On Fri, 14 Dec 2007, Metz, Thomas (IRRI) wrote:
> Under Windows, I have used RODBC to connect to Excel spreadsheets as per
> the example below:
>
> library(RODBC);
> connect = odbcConnectExcel("testdata.xls");
> query = "SELECT [data$.ethn], [data$.sex], [data$.age],
>[data$.height
This question is nothing to do with RODBC.
You need to study the rw-FAQ Q2.9: I believe you should be able to get up
to (almost) 3GB on that system. (BTW, you seem confused about units: I
hope you have 3GB of RAM, where G means 1024^3.)
On Thu, 13 Dec 2007, Thomas Pujol wrote:
> I am using R
or just try looking in the annotate package from Bioconductor
Gabor Grothendieck wrote:
> On Dec 13, 2007 9:03 PM, Farrel Buchinsky <[EMAIL PROTECTED]> wrote:
>> I would like to track in which journals articles about a particular disease
>> are being published. Creating a pubmed search is trivial
Under Windows, I have used RODBC to connect to Excel spreadsheets as per
the example below:
library(RODBC);
connect = odbcConnectExcel("testdata.xls");
query = "SELECT [data$.ethn], [data$.sex], [data$.age],
[data$.height], [data$.weight],
[label$.label]
On Dec 13, 2007, at 9:16 PM, Farrel Buchinsky wrote:
> I am afraid not! The only thing I know about Python (or Perl, Ruby
> etc) is that they exist and that I have been able to download some
> amazing freeware or open source software thanks to their existence.
> The XML package and specifical
On Dec 13, 2007 9:03 PM, Farrel Buchinsky <[EMAIL PROTECTED]> wrote:
> I would like to track in which journals articles about a particular disease
> are being published. Creating a pubmed search is trivial. The search
> provides data but obviously not as an R dataframe. I can get the search to
> ex
I am afraid not! The only thing I know about Python (or Perl, Ruby etc) is
that they exist and that I have been able to download some amazing freeware
or open source software thanks to their existence.
The XML package and specifically the xmlTreeParse function looks as if it is
begging to do the ta
On Dec 13, 2007, at 9:03 PM, Farrel Buchinsky wrote:
> I would like to track in which journals articles about a particular
> disease
> are being published. Creating a pubmed search is trivial. The search
> provides data but obviously not as an R dataframe. I can get the
> search to
> export t
> I am trying to detrend and transform variables to achieve
> normality and stationarity (for time series use, namely spectral
> analysis). I am using the boxcox transformations.
>
>
>
> As my dataset contains zeros, I found I need to add a constant to
> it in order to run "boxcox".
If your data
Kevin,
I guess this may not have a one line solution (I may be wrong). I would try
something like this:
-Get all the unique (use 'unique') values in 'temp'
-Loop through these values and extract the subset of data and perform the
ANOVA on this subset.
As an example, code may go something like thi
I would like to track in which journals articles about a particular disease
are being published. Creating a pubmed search is trivial. The search
provides data but obviously not as an R dataframe. I can get the search to
export the data as an xml feed and the xml package seems to be able to read
it.
Hi,
Hope this does not sound too ignorant .
I am trying to detrend and transform variables to achieve normality and
stationarity (for time series use, namely spectral analysis). I am using the
boxcox transformations.
As my dataset contains zeros, I found I need to add a constant to it
I am using RODBC and "odbcConnect". I have successfully used odbcConnect to
extract "modest" amounts of data from SQL. For convenience, (and maybe speed?)
I wish, if possible, to extract larger amounts of data in a single query.
(I am running R2.6.0 under a machine running Windows Small B
Yea that worked !!
Did
paste(SELECT * FROM MET WHERE molid=", molid, "AND mode=", "\'" mode,
"\'" , sep="")
I guess all I needed was the sep="" which I totally forgot about.
Cheers,
Paul
Eric wrote:
> Does this work for you?
>
> "SELECT * FROM data WHERE molNUM = 165 AND mode = '+'"
>
> Possibl
Go to your Rprofile file (in the etc directory) and add the following line:
options(htmlhelp=TRUE)
I hope this helps
Francisco
Knut Krueger wrote:
> I forgot how to switch between Windows helpfiles and Browser helpfiles.
> f.e ?lm should open the browser.
