Ryacas provides an interface to yacas from R. So my suggestion is to search the
web to see if yacas can solve your specific problem.
Regards
Søren
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of jpm miao
Sent: 16. august 2012
Worked!
Thanks.
David.
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__
R-help@r-project.org mailing list
Dear All,
I would be very grateful if someone could answer the following question:
I am trying to randomly generate values from the zero-truncated negative
binomial distribution. I already found rposnegbin in the package VGAM. However,
I need a detailed code since I am trying to export it to
Hi all,
I have a 4 by 100 matrix. I wish to extract each column and make it into a 2
by 2 matrix.
I also want to assign names for each 2X2 matrix.
For example
set.seed(2)
a=matrix(rnorm(400),ncol=100)
a1=matrix(a[,1],ncol=2)
a2=matrix(a[,2],ncol=2)
.
.
.
a100=matrix(a[,100],ncol=2)
Any
John/Michael, thanks for your comments.
I get the following error when I run my code (the code is attached as is a
section of the data set):
Error in family$linkfun(mustart) : Value 1.125 out of range (0, 1)
If anyone can please provide assistance that would be much appreciated - I'm
not sure
David Reiner david.rei...@xrtrading.com
on Wed, 15 Aug 2012 09:59:10 -0500 writes:
As a mathematician, I have to correct the subject line
to 'Distributive Matrix operations' -- David
Yes, indeed;
that was *really* disturbing to me as well, about this thread!
{in addition to the
set.seed(2)
a=matrix(rnorm(400),ncol=100)
#use a list
aList-list()
for(i in 1:dim(a)[2]){
aList[[i]]-matrix(a[,i],ncol=2)
}
#get lots of variables
for(i in 1:dim(a)[2]){
assign(paste(a,i,sep=),matrix(a[,i],ncol=2))
}
On 16.08.2012, at 08:07, bantex wrote:
Hi all,
I have a
Hello
This is a problem I encountered repeatedly and I found no answer that made
me really happy. I hope it is not too trivial.
I would like to give the concentration of a substance in a plot title:
5 ug/ml substance
the '5' would be a variable and the ug should be micrograms (with greek
Yin Aphinyanaphongs yinnersp...@gmail.com
on Tue, 14 Aug 2012 09:13:21 -0400 writes:
I am using the reshape package to convert a series of
values into a binary matrix. The binary matrix is very
sparse with many zeros and I'd like to use cast to
generate a sparse matrix
Hi Dominik,
You can try
x - 5
plot(rnorm(50), main=bquote(.(x) * mu * g/m^3 * substance))
Regards,
- Jon
On Thu, Aug 16, 2012 at 3:37 PM, Dominik Refardt
dominik.refa...@gmail.com wrote:
Hello
This is a problem I encountered repeatedly and I found no answer that made
me really happy. I
-Original Message-
I have a 4 by 100 matrix. I wish to extract each column and
make it into a 2 by 2 matrix.
#make a toy 4x100 matrix:
m - matrix(rnorm(400), ncol=100)
#use lapply after on-the-fly conversion to a list-like object:
a.list - lapply(as.data.frame(m),function(x)
I would like to give the concentration of a substance in a plot title:
5 ug/ml substance
Examples of including a variable in text are given in the ?plotmath page, under
## How to combine math and numeric variables.
For your case,
plot(1:10)
conc=5
title(main=bquote(.(conc)~mu*g/ml
Thanks to you both. I'm not sure all.equal() was right for this situation, is
it? If the differences were to the right of the decimal and differed as a
function of precision, then I see all.equal() being the right function. But,
the differences in the reproducible code were to the left of the
On Aug 16, 2012, at 09:37 , Dominik Refardt wrote:
Hello
This is a problem I encountered repeatedly and I found no answer that made
me really happy. I hope it is not too trivial.
I would like to give the concentration of a substance in a plot title:
5 ug/ml substance
the '5' would
That worked but how do you get the location of one particular tweet.
You are teetering on the edge of potential stalking here,
since locating a specific tweet implies locating a specific
individual. In some countries that can be a problem.
