Thanks all, that worked!
Yes, it should be
for (i in 1:length(a)) a[i]- scalar * a[i] * i
And now is...
a - a * scalar * seq_along(a)
That is almost as cool as the PERL programming language :-)
2012/6/4 Rui Barradas ruipbarra...@sapo.pt
Hello,
Just learning the alphabet? If yes, there's a
Dear R-listers,
I am giving part of my R code :
###
n=15
m=1
library(partitions)
library(gregmisc)
library(combinat)
x = t(restrictedparts(n-m,m))
l = length(x[,1])
for(u in 1:l){
A= unique(matrix( unlist(permn(x[u,])), ncol=m,
Hi Trinh,
Please check ?dose.p() from library MASS.
Since you wanted the rate, the code was a bit modified from the example.
datPr
X Event Trial
1 1210.0 8 8
2 121.0 6 8
3 60.5 6 8
fm-glm(Event/Trial~X,data=datPr,family=quasibinomial(link=probit))
Thank you ! Rui Barradas .
Yes, the arrow character can not be recoginised by R.
the character will be treat as a stop command...
So I replace all the character with space, then it works..
Just want to know is there any way for R to handle it.
Not sure does eveybody able to view the
I have been try to work with Merror n I keep getting this message:
Error in solve.default(H) :
system is computationally singular: reciprocal condition number = 0
can some one please tell me what it mean and how to solve this.
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On Jun 5, 2012, at 08:42 , climmi wrote:
Thank you ! Rui Barradas .
Yes, the arrow character can not be recoginised by R.
the character will be treat as a stop command...
So I replace all the character with space, then it works..
Just want to know is there any way for R to handle
Dear Gunter Berton and all,
As
you can see, my data has three lists each containing 366 entries. I converted
them into the matrix. I now want to approx./interpolate 366 entries over 365
intervals, Which means I want to have a matrix with 365 entries.
I
used
approx(matrix,
Dear all,
I have been trying to convert coordinates from longitude/latitude to utm
but I got an error. As soon as the longitude coordinate is greater than 90,
I get the folloowing error message: error in pj_transform: latitude or
longitude exceeded limits
Here is what I did:
I believe that dozens of people asked this question before you.
If you search this error in google, you can find your answer!
Ozgur
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Hi,
Please let me whether there is a package which can parse linux
configuration file '.conf. If there is no specific package can any of the
generic package handle this?
Thanks Regards,
Thomas
[[alternative HTML version deleted]]
__
UTM needs a +zone and +south marker for the southern hemisphere, so
your PROJ.4 specification is incomplete, probably assuming +zone=1
which is completely wrong for longitude=126.
http://en.wikipedia.org/wiki/File:Utm-zones.jpg
Longitude 126.59 suggests zone 51 or 52, but it really depends on
Dear useRs, Simon,
@Simon: I'll send the data offline.
The command summary(...,freq=F) yields the same result as before. Why
did the default change from freq=F in version 1.7-13 to freq=T in
version 1.7-17? Especially since the original default appeared to be
better.
I noticed that some of my
Hi,
Try
pchisq(q,df)
available at help(Chisquare)
Ozgur
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On 04/06/2012 17:25, Fredrik Karlsson wrote:
Dear list,
I have non-ascii labels that I want to include somethow in a .eps file for
inclusion into a LateX document. Is this possible?
What I see is that the (attached) png file looks ok but the .eps renders
all non-ascii labels at the bottom as
Thanks again but my errors are still here. Is it maybe coming from the next
fonction (I combinate these 2 functions but I thought it was coming from the
first one):
process.all - function(df.list, mat){
f - function(station)
na.fill(df.list[[ station ]], df.list[[
Dear R-Users, I'd like to have some tips about printing graph.
I use the command par to print more graphs in one window:par(mfrow=c(6,1));
par(oma=c(2.5, 2.5, 2.5, 2.5)); par(mar=c(0.5,4, 0.5, 0.5))
But this command doesn't run with complex graphic command (i.e. xyplot,
ternaryplot).How can
Hello,
Yes, there are ways for R to handle it. You could use readChar/writeChar
after openning the file in binary mode.
Something like,
len - file.info(test.txt)$size
#len
fc - file(test.txt, open=rb)
s - readChar(fc, nchars=len, useBytes = TRUE)
close(fc)
#grep(\032, s)
s - gsub(\032, , s)
Hello,
Try
apply(mat, 2, approx, method=”linear”, n=365)
This reads apply to each column (dimension = 2) of mat the function
approx with extra args method and n.
