Hi,
I have a dataset where each case is characterized by a histogram. I
would like to cluster these cases using a sensible distance measure,
possibly relative entropy? Is there a way I can use R facilities to do
this (hclust etc.). I couldn't find an alternative to dist that would
compute
Hi,
I what to solve this problem:
alfab - ABCEDFG#[1] ABCEDFG
chara - strsplit(alfab, )#[1] A B C E D
F G
Then I do some changes before I want the character together again, say,
remove two letters.
Now, chara is A B E D G.
Is there any opposite
Ragnhild Sørum wrote:
Hi,
I what to solve this problem:
alfab - ABCEDFG#[1] ABCEDFG
chara - strsplit(alfab, )#[1] A B C E
D F G
Then I do some changes before I want the character together again, say,
remove two letters.
Now, chara is A B E D G.
?paste
On Fri, 7 Nov 2003, Ragnhild Sørum wrote:
I what to solve this problem:
alfab - ABCEDFG#[1] ABCEDFG
chara - strsplit(alfab, )#[1] A B C E D
F G
Then I do some changes before I want the character together again, say,
remove two
Hi all;
could you remind me what is the function to change the precision of the
operations done in R? I can't remember nor find it.
Best regards
Javier
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Gabor == Gabor Grothendieck [EMAIL PROTECTED]
on Thu, 6 Nov 2003 15:33:04 -0500 (EST) writes:
Gabor Its also possible to avoid these intricacies by not
Gabor using an intermediate text representation, i.e. csv,
Gabor in the first place.
Gabor The following R code uses the
Pascal A. Niklaus wrote:
Hi all,
I'm not sure how to correctly analyse the following data with glm, and
hope for some advice from this list, ideally showing how to specify
the model in R and perform the tests, and also for suggestions of
literature.
The data structure is like this:
- 20
javier garcia - CEBAS wrote:
Hi all;
could you remind me what is the function to change the precision of the
operations done in R? I can't remember nor find it.
There isn't one. R does all its calculations with system defined
double-precision or integers (on most platforms where R is used,
Hi,
The R FAQ suggests that we can cite Ihaka and Gentleman (1996) in
publications. Does anyone have an electronic copy of this that could
be (legally) made publically available -- my library here doesn't take
this journal (Journal of Computational and Graphical Statistics).
Otherwise, I can
Hi,
I'm trying to work out how the nlme function estimates the variances of
the fixed effects parameters, so I tried to look at the code for these
functions: summary.nlme, summary.lme, MEestimate.
MEestimate
Error: Object MEestimate not found
summary.nlme
Error: Object summary.nlme not found
What's going on?
Namespaces. See `Writing R Extensions'.
?getS3method
?getAnywhere
will show these to you.
It is probably easier to read the source code for the package, though.
On Fri, 7 Nov 2003, Simon.Bond wrote:
Hi,
I'm trying to work out how the nlme function estimates the
On Fri, 7 Nov 2003 10:32:44 +0100, Martin Maechler wrote:
1) dataload is *not* free software in the sense of the
Free Software Foundation (which has existed for a much
longer time than MS windows!):
Actually they're contemporaneous. MS Windows version 1.0 came out in
1985, the same
While I don't disagree with what you say, the purpose of this
is to interface to Excel which is even less free (you
have to pay for Excel but not for dataload) so
perhaps the status of the glue used between R and Excel
is not as important.
From an expediency viewpoint, I found that
I am trying to plot positions on a grid where the x and y axis equate to
longitudinal and latitudinal co-prdinates respectively. As these
co-ordinates are southings and westings, i need the origin of my graph to
contain the two highest values of each co0ordinate, with the values
decreasing in both
Assuming the start tag contains the word start and the end tag
contains the word end you could do this:
lines - readLines( input.txt )# lines is a vector of lines
g - grep( start|end, lines ) # positions of tags
lines - lines[ seq( g[1]+1, g[2]-1 ) ]
mydata - read.table(
Jason Turner [EMAIL PROTECTED] writes:
javier garcia - CEBAS wrote:
Hi all;
could you remind me what is the function to change the precision of
the operations done in R? I can't remember nor find it.
There isn't one. R does all its calculations with system defined
Prof Brian Ripley [EMAIL PROTECTED] writes:
What's going on?
Namespaces. See `Writing R Extensions'.
?getS3method
?getAnywhere
will show these to you.
It is probably easier to read the source code for the package, though.
