Hi All,
I have some data I need to write as a file from R to use in a different
program.
My data comes as a numeric matrix of n rows and 2 colums, I need to transform
each row as a two rows 1 col output, and separate the output of each row with a
blanck line.
Foe instance I need to go from
Hi All,
I am using the function fractions() and cognates from MASS. I would
like to be able to tell if some calculations I am doing on some
rationals are transformed into floats and then retransformed into
rationals.
For instance I suspect that
as.fractions(1/8) + as.fractions(1/4)
does
Hi All,
I have a n x m matrix. The n rows are individuals, the m columns are variables.
The matrix is in itself a collection of 1s (if a variable is observed for an
individual), and 0s (is there is no observation).
Something like:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]1011
Dimitris Rizopoulos wrote:
you could also have a look at function lca() from package `e1071' that
performs a latent class analysis, e.g.,
fit1 - lca(data, 2)
I tried but I got:
lca(data, 2)
Error in matrix(0, 2^nvar, nvar) : matrix: invalid 'nrow' value (too large or
NA)
In addition:
Hi All,
a colleague wants to calculate the Fisher information matrix for a model he
wrote (not in R). He can easily get the neg-log-likelihood and the best fit
parameters at the minimum. He can also get negLLs for other parameter values
too.
Given these data, is there a way in R to calculate
Hi All,
is there a function in R that allows me to work with fractions without
transforming them to floats (or whatever) in between?
Something that would calculate something like:
(1/2 + 1/8) * 1/2 = 5/16
without ever transforming to 0.5 and 0.125?
Best,
Federico
--
Federico C. F. Calboli
Henrique Dallazuanna wrote:
require(MASS)
?as.fractions
as.fractions(1/2+1/8)
I thought that as.fractions did transform the fractions *first* into floats and
*then* found the rational approssimation (a passage I'd rather avoid):
fractionspackage:MASSR
Hi All,
I am writing a fucntion where I would like to use fractions for all the
(numerous) passages.
Is there a way of creating an environment *within a fucntion* so that all the
numbers/calculations are fractions?
Best,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public
Muhammad Subianto wrote:
library(MASS)
as.fractions(c(0, 0.15, 0.827, .06, 0.266))
[1] 0 3/20 62/75 1/15 4/15
Seems to make things a bit too slow, even though I get a good increase in
precision.
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public
Hi All,
I have the following problem: I have a vector
x = rep(0,15)
x[1:2] = 1
x
[1] 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
I need to be able to call that vector 'x' so that if condition 'A' is true,
only
the first value is kept 'as is' and all the others are put to 0
if(A == T)
function(x) with x
Hi,
is there a way of telling Emacs + ESS to show words that are already
a function in R (such as 'length') is a different colour/font?
Best,
Federico
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel
Hi All,
can I have a plot where the symbol for the dots is smaller than pch
=20 but bigger than pch = '.'?
Best,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44
Liaw, Andy wrote:
I don't see why making copies of the columns you need inside the loop is
better memory management. If the data are in a matrix, accessing
elements is quite fast. If you're worrying about speed of that, do what
Charles suggest: work with the transpose so that you are
Charles C. Berry wrote:
This is a bit different than your original post (where it appeared that
you were manipulating one row of a matrix at a time), but the issue is
the same.
As suggested in my earlier email this looks like a caching issue, and
this is not peculiar to R.
Viz.
Hi All,
I would like to ask the following.
I have an array of data in an objetct, let's say X.
I need to use a for loop on the elements of one or more columns of X and I am
having a debate with a colleague about the best memory management.
I believe that if I do:
col1 = X[,1]
col2 = X[,2]
Charles C. Berry wrote:
Whoa! You are accessing one ROW at a time.
Either way this will tangle up your cache if you have many rows and
columns in your orignal data.
You might do better to do
Y - t( X ) ### use '-' !
for (i in whatever ){
do something using Y[ , i ]
}
My
Ndoye Souleymane wrote:
Hi,
Let me suggest you to save your spss file in txt and use the read.table
function to load your file in R.