> Maybe anybody could give me a hint?
>
Hi Monica,
Try
sapply(as.list(str),function(x) unlist(strsplit(x,"
"))[1])
--- Monica Pisica <[EMAIL PROTECTED]> wrote:
>
> Hi everyone,
>
> I have a vector of strings, each string made up by
> different number of words. I want to get a new
> vector which has only the first word of each str
The author is actually "Ye", and not "Yi". It is titled "On Measuring and
correcting the effects of data mining and model selection" by Jianming Ye,
JASA(1998).
Here is link from JSTOR:
http://www.jstor.org/view/01621459/di015668/01p00145/0?currentResult=0162145
9%2bdi015668%2b01p00145%2b0%2cFF1
Hi Livia,
Below is my solution. I believe that other list
members will post a better one.
Regards,
Moshe.
> rm(list=ls())
>
eval(parse(text=paste(paste("x",1:5,sep=""),paste("<-1.",1:5,";",sep=""
>
eval(parse(text=paste(paste("y",1:5,sep=""),paste("<-2.",1:5,";",sep=""
>
eval(parse(text
On Thursday 13 December 2007, Frank E Harrell Jr wrote:
> Andrew Park wrote:
> > Hi there
> >
> > In rpart, one can get a graph of R-squared (using rsq.rpart (fit)), in
> > which the x axis is the number of splits, and which contains two lines -
> > an "apparent" R squared and an Rsquared based on
On Thursday 13 December 2007, Judith Flores wrote:
> Dear R-experts,
>
>I need to run a R code on different computers (MACs
> and PCs) on different monitos sizes. The code
> generates graphs that are displayed in the same page.
> But there are some elements of the graphs that appear
> at the ri
Dear R-experts,
I need to run a R code on different computers (MACs
and PCs) on different monitos sizes. The code
generates graphs that are displayed in the same page.
But there are some elements of the graphs that appear
at the right position on a PC, but they don't appear
at all when I run it
On Dec 13, 2007 4:15 PM, Ilona Leyer <[EMAIL PROTECTED]> wrote:
> Dear All,
> I want to analyse treatment effects with time series
> data: I measured e.g. leaf number (five replicate
> plants) in relation to two soil pH - after 2,4,6,8
> weeks. I used mixed effects models, but some analyses
> didn
Dear All,
I want to analyse treatment effects with time series
data: I measured e.g. leaf number (five replicate
plants) in relation to two soil pH - after 2,4,6,8
weeks. I used mixed effects models, but some analyses
didn´t work. It seems for me as if this is a randomly
occurring problem since so
Hi all,
I have a quick question, I want to send a select statement to the
mysql server. With the code below I connect and ask my select statement.
As you can see from the code I want to have a mode = "+". The '+' needs
to be free text and hence the "\"" around it. But it doesn't work. If i
Hi,
Is there a R library (code) to calculate Average Precision?
I have been using ROCR to calculate other measure like AUC, F-measure etc.
But ARP (average precision) was not available although it provides
precision/recall.
I am almost sure that someone already implemented it.
wan
_
Andrew Park wrote:
> Hi there
>
> In rpart, one can get a graph of R-squared (using rsq.rpart (fit)), in which
> the x axis is the number of splits, and which contains two lines - an
> "apparent" R squared and an Rsquared based on the x error.
>
> I would like to caclulate these R-squared value
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of tom soyer
> Sent: Thursday, December 13, 2007 1:27 PM
> To: r-help@r-project.org
> Subject: [R] counting weekday in a month in R
>
> Hi,
>
> I am trying to count weekday of the month using R. For
> e
Hi there
In rpart, one can get a graph of R-squared (using rsq.rpart (fit)), in which
the x axis is the number of splits, and which contains two lines - an
"apparent" R squared and an Rsquared based on the x error.
I would like to caclulate these R-squared values, but cannot work out from the
You may want to check package dlm and, possibly, dse.
In dlm you can cast a VARMA model in state space form (dlmModARMA) and
estimate unknown parameters by maximum likelihood (dlmMLE).