When reading the source code for the nlme package is the
On Fri, 2003-11-07 at 04:40, Jason Turner wrote:
javier garcia - CEBAS wrote:
Hi all;
could you remind me what is the function to change the precision of the
operations done in R? I can't remember nor find it.
There isn't one. R does all its calculations with system defined
I didn't realise version 1 of the R programming language was quite so old:
http://www.jbum.com/idt/r.html
Does anyone still have hardware that runs it?
http://www.jbum.com/idt/hyperboreans.html
Baz
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Hi,
I use a little script¹ to make a chi-square-test on 162
factors (it makes no difference if I take the numeric variant
of the factors). At factor nr. 4 is stops with an error:
[1] v1= V7.KARTM v11= V7.KAR1M
Error in chisq.test(d1, d2) : x and y must have at least 2 levels
But x and y /have/
Need to wrap the xyplot() in print(), as in
print(xyplot(...))
Andy
From: Dr. Peter Schlattmann
Dear all,
I am trying to use xyplot inside a function
plotme-function(dataframe)
{
xyplot(x~y|z,data=dataframe)
}
x,y,z are members of the data frame.
When calling function
On Fri, 2003-11-07 at 07:54, Barry Rowlingson wrote:
I didn't realise version 1 of the R programming language was quite so old:
http://www.jbum.com/idt/r.html
Does anyone still have hardware that runs it?
http://www.jbum.com/idt/hyperboreans.html
Baz
I think that you just answered
If you want to preserve the aspect ratio of x and y axis, use asp=1
within the plot function.
To reverse the direction of the x and y axes, if I understand
correctly what you are trying to do, I am not sure. There may be more
efficient ways. What I would do is something along the following
Christoph Bier [EMAIL PROTECTED] writes:
Hi,
I use a little script¹ to make a chi-square-test on 162 factors (it
makes no difference if I take the numeric variant of the factors). At
factor nr. 4 is stops with an error:
[1] v1= V7.KARTM v11= V7.KAR1M
Error in chisq.test(d1, d2) : x and
Marc Schwartz [EMAIL PROTECTED] writes:
On Fri, 2003-11-07 at 07:54, Barry Rowlingson wrote:
I didn't realise version 1 of the R programming language was quite so old:
http://www.jbum.com/idt/r.html
Does anyone still have hardware that runs it?
On Fri, 2003-11-07 at 07:55, Christoph Bier wrote:
Hi,
I use a little script to make a chi-square-test on 162
factors (it makes no difference if I take the numeric variant
of the factors). At factor nr. 4 is stops with an error:
[1] v1= V7.KARTM v11= V7.KAR1M
Error in chisq.test(d1,
Greetings.
There seems to be a problem with the P-value computation in the
cor.test with method=spearman. In R1.8.0 (MS Windows) I
seem to be getting intermittently nonsense P-values, but the rho's
are OK. I can get this reproducibly with the toy example attached
where the first use is OK
Hi,
I found two sorts of functions to use the SOM algorithm, one in the package
som and the other one in the package class, but none of them deal with
the missing values.
The som function in package som gives results with option init=random,
but the value of the code vector has missing values
It works fine for me, on a Sun.
x-1:100
y-rep(c(2,3,4,5),25)
cor.test(x,y,m=p)
Pearson's product-moment correlation
data: x and y
t = 0.3837, df = 98, p-value = 0.702
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.1588952 0.2333745
On Fri, 2003-11-07 at 08:45, Steve Roberts wrote:
Greetings.
There seems to be a problem with the P-value computation in the
cor.test with method=spearman. In R1.8.0 (MS Windows) I
seem to be getting intermittently nonsense P-values, but the rho's
are OK. I can get this reproducibly
As Peter said, the standard chi-square for a 2x2 table won't work for
your data. However, if you tabulate the numbers first and then ask for
the chi-square, you get a warning, not an error:
contable - array(c(42, 0,0,0), dim=c(2,2))
chisq.test(contable)
Pearson's Chi-squared test
Marc Schwartz wrote:
[answer to my problem]
Thanks!
BTW to the german speaking readers: What's the R pendant to or
translation for Konfigurationsfrequenzanalyse (KFA)?
I am not a German speaker, however:
Analysis of Configuration Frequencies (CFA)?
Sounds reasonable ;-).
There is a
I am trying to work through CircStats (R 1.8, CircStats just downloaded
last week). circ.plot just does not work. I am plotting about 800 points
and they come out very different (and apparently correct) in rose.diag,
but circ.plot does not work (high angles or low negative angles are not
Peter Dalgaard wrote:
Well, the error message might be slightly beside the point, but the
issue would seem to be that there are no ja's inside either vector.
