one solution is:
img1-matrix(1:5)
img2-matrix(2:5)
col-1:5 # col-c(green,yellow,...)
image(img1,col=col[sort(unique(img1))])
image(img2,col=col[sort(unique(img2))])
On 12/26/06, Milton Cezar Ribeiro [EMAIL PROTECTED] wrote:
Dear All,
How can I define a color sequence for each image value? I
On Sun, 24 Dec 2006, John Fox wrote:
If I remember Freedman's recent paper correctly, he argues that sandwich
variance estimator, though problematic in general, is not problematic in the
case that White described -- an otherwise correctly specified linear model
with heteroscedasticity
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cG9zZSBJIGhhdmUgYSB2ZWN0b3IgeCwgYW5kIEkgd2FudCB0byBjYWxjdWxhdGUgeVtpXT14
W2krMV0teFtpXSwgaXQgaXMgdmVyeSBlYXN5LiBJIGp1c3QgbmVlZCB0byB3cml0ZSB5PC14
WzI6bGVuZ3RoKHgpXS14WzE6bGVuZ3RoKHgpLTFdLiANCiANCk5vdyBpZiBJIGtub3cgeSwg
Dear Users,
I am new to R. I use write() to write my data in .txt format. I'd like
to write to a disc any kind of data in a .dat S-PLUS format.
Please help.
SM
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PLEASE
On Dec 26, 2006, at 12:07 AM, Pedro Mardones wrote:
I'm wondering if there is any 'efficient' approach for selecting a
sample of 'every nth rows' from a dataframe. For example, let's use
the dataframe GAGurine in MASS library:
length(GAGurine[,1])
[1] 314
# select an 75% of the dataset,
Hi Everyone,
I am stuck with a simple problem. Suppose I have a vector x, and I want
to calculate y[i]=3Dx[i+1]-x[i], it is very easy. I just need to write
y-x[2:length(x)]-x[1:length(x)-1].
Now if I know y, and want to know the vector x defined by
x[i]=3Dx[i-1]+y[i-1] for all i, how can I do
x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop?
It looks like
x - cumsum(y)
What does 3D mean?
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PLEASE do read the posting guide
Hi Richard,
3D is automatically generated by the mailing list software, probably
because I had ] followed by =3D without a space in the original post.
What I meant was to calculate x[i] =3D x[i-1] + y[i-1]
For example, if X - 1:10
Then I want the vector Y to be 1, 3, 6, 10, 15, 21, 28, 36,
x[i]=3Dx[i-1]+y[i-1] for all i, how can I do this without a loop?
It looks like
x - cumsum(y)
What does 3D mean?
The =3D is probably an encoding error - it should just be =.
In general to vectorise an iterative problem, you will need to solve
the recurrence relation
I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below.
The mailing list software just keep adding 3D's. Sorry.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
Sent: Tuesday, December 26, 2006 10:03 AM
To: Richard M. Heiberger;
In your case, the recurrence relationship for x can be solved easily:
Notice that
sum{i=1,n}(x[i]-x[i-1]) = x[n] - x[0]
and therefore
x[n] = x[0] + sum{i=1,n}(y[i-1] for n=1, N, with the appropriate initial
condition for i=0, (x[0],y[0]).
Thus cumsum on y will give you a direct answer.
What you describe is called stratified sampling. It was discusssed last
month (and other times) on this list:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/90220.html
Using
RSiteSearch(stratified sampling)
will produce many hits to relevant articles and packages.
On Mon,
Yes, this solves my problem. Thanks for your help.
-Original Message-
From: Christos Hatzis [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 26, 2006 10:58 AM
To: Geoffrey Zhu; r-help@stat.math.ethz.ch
Subject: RE: [R] vectorizing an iterative process.
In your case, the recurrence
I am looking for hints on how to estimate ratings for competitors
in an ongoing pairwise competition using R... my particular area of
interest being the game of Go, but the idea of identifying ratings
(on a continuous scale) rather than relative rankings seems easily
generalized to other
Geoffrey Zhu wrote:
I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below.
The mailing list software just keep adding 3D's. Sorry.
Rather, I suspect that *your* mailer is sending in Quoted-Printable,
without setting the appropriate headers. Take a look at
Have you considered Bradley-Terry models? RSiteSearch(bradley,
functions) just returned 31 hits for me.
Hope this helps.
Spencer Graves
Jeff Newmiller wrote:
I am looking for hints on how to estimate ratings for competitors
in an ongoing pairwise competition using R...
There is a substantial literature on 'statistics in sports' and pairwise
comparisons are of obvious interest. Here is a starting point:
http://www.amstat.org/sections/sis/
You might browse the newsletters posted there.