Location is currently disabled by default in
Thanks to you both. I'm not sure all.equal() was right for this situation, is
it? If the differences were to the right of the decimal and differed as a
function of precision, then I see all.equal() being the right function. But,
the differences in the reproducible code were to the left of
Hi
Without some data it is difficult to say where is difference. However from code
below it seems to me that it is an adaptation from my attempt to produce colour
coded points on biplot.
http://tolstoy.newcastle.edu.au/R/e4/help/08/08/20012.html
fit-princomp(iris[,1:4], cor=T)
biplot(fit,
Hi
After importing data from Excel through ODBC.
In the inclusion dataset, class(inclusion$Value) is coming as factor.
After filtering the data, length(inclusion$Value == 0),the answer is
coming as 4879, but actually Value contains only 225 rows. So how can I
How do you know?
What does
Dear expeRts,
I have a master file master.tex containing the preamble and which inputs (via
\input{chapter01}, \input{chapter02}, ...) chapters. The chapters are .Rnw
files. My goal is to use patchDVI::SweavePDF to compile the chapters (say,
chapter.Rnw) individually (each chapter starts with
Sensitivity and specificity are improper scoring rules so beware. They are
optimized by a bogus model.
Frank
Diana Marcela Martinez Ruiz wrote
Hello,
As obtained from a table svyglm clasificaion, sensitivity and
specificity. The funtion ConfusionMatrix () of the library (caret)
gives
Thanks a lot. However it's not completely what I want. There is space
missing between the variable and the mu. Can this be added?
On Thu, Aug 16, 2012 at 12:09 PM, Miguel Manese jjon...@gmail.com wrote:
Hi Dominik,
You can try
x - 5
plot(rnorm(50), main=bquote(.(x) * mu * g/m^3 *
Hi,
I altered .internal file as Ligges indicated, but I have a problem when I make
R CMD build pkg:
ERROR
packaging into .tar.gz failed
The question is the following: what is the problem? I have installed Rtools,
Must I do anything else?.
Obs.: I work under Windows 7, but the package
I forget one question:
Where do I indicate this path? :
PATH=c:\Rtools\bin;c:\Rtools\gcc-4.6.3\bin;c:\MiKTeX\miktex\bin;
c:\R\R-2.15\bin\i386;c:\windows;c:\windows\system32
Regards,
Eva
--- El jue, 16/8/12, Eva Prieto Castro evapcas...@yahoo.es escribió:
De: Eva Prieto Castro
Hi,
I want to plot ROC curve for my detection algorithm which detects
features in different images at two different thresholds.
6 different images used and obtained tp, fp and fn. No tn in my case.
in first threshold run i obtained 6 values of tp,fp and fn. In second
threshold run agian i
Thanks a lot. Both
plot(1:10)
conc=5
title(main=bquote(.(conc)~mu*g/ml substance))
and
plot(0,0)
conc - 5
title(main=bquote(.(conc)* *mu*g/ml* substance))
do exactly what I need.
On Thu, Aug 16, 2012 at 1:33 PM, peter dalgaard pda...@gmail.com wrote:
On Aug 16, 2012, at 09:37 , Dominik
Thanks guys for the help. I finally know what to do know :)
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
Rich,
The cenboxplot function uses cenros to estimate the censored values. The
cenros function requires at least 2 uncensored observations to be able to
do the regression. The cenros function does issue a warning when there are
more than 80% censored data, but that is suppressed in
Hello,
I'm most grateful for your time to read this.
I have a uber size 30GB file of 6 million records and 3000 (mostly
categorical data) columns in csv format. I want to bootstrap subsamples for
multinomial regression, but it's proving difficult even with my 64GB RAM
in my machine and twice
Hi,
Could you indicate the way you resolved?. I am having the same problem you
had, and I can't resolve it. What do you mean when you say that folders have
be in front?.
Thanks in advance.