Best,
Giovanni
> Date: Thu, 13 Dec 2007 11:17:47 -0800 (PST)
> From: creepa1982 <[EMAIL PROTECTED]>
> Sende
Hi,
I am trying to count weekday of the month using R. For example, 1/4/2001
is the 4th weekday of Jan, and 1/5/2001 is the 5th weekday of the month, and
1/8/2001 is the 6th weekday of the month, etc. I get as far as extracting
the weekdays from a sequence of dates (see below). But I have not yet
Deepayan:
Very nice, thanks for introducing me to a new resource. I will include the
entire, functioning example in the event others may find it useful.
--Seth
## Sample code for overlaying data points on a contour graph, using
xyplot and contourplot ##
library(lattice)
mo
Hi Seth,
An alternative would be to use ggplot2, http://had.co.nz/ggplot2:
model <- function(a,b,c,X1,X2) {
(exp(a + b*X1 + c*X2)) / (1 + exp(a + b*X1 + c*X2))
}
g <- expand.grid(X1 = seq(0.40, 0.8,0.01), X2 = seq(0.03,0.99,0.03))
a <- -37.61
b <- 34.88
c <- 28.44
g$z<- model(a, b, c, g$X1,g$
On 12/13/07, Seth W Bigelow <[EMAIL PROTECTED]> wrote:
>
> Friends: I wish to overlay data points on a contour graph. The following
> example produces a nice contour plot, but I have not mastered the concept
> of using panel functions to modify plots. Can someone show me how to
> overlay the data
I should say that the name of this chart varies even among
Spanish-speaking countries. In Argentina is "diagrama de torta" which
is something like "cake-chart".
Julian
ahimsa campos-arceiz wrote:
> Two non-eatable examples from Spain and Japan:
>
> in Spanish we call them "diagrama de sector
Hello R help-ers,
I have a basic question, but I have been playing with it for a while and
haven't quite gotten a hang of how to get it working. I want to perform
multiple one-way ANOVAs on subsets of data and am not sure how to do it in
an automated way. I am thinking of doing something simila
Friends: I wish to overlay data points on a contour graph. The following
example produces a nice contour plot, but I have not mastered the concept
of using panel functions to modify plots. Can someone show me how to
overlay the data points (given after contour plot statement) on the
contourplot?
Thank you for the reply. With the improved formula, mvpart worked like a charm.
Sincerely,
Jeff
Jeff H Bardwell, M.S.
Biology Department
ENV 1101 Lab Coordinator
Goebel 115, OH: Thu 1pm-4pm
710-6596 (e-mail preferred)
From: Gavin Simpson [mailto:[EMAIL PROTEC
Hi all,
does anyone know of a package/function for fitting Vector Autoregressive
Moving Average models? I looked through most of the packages available but
could only find functions to fit a VAR.
Any help would be appreciated!
Benjamin
--
View this message in context:
http://www.nabble.com
Hello
Is there an easy way, i.e. a function in a package, to calculate the
area under the curve (AUC) for drug serum levels?
Thanks for any advice
--
Armin Goralczyk, M.D.
--
Universitätsmedizin Göttingen
Abteilung Allgemein- und Viszeralchirurgie
Rudolf-Koch-Str. 40
39099 Göttingen
--
Dept. of
> Yes, I know this works. I just give it for example to show what I
> would like to achieve. In fact, I have many variables, you can
> imagine it is very annoying to construct the data frame by mannually
> assigning the variables and binding them. Therefore, I would like to
> do it automaticlly
Hello,
I'm new to R, and I would like to know if there is a way to smooth a curve
using a Gaussian smoothing with R.
Thank you very much,
TDB
--
View this message in context:
http://www.nabble.com/Gaussian-Smoothing-tp14321313p14321313.html
Sent from the R help mailing list archive at Nabb
Dear all,
I run the GAMs (generalized additive models) in gam(mgcv) using the
following codes.
m.gam
<-gam(mark~s(x)+s(y)+s(lstday2004)+s(ndvi2004)+s(slope)+s(elevation)+disbinary,family=binomial(logit),data=point)
And two repeated warnings appeared.
Warnings$B!'(B
1: In gam.fit(G, family = G
On Thu, 13 Dec 2007, Tim Coote wrote:
> I've hunted around to try to work this out and cannot find anything
> aposite, although there are exhortations to read News files and the fine
> manual in response to similar queries. Hope I haven't missed anything
> obvious.
>
> I want to read in csv files
I've hunted around to try to work this out and cannot find anything
aposite, although there are exhortations to read News files and the fine
manual in response to similar queries. Hope I haven't missed anything
obvious.