You might enjoy:
Bridging Different Eras in Sports by Scott M.
dear R experts:
This is just a minor, minor nuisance, but I thought I would point it out:
dataset - read.table(file=pipe(cmdline), header =T,
+ na.strings=c(NaN, C,I,M, E), sep=,,
as.is=T, nrows=);
Error: cannot allocate vector of size 781249 Kb
If I extend nrows by a
I would start with elimination-by-aspects models:
?eba
I would read the Tversky 1972 paper (cited on the help page for the
eba() function), which is brilliant.
Jeff Newmiller wrote:
I am looking for hints on how to estimate ratings for competitors
in an ongoing pairwise competition using
One approach that is already coded in R is the Bradley-Terry model
(found in the BradleyTerry package of all places).
This could be a good place to start if you want something quick, others
have given you references if you want more detail and/or control.
--
Gregory (Greg) L. Snow Ph.D.
Hello,
I am hoping someone can clarify why I might obtain a quite different value
in R SPSS for a McNemar test I ran.
Firstly, here is the R syntax output
R OUTPUT
mctest - as.table(matrix(c(128,29,331,430),
+ ncol =2, dimnames = list(group=c(preMHT,postMHT),
+ assault=c(yes,no
Bob Green wrote:
Hello,
I am hoping someone can clarify why I might obtain a quite different value
in R SPSS for a McNemar test I ran.
Firstly, here is the R syntax output
R OUTPUT
mctest - as.table(matrix(c(128,29,331,430),
+ ncol =2, dimnames = list(group=c(preMHT,postMHT),
+
On Mon, 25 Dec 2006, Achim Zeileis wrote:
On Sun, 24 Dec 2006, John Fox wrote:
If I remember Freedman's recent paper correctly, he argues that sandwich
variance estimator, though problematic in general, is not problematic in the
case that White described -- an otherwise correctly specified
Peter,
Thanks for your reply. To perform the analysis in R, I used the table
format suggested in the book by Everitt Hothorn, whereas in SPSS the
analysis was performed directly from the 2 variables, rather than using
count data.
There is still something I don't understand - I tried to
Bob Green wrote:
Peter,
Thanks for your reply. To perform the analysis in R, I used the table
format suggested in the book by Everitt Hothorn, whereas in SPSS the
analysis was performed directly from the 2 variables, rather than using
count data.
You still need the right table.
Hello,
I have a longitudinal data with about 30 subjects. I used xyplot() to plot
the longitudinal data. One problem is that xyplot() recycles the color of
auto.key so that every 7th subject has the same color (symbol if setps() was
used). Is there a way so that every subject will have a unique
Set the colors yourself using the par.settings= argument of xyplot or
trellis.par.set. See ?xyplot and ?trellis.par.set for more info.
If you follow the instructions on the last line of every post to r-help you
will get more detailed answers.
On 12/26/06, Osman Al-Radi [EMAIL PROTECTED] wrote:
Frank,
You might want to use a package that is associated with some textbook that
can give you guidance and examples. A lot of people would do this with
WinBUGS, which is usually run separately although it has some kind of R
interface. I'm sure there are lots of nice texts about that
Hi all,
I am using the zoo package to plot time series. I have a problem with
formatting the axes.
my zoo object (z) looks like the following.
c1
1992-01-10 21
1992-01-17 34
1992-01-24 33
1992-01-31 41
1992-02-07 39
1992-02-14
Try this:
# test data
library(zoo)
z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40),
index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086,
8093, 8100, 8107), class = Date), class = zoo)
z
# plot without X axis
plot(z, xaxt = n)
# unlabelled tick at each point
axis(1,
Dear Gabor,
Thank you for your quick reply.
This solution works for my univariate zoo class time series. I first tried
it for a timeseries with 4 columns of data, it did not plot the labels nor the
ticks, I tried it on a one dim timeseries (one column zoo class data as the
example
in the
Hi Sebastian
try
save(x,y, file = filename.txt, ascii = TRUE)
where x, y are your R objects. You should read
?save
good luck
AA.
- Original Message
From: Sebastian Michalski [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Monday, December 25, 2006 7:36:50 AM
Subject: [R] writing to
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's rows into different data frames
according to the levels of 'aa' column,
try ?assign
On 12/26/06, jingjiangyan [EMAIL PROTECTED] wrote:
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's
Please read the last line of every message to r-help and follow that.
On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Dear Gabor,
Thank you for your quick reply.
This solution works for my univariate zoo class time series. I first tried
it for a timeseries with 4 columns of data, it did not
In the following the components of ss are the data frames in question:
ss - split(df, df$aa)
On 12/26/06, jingjiangyan [EMAIL PROTECTED] wrote:
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6
Peter,
I now see the original E H table was based on matched pairs not the raw
counts. I now understand this much better and have the syntax generates
results that correspond with your results (and SPSS),
Thanks again,
Bob
__
Spencer Graves wrote:
Have you considered Bradley-Terry models? RSiteSearch(bradley,
functions) just returned 31 hits for me.
Hope this helps. Spencer Graves
Thanks to everyone who responded... this was very helpful. I have a bit of
reading and investigation to do.
I think
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