That is what I use to do.
The problem here is that the files are old data that were made with an ancient
version of spss and cannot be changed to txt (or
Hi All,
does anyone ever import old SPSS files in a sl3 format?
read.spss('file.sl3') does not seem to work... it's not recognised as
a supported SPSS format at all.
Best,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk
Hi All,
just a quick (?) question while I wait my code runs...
I'm comparing the identity of the lines of a dataframe, doing all possible
pairwise comparisons. In doing so I use identical(), but that's by the way. I'm
doing a (not so) quick and dirty check, and subsetting the data as
Hi All,
I have two numerical matrices of 2 columns and many rows.
The two coulumns of matrix (1) form a number of 'pairs' of numbers, e.g:
[,1] [,2]
[1,]10
[2,]34
[3,]34
[4,]58
[5,]10
[6,]10
[7,]67
Matrix (2) contains the *unique*
Hi All,
I have a vector of data, a vector of bin breakpoints and I want to put my data
in the bins and then extract fanciful informations like the mean value of each
bin.
I know I can write my own function, but I would have thought that R should have
somewhere a function that took as
Hi All,
I have a data frame of three columns, all of which names. The names in the
three
cols overlap up to a point, so I might have *Harry* in all three columns,
*Tom* in cols 1 and 3 and *Bob* in col 3 only.
harry paulbob
anita harry tom
frank jackharry
tom peteben
Prof Brian Ripley wrote:
Are the columns factors or character? I'll try to write code that copes
with both:
nm - unique(c(as.character(col1), as.character(col2), as.character(col3)))
DF[] - lapply(DF, function(x) match(x, nm))
Cheers,
it worked.
Federico
--
Federico C. F. Calboli
Hi All,
I have two data frames. The first contains data about a number of individuals,
coded in the first column with a name, in an order I find convenient.
The second contains different data about the same indivduals, in a different
order. Both data frame have the individual names in the
Hi All,
I would like to know, is there a *ballpark* figure for how many
parameters the minimisation routines can cope with?
I'm asking because I was asked if I knew.
Cheers,
Federico
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Hi All,
I have a non-square matrix of 0s/1s. The rows are individuals, and the colums
factors. The code is 1 if a factor is present, 0 if not.
I would like to rearrange the order of the rows to find if there are any
diagonal blocks. Is there a fucntion that would do that already in R?
Cheers,
Hi All,
I am using plot(x, type = 'l') for some plotting, but I would like rounded
edges
rather than jagged edges in the plot (purely for aestetic reasons).
How could I achieve that?
Cheers,
Federico
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College,
Dimitris Rizopoulos wrote:
probably you want to use the `lend' argument of ?par(); I hope it helps.
Does not seem to work in my case.
F
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594
Prof Brian Ripley wrote:
It I understand you aright, that is done by par(lend) but the default is
round. So maybe your graphics device (unstated) on your OS (unstated)
does not support this.
graphics device = X11 (xserver-xorg)
OS = Debian GNU/Linux, Kernel 2.4.27-2-686-smp
We need
Hi All,
I need to write as text files 1000 ish variation of the same data frame,
once I permute a row.
I would like to use the function write.table() to write the files, and
use a loop to do it:
for (i in 1:1000){
bb8[2,] = sample(bb8[2,])
write.table(bb8, quote = F, sep = '\t', row.names =
, so
I get back close to square one annoyingly often.
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t
Hi All,
I am (almost) successfully using apply() to apply a function recursively
on a data matrix. The function is question is as.genotype() from the
library 'genetics'
apply(subset(chr1, names$breed == 'lab'),2,as.genotype,sep =)
Unfortuantely apply puts it's results into a matrix object
Hi All,
is there any general advice about speeding up recursive functions
(not mentioning 'don't use them')?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20
need as.matrix to behave in a consisten
fashions, so I get
as.matrix(x, whatever)
[,1] [,2]
x12
and
as.matrix(rbind(x,y), whatever)
[,1] [,2]
x12
y34
I tried byrow =T, does not make a thing.