I want to read in csv files that contain dates, or date times. If I read
the
>cbcname1 <- rnorm(100,0,1)
>cbcname2 <- rnorm(100,0.5,1)
>y <- rnorm(100,0.6,1)
>lm1 <- lm(y~cbcname1+cbcname2)
>
>I have finished the linear regression until here. Now I would
> like to predict for new data. This is the question part.
>
>data <- data.frame(cbcname1=0.1,
Hello everybody, i had a big issue with RMySQL,
working on an installation of R on multiple nodes via net file-system.
The point is that we installed Xampp as suite for mysql, perl, apache
and php on every node, and we need to use R scripts with mysql.
Xampp doesn't compile anything, it gives you
> Thank you very much for your reply.I came across with another
> problem for assigning the variables.
> in order to achieve cbcname1 = 0.1, I would use sth like:
>
> paste("cbc","name1","=",function(),sep="")
>
> But it seems incorrect. Any advice?
This sounds like a different problem
Dear useRs,
the seqinR package contains utilities to import and analyze biological
sequence data. For a general introduction see this document:
http://pbil.univ-lyon1.fr/software/SeqinR//vignette.pdf
Please do not use r-help for questions about seqinR or r-bugs
for bug report about seqinR. Use in
> I get some data in the following format and I would like to combine them
to
> form a dataframe.
>
> The data is like:
>
> cbcname1 = 0.1,
> cbcname2= 0.2,
> cbcname3=0.3,...
>
> name1, name2, name2 are just some random names. I would like to achieve
sth
> like:
> (cbcname1=0.1, cbcname2=0.2,
Dear readers,
Is it possible to compute a Model I regression (Sokhal & Rolf 1995) in R
where there is more than one value of Y for each value of X?
Thanks, John Lowen
--
View this message in context:
http://www.nabble.com/More-than-one-value-of-Y-for-each-value-of-X-%28Model-I-regression%29-
> great! I knew you would have thought this through. That's perfect. As
> always there's the trade-off between writing code and documenting the
> code already written. In this case the trade-off turned toward the
> code part I guess.
>
> Autodetection of strings by aes would be even greater but tha
On 2007-December-13 , at 15:56 , hadley wickham wrote:
> Hi Jiho,
>
> The key to solving this problem is to use aes_string instead of aes.
> Instead of the complicated munging that aes does to get the names of
> the variables, aes_string works directly with strings, so that:
>
> aes_string(x = "m
Thanks for the reply..though I am trying to still work around the problem of
overlapping labels
--
Chandrasekhar Ganduri
320 West Union St, APT # 105
Athens, OH - 45701
Ph: 740-274-9170
[[alternative HTML version deleted]]
__
R-help@r-project.
I deleted the file zzz.R from the R subdirectory which contained the following:
## startup file for clValid package
#.First.lib <- function(libname, pkgname, where) {
# if( !require(methods) ) stop("we require methods for package clValid")
# where <- match(paste("package:", pkgname, sep=""), sea
> Is the function "current.grobTree()" implemented in ggplot2? According to
> the draft ggplot2 book, on page 43, we can get a list of all grobs with
> current.grobTree(). But when I try that, I get 'Error: could not find
> function "current.grobTree"'.
Ooops - the book is now out of date. Paul M
Hi Jiho,
The key to solving this problem is to use aes_string instead of aes.
Instead of the complicated munging that aes does to get the names of
the variables, aes_string works directly with strings, so that:
aes_string(x = "mpg", y = "wt") == aes(x = mpg, y = wt)
So your function would look l
> PS: If you care to explain: why do all parameters in the code below
> have a "." before the name, except precisely "limits"? I know it has
> to do with "proto", but could not find out why this one was different.
Not really - I started off with the convention that variables should
start with a .