Eva
--
View this message in context:
You could also use a 3d array instead of a list:
m - matrix(1:400, ncol=100, byrow=TRUE)
a - array(m, dim=c(2, 2, 100), dimnames=list(row=c(1, 2),
col=c(1, 2), tbl=c(paste0(a, 1:100
a[,,1] # First table by index
col
row 1 2
1 1 201
2 101 301
a[,,a1] # First table by
On Thu, 16 Aug 2012, David L Lorenz wrote:
The cenboxplot function uses cenros to estimate the censored values. The
cenros function requires at least 2 uncensored observations to be able to
do the regression. The cenros function does issue a warning when there are
more than 80% censored data,
plot(rnorm(50), main=bquote(.(x)* * mu * g/m^3 * substance))
John Kane
Kingston ON Canada
-Original Message-
From: dominik.refa...@gmail.com
Sent: Thu, 16 Aug 2012 12:50:57 +0200
To: jjon...@gmail.com
Subject: Re: [R] Variables and greek letters in a plot title
Thanks a lot.
Zuki,
You did not provide any information on what table$attrib is. Using a log
scale for the y-axis doesn't make much sense to me, since the y-axis is
the cumulative proportion of observations, from 0 to 1. Perhaps you are
interested in the log scale for the x-axis, instead? In any case,
I suggest that you re-post your question, but this time make it easy for
readers to help you by providing a simple reproducible example of your
problem. Provide some fake data, or a small subset of your data using,
for example, the function dput(). Trim the fat from your code. Get rid
of
If you don't know what 1:100 is, you should (re)read the Introduction to R
document that comes with the software. You can also try typing expressions like
this alone at the command line to see what they are.
If you don't know what the first argument to the ecdf function is, you need to
learn
A quick hack to give you space between the line and point (if you only
used solid lines) is to specify lty='ff' to the legend function.
If you want more control then look at setting trace=TRUE and
plot=FALSE and looking at the printed outcome and the return value
from legend. This does not plot
Why not put this into a database, and then you can easily extract the
records you want specifying the record numbers. You play the one time
expense of creating the database, but then have much faster access to
the data as you make subsequent runs.
On Thu, Aug 16, 2012 at 9:44 AM, Tudor Medallion
The read.csv.sql function in the sqldf package may make this approach
quite simple.
On Thu, Aug 16, 2012 at 10:12 AM, jim holtman jholt...@gmail.com wrote:
Why not put this into a database, and then you can easily extract the
records you want specifying the record numbers. You play the one
To clarify:
Is TN = 0 or do you not know TN (N)?
On 16.08.2012, at 11:51, vjyns wrote:
Hi,
I want to plot ROC curve for my detection algorithm which detects
features in different images at two different thresholds.
6 different images used and obtained tp, fp and fn. No tn in my case.
I am building several pedigrees and would like to construct the graphs
automatically.
I used the following program:
require(kinship2)
pedig = with(Dados, pedigree(id=iid, dadid=fid, momid=mid, sex=sex,
famid=famid, missid=0))
for(famid in 1:15){
pedig[famid] - pedig['famid']
Hi
I have two vectors of integers and I am plotting a box plot that contains
both vectors. I have done the following:
vec1 -scan(file1)
vec2 -scan(file2)
boxplot(vec1, vec2, range=0, horizontal=TRUE, names=c(One,Two),
boxwex=0.15)
I managed to get both boxplots on the same figure but there
Check out:
http://rtutorialseries.blogspot.com/2011/02/r-tutorial-series-two-way-repeated.html
On 8/15/2012 11:32 AM, Diego Bucci wrote:
Hi,
I performed an ANOVA repeated measures but I still can't find any good news
regarding the possibility to perform multiple comparisons.
Can anyone help
On Aug 16, 2012, at 12:03 AM, Dinuk Jayasuriya wrote:
John/Michael, thanks for your comments.
I get the following error when I run my code (the code is attached
as is a section of the data set):
No, it's not (anymore at least). It needs to be a mime-text formated
file at the time it
Dear All,
I am evaluating the value of loglikelihood and it ends up with the sum of
tiny numbers.