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department
Hi All,
I found a solution for my question:
I have the following adjacency matrix for a directed graph:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]00000000
[2,]00000000
[3,]10000000
[4,]
All in all I have 3 paths (not all of the same length) between (8)
and (1).
Is there already a function in R (whatever the library) that will
list the nodes touched in all those three paths (i.e. 8 - 6 - 1; 8 -
7 - 4 - 3 - 1; 8 - 7 - 5 - 3 - 1)?
Regards,
Federico Calboli
--
Federico C. F
Hi everyone,
I would like to find out when and where exactly I get the following
warning in a piece of code I've written:
Error in [-(`*tmp*`, iseq, value = numeric(0)) :
nothing to replace with
The code is a for () loop performing a somewhat trivial calculation,
modulated by a number
Hi All,
is there an equivalent of the Unix command 'less' (or 'more'), so I can
look at what's inside a data.frame or a matrix without having it printed
out on console?
I am using R on Debian Linux and Mac OS 10.4.5
Cheers,
F
--
Federico C. F. Calboli
Department of Epidemiology and Public
that is has a subset each matrix in the
list subsetting so I get the 2nd and 3rd row of each (and all columns).
How could I do that (apart from looping)?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place
On Thu, 2006-02-09 at 09:22 +0100, Christoph Buser wrote:
Dear Frederico
From your example it is not clear to me what you like to obtain:
Please have a look on the slightly changed example here (I
changed two values to show a potentially undesired side effect of
your coding.
test -
HI All,
I have a data frame such as:
test
x y p d
[1,] 1 0 10 21 0
[2,] 2 3 11 12 0
[3,] 3 4 12 23 0
[4,] 3 5 13 24 0
and I want to perfor some operations on the first two coulums,
conditional on the uneqaulity values on the 3rd and 4th columns.
For instance:
j = 3
test[test[,1] ==
,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
Hi All,
I have the following problem, that's driving me mad.
I have a dataframe of factors, from a genetic scan of SNPs. I DO have
NAs in the dataframe, which would look like:
V4 V5 V6 V7 V8 V9 V10
1 TT GG TT AC AG AG TT
2 AT CC TT AA AA AA TT
3 AT CC TT AC AA NA TT
4 TT
Hi All,
a colleague asked me if R has a function producing a Partial Rank
Correlation Coefficient, sensu Blower and Dowlatabadi 1994 [1].
I personally would not have a clue, and I could not find something like
that on the search page... although I would not be surprised if it's
there under a
. Would this be at all possible?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44 (0)20 75943193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
] .
That's exactly my situation, and is exactly what I want to do.
After taking out the typo (and bug) drow[i]0 the code seems to
work fast enough... I'll tinker a bit with it, but it could be good
enough as it is.
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology
Hi All,
is there any chance of vectorising the two ifelse() statements in the
following code:
for(i in gp){
new[i,1] = ifelse(srow[i]0, new[srow[i],zippo[i]], sample(1:100, 1,
prob =Y1, rep = T))
new[i,2] = ifelse(drow[i]0, new[drow[i]0,zappo[i]], sample(1:100,
1, prob =Y1, rep = T))
}
In my absentmindedness I'd forgotten to CC this to the list... and
BTW, using gc() in the loop increases the runtime...
My suggestion is that you try to vectorize the computation as much
as you
can.
From what you've shown, `new' and `ped' need to have the same
number of
rows, right?
On 4 Jul 2005, at 12:41, Uwe Ligges wrote:
Federico Calboli wrote:
In my absentmindedness I'd forgotten to CC this to the list...
and BTW, using gc() in the loop increases the runtime...