Three notes.
1. Your data does NOT have three list, it IS one list with three vectors.
2. 'matrix' is a function so choose
Hi,
I want to use N gram approach for assessing words association and their
placing in a sentence. I have tried textcat package and the syntaxes but it
deals with language classificaiton rather than a user defined
classification. Can someone help me with this?
Basically, I have these two
Thank you for your commentaries and suggestions.
Site 0 and site 1 are interpretable like events.
In fact these data come from a simultaneous observations of individuals in
two different sites (so they are independent observations: while one
individual is observed in one site it can't be in
Hi,
x-matrix(0,80,ncol=1)
will create x matrix with all elements 0 (to be filled by the sums that you
need)
sum(y[y[,4]==1,5])
will calculate the sum of 5th column of y with 4th column=1
Similarly,
sum(y[y[,4]==80,5])
will calculate the sum of 5th column of y with 4th column=80.
You can
Hi,
How to calculate chi sqaure p value for given statistical value and degrees
of freedom.
Input :
x = statistical value
d = degrees of freedom
output:
p value = ?
Regards
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Dear Rui,
i am greatful for everything you did. bret advised me to use dput(), which i
did. yes! i forgot to complement him. i literally feel sorry for that. i hope
you wont mind and continue extending your help.
regards and love for every one
eliza botto
waters inn
Date: Tue, 5 Jun 2012
On 05.06.2012 00:36, Erdal Karaca wrote:
Thanks all, that worked!
Yes, it should be
for (i in 1:length(a)) a[i]- scalar * a[i] * i
And now is...
a- a * scalar * seq_along(a)
That is almost as cool as the PERL programming language :-)
Almost?
Uwe Ligges
2012/6/4 Rui
Do we need an Obfuscated R contest?
On Tue, Jun 5, 2012 at 8:17 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 05.06.2012 00:36, Erdal Karaca wrote:
Thanks all, that worked!
Yes, it should be
for (i in 1:length(a)) a[i]- scalar * a[i] * i
And now is...
a- a * scalar *
On 05.06.2012 14:46, jim holtman wrote:
Do we need an Obfuscated R contest?
I already know potential winners
Uwe
On Tue, Jun 5, 2012 at 8:17 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 05.06.2012 00:36, Erdal Karaca wrote:
Thanks all, that worked!
Yes, it
Hi all,
I would like to predict some values for an nls regression function
(functional response model Rogers type II). This is an asymptotic function
of which I would like to predict the asymptotic value
I estimated the paramters with nls, but can't seem to get predictions for
values of m
Dear R users,
currently, i am working on the canonical correlation analysis (R package
vegan or CCA). I have a trouble to calculate the pvalue of the association
between each dependent variable and each independent variable, also i want
to calculate the pvalue of the association between a
On Jun 5, 2012, at 4:52 AM, lincoln wrote:
Thank you for your commentaries and suggestions.
Site 0 and site 1 are interpretable like events.
In fact these data come from a simultaneous observations of
individuals in
two different sites (so they are independent observations: while one
You might try R-Sig-Geo for the reasons, but you need to include the UTM
zone to convert:
SP-SpatialPoints(cbind(126.59,-14.30),proj4string=CRS(+proj=longlat))
SP
SpatialPoints:
coords.x1 coords.x2
[1,]126.59 -14.3
Coordinate Reference System (CRS) arguments: +proj=longlat
Joachim Audenaert Joachim.Audenaert at pcsierteelt.be writes:
Hi all,
I would like to predict some values for an nls regression function
(functional response model Rogers type II). This is an asymptotic function
of which I would like to predict the asymptotic value
I estimated the
I am trying to use ggplot() to produce a graph and am getting warnings that
I don't understand. This is happening from within RStudio, but also happens
if I start Windows R GUI. Can anyone help me understand the warning and
how to make sure the right fonts are designated? The relevant code
Hello,
I believe the error is in function 'g'. If I'm right, follow these steps
1. Just before the first if include
flag - TRUE
2. Just before for(y in ord) include
flag - FALSE
3. Just before break include
flag - TRUE
3. Change the return value form simply x to
if(flag) x else NA
The code
Hi All,
Here is the problem: I'm trying to generate a number of Fourier Descriptors
figures for an experiment.
