Hi,
I'm trying to figure out how to use capturing parenthesis in regular
expressions in R. (Doing this in Perl, Java, etc. is fairly trivial,
but I can't seem to find the functionality in R.)
For example, given the string:10 Nov 13.00 (PFE1020K13)
I want to capture the first to digits
:
On Thu, 4 Nov 2010, Noah Silverman wrote:
Hi,
I'm trying to figure out how to use capturing parenthesis in regular
expressions in R. (Doing this in Perl, Java, etc. is fairly trivial,
but I can't seem to find the functionality in R.)
For example, given the string:10 Nov 13.00
Hi,
I have a process in R that produces a lot of output. My plan was to
build up a matrix or data.frame row by row, so that I'll have a nice
object with all the resulting data.
I started with:
results - matrix(ncol=3)
names(results) - c(one, two, three)
Then, when looping through the data:
That was a typo.
It should have read:
results[results$one 100,]
It does still fail.
There is ONE column that is text. So my guess is that R is seeing that
and assuming that the entire data.frame should be factors.
-N
On 11/10/10 11:16 PM, Michael Bedward wrote:
Hello Noah,
If you set
That makes perfect sense.
Since I need to build up the results table sequentially as I iterate
through the data, how would you recommend it??
Thanks,
-N
On 11/11/10 12:03 AM, Michael Bedward wrote:
All values in a matrix are the same type, so if you've set up a matrix
with a character
Still doesn't work.
When using rbind to build the data.frame, it get a structure mostly full
of NA.
The data is correct, so something about pushing into the data.frame is
breaking.
Example code:
results - data.frame()
for(i in 1:n){
#do all the work
#a is a test label. b,c,d are
Langfelder
Sent: Thursday, November 11, 2010 12:25 PM
To: Noah Silverman
Cc: r-help@r-project.org
Subject: Re: [R] Populating then sorting a matrix
and/or data.frame
On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
n...@smartmediacorp.com
wrote:
Still doesn't work.
When using rbind to build
That makes perfect sense. All of my numbers are being coerced into
strings by the c() function. Subsequently, my data.frame contains all
strings.
I can't know the length of the data.frame ahead of time, so can't
predefine it like your example.
One thought would be to make it arbitrarily long
David,
Great solution. While a bit longer to enter, it lets me explicitly
define a type for each column.
Thanks!!!
-N
On 11/11/10 4:02 PM, David Winsemius wrote:
On Nov 11, 2010, at 6:38 PM, Noah Silverman wrote:
That makes perfect sense. All of my numbers are being coerced
Hi,
I'm trying to use the solve() function in R to invert a matrix. I get
the following error, Lapack routine dgesv: system is exactly singular
However, My matrix doesn't appear to be singular.
[,1] [,2] [,3] [,4]
[1,] 0.99252358 0.93715047 0.7540535 0.4579895
Hi,
I have a process (not in R) that records events with a time stamp. So,
I have a huge series of maybe 100,000 time stamps.
I'd like to break it up into hourly (Or daily) intervals and then count
how many events occurred in each interval. That way I can graph it.
Ideally, converting the
] -2.132894e-08 2.128452e-08
When dealing with numerical matrices you have to be prepared for the
unexpected.
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Noah Silverman
Sent: Sunday, 21 November 2010 2
Hi,
I have a series of data (about 80,000 pairs of x,y).
Plotting it shows a great chart. However, R has randomly chosen about 6
labels for my x axis. Now, clearly I can't show them all, but would
like some control over the granularity of what is displayed. I can't
find anything in the
I think I've come across a bug in the command line switches.
From R --help
--vanillaCombine --no-save, --no-restore, --no-site-file,
--no-init-file and --no-environ
--slave Make R run as quietly as possible
-q, --quiet
Hi,
I'm experimenting with a few learners that require a matrix as their
input. (Currently svmpath, vbmp, etc.)
I currently have a dataframe with 50 columns and 20,000 rows.
I tried using:
x - as.matrix(my_data.frame)
If I then as, is.matrix(x), I get TRUE.
However everywhere I've tried
:
On Fri, Oct 16, 2009 at 01:33:14AM -0700, Noah Silverman wrote:
Hi,
I'm experimenting with a few learners that require a matrix as their
input. (Currently svmpath, vbmp, etc.)
I currently have a dataframe with 50 columns and 20,000 rows.
I tried using:
x- as.matrix(my_data.frame)
If I
Hi,
I have a data.frame that I need to scale.
I've been using the scale function and it works nicely.
Some of the libraries I'm testing won't accept negative values for data,
so I need to find a way to scale the data from 0 to 1
Any ideas?
Thans!
