standard normal
distribution for the random effect, which doesn't satisfy my case, i.e. my
random effect also follows a normal distribution, but observations ov
Hello group
Im trying to plot 3d with scatterplot packages, i got error say "
length(color) must be equal length(x) or 1 " may data has dimensions
(lon,lat,lev,time) ,the time in month i want to
calculate the monthly mean for the time how can i make that , is there any
function doi
Agree with Greg's point. In fact it does not make logical sense in many
cases. Similar to the use of the "statistically unreliable" reliability
measure Cronbach's alpha in some non-statistical fields.
--
View this message in context:
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On 2010-08-31 13:49, Greg Snow wrote:
Look at the pairs2 function in the TeachingDemos package, this lets you produce
smaller portions of the total scatterplot matrix at a time (with bigger plots),
you could print the smaller portions then assemble the full matrix on a large
wall, or just use
Greg, thanks for the suggestion. That's useful to know for future work.
It's not so good in this case, because I'm making the plots for a
colleague who doesn't know R, and it would be a bother for me to have to
send him several files and him reassemble them. What I did was to use
pairs.panels,
William,
Thanks. I adapted your example by doing:
library(psych)
pdf(file="myfile.pdf",width=30,height=30)
pairs.panels(data,gap=0)
dev.off()
The R part worked. I could see it doing so when I replaced the call to
'pdf' by
windows(width=30,height=30)
. The remaining problem was that A
When "pairs" draws plots, "lower.panel" invokes "f.xy". Maybe there is
something in "f.xy" incompatible with "pairs". You can read the code of
"pairs" to see what happens.
"pairs" has two methods, as you can see in the help message (?pairs).
According to your code, pairs is supposed to invoke
Is there a complete list of these very handy and power functions in the base
R?
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
R-he
Qiu, Weiyu ualberta.ca> writes:
>
> Hi,
>
> I'm doing a generalized linear mixed model, and I currently use an
> R function called "glmm". However, in
> this function they use a standard normal distribution for the random
> effect, which doesn't satisfy my
> case, i.e. my random effect also
This data are kilojoules of energy that are consumed in starving fish over a
time period (Days). The KJ reach a lower asymptote and level off and I
would like to use a non-linear plot to show this leveling off. The data are
noisy and the sample sizes not the largest. I have tried selfstarting
we
Hi,
I'm doing a generalized linear mixed model, and I currently use an R function
called "glmm". However, in this function they use a standard normal
distribution for the random effect, which doesn't satisfy my case, i.e. my
random effect also follows a normal distribution, but observations ov
On 02-Sep-10 18:01:55, Greg Snow wrote:
> Just to add to Ted's addition to my response. I think you are moving
> towards better understanding (and your misunderstandings are common),
> but to further clarify:
> [Wise words about P(A|B), P(B|A), P-values, etc., snipped]
>
> The real tricky bit abo
Suggestion: use the power of R.
If x and y are independent then sorting y based on x is meaningless.
If sorting y based on x is meaningful, then they are not independent.
Trying to force non-independent things to pretend that they are independent
just causes future headaches.
Part of the grea
Be happy, don't do post-hoc power analyses.
The standard "post-hoc power analysis" is actually counterproductive. It is
much better to just create confidence intervals. Or give a better
description/justification showing that your case is not the standard/worse than
useless version.
--
Gregor
Totally agreed.
I made a mistake in calling the categorization a GAM. If we apply a step
function to the continuous age we get a limited range ordinal variable.
Categorizing is creating several binary variables from the continuous (with
treatment contrasts).
Stephen B
-Original Message
On Thu, Sep 2, 2010 at 11:09 AM, Ivan Allaman wrote:
>
> Good morning gentlemen!
>
> How using a weighted model in nls2? Values with the nls are logical since
> values with nls2 are not. I believe that this discrepancy is due to I did
> not include the weights argument in nls2.