>Hi,
>
>I would like to use R to estimate the following model:
>
>X(t) = a + b1*X(t-1) + b2*X(t-2) + c1*Y(t) + c2*Y(t-1) + c3*Y(t-2)
>
>Is there any R function that performs this type of estimation? I know
>that if I only have one time series (i.e. lagged value of X) on the
>right hand side then th
If you don't need a complex error structure then the dyn package
(and also the dynlm package) can do it. Using R's builtin
EuStockMarkets time series:
library(dyn)
z <- as.zoo(EuStockMarkets)
mod1 <- dyn$lm(DAX ~ lag(DAX, -(1:2)) + lag(FTSE, -(0:2)), z)
mod1
# compare to model without FTSE
mod2
There is probably a better way but something like this
should do it for your second version.
aa <- 1:20
bb <- c(5,10,15,20 )
cc <- c(25, 50, 75, 100)
plot(aa)
axis(side=3, labels=cc, at=bb)
--- Ronaldo Reis Junior <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I try to make a plot like this:
>
> Y |
>
Hello everyone,
I get some data in the following format and I would like to combine them to
form a dataframe.
The data is like:
cbcname1 = 0.1,
cbcname2= 0.2,
cbcname3=0.3,...
name1, name2, name2 are just some random names. I would like to achieve sth
like:
(cbcname1=0.1, cbcname2=0.2, cbcnam
Folks,
Thanks a lot for a very speady answer. I usually read the list quite a bit
because sometimes others put questions i may be interested in. One obs. i made
was that lots of neat tricks can be made with apply, sapply and tapply. But
even if i read all the help on these functions , and the
I forgot how to switch between Windows helpfiles and Browser helpfiles.
f.e ?lm should open the browser.
Maybe anybody could give me a hint?
Regards Knut
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read th
On Thu, 13 Dec 2007, Monica Pisica wrote:
>
> Hi everyone,
>
> I have a vector of strings, each string made up by different number of
> words.
You need to define 'word' and 'first'. Your solution says the first word
of " aa" is "", which is not what most people would think.
> I want to get a
Good afternoon Monica,
Relying on regular expressions, substituting nothing "" for everything
starting with a space until the end of the "line" (i.e. with a dollar sign)
str1 <- sub(" .*$", "", str)
Regards,
Sean
Monica Pisica wrote:
>
>
> Hi everyone,
>
> I have a vector of strings, eac
Monica Pisica wrote:
> Hi everyone,
>
> I have a vector of strings, each string made up by different number of words.
> I want to get a new vector which has only the first word of each string in
> the first vector. I came up with this:
>
> str <- c('aaa bbb', 'cc', 'd eee aa', 'mmm o n')
> st
> I try to make a plot like this:
>
> Y |
> |
> | o
> | oo
> | o o
> | o o
> |o o o
> |
>0 10 20 30 40 50 60 70 80 90 100 (A)
> 100 90 80 70 60 50 40 3
Hi,
I would like to use R to estimate the following model:
X(t) = a + b1*X(t-1) + b2*X(t-2) + c1*Y(t) + c2*Y(t-1) + c3*Y(t-2)
Is there any R function that performs this type of estimation? I know
that if I only have one time series (i.e. lagged value of X) on the
right hand side then there are R
one way is the following:
str <- c('aaa bbb', 'cc', 'd eee aa', 'mmm o n')
sapply(strsplit(str, " "), "[", 1)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven,
Hi,
if you only want the first word, then this should do the trick :
R> sub( " +.*", "", str )
[1] "aaa" "cc" "d" "mmm"
Cheers,
Romain
Monica Pisica wrote:
> Hi everyone,
>
> I have a vector of strings, each string made up by different number of words.
> I want to get a new vector which
See the data imput export manual on CRAN
--- [EMAIL PROTECTED] wrote:
> Hello everyone,
>
> I'll to request some input on what is available for
> use as an R/Excel
> interface; any help will be appreciated.
>
> Tony.
> [[alternative HTML version deleted]]
>
> ___
Hi everyone,
I have a vector of strings, each string made up by different number of words. I
want to get a new vector which has only the first word of each string in the
first vector. I came up with this:
str <- c('aaa bbb', 'cc', 'd eee aa', 'mmm o n')
str1 <- rep(1, length(str))
for (i in
Something like this?