All I need is that they are created within a loop and saved with sequential
names (e.g., s1_1.png, s1_2.png etc..) in my directory.
I created a nested loop with a counter for the
I'm not the one who is choosing to use gradient background. It's our company
policy.
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Hi all, I'm new to using R, and apologize for simplicity of this
question.
I'm using a data set with over 60,000 observations, Two variables are
patient ID, and cost incurred by the patient. I'd like to generate
frequency/table by patient and cost IF the total cost is over 2000.
Right now I'm
Hi all,
I'm struggling with nls. How do you know if your model is significant? For a
lm, you get a p-value, but you don't get it for a nls. Is there a way to
calculate it?
For a lm I use this:
a-summary(lm(model ~obs))
f.stat-a$fstatistic
Hi Sandy,
I was wondering if you ever recieved an answer regarding the use of multiple
comparisons for gls models?
I have also been searching various forums and books to see if there are any
methods I could use and have only found people, such as yourself, asking the
same question.
Many thanks
I'm having a very frustrating problem, trying to find the inverse distance
squared weighted interpolants of some weather data.
I have a data frame of weights, which sum to 1. I have attached the weights
data. I also have a data frame of temperatures at 48 grid points, which I
have also attached.
http://r.789695.n4.nabble.com/file/n4632406/temp3880.csv temp3880.csv
http://r.789695.n4.nabble.com/file/n4632406/weight3880.csv weight3880.csv
Here are the files I promised to upload.
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1. Before you post to this list again, please read An Introd to R --
or other basic R tutorial. Intro ships with every R installation.
There's a reason for this -- to avoid badgering this list with basic R
queries that minimal homework could answer.
2. However, see also ?table and links therein.
I suggest you consult a local statistician. You could also post to a
statistical help list like stats.stackexchange.com. Your query has
nothing to do with R, but is rather about the meaningfulness (or lack
thereof) of statistical significance in nonlinear modeling.
-- Bert
On Tue, Jun 5, 2012 at
Hello,
Try
Total - aggregate(cost~patient, data=x)
Total[Total$cost 1000, ]
As for writing to a csv file, see ?write.csv
Hope this helps,
Rui Barradas
Em 05-06-2012 16:34, mkm1616 escreveu:
Hi all, I'm new to using R, and apologize for simplicity of this
question.
I'm using a data set
I need a nudge in the right direction to get started using NADA. I bought
Helsel's second addition and am currently reading it; NADA is installed in
R.
My data has been restructured with a couple of awk scripts. The data frame
structure now has a flag if the quantity is censored (ceneq1
Superb. It works. The results loo great. I just tweaked the function a little
bit to suit my needs though. I passed merges and vcount separately as
fgc$merges and vcount(g). The vcount attribute was not there in fgc object.
Also, I removed the last row from the merges matrix before returning
On 05/06/2012 11:17, Francesco Nutini wrote:
Dear R-Users, I'd like to have some tips about printing graph.
I use the command par to print more graphs in one window:par(mfrow=c(6,1));
par(oma=c(2.5, 2.5, 2.5, 2.5)); par(mar=c(0.5,4, 0.5, 0.5))
But this command doesn't run with complex
I'm using the Match package to do propensity score matching. Here's some
example code that shows the problem that I'm having (much of this code is
taken from the Match package documentation):
*data(lalonde)
glm1 - glm(treat~age + I(age^2) + educ + I(educ^2) + black +
hisp + married
Update:
The IT people agreed to test R separately. R is now approved and RStudio
is not.
The folks at RStudio are baffled as to why all those registry entries
are being
recorded. They directed me to the source code which details the known
accesses
to the registry during installation. I have
One thing to note is that there are more than one model that can be
called exponential. Two of the common ones are:
y = exp( a + b*x + error )
y = exp( a + b*x ) + error
The common way to fit the first is to take the log of both sides and
just fit a linear model with log(y), I expect (but am
On Jun 5, 2012, at 11:50 AM, aledanda wrote:
Hi All,
Here is the problem: I'm trying to generate a number of Fourier
Descriptors
figures for an experiment.
Somehow I'm guessing that this will involve load an unnamed package.
Yep:
?create.fourier.descriptor
No documentation for
Hello,
The files you've uploaded are the weights file and the results file, not
the original temp.csv.