>
Just to follow u
Hi Marcio,
Is this what you want?
x <- c(2,6,8,8,1)
y <- c(1,6,3,5,4)
o <- order(x)
# If you want each vector order by x
x[o]
y[o]
You can also use sort(), but then each vector would be sorted by
itself, not both by x.
HTH,
Josh
On Thu, Sep 2, 2010 at 1:48 PM, Mestat wrote:
>
> Hi listers,
Hi listers,
I could order a data that like this:
x<-c(2,6,8,8,1)
y<-c(1,6,3,5,4)
o<-order(x)
frame<-rbind(x,y)[,o]
But, I would like to know if there is a way to order my data without setting
up a data frame. I would like to keep independent vectors x and y.
Any suggestions?
Thanks in advance,
Mar
Thanks Peter
Actually I should have specified. These are not actually the functions I
ultimately want to solve can't be rearranged explicitly and substituted. I
do need a way to solve simultaneously.
Ben
--
View this message in context:
http://r.789695.n4.nabble.com/Simultaneous-equations-tp25
I am aware this is fairly simple, but is currently driving me mad! Could
someone help me out with conducting a post-hoc power analysis on a Wilcoxon
test. I am being driven slightly mad by some conflicting advice!
Thanks in advance,
Lewis
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On Thu, 2 Sep 2010, stephenb wrote:
sorry to bump in late, but I am doing similar things now and was browsing.
IMHO anova is not appropriate here. it applies when the richer model has p
more variables than the simpler model. this is not the case here. the
competing models use different variabl
On 09/02/2010 09:25 PM, benhartley903 wrote:
>
> Dear all,
> I am relatively new to R and have had some difficulty in understanding the
> user manual for a package that I have downloaded to evaluate non-linear
> simultaneous equations.
> The package is called systemfit.
> Does anyone have any e
benhartley903 wrote:
>
> Dear all,
> I am relatively new to R and have had some difficulty in understanding the
> user manual for a package that I have downloaded to evaluate non-linear
> simultaneous equations.
> The package is called systemfit.
> Does anyone have any experience of this pack
Dear all,
I am relatively new to R and have had some difficulty in understanding the
user manual for a package that I have downloaded to evaluate non-linear
simultaneous equations.
The package is called systemfit.
Does anyone have any experience of this package?
What I am trying to do is solve
Tal wrote:
> A friend recently brought to my attention that vector assignment actually
> recreates the entire vector on which the assignment is performed.
...
I brought this up in r-devel a few months ago. You can read my posting,
and the various replies, at
http://www.mail-archive.com/r-de...
Perhaps even more to the point, "covariate adjustment" and
"classification" should not be separate. One should fit the
appropriate model that does both.
-- Bert
On Thu, Sep 2, 2010 at 11:34 AM, Ben Bolker wrote:
> On 10-09-02 02:26 PM, James Nead wrote:
>> My apologies - I have made this more co
David,
The original poster was not looking at distributions and testing distributions,
I referred to the distribution of the p-value to help them understand (in
reference to the paper mentioned).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.o
On 10-09-02 02:26 PM, James Nead wrote:
> My apologies - I have made this more confusing than it needs to be.
>
> I had microarray gene expression data which I want to use for
> classification algorithms. However, I want to 'adjust' the data for
> all confounding factors (such as age, experiment nu
My apologies - I have made this more confusing than it needs to be.
I had microarray gene expression data which I want to use for classification
algorithms. However, I want to 'adjust' the data for all confounding factors
(such as age, experiment number etc.), before I could use the data as inpu
On Sep 2, 2010, at 2:01 PM, Greg Snow wrote:
The real tricky bit about hypothesis testing is that we compute a
single p-value, a single observation from a distribution, and based
on that try to decide if the process that produced that observation
is a uniform distribution or something e
James Nead yahoo.com> writes:
>
> Sorry, forgot to mention that the processed data will be used as input for a
> classification algorithm. So, I need to adjust for known effects before I can
> use the data.
>
> > I am trying to adjust raw data for both fixed and mixed effects.