I.e. it first reduces each factor to those levels that are actually
present, then checks whether there are at least two levels.
Thanks for
Hello,
Can anyone tell me how to label individual bars on a barplot? I want to put an * or
letter ABOVE the bar to denote statistical significance. Is this possible and how?
Thanks,
Suzanne
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McAfee VirusScan Online from the
Spencer Graves wrote:
As Peter said, the standard chi-square for a 2x2 table won't work for
your data. However, if you tabulate the numbers first and then ask for
the chi-square, you get a warning, not an error:
Thanks for the solution of my problem!
Best regards,
Christoph
On Fri, 2003-11-07 at 09:46, Suzanne E. Blatt wrote:
Hello,
Can anyone tell me how to label individual bars on a barplot? I want
to put an * or letter ABOVE the bar to denote statistical
significance. Is this possible and how?
Thanks,
Suzanne
You might want to take a look at the most
On Fri, 7 Nov 2003, Christoph Bier wrote:
Peter Dalgaard wrote:
Well, the error message might be slightly beside the point, but the
issue would seem to be that there are no ja's inside either vector.
I.e. it first reduces each factor to those levels that are actually
present, then
Hi!
On Fri, Nov 07, 2003 at 10:46:27AM -0500, Suzanne E. Blatt wrote:
Can anyone tell me how to label individual bars on a barplot? I want
to put an * or letter ABOVE the bar to denote statistical
significance. Is this possible and how?
You can use text() to put arbitrary strings in
Hi,
I would like to know if anyone knows how to draw a plot with two Y-axises and
one X-axis? When you have two sets of y values that do not have the same
scale, but correspond to the same x value, I would like to plot them on one
graph.
Could you please help me?
Thank you
Soyoko
Stephen Ellner schrieb:
I think an additional problem is that the derivatives function
must be 'told' where to 'look' in y for the values of R, C, and P
that are used to compute Rprime,Cprime, Pprime - it has no way of
'knowing' that the model's state vector y consists of
the variables (R,C,P).
Dear All,
I am new to R and am trying to learn how to create functions using R.
Below is code which calculates Lin's Concordance Coefficient. After
I calculate the coefficient I want to create a scatter plot which
annotates the coefficient along with preceding text onto the plot.
The below code
You want to use paste() inside text(), not cat().
Andy
From: Craig H. Ziegler [mailto:[EMAIL PROTECTED]
Dear All,
I am new to R and am trying to learn how to create functions
using R. Below is code which calculates Lin's Concordance
Coefficient. After I calculate the coefficient I
Try using paste() instead of cat() in
text(20,60,cat(Lin's Concordance Coefficient = ,lincc))
The last expression in your function will be returned as the value of
the function; that is the best way to get results out of a function
and available for use outside the function.
Hi,
On Fri, 7 Nov 2003, umeno wrote:
I would like to know if anyone knows how to draw a plot with two Y-axises and
one X-axis? When you have two sets of y values that do not have the same
scale, but correspond to the same x value, I would like to plot them on one
graph.
I think both
umeno wrote:
Hi,
I would like to know if anyone knows how to draw a plot with two Y-axises and
one X-axis? When you have two sets of y values that do not have the same
scale, but correspond to the same x value, I would like to plot them on one
graph.
Could you please help me?
Please look
Hi,
I am trying to figure out how to lable the second y-axis after the following
codes:
plot(x, y,
xlab=time,
ylab=pay1
)
par(new=TRUE)
plot(x,y2,
ann=FALSE,
xaxt=n,
yaxt=n,
pch=7
)
axis(4)
Then, I want to label the second y axis
I have x,y data and would like to compute the average of y at each level
of x.
x
[1] 0.006110 0.007027 0.007027 0.007027 0.008081 0.008081 0.008081 0.008081
[9] 0.008081 0.008081 0.008081 0.009293 0.009293 0.009293 0.009293 0.009293
[17] 0.009293 0.009293 0.009293 0.010686 0.010686 0.010686
Try (untested):
p1 - tapply(y, x, function(z) mean(z == 0))
HTH,
Andy
From: Bill Simpson [mailto:[EMAIL PROTECTED]
I have x,y data and would like to compute the average of y at
each level
of x.
x
[1] 0.006110 0.007027 0.007027 0.007027 0.008081 0.008081
0.008081 0.008081 [9]
Hi,
On Fri, 7 Nov 2003, umeno wrote:
Then, I want to label the second y axis pay2. I tried title(ylab=pay2),
but it put this lable on the first y-axis. Does anyone know how to move this
to the second y-axis?