So this is untested but it seems you have a standard matrix multiply
problem.
temp3880W - temp[, 3:50] %*% weight3880
Hope this helps,
Rui Barradas
Em 05-06-2012 15:48, alonis10
I am having a problem visualizing what you are doing here.
What I see is a temp file of 760 elements. From my point of view and reading
in your data it would have a have dim(760, 1) but your code seems to suggest
that it is not a single vector.
Again I must be missing something completely
A client has inquired about producing a decision tree from data which could
include:
- ID of brand purchased
- Importance ratings (1-10 scale) for a number of relevant attributes
(price, strength, recommended by a friend, etc.) In other words, a rating
of how important each attribute is in the
Post on a statistics or data mining discussion site (e.g.
stats.stackexchange.com). There are tons of different algorithms for
fitting decision trees, and you are more likely to get an informative
discussion on their relative strengths and weaknesses there than here.
This is really NOT an R
Hi all,
How do I obtain the current active path of a function that's being called?
That's to say, I have several source files and they all contain definition
of function A.
I would like to figure out which function A and from which file is the one
that's being called and is currently active?
I read your website but still don't know the difference between the three
formulas...
Thank you!
On Mon, Jun 4, 2012 at 11:14 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:
Hi Michael,
This is far from exhaustive (I wrote it as an introduction some years
ago) but you may find it useful to
Hi
Here's one way to approach it ...
ggplot(data.frame(x=1:10, y=1:10)) +
geom_polygon(aes(x=xx, y=yy), fill=grey70,
data=data.frame(xx=c(0, 0, 4, 4), yy=c(0, 11, 11, 0))) +
geom_point(aes(x=x, y=y))
... (supply a separate data source for the background region polygon).
Hi,
On 6 June 2012 08:58, Paul Murrell p.murr...@auckland.ac.nz wrote:
Hi
Here's one way to approach it ...
ggplot(data.frame(x=1:10, y=1:10)) +
geom_polygon(aes(x=xx, y=yy), fill=grey70,
data=data.frame(xx=c(0, 0, 4, 4), yy=c(0, 11, 11, 0))) +
geom_point(aes(x=x, y=y))
Define: one (who?)
The individual analyzing the data.
Define: usually
More than 50% of the time.
Define: good
A decision tree including branches on at least 2 variables.
Define: data of this kind
I believe I may have specified this in my original post:
- ID of brand purchased
-
Well that answers the question as answer. So while you are working
within the system to get your company to change the policy you can use
rasterImage to add a gradient background to the plot, then use points
or lines or other functions to put the parts of interest back on top
of the gradient (if
The R book by Michael Crawley has some discussion the type of R syntax you are
looking for in the chapter on statistical modeling. As for the formulae you
gave...
lm(y ~ x*w - 1) fits an interaction between x and w without an intercept,
along with the main effects for x and w
lm(y ~ x:w - 1)
There are several ways that a function can come into being within R,
it can be sourced from a file like in your case, but it could also be
typed in by hand at the command prompt, or created by another
function, etc. So R does not in general keep links to the files from
which the file was
Can you provide an example of the code file that you use to call the different
functions? Without that it might be hard for people to answer your question.
Offhand, I'd say that it is whatever version of function A was called last. If
you are loading functions into the workspace, they are
You mention 3 models. In all of them, the '-1' simply removes
the intercept term; help('formula') explains the use of '-'
in general.
1. lm(y~ x * w - 1) is clearly explained in help('formula');
2. lm(y~ x:w - 1) ditto (and this is a model to avoid);
3. lm(y~ x/w - 1) this is equivalent to
Thank you!
But could you please explain why the 2nd formula is a model to avoid?
Thanks again!
On Tue, Jun 5, 2012 at 6:18 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
You mention 3 models. In all of them, the '-1' simply removes
the intercept term; help('formula') explains the use of '-'
in
Thanks so much for your help!
I'd like but however I couldn't provide the code since they are not in
public domain...
But lets imagine I inherited a big pile of R projects/codes from other
people and there are lots of sources in the programs.
And there are many definitions of function A in the
OY!
Have you had any courses in linear models/regression? If so, it's the
so-called effects hierarchy principle: generally one does not
expect interactions -- 2nd order effects -- when main effects -- first
order effects -- are absent.
If not, it's difficult to explain, but it certainly has
On Jun 5, 2012, at 7:51 PM, Michael wrote:
Thanks so much for your help!
I'd like but however I couldn't provide the code since they are not in
public domain...