Hi,
This is probably going to be one of those, It depends what you want
kind of answers, but I'm very curious to see if the group has an opinion
or some general suggestions.
The actual experiment is too complicated for a quick e-mail, but I'll
summarize well enough(hopefully) to get the
Hi,
This is probably going to be one of those, It depends what you want
kind of answers, but I'm very curious to see if the group has an opinion
or some general suggestions.
The actual experiment is too complicated for a quick e-mail, but I'll
summarize well enough(hopefully) to get the
Hi,
I have a process using svm from the e1071 library. it works.
I want to try using the KSVM library instead. The same data used wiht
e1071 gives me an error with KSVM.
My data is a data.frame.
sample code:
svm_formula - formula(y ~ a + B + C)
svm_model - ksvm(formula, data=train_data,
Hi,
It looks like you are potentially dealing with two separate issues.
1) Access - Mysql has very find grained permissions as to who can access
what and from where. You need to make sure that your username in mysql
is allowed to access the database/tables from your location.
2) Corruption
Hi,
I've search rseek.org high and low and can't seem to find an answer to this.
I want to maximize likelihood for a set of training data, but the data
is grouped. (Think multiple trials.)
It would probably be possible to do this with some nested for loops
manually, but would be painfully
Hi,
I'm trying to normalize some data.
My data is organized by groups. I want to normalize PER GROUP as
opposed to over the entire data set.
The current double loop that I'm using takes almost an hour to run on
about 30,000 rows of data in 2,500 groups.
I'm currently doing this:
/ data$sum
data
.. or even
transform(data, norm=ave(y, group, FUN = function(x) x/sum(x)))
I hope it helps.
Best,
Dimitris
Noah Silverman wrote:
Hi,
I'm trying to normalize some data.
My data is organized by groups. I want to normalize PER GROUP as
opposed to over
David,
That helps me a lot. Thanks!!!
-N
On 4/12/10 9:06 PM, David Winsemius wrote:
dat - as.data.frame(matrix( rnorm(200), 100 , 2)) # bivariate normal
n=100
ab - matrix( c(-5,-5,5,5), 2, 2) # interval [-5,5) x [-5,5)
nbin - c( 20, 20) # 400 bins
bins - bin2(dat, ab, nbin) # bin
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit by definition returns a
value between 0 and 1 a a probability. Can anyone suggest why I'm
seeing results outside of the {0,1} range??
The call in R is:
m - clogit(score ~
Thanks David,
That explains a lot. I appreciate it.
--
Noah
On 4/20/10 3:48 PM, David Winsemius wrote:
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by group.
My understanding was that a conditional logit
On 4/20/10 4:22 PM, Noah Silverman wrote:
Thanks David,
That explains a lot. I appreciate it.
--
Noah
On 4/20/10 3:48 PM, David Winsemius wrote:
On Apr 20, 2010, at 5:59 PM, Noah Silverman wrote:
Hi,
I'm calculating a conditional logit on some data stratified by group.
My
Hi,
An example data set is:
grouplevelcolor
A1blue
A1Red
B1blue
B2Red
A2Red
B2Red
B2blue
B2blue
A2blue
A2Red
I'd like to
Hi,
I'm looking for an easy way to discretize factors in R
I've noticed that the lm function does this automatically with a nice
result.
If I have
group - c(A, B,B,C,C,C)
and run:
lm(result ~ x1 + group)
The lm function has split the group into separate binary variables {0,1}
before
to lm, but not actually doing any regression?
Thanks again!
-N
On 5/15/10 11:17 AM, Thomas Stewart wrote:
Maybe this?
group - factor(c(A, B,B,C,C,C))
model.matrix(~0+group)
-tgs
On Sat, May 15, 2010 at 2:02 PM, Noah Silverman
n...@smartmediacorp.com mailto:n...@smartmediacorp.com wrote
1 0
4 4 3 0 0 1
5 5 4 0 0 1
6 6 5 0 0 1
Any ideas?
-N
On 5/15/10 11:02 AM, Noah Silverman wrote:
Hi,
I'm looking for an easy way to discretize factors in R
I could, but with close to 100 columns, its messy.
On 5/16/10 11:22 AM, Peter Ehlers wrote:
On 2010-05-16 11:06, Noah Silverman wrote:
Update,
I have it working, but now its producing really ugly labels. Must be a
small adjustment to the code. Any ideas??
##Create example data.frame
Hello,
I need to build a list of lists
We have 20 groups we are generating MCMC samples for. There are 10
coefficients, and 1 MCMC iterations.
I would like to store each iteration by-group in a list. My problem is
with the first iteration.