> The data tha
Just to add to Ted's addition to my response. I think you are moving towards
better understanding (and your misunderstandings are common), but to further
clarify:
First, make sure that you understand that the probability of A given B, p(A|B),
is different from the probability of B given A, p(B
Sorry, forgot to mention that the processed data will be used as input for a
classification algorithm. So, I need to adjust for known effects before I can
use the data.
thanks
From: Bert Gunter
Cc: r-help@r-project.org
Sent: Thu, September 2, 2010 12:46:13
Hi Bert,
Thanks for the reply.
Height, was just as an example. Perhaps, I should have said 'x' instead of
height.
Essentially, what I want to do is adjust the data for known effects. After I've
done this, I can conduct further analysis on the data (for example, if another
variable 'z' has an
Why should height be a random effect?
I suspect you may need a tutorial on what a random effect in a mixed
model is. I see no obvious reason why one would cluster on height.
Perhaps if you clarify, it may become obvious either what your
concerns are (and that your model is correct) or that you
mis
Thomas,
Thanks a lot. This solves my problem.
Wayne (Yanwei) Zhang
Statistical Research
>CNA
-Original Message-
From: Thomas Lumley [mailto:tlum...@u.washington.edu]
Sent: Thursday, September 02, 2010 11:24 AM
To: Zhang,Yanwei
Cc: r-help@r-project.org
Subject: Re: [R] Help on glm an
On Thu, 2 Sep 2010, Zhang,Yanwei wrote:
Dear all,
I'm trying to use the "optim" function to replicate the results from the "glm" using an example
from the help page of "glm", but I could not get the "optim" function to work. Would you please
point out where I did wrong? Thanks a lot.
The fo
Hi,
Sorry for a basic questions on linear models.
I am trying to adjust raw data for both fixed and mixed effects. The data that
I
output should account for these effects, so that I can use the adjusted data
for
further analysis.
For example, if I have the blood sugar levels for 30 patients,
Dear all,
I'm trying to use the "optim" function to replicate the results from the "glm"
using an example from the help page of "glm", but I could not get the "optim"
function to work. Would you please point out where I did wrong? Thanks a lot.
The following is the code:
# Step 1: fit the glm
Good morning gentlemen!
How using a weighted model in nls2? Values with the nls are logical since
values with nls2 are not. I believe that this discrepancy is due to I did
not include the weights argument in nls2.
Here's an example:
MOISTURE <- c(28.41640, 28.47340, 29.05821, 28.52201, 30.9
Hi Dunia,
You need to convert the character strings to Dates.
time1 <- as.Date(c("21/04/2005","23/05/2005","11/04/2005"), "%d/%m/%Y")
time2 <- as.Date(c("15/07/2009", "03/06/2008", "15/10/2005"), "%d/%m/%Y")
time2-time1
Best,
Ista
On Thu, Sep 2, 2010 at 10:32 AM, Dunia Scheid wrote:
> Hello a
sorry to bump in late, but I am doing similar things now and was browsing.
IMHO anova is not appropriate here. it applies when the richer model has p
more variables than the simpler model. this is not the case here. the
competing models use different variables.
you are left with IC.
by transfo
Hi,
Have you gotten help on this?
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of ma...@unizar.es
Sent: Saturday, August 28, 2010 5:08 PM
To: r-help@r-project.org
Subject: [R] star models
Hi,
I am traying to implement an STAR
Hello all,
I've 2 strings that representing the start and end values of a date and
time.