But lets imagine I inherited a big pile of R projects/codes from other
people and there are lots of sources in the programs.
And
I think your best option is to invest time in learning how to build an R
package, and then
package each project. Conflicting names are not a problem when you have
them inside package
namespaces. See the extensions document in R in the R distribution
R-2.15.0/doc/manual/R-exts.html
One way to
On Jun 5, 2012, at 8:35 PM, arun wrote:
Hi Dave,
I am interested in your suggestion. But, look at this scenario.
func-function()x
comment(func)-test
comment(func)
[1] test
func-function()y
comment(func)-test2
comment(func)
[1] test2
func-function()x
comment(func)
NULL
Folks,
Can you point me to any examples of using the GSL library to generate
correlated uniform random variables?
I want to generate correlated defaults among portfolios of loans.
Currently I generate simulations by using rmvsnorm (and then inverting using
the standard normal distribution
Thank you David Michael for your answers.
It makes more sense now!
Cheers,
Boris
2012/6/6 David L Carlson dcarl...@tamu.edu
You might try R-Sig-Geo for the reasons, but you need to include the UTM
zone to convert:
SP-SpatialPoints(cbind(126.59,-14.30),proj4string=CRS(+proj=longlat))
SP
Hi,
Actually, pchisq(q,df) calculates the cumulative distribution function by
default.
To calculate the p-value, you can use either
1-pchisq(q,df)
or
pchisq(q,df,lower.tail=FALSE)
PS: I checked, the p value yielded by R and the calculator for which you
give a link, for some q and df values,
Hi,
I have say 1000 different Independent Variables which have each been
classified into diiferent clusters using some clustering method.
I want to show these 1000 varaible in a report, but want some criterion to rank
them so i can show the most significant ones (based on the cluster groups)
Hello,
Better yet is function aggregate.
# Create some data matrix
Y - matrix(c(sample(80, 507, TRUE), sample(-1:1, 4*507, TRUE)), ncol=5)
X - aggregate(Y[, 5]~Y[, 1], Y, sum) # In your case Y[, 4] not Y[, 1]
Hope this helps,
Rui Barradas
Em 05-06-2012 09:53, Özgür Asar escreveu:
Hi,
Dear all,
I just wrote a script to have each of my three mcmc chins running on a
different computer core to improve computation speed. To do it I use the
function sfLapply from the package snowfall.
Before using parallell computation, I did my diagnostic analyses just after
the model updating,
Hi,
I am looking at the change in N concentration in plant roots over 4 time
points and I have fit a spline to the data using ns and lme:
fit10 - lme( N~ns(day, 3), data = rcn10G)
I may want to adjust the model a little bit, but for now, let's assume it's
good. I get output for the fixed
Was it really necessary to reply with sarcasm i.e. badgering the list.
The point of an open community is to seek advice. i mentioned I was
new to R and apologized to the community. A simple response with some
guidance would be helpful.
I'm looking at other posts, there are some other simple
Hi,
Try this example:
set.seed(1)
dat2-data.frame(patient=rep(c(1:20),5),cost=rnorm(100,15,1))
agg1-aggregate(dat2,by=list(dat2$patient),FUN=sum)
agg2-agg1$cost 75
agg3-data.frame(agg1[agg2,2:3])
A.K.
- Original Message -
From: mkm1616 mkm1...@gmail.com
To: r-help@r-project.org
Cc:
Here goes bert again with etiquette lessons.
Why are you so crabby?
On Jun 5, 11:16 am, Bert Gunter gunter.ber...@gene.com wrote:
I suggest you consult a local statistician. You could also post to a
statistical help list like stats.stackexchange.com. Your query has
nothing to do with R, but
Hello R-help.
I believe I may be coding the offset incorrectly in my formula. I am
testing significance between different group pairings using the VGAM
package. My data are zero-truncated and poisson-distributed and
overdispersed, so I'm using the posnegbinomial function in vglm (code
Thank you Rui. Appreciate the help.
Apologies for badgering the msg board. lol
On Jun 5, 11:28 am, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Try
Total - aggregate(cost~patient, data=x)
Total[Total$cost 1000, ]
As for writing to a csv file, see ?write.csv
Hope this helps,
Rui
Hi guys,
I'm a new to R and following along with Tutorials using this book:
http://www.amazon.com/Practical-Statistical-Analysis-Non-structured-Applications/dp/012386979X
In one of them, they use the twitteR package and describe the following
function (see below). From what I can tell from the
Hi,
Oops! I selected the wrong columns in the previous reply.