Below is an example: suppose I would like to calculate sum_i (log (sum_j x
[i, j] )), the index of log (x) is in the range, say (-2000, 0). I am aware
that exp(-744.5) will be expressed as 0 in 32
Hello,
Without data, I've made up my own. Try the following.
x - rnorm(100)
y - rnorm(100)
bp - boxplot(x, y, range = 0, names=c(One,Two))
bx - bxp(bp, horizontal=TRUE, boxwex = 0.15, at = c(1, 1 + 2*0.15))
bx
Note that in the call to boxplot you only need the arguments that affect
the
I had noticed this oversight a while ago. In an updated version of the metafor
package (hopefully to be released in the near future), the argument will be
called maxiter (as intended). Using maxit will then work as well, due to
partial matching.
Thanks for bringing this to my attention though.
Hi,
I have a table in which one column has the name of the objects as shown below.
Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
hi,
i'm trying to understand r data structures. i see that vectors, matrix,
factors and arrays have a dimension.
there seems to be no mention of dimensionality anywhere for lists or
dataframes. can i consider lists and frames to be of fixed dimension 2?
thanks,
jay s
Hi,
On Thu, Aug 16, 2012 at 2:49 PM, Schumacher, Jay S j...@neo.tamu.edu wrote:
hi,
i'm trying to understand r data structures. i see that vectors, matrix,
factors and arrays have a dimension.
Out of curiosity, where do you see that vectors and factors have a
dimension? I mean -- I
Hi R community
I copied a bit of my R code that gets some data from a database. You
won't be able to run the code, but I am a beginner so you will
probably understand what going on.
I would like to make a variable I can refer to inside the sqlQuery.
Instead of writing the start date and time (ex
On Thu, Aug 16, 2012 at 01:41:17PM -0400, Jie wrote:
Dear All,
I am evaluating the value of loglikelihood and it ends up with the sum of
tiny numbers.
Below is an example: suppose I would like to calculate sum_i (log (sum_j x
[i, j] )), the index of log (x) is in the range, say (-2000, 0).
Sapana Lohani lohani.sap...@ymail.com wrote on 08/16/2012 01:41:23 PM:
Hi,
I have a table in which one column has the name of the objects as shown
below.
You don't provide example data, so I'm not sure what you mean by a
table. I will assume that you mean a data frame.
Name
Perhaps the sprintf function is what you are looking for. It is one
way to insert information from a variable into a string. A couple of
other options are paste, paste0, and the gsubfn package, but I think
sprintf will be simplest for what you are asking.
On Thu, Aug 16, 2012 at 1:30 PM,
?strsplit
-- Bert
On Thu, Aug 16, 2012 at 11:41 AM, Sapana Lohani lohani.sap...@ymail.com wrote:
Hi,
I have a table in which one column has the name of the objects as shown below.
Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope
Hi,
Try this:
dat1-readLines(textConnection(Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Hello,
Try the following.
x - c( Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15),
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15),
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10),
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) )
Hi,
Forgot the last part.
colnames(dat3New)-c(name,slope,percentage)
dat3New
name slope percentage
1 Budlamp-Woodcutter Complex 15 to 60% slope (60/25/15)
2 Budlamp-Woodcutter Complex 15 to 60% slope (60/25/15)
3
On Aug 16, 2012, at 11:49 AM, Schumacher, Jay S wrote:
hi,
i'm trying to understand r data structures. i see that vectors,
matrix, factors and arrays have a dimension.
there seems to be no mention of dimensionality anywhere for lists
or dataframes. can i consider lists and frames to
On Thu, Aug 16, 2012 at 9:01 AM, Diana Marcela Martinez Ruiz
dianamm...@hotmail.com wrote:
Hello,
As obtained from a table svyglm clasificaion, sensitivity and specificity.