For example,
time1 <- c("21/04/2005","23/05/2005","11/04/2005")
time2 <- c("15/07/2009", "03/06/2008", "15/10/2005")
as.difftime(time1,time2)
Time differences in secs
[1] NA NA NA
attr(,"tzone")
[1] ""
How ca
..i'd like to add that i actually wanted to test the location of differences
of paired samples coming from an non-normal asymetric distribution. the
alternative hypothesis was that negative differences are more often than in
0.5 of all cases. thus i tested
(x=nr.diff.under.0,n=all.diffs,0.5,altern
if I try as.dist I get the following error:
On Thu, 2010-09-02 at 09:57 -0400, Sarah Goslee wrote:
> > mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3))
> > as.dist(cor(mydata))
>x1 x2
> x2 -0.5960396
> x3 0.3973597 0.500
print(xtable(as.dist(cor(mydata)),digits=3)
try lower.tri
and see
??lower.tri
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147
oops sorryreally careless. thanks !
On Thu, Sep 2, 2010 at 10:03 AM, Alain Guillet
wrote:
> Hi,
>
> Are you sure you don't want to do ks.test(y, punif, min=0, max=1,
> alternative="greater") instead of what you tried?
>
> Alain
>
>
> On 02-Sep-10 15:52, Samsiddhi Bhattacharjee wrote:
>>
>>
Hi,
Are you sure you don't want to do ks.test(y, punif, min=0, max=1,
alternative="greater") instead of what you tried?
Alain
On 02-Sep-10 15:52, Samsiddhi Bhattacharjee wrote:
ks.test(y, runif, min=0, max=1, alternative="greater")
--
Alain Guillet
Statistician and Computer Scientist
Like this?
> mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3))
> as.dist(cor(mydata))
x1 x2
x2 -0.5960396
x3 0.3973597 0.500
Sarah
On Thu, Sep 2, 2010 at 9:51 AM, Olga Lyashevska wrote:
> Dear all,
>
> mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3))
> cor(
I was just testing out ks.test()
> y <- runif(1, min=0, max=1)
> ks.test(y, runif, min=0, max=1, alternative="greater")
One-sample Kolmogorov-Smirnov test
data: y
D^+ = 0.9761, p-value < 2.2e-16
alternative hypothesis: the CDF of x lies above the null hypothesis
>
It seems that eve
Dear all,
mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3))
cor(mydata)
x1 x2x3
x1 1.000 -0.5960396 0.3973597
x2 -0.5960396 1.000 0.500
x3 0.3973597 0.500 1.000
I wonder if it is possible to fill only lower triangle of this
correlation matr
I think I found a simple solution, although it requires some tinkering to
find the x,y coordinates of the plot region to place the text...
barplot ( )
text(x= 2.9, y = 0.43, srt=90, labels = "H", cex = 1.5, col="blue") #srt
rotates the "H" character, so that it resembles an error bar
text(x=3.5
thanks a lot for the elaborations.
your explanations clearly brought to me that either
binom.test(1,1,0.5,"two-sided") or binom.test(0,1,0.5) giving a
p-value of 1 simply indicate i have abolutely no ensurance to reject H0.
considering binom.test(0,1,0.5,alternative="greater") and
binom.
Dear Thierry,
Thanks for the quick answer. I'm moving this to r-sig-mixed-models
(but also posting on r-help to notify).
I reserved "Mixed-effects models in S and S-PLUS" by Pinheiro and
Bates, New York : Springer, 2000. Do you know any other good
references?
Cheers, Mikkel.
2010/9/2 ONKELINX,
The absence of stepwise methods works to your advantage, as these
yield invalid statistical inference and inflated regression
coefficients.
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Thu, 2
Hello all
I wonder what can i use to cluster vectors which composed of several
factors.
lets say around 30 different factors compose a vector, and if the factor is
present then it encoded as 1, if not presented then it will be encoded as 0.
I was thinking of using hierarchical clustering, as i kn
Dear Mikkel,
You need to do some reading on terminology.
In your model the fixed effects are channel 1, 2 and 3. samplenumber is
a random effect and the error term is an error term
The model you described has the notation below. You do not need to
create the grouped data structure.
lme(channel0
Hi:
One option to search for functions in R is to download the sos package:
library(sos)
findFn('Anderson-Darling')
On my system (Win 7), it found 51 matches. The two likeliest packages are
nortest and ADGofTest. To answer your question, nortest still exists on
CRAN. I can't comment on Ubuntu, b
Hi:
Thanks, David and Thierry. I knew what I did was inefficient, but I'm not
very adept with cast() yet. Thanks for the lesson!