If the data size increases, you cannot expect linear run time
behaviour, e.g. because gc
On 4 Jul 2005, at 15:15, Peter Dalgaard wrote:
Your original code got lost in the threading, but that order of
magnitude suggests that you have N^2/2 behaviour somewhere. The
typical
culprit is code like
x - numeric(0)
for (i in 1:N){
newx -
x - c(x, newx)
}
in which the
Hi All,
I'd like to ask for a few clarifications. I am doing some calculations
over some biggish datasets. One has ~ 23000 rows, and 6 columns, the
other has ~62 rows and 6 columns.
I am using these datasets to perform a simulation of of haplotype
coalescence over a pedigree (the datestes
a gross figure of the
number of functions in R considering the base install and all the
libraries available?
Apart from graphics and lattice, are there any more packages producing
eye catching graphics (possibly with a survival analysis/epidemiological
bend)?
Cheers,
Federico Calboli
--
Federico C
= made up of a number of repetitions of all the elements of
vector1
vector3 = a vector of NAs that is meant to get the result of the
counting
My problem is that vector1 is about 6 terms, and vector2 is
62... can anyone suggest a faster code than the one I wrote?
Cheers,
Federico
On 20 Jun 2005, at 21:24, Erin Hodgess wrote:
Hello, Federico!
I'm a bit confused about your question, please:
What sorts of things are in Vector1, please?
numbers (as in numeric) that code individuals
Why are you counting NAs in Vector3, please?
I am counting how many times the code
On Tue, 2005-05-24 at 18:05 +0200, Guillaume Chapron wrote:
Hello,
I'm running a PB G4 with Mac OS 10.4.1. I have downloaded the latest
version R-2.1.0a.dmg. It appears that R does not work. It launches
itself, but the window never gets ready, there is written Loading R...
and a small
above or should I actually write the statements of F1 to make
the parsing faster?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
of any difference in the tarball built by R 2.0.1 and
2.1.0).
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t
for Debian from the latest .deb files for Sarge on x86 arch.
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t
On Fri, 2005-04-15 at 14:30 -0400, Frank Duan wrote:
Hi R people,
I met a naive prolem. Could anyone give me a hint how to create such a
vector with entries: one, two, three, ...?
rvect - c(one, two, three)
rvect
[1] one two three
Is it what you want?
F
--
Federico C. F. Calboli
... what exacly is glm doing with the weight vector?
In any case, how would I go about weighting my individuals in a logistic
regression?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2
: non-integer #successes in a binomial glm! in: eval(expr, envir,
enclos)
##
Why such inconsistency?
Regards,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place
or other reference apart from the green
book and the VR S-programming?
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t
On Tue, 2005-04-12 at 17:45 +0800, Feng Chen wrote:
Maybe you can try this:
http://cran.r-project.org/doc/manuals/fullrefman.pdf
I was thinking of something that would not limit itself to defining all
possible functions and that I do not have already on my HD...
F
--
Federico C. F. Calboli
On Mon, 2005-04-11 at 12:52 +0200, Clark Allan wrote:
HI
sorry to be a nuisance to all!!!
how can i see the code of a particular function?
e.g. nnet just as an example
for some function just type the function name and you get the code:
ls
function (name, pos = -1, envir =
On Wed, 2005-04-06 at 09:11 +0100, michael watson (IAH-C) wrote:
OK, so I tried using lm() instead of aov() and they give similar
results:
My.aov - aov(IL.4 ~ Infected + Vaccinated + Lesions, data)
My.lm - lm(IL.4 ~ Infected + Vaccinated + Lesions, data)
Incidentally, if you want
On Wed, 2005-04-06 at 21:40 +0900, Kum-Hoe Hwang wrote:
Howdy, R gurus
I 'd like to know hwo to calculate or estimate SS of Type I and Type III in
ANOVA or other anaysis in R.
Thanks,
If memory seves me well, try Anova in the car package
F
--
Federico C. F. Calboli
Department of
of what's in xxx$p.value, even if HWE.exact failed
to calculte it. Dump anything in there!