> x<-0:100
> y<-dnorm(x,50,20)
> plot(x,y)
> axis(3,seq(0,100,by=20),labels=seq(100,0,by=-20))
On 12/13/07, Ronaldo Reis Junior <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I try to make a plot like this:
>
> Y |
> |
> | o
> | oo
> | o
On 12/13/2007 8:16 AM, Ronaldo Reis Junior wrote:
> Hi,
>
> I try to make a plot like this:
>
> Y |
> |
> | o
> | oo
> | o o
> | o o
> |o o o
> |
>0 1
Hi,
I try to make a plot like this:
Y |
|
| o
| oo
| o o
| o o
|o o o
|
0 10 20 30 40 50 60 70 80 90 100 (A)
100 90 80 70 60 50 40 30 20 10 0 (B)
o
try something like this:
combn(LETTERS[1:3], 2, paste, collapse = "")
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(
Hi all,
Suppose I have a population of 3 alphabets : A, B, C. From this population,
number of ways that any 2 can be chosen is 3 i.e. AB, AC, and BC.
Is there any R function to generalize this process, for any number of
alphabets/numbers and for any sub-sample size?
Thanks and r
Hi Hadley,
Thanks for the help.
I found the problem. In the code you sent me earlier to correct the
scale_y_continuous, (below) the parameter is "limits", rather than
"limit", as you write it. If I run my code without introducing the
change you sent me, it works, but if I first correct it, th
A smooth of `disbinary' doesn't really make sense. It should enter the
model parametrically. The problem is easy to see in 1D. Consider the
model
y_i = f(x_i) + error_i
where `f' is a smooth function. If x_i only takes the values 0 and 1
then you really don't have much of a basis for estimating
Tiago R Magalhães wrote:
> Dear members of the mailing list,
>
> I want to fetch warnings() from a series of prop.test calls. I want to
> get "none" if no warning is issued, and "warning" if there is a problem.
>
> I have looked (and relooked) at options(warn) and warning(), warnings()
> and 'last.
Dear members of the mailing list,
I want to fetch warnings() from a series of prop.test calls. I want to
get "none" if no warning is issued, and "warning" if there is a problem.
I have looked (and relooked) at options(warn) and warning(), warnings()
and 'last.warning' but to no avail. I am a biol
Öhagen Patrik wrote:
>
> Dear List,
>
> Please put me on the right track. I have been searching for the R functions
> for measures of agreement. Where should I look?
>
Hi Patrik,
Tried the irr package?
Jim
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R-help@r-project.org mailing list
https:
On Wed, 12 Dec 2007, Spencer Graves wrote:
>
> How can I identify the problem generating a warning in R CMD check
> for "Rd files with unknown encoding"?
>
> Google identified an email from John Fox with a reply from Brian
> Ripley about this last 12 Jun 2007.
But not on this list:
htt
At 12:17 06/12/2007, Michael Dewey wrote:
>I have a series of question about methods and generics.
To avoid extra clutter I have deleted the rather long question which
is obtainable from
https://www.stat.math.ethz.ch/pipermail/r-help//2007-December/147754.html
I received a very helpful response
Follow up.
On 2007-December-13 , at 10:45 , jiho wrote:
> foo1 <- function(uv="u")
> {
> # solution 1: do not use the data argument at all
> # (forces the use of qplot, could be more elegant)
> B = A[A$y<=5,]
> qplot(B$x, B$y, fill=B[[uv]], geom="tile")
> }
---> act
Hi everyone, Hi ggplot users in particular,
ggplot makes it very easy to plot things given their names when you
use it interactively (and therefore can provide the names of the
columns).
qplot(x,foo,data=A) where A has columns (x,y,foo,bar) for example
but I would like to use this fro
I´m sorry for no explain this before. It´s nls2 from INRA not from CRAN that i
was unsecesfull to install.
Mensaje citado por Patrick Connolly <[EMAIL PROTECTED]>:
> On Wed, 05-Dec-2007 at 01:40PM +0100, Florencio González wrote:
>
> |>
> |> Hi Thanks for your suggestion, I'm trying to install t
cut(data, breaks=n)
splits the data in n bins of (approximately) the same size.
The used size is obtained by:
max(data) - min(data)
n
> x=rnorm(x)
> cut(x,breaks=3)
[1] (1.79,9.97] (-6.39,1.79] (9.97,18.2] (9.97,18.2] (-6.39,1.79]
[6] (
melissa cline wrote:
> Hello,
>
> I'm trying to bin a quantity into 2-3 bins for calculating entropy and
> mutual information. One of the approaches I'm exploring is the cut()
> function, which is what the mutualInfo function in binDist uses. When it's
> called in the format cut(data, breaks=n),
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