The funtion ConfusionMatrix () of the library (caret)
gives these results but not how to apply it to svyglm.
predict()
It would be helpful to distinguish between a formal dimension attribute,
and a (personal) conceptual model of whether or not any particular R
object, or type of object, has dimension. Mention of data frames having
dimension can be found in the help page for the dim() function.
foo - 1:10
are these correct/accurate/sensible statements:
a vector is a one dimensional object.
a matrix is a two dimensional object.
a list is a one dimensional object.
i'm working from this web page:
http://www.agr.kuleuven.ac.be/vakken/statisticsbyR/someDataStructures.htm
I sometimes do this sort of thing with tricks like this:
sql - select * from mytable where dt = 'ADATE'
dbGetQuery( con, gsub('ADATE', '2012-06-12 23:14', sql) )
Or if mydates is a vector of dates stored as a POSIXt object:
for (id in mydates) {
dbGetQuery( con, gsub('ADATE',
On Thu, Aug 16, 2012 at 4:50 PM, Schumacher, Jay S j...@neo.tamu.edu wrote:
are these correct/accurate/sensible statements:
a vector is a one dimensional object.
a matrix is a two dimensional object.
a list is a one dimensional object.
i'm working from this web page:
yes, thank you, conceptual model (rather than formal dimension attribute)
is where i'm coming from at this point.
It would be helpful to distinguish between a formal dimension attribute,
and a (personal) conceptual model of whether or not any
On Aug 16, 2012, at 1:50 PM, Schumacher, Jay S wrote:
are these correct/accurate/sensible statements:
a vector is a one dimensional object.
a matrix is a two dimensional object.
a list is a one dimensional object.
i'm working from this web page:
I don't disagree with Michael, but I would add that to me it also depends.
If one thinks in terms of subsetting an object (for objects that can be
subsetted)
To subset a vector, one supplies *one* value for the index:
myvector[3]
myvector[ 2:5 ]
are valid statements.
Similarly for a list
On Thu, Aug 16, 2012 at 5:44 PM, MacQueen, Don macque...@llnl.gov wrote:
Whereas for a matrix or data frame, one must supply *two* index values
(even if one of them may be omitted)
mydf[ 1 , 3 ]
mydf[ , 5 ]
mymat[ 2:5 , ]
mymat[ 3 , 4:6 ]
are valid statements.
Not quite:
Dear List
I'm trying to install a package not present in cran named A2R (
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=79)
After running the demo script I retrieve the following error:
cannot change value of locked binding for '._a2r_counter'
Please could someone give to me a
Hi, I'm having trouble installing and using the 'foreign' package in R on a
UNIX machine.
sessionInfo()
R version 2.11.1 (2010-05-31)
sparc-sun-solaris2.10
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
On Aug 16, 2012, at 4:33 PM, Kevin Goulding wrote:
Hi, I'm having trouble installing and using the 'foreign' package in
R on a
UNIX machine.
sessionInfo()
R version 2.11.1 (2010-05-31)
sparc-sun-solaris2.10
locale:
[1] C
attached base packages:
[1] stats
Dear all,
I have a question on applying a function to the data according to factor
levels.
For example, for the data below, what is the best way to apply a function
to
values according to different levels of samples (1,2,3,4,5)?
values ind sample
1 0.03325 1 1
2 0.03305 1 1
?tapply
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of li li
Sent:
The package hasn't been modified since version 0.0-4 in August 2005. Your
only hope is to try to contact the author:
Package: A2R
Type: Package
Title: Romain Francois misc functions
Version: 0.0-4
Date: 2005-08-05
Author: Romain Francois francoisrom...@free.fr
Maintainer: Romain Francois
Hello,
Or ?aggregate, depending on the wanted output. With the dataset provided
by the op, apply function mean.
dat - read.table(text=
values ind sample
1 0.03325 1 1
2 0.03305 1 1
3 0.03185 1 1
[...etc...]
44 0.03060 5 4
45 0.06605 5 5
, header = TRUE)
Dear R users,
I work with a descrete variable (sclae 0 - 27) which is highly skwed to the
right (many zeros and small numbers). I measure this variable on a control and
intervention cohort 5 times a year. When I analyze analyze this varoable at
each time point separately and use GLM with
Hi guys,
I am trying to write a function that allows me to perform specific
transformations to each element of
a 2 by 2 matrix to generate a 3 by 3 matrix.