Try this:
set.seed(1)
dat2-data.frame(patient=rep(c(1:20),5),cost=rnorm(100,15,1))
agg1-aggregate(dat2,by=list(dat2$patient),FUN=sum)
agg2-agg1$cost 75
agg3-data.frame(agg1[agg2,c(1,3)])
names(agg3)-c(patient,cost)
agg3
FYI - here are the errors (I can get rid of the first one by removing the
progress parameter - which I understand controls whether a progress bar is
displayed). The second one stops at the gsub call.
Maybe if someone could give me a properly formatted example I could work
from instead?
Thanks!
Hi there!
I have two date columns in a dataframe I need to selectively collapse based
on missing values and which date comes first (imported from an text file).
This is what I did:
RHSSP$CT- as.POSIXct(RHSSP$CT,
format='%m/%d/%y %H:%M')
RHSSP$MRI
This is precisely what I needed; I can't believe how simple it is. Thanks!
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__
Hi All,
I am trying to fit a piecewise lasso regression, but package Segmented does not
work with Lars objects.
Does any know of any package or implementation of piecewise lasso regression?
Thanks,
Lucas
__
R-help@r-project.org mailing list
Hi Dave,
I am interested in your suggestion. But, look at this scenario.
func-function()x
comment(func)-test
comment(func)
[1] test
func-function()y
comment(func)-test2
comment(func)
[1] test2
func-function()x
comment(func)
NULL
func-function()y
comment(func)
NULL
Does it imply
Hi all
Dear all,
I need to compute boundaries for 3 stages group sequential design with 2
variables using Simpson’s integration method.
There is a r program written by Michael A. Proschant et Al. (2006)
(Statistical Monitoring of Clinical Trials: A Unified Approach) – see file
attached – to
How do I create a legend with ggplot?
I think should be getting the FuelTypeNum in the legend.
Plot:
http://r.789695.n4.nabble.com/file/n4632471/Rplot.jpeg
My code is:
ggplot(data=tempTable, aes(x=Bands8, y=SubPercent, fill=FuelTypeNum)) +
geom_bar(position=stack, stat=identity) +
Hi,
My input is chi square statistical value and degrees of freedom. But i m
getting different p values with the above formula.
I double checked my values with the below calculator.
http://vassarstats.net/tabs.html#csq
Pls help me out.
--
View this message in context:
I'm attempting to change a data set by compressing rows into columns.
Currently there are several rows that all have information about one
patient, but at different cycles. I'm trying to make each patient only
have one row in the data set.
Does anyone know a good way to combine data sets by
You can use the debug, fix, or edit functions to insert break points
into the version of the function in memory without needing to edit the
original source code.
On Tue, Jun 5, 2012 at 5:51 PM, Michael comtech@gmail.com wrote:
Thanks so much for your help!
I'd like but however I couldn't
See also ?trace ,?debugger, ?traceback and links therein.
-- Bert
On Tue, Jun 5, 2012 at 9:48 PM, Greg Snow 538...@gmail.com wrote:
You can use the debug, fix, or edit functions to insert break points
into the version of the function in memory without needing to edit the
original source code.
Does
?predict.ns
not do what you want without having to explicitly manipulate the spline basis?
-- Bert
On Tue, Jun 5, 2012 at 1:56 PM, Ranae ranae.diet...@gmail.com wrote:
Hi,
I am looking at the change in N concentration in plant roots over 4 time
points and I have fit a spline to the data
Hi,
A practical way migh be
out-rep(matrix(arg),4)-ma
out[2,]--out[2,];out[5,]--out[5,]
out
[,1] [,2] [,3] [,4]
[1,] 0.8 0.4 0.80 1.0
[2,] 1.0 0.0 1.00 0.3
[3,] 0.7 0.9 0.65 1.0
[4,] 1.0 0.5 0.60 0.9
[5,] 0.5 1.0 0.00 1.0
Best
Ozgur
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Dear Professor Ripley,
Thank you for your prompt answer. Indeed, cairo_ps solved the problem for me.
Thank you!
Fredrik
5 jun 2012 kl. 11:51 skrev Prof Brian Ripley rip...@stats.ox.ac.uk:
On 04/06/2012 17:25, Fredrik Karlsson wrote:
Dear list,
I have non-ascii labels that I want to
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