Is there any way of forcing the loop to carry on?
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2
On Tue, 2005-04-05 at 15:51 +0100, michael watson (IAH-C) wrote:
So, what I want to know is:
1) Given my unbalanced experimental design, is it valid to use aov?
I'd say no. Use lm() instead, save your analysis in an object and then
possibly use drop1() to check the analysis
2) Have I used
On Mon, 2005-04-04 at 10:11 +0200, Giorgio Corani wrote:
Dear All,
As far I as I have understood reading both your past posting and the
documentation, in order to have the command-line completion facility, I
have to run R within emacs.
However, as I try to start R within emacs as
On Fri, 2005-04-01 at 20:27 +0200, Hartmut Weinrebe wrote:
Having searched and searched I still haven't found what's the problem with my
data (I've attached the relevant file).
Every time I tried to use the CANCOR-Function I got error messages.
So I turned to check my two sets of variables
On Sat, 2005-03-26 at 17:44 +, Hess, Stephane wrote:
Dear all,
I have 5 plots that I would like to include in a single figure, spread over
two rows. If I use mfrow=c(2,3), and produce my plots one after the other, I
will end up with three plots in the first row, and 2 in the second
On Wed, 2005-03-23 at 11:58 -0500, Shaw, Philip (NIH/NIMH) wrote:
Hi
I am struggling with nested random effects and hope someone can help.
I have individuals (ID) who are nested within families (FAM). I want to
model an outcome variable, and take account of the intercorrelation of
On Wed, 2005-03-23 at 10:04 -0800, Berton Gunter wrote:
An interaction random effect/fixed effect is noted as
random ~1|random/fixed
in your case random =~1|ID/FAM (but I don't uderstand why indiviuals
withing families are fixed and and families are random, but there you
go).
,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
On Tue, 2005-03-08 at 15:57 -0500, Wiener, Matthew wrote:
Remove the (now empty, because you deleted all objects) file .RData from
the directory.
Hope this helps,
Thanks, it did fix the problem.
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public
obtained this way? I
don't.
So I'd argue that the lack of a GUI is a good thing, because it forces
the users to think a bit more about what they want to do, and gives more
control on what is going on.
Best,
Federico Calboli
--
Federico C. F. Calboli
Dipartimento di Biologia Evoluzionistica
that will calculate the a
phylogenetic tree from such data? I installed ape, but, despite
reading the docs, I cannot find a function that would calculate a tree
from data like mine (my sight may be getting worse though). Any
suggestion is welcome.
Regards,
Federico Calboli
--
Federico C. F
[ This was posted to the R-packages list (which I moderate),
but definitely doesn't belong there. Martin Maechler
]
--- start of forwarded message ---
From: Federico Calboli [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: how to upload
Date: 15 Jul 2004 18:30:26 +0100
Dear All,
I
a nice
table will come later). I know it is possible to print tables through
LaTeX and the Design/Hmisc libraries, although I would not have a clue
about printing it together with graphs, but I'd like something quicker
if at all possible.
Regards,
Federico Calboli
On Sat, 2004-05-22 at 03:58, Deepayan Sarkar wrote:
Exactly what do you wish to square and sum ? If it's the 'errors' (which
in this context is ambiguous), extract them (see ?residuals.lme),
square them and sum them. But what are you planning to do with this
after you get it ?
Make a oh
not ranting too much.
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 209 4286
f.calboli at ucl.ac.uk
fcalboli at alma.unibo.it
to fit a second (and wrong in a way) model for something so
trivial.
I know that the issue has been dealt with before, but I could not find a
clear cut answer searching through the archives.
Regards
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di
From what I gather you have
2 treatments
6 samples nested in treatment
12 replicates in sample.
I do not know what your dependent veriable is, but I would:
code the two treatments as T1 and T2
code the 6 samples as a, b, c, d, e, f so the first three samples
belonging to T1 are not confused
I have R 1.9.0 on sid, apt-getting stuff from cran... I too noticed that
cran has R 1.8.1 for Debian, but as I got R 1.9.0 I just ignored the
differences in labelling.