Input into function-2 by 2 matrix
Output from function -3 by 3 matrix
For example:
# a is my original 2 by 2 matrix
#b is my new 3 by 3
HI,
Slightly different way to do this:
dat1-readLines(textConnection(Name
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15)
Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10)
Terrarossa-Blacktail-Pyeatt Complex - 1 to
Hello,
I've been trying to setup a site library that allows the users to manage
the R packages themselves, but am having an issue with permissions. As
seen below, when installing a package using install.packages, the umask
used is always 022. Instead, I would like it to be 002, allowing any user
Hi,
Try this:
dat1-read.table(text=
no. values ind sample
1 0.03325 1 1
2 0.03305 1 1
3 0.03185 1 1
4 0.03515 1 1
-
-
42 0.01085 5 2
43 0.01660 5 3
44 0.03060 5 4
45 0.06605 5 5
Got it. Thanks so much.
Hannah
2012/8/16 arun smartpink...@yahoo.com
Hi,
Try this:
dat1-read.table(text=
no. values ind sample
1 0.03325 1 1
2 0.03305 1 1
3 0.03185 1 1
4 0.03515 1 1
-
-
42 0.01085 5
Dear all,
I am trying to use R to fit mixed models.
Take the following example, where ind is a random effect and
sample is fixed. I wanted to fit
Model 1: values = ind + sample
Model 2: values =ind * sample
Model 3: values=ind(sample) + sample
Tried to use the below for mod1, but it
Hi I am new to R so am struggling with the commands here
I have one column in a table that looks like
slope (60/25/15)
slope (90/10)
slope (40/35/15/10)
slope (40/25/25/10)
I want to have 4 columns with just the number inside the parenthesis. when
there is no number that cell can have 0. I
Hi guys,
After a long while I came up with this :
set.seed(2)
a-matrix(rnorm(4),ncol=2)
abc-function(a)
{
b=matrix(NA,nrow=3,ncol=3)
b[1,1]=a[1,1]+1
b[1,2]=a[2,1]*a[2,2]
b[1,3]=a[2,2]+a[1,1]
b[2,1]=a[1,1]-5
b[2,2]=sqrt(a[2,2])
b[2,3]=a[1,1]/a[2,2]
b[3,1]=a[2,2]-3
hi,
TN=0 in all cases, with this how can i plot the ROC, i need help in this
regard.
thank you.
Date: Thu, 16 Aug 2012 09:47:20 -0700
From: ml-node+s789695n4640509...@n4.nabble.com
To: vijay...@outlook.com
Subject: Re: no true negative data, need roc curve
To clarify:
Is TN = 0
To Whom It May Concern,
In attempting to install the e1071 package, I get the message below after
selecting the mirror site.
-- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository
http://mirror.fcaglp.unlp.edu.ar/CRAN/bin/windows/contrib/2.15
1. Post on R-sig-mixed-models, not here.
2. Models 2 and 3 make no sense (to me, anyway). What do you think
they mean? (Don't answer here -- explain on the mixed models list).
-- Bert
On Thu, Aug 16, 2012 at 9:13 PM, li li hannah@gmail.com wrote:
Dear all,
I am trying to use R to fit
TN=0 in all cases, i had only tp, fp and fn for 6 images (two sets).
suggest me how plot the roc curve.
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Hi Diego,
I am struggeling with this question also for some time and there does
not seem to be an easy and general solution to this problem. At least I
haven't found one.
However, if you have just one repeated-measures factor, use the solution
describe by me here:
Hello,
Try the following.
x - c(slope (60/25/15),
slope (90/10),
slope (40/35/15/10),
slope (40/25/25/10))
dat - data.frame(X = x)
lst - unlist(lapply(dat, function(.x) gsub(slope \\((.*)\\)$, \\1,
.x)))
lst - strsplit(lst, /)
keep - seq_len(max(unlist(lapply(lst, length
vec - rep(0,
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