The time is less than half without the fun = mean statement, too. On my
system, the timing of David's call was 2.06 s elapsed; with Thierry's, it
was 4.88 s. Both big im
Dear R-community,
I'm analysing some noise using the nlme-package. I'm writing in order
to get my usage of lme verified.
In practise, a number of samples have been processed by a machine
measuring the same signal at four different channels. I want to model
the noise. I have taken the noise (the s
On 09/02/2010 08:50 PM, Jens Oldeland wrote:
> ...
> I am looking for a way to add labels, i.e. absolute values, into a
> stacked bar chart using the basic plot functions of R. The labels
> should be inside the stacked bars.
barpos<-barplot(height,beside = FALSE,
horiz=TRUE,col = c(2, 3))
librar
Hi,
I'm trying to run an anderson-darling test for normality on a given variable
'Y':
ad.test(Y)
I think I need the 'nortest' package, but since it does not appear in any of
the Ubuntu repositories for 2.10.1, I am wondering if it goes by the name of
something else now?
Thanks
--
View this me
It is dependent on another dll but it did not give compilation errors. It
seemed to link fine at that point. Why does it have a problem at this stage?
From: "Duncan Murdoch"
To: raje...@cse.iitm.ac.in
Cc: "r-help"
Sent: Thursday, September 2, 2010 4:05:14 PM
Subject: Re: [R] R CMD check P
Dear r-help,
I am using CAH. I would cut my dendogram. What is the command in R that
allows draw a graph of the semi-partial R-squared ?
Best Regards
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On 02/09/2010 6:46 AM, raje...@cse.iitm.ac.in wrote:
It is dependent on another dll but it did not give compilation errors. It seemed to link fine at that point. Why does it have a problem at this stage?
Windows needs to be able to find the other DLL at load time. It will
find it if it's in t
Hi,
I am looking for a way to add labels, i.e. absolute values, into a
stacked bar chart using the basic plot functions of R. The labels should
be inside the stacked bars.
For example,
### I have this dataset
height = cbind(x = c(465, 91) / 465 * 100,
y = c(840, 200)
picking up on Thierry's example, I don't think you need any function because
you are just reshaping
(not aggregating). Therefore:
bigtab2 <- cast(data = big, study + subject + cycle + day ~type, value =
"obs")
> head(bigtab2)
study subject cycle day ALB ALP ALT AST
1 1 1 1 1
On 02/09/2010 2:29 AM, raje...@cse.iitm.ac.in wrote:
Hi,
I've built my own package in windows and when I run R CMD check Package-Name I
get,
* install options are ' --no-html'
* installing *source* package 'AceTest' ...
** libs
making DLL ...
g++ ...etc.
installing to
... done
** R
** p
Dear Dennis,
cast() is in this case much faster.
> system.time(bigtab <- ddply(big, .(study, subject, cycle, day),
function(x) xtabs(obs ~ type, data = x)))
user system elapsed
35.360.12 35.53
> system.time(bigtab2 <- cast(data = big, study + subject + cycle + day
~type, value = "ob
On 09/02/2010 04:56 AM, josef.kar...@phila.gov wrote:
I have a simple barplot of 4 mean values, each mean value has an
associated 95% confidence interval drawn on the plot as an error bar.
I want to make a legend on the plot that uses the error bar symbol, and
explains "95% C.I."
How do I show th
Hi:
I did the following test using function ddply() in the plyr package on a toy
data frame with 5 observations using five studies, 20 subjects per
study, 25 cycles per subject, five days per cycle and four observations by
type per day. No date-time variable was included.
# Test data frame
bi
Hi ,
I use library(gee),library(geepack),library(yags) perform GEE data analysis
, but all of them cannot do variable selection!