BTW, it is R that tells me it is R 1.9.0 when I fire it up. And has the
stats library that I did not remember in R 1.8.*
,
according to your coding of the data, this stem is not always necessary.
HTH
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 209 4286
f.calboli at ucl.ac.uk
fcalboli
)?
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 209 4286
f.calboli at ucl.ac.uk
__
[EMAIL PROTECTED] mailing list
the desired symbol (any
combination of keys seems to fail...)
Can anyone advice how to produce the ~ symbol, short of a copy/paste
from MS Word?
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel
On Tue, 2004-03-30 at 11:48, Henrik Bengtsson wrote:
Press-Alt + Num1 + Num2 + Num6 + Release-Alt where Num1, Num2
Num6 are 1, 2 6 on the *numeric keyboard* (not the ones above the
letter keys) will produce ASCII character 126 (~, tilde) on my WinXP
Pro in both Rterm and Rgui for R v1.8.1.
On Tue, 2004-03-30 at 12:38, Erich Neuwirth wrote:
Before using the sequence
Press-Alt + Num1 + Num2 + Num6 + Release-Alt
one has to make sure that the numeric keyboard is in NumLock mode,
not in cursor mode.
If it is in cursor mode, this trick will not work
(at least it does not on my
is that the colour used for line1
at temperature 18C is not the same for line1 at temperature 28C,
which I find extremely confusing.
How can I specify to keep the same colour for the same line across the
plot?
regards,
Federico Calboli
--
=
Federico C. F. Calboli
am doing is more of an
advantage, IMHO, rather than a shortcoming... from a usability
standpoint I never felt I had any problem I could not deal with after
some thinking, or asking the list.
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento
that doing
things with R is going to be a better (= more sensible) option.
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 251 208
f.calboli at ucl.ac.uk
can ONLY be performed on a distance matrix,
and I can therefore reasonably expect that some form of transformation
to a distance matrix has been performed by Statistica prior to the MDS.
It would at least be a first step to understand what exactly Statistica
did with the data.
Regards,
Federico
on the standard plot function, but I
was quite unsucessful. Can anyone give advice? I am happy to give a toy
example if needed.
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39
. The following should do what you asked for:
plot(mod.pro$X, asp=1, pch=1)
points(mod.pro$Yrot, pch=2)
segments(mod.pro$X[,1], mod.pro$X[,2], mod.pro$Yrot[,1],
mod.pro$Yrot[,2])
The above solves my problem. Thanks for your help.
Best regards,
Federico Calboli
Dear All,
how do I compute the left eigenvector of a matrix? I gather that eigen
computes the right eigenvectors...
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
PLEASE NOTE NEW ADDRESS
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel
could be happy just
comparing the two diagonals.
regards,
Federico Calboli
--
=
Federico C. F. Calboli
PLEASE NOTE NEW ADDRESS
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 251 208
f.calboli at ucl.ac.uk
in other problems?
Again, many thanks to all for the invaluable help.
Regards,
Federico Calboli
--
=
Federico C. F. Calboli
PLEASE NOTE NEW ADDRESS
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 251 208
11
[2,]111
[3,]111
I would have expected the identity matrix here.
I find the same result with every other square matrix I used.
BTW, I am using R 1.8.1 on Linux Mandrake 9.1
Cheers,
Federico Calboli
--
=
Federico C. F. Calboli
,
Federico Calboli
--
=
Federico C. F. Calboli
PLEASE NOTE NEW ADDRESS
Dipartimento di Biologia
Via Selmi 3
40126 Bologna
Italy
tel (+39) 051 209 4187
fax (+39) 051 251 208
f.calboli at ucl.ac.uk
__
[EMAIL PROTECTED
1 - 100 of 122 matches
Mail list logo