?mtext will do what you want.
For example, here is a few codes from Venables
Douglas Bates wrote:
Jason Turner [EMAIL PROTECTED] writes:
...
There isn't one. R does all its calculations with system defined
double-precision or integers (on most platforms where R is used, these
are both 32 bits). See help(is.single)
Usually the double-precision representation is 64 bits
Hello,
Thanks for the suggestion below. I did one of my figures and it worked perfectly. I
tried it on my second figure and got double-labelling on the first bar. I used the
same code to create both figures, and the same code to label. I've attached the code
for one of the plots (there
one way that works is:
sapply(split(y,x), mean)
Bill
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Should I be able to use axis() on a barplot? i have a data.frame, the first
3 values of which are:
c[1:3,]
median mean
A156.5 58.5
A61 73.0 73.0
A62 63.0 63.0
str(c)
`data.frame': 19 obs. of 2 variables:
$ median: num 56.5 73 63 161 51 55 44.5 22 54 49 ...
On Fri, Nov 07, 2003 at 03:16:13PM -0500, Suzanne E. Blatt wrote:
Thanks for the suggestion below. I did one of my figures and it
worked perfectly. I tried it on my second figure and got
double-labelling on the first bar. I used the same code to create
both figures, and the same code to
On Fri, 2003-11-07 at 14:29, Siddique, Amer wrote:
Should I be able to use axis() on a barplot? i have a data.frame, the first
3 values of which are:
c[1:3,]
median mean
A156.5 58.5
A61 73.0 73.0
A62 63.0 63.0
str(c)
`data.frame': 19 obs. of 2
Thanks!
That did it!!
Suzanne
Philipp Pagel [EMAIL PROTECTED] wrote:
On Fri, Nov 07, 2003 at 03:16:13PM -0500, Suzanne E. Blatt wrote:
Thanks for the suggestion below. I did one of my figures and it
worked perfectly. I tried it on my second figure and got
double-labelling on the first
Hello
I am writing to you regarding your interest in Copula/copulae and it´s
usage. I am currently studying Copula for my Master thesis and also
therefore have a large interest in it. Now I am looking for information
regarding copula as well as Copula and R.
Code, information how to calculate
Dear R users,
I'm trying to use rpart() to build a classification tree on a big dataset.
The number of samples is n=100 and the number of variables is p=1.
At first I stored all the data in a data.frame and got a stack overflow
error; then I changed the data into a matrix and the problem
Try something like this (suppose x is the matrix of predictors in the
training set, and xtest is the same for the test set):
my.rp - rpart(y ~ x, ...)
test.pred - predict(my.rp, newdata=data.frame(x=I(xtest)))
Make sure the name of the variable in the data frame given to newdata
matches the name
Original Message
Subject: Sun Solaris 8 compile problem.
Date: Thu, 06 Nov 2003 23:03:36 -0700
From: Jack Matthews [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
I am new to the list was wondering if anyone can help me find the reason
and remedy for compiling R on an ultrasparc 60
On Fri, Nov 07, 2003 at 06:19:26PM -0700, Jack Matthews wrote:
I am new to the list was wondering if anyone can help me find the reason
and remedy for compiling R on an ultrasparc 60 solaris 8 with gcc 3.2.2
in 64 bit mode? Fails in checking for a fortran library.
RTFM, in this case the
On 7 Nov 2003 at 8:19, Thomas Lumley wrote:
Just to make the point more clear:
fisher.test( matrix(c(42,0,0,0),2,2))
Fisher's Exact Test for Count Data
data: matrix(c(42, 0, 0, 0), 2, 2)
p-value = 1
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence
On 7 Nov 2003 at 9:47, Bill Simpson wrote:
Since your y is 0-1 the following works:
tapply(x, y ,mean)
aggregate(x, y, mean)
gives the output in a different format.
Kjetil Halvorsen
I have x,y data and would like to compute the average of y at each level
of x.
x
[1] 0.006110 0.007027
That's not a sensible thing to do. Supply predict.rpart with a data frame
that contains just the variables rpart selected.
R does have limits, and attempting to use 10,000 variables is hitting
them, But surely any statistician is aware of the dangers of selecting
from 1 variables on just
Actually not, 3.2.2 works in 64-bit mode (although I would recommend a
later version of gcc).
The issue seems more basic: the environment does not have a working
compiler. A likely reason is the the 64-bit shared libraries are not in
the LD_LIBRARY_PATH, or not high enough up it. That
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