Both step and stepAIC can do variable selection based on AIC criterion under
linear regression and glm,
but they cannot work when model is based on GEE.
I want to a
You state: "in reverse the p-value of 1 says that i can 100% sure
that the estimate of 0.5 is true". This is where your logic about
significance tests goes wrong.
The general logic of a singificance test is that a test statistic
(say T) is chosen such that large values represent a discrepancy
betw
Hi,
I didn't find any post on this subject so I ll ask you some advices.
Let's say that I have two library trees.
Number 1 is the default R library tree on path1
Number 2 is another library tree on a server with all packages on path2.
When I set library(aaMI,lib.loc=paths2) it loads the package
Dear Greg,
First convert your data.frame to a long format using melt(). Then use
ddply() to calculate the averages. Once you get at this point is should
be rather straightforward.
library (ggplot2)
v1 <- c(1,2,3,3,4)
v2 <- c(4,3,1,1,9)
v3 <- c(3,5,7,2,9)
gender <- c("m","f","m","f","f")
d.da
On 02-Sep-10 07:16:54, khush wrote:
> Hi,
>
> I am using vim to edit my files...I do not know what has been
> pressed by me as my R code get converted to such a crap...
>
> %PDF-1.4
> %<81>â<81>ã<81>Ï<81>Ó\r
> 1 0 obj
> <<
> /CreationDate (D:20100902122215)
> /ModDate (D:20100902122215)
i test the null that the coin is fair (p(succ) = p(fail) = 0.5) with
one trail and get a p-value of 1. actually i want to proof the
alternative H that the estimate is different from 0.5, what certainly
can not be aproven here. but in reverse the p-value of 1 says that i
can 100% sure that t
On Sep 2, 2010, at 09:16 , khush wrote:
> Hi,
>
> I am using vim to edit my files...I do not know what has been pressed by me
> as my R code get converted to such a crap...
>
> %PDF-1.4
> %<81>â<81>ã<81>Ï<81>Ó\r
> 1 0 obj
> <<
> /CreationDate (D:20100902122215)
> /ModDate (D:2010090212
Hello,
I have a data.frame with the following format:
> head(clin2)
Study Subject Type Obs Cycle Day Date Time
1 A001101 10108 ALB 44.098 1 2004-03-11 14:26
2 A001101 10108 ALP 95.098 1 2004-03-11 14:26
3 A001101 10108 ALT 61.098 1 2004-0
Hi,
I am using vim to edit my files...I do not know what has been pressed by me
as my R code get converted to such a crap...
%PDF-1.4
%<81>â<81>ã<81>Ï<81>Ó\r
1 0 obj
<<
/CreationDate (D:20100902122215)
/ModDate (D:20100902122215)
/Title (R Graphics Output)
/Producer (R 2.11.1)
/Creator (R)
how t
Hi
I'm sorry, but I don't think that coherence is the same as the cross
spectrum. People use coherence since it is much easier to deal with. I know
how by using R to plot and calculate the coherence and phase, but what I
didn't know is how to calculate the cross spectrum by using R.
Regards
Aft
On Thu, Sep 2, 2010 at 7:17 AM, Yi wrote:
> Hi, folks,
>
> runif (n,min,max) is the typical code for generate R.V from uniform dist.
>
> But what if we need to fix the mean as 20, and we want the values to be
> integers only?
It's not clear what you want. Uniformly random integers with expected
Hi Dejian,
You're right on this! Do you know how to pass those two argument into
lower.panel? Thanks!
...Tao
From: Dejian Zhao
To: r-help@r-project.org
Sent: Tue, August 31, 2010 6:10:16 PM
Subject: Re: [R] "pairs" with same xlim and ylim scale
I think you
Hi,
I have to compute the singular value decomposition of rather large
matrices. My test matrix is 10558 by 4255 and it takes about three
minutes in R to decompose on a 64bit quadruple core linux machine. (R is
running svd in parallel, all four cores are at their maximum load while
doing this.) I
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