Here is a toy example:
Chain - list()
for (j in
]], coef)
If it does, this has the additional advantage that it tends to be
faster to initialize the list at size rather than expanding it as
needed.
HTH,
Josh
On Sun, May 30, 2010 at 2:52 PM, Noah Silverman n...@smartmediacorp.com
wrote:
Hello,
I need to build a list of lists
We
)
}
}
On Sun, May 30, 2010 at 6:05 PM, Noah Silverman n...@smartmediacorp.com
wrote:
That would be great, except I just realized I made a typo when sending
my code.
I'm tracking 20 coefficents for 10 groups. So I need a top list of 10
groups. Then each of the 10,000 samples for each of the 20
We're running Monte Carlo repeated measures for several groups.
The goal is to determine the number of time each group has the highest
score.
A toy example:
[,1] [,2] [,3]
0.1 0.2 0.3
0.1 0.2 0.3
0.1 0.2 0.3
0.1 0.3 0.2
0.1 0.3 0.2
0.2 0.3 0.1
For this example:
Hi,
Working on a report that is going to have a large number of graphs and
summaries. We have 80 groups with 20 variables each.
Ideally, I'd like to produce ONE page for each group. It would have two
columns of 10 graphs and then the 5 number summary of the variables at
the bottom.
So, perhaps
Lattice looks nice, but how can I put some summary text at the bottom?
On 5/31/10 11:27 AM, RICHARD M. HEIBERGER wrote:
Use lattice.
require(lattice)
?lattice
?xyplot
__
R-help@r-project.org mailing list
I'm running a long MCMC chain that is generating samples for 22 variables.
I have each run of the chain as a row in a matrix.
So: Chain[,1] is the column with all the samples for variable one.
Chain[,2] is the column with all the samples for variable 2, etc.
I'd like to fit all 22 on a single
Hi,
I used the term run, as each iteration of the Gibbs sampler produces
22 variables (coefficients for Beta in a regression model)
The example wont work
On 6/1/10 5:54 AM, Ben Bolker wrote:
Noah Silverman noah at smartmediacorp.com writes:
I'm running a long MCMC chain
AM, Noah Silverman wrote:
Hi,
Working on a report that is going to have a large number of graphs and
summaries. We have 80 groups with 20 variables each.
Ideally, I'd like to produce ONE page for each group. It would have two
columns of 10 graphs and then the 5 number summary
Hi,
I want to add a box at the bottom of a lattice window (device/page?).
Lattice has drawn a nice group of panels with all the plots I need. How
do I add my own summary text at the bottom (several lines worth?)
__
R-help@r-project.org mailing list
On Tue, Jun 1, 2010 at 9:37 AM, Noah Silverman
n...@smartmediacorp.com mailto:n...@smartmediacorp.com wrote:
Hi,
I used the term run, as each iteration of the Gibbs sampler produces
22 variables (coefficients for Beta in a regression model)
The example wont work
, 2010 at 10:51 AM, Noah Silverman
n...@smartmediacorp.com mailto:n...@smartmediacorp.com wrote:
You are correct,
I initially missed the as.mcmc step. Without it, R doesn't want
to squeeze so many plots onto a page.
I've had that problem before with lattice...which makes me wonder
::splitTextGrob
function might help.
HTH,
baptiste
On 1 June 2010 19:37, Noah Silverman n...@smartmediacorp.com wrote:
Hi,
I want to add a box at the bottom of a lattice window (device/page?).
Lattice has drawn a nice group of panels with all the plots I need. How
do I add my own summary text
- matrix(runif(2200),ncol=22)
m - as.mcmc(x)
p = xyplot(m, layout = c(2, 11))
pdf(,height=15)
arrange(p, tableGrob(as.matrix(summary(iris)), theme=theme.white()),
heights= unit(c(3,1),null))
dev.off()
HTH,
baptiste
On 1 June 2010 20:52, Noah Silverman n...@smartmediacorp.com wrote
Hi,
I'm working on a project using the kernlab library.
For one phase, I want the decision values from the SVM prediction, not
the class label. the e1071 library has this function, but I can't find
the equivalent in ksvm.
In general, when an SVM is used for classification, the label of an
I'm reading Gellman's book Data Analysis Using Regression and
Multilevel-Hierarchical Models
In Chapter 7 (and later), he makes frequent referent to a function names
sim.
I can't find the function anywhere, not in my standard R install, or in
any of the packages.
Doe anyone have a
Hi,
I have a dataset where the results are coded (yes, no) We want to
do some machine learning with SVM to predict the yes outcome
My problem is that if I just use the as.factor function to convert, then
it reverses the levels.
--
x - c(no, no, no, yes, yes, no, no)
Hi,
I have a dataset where the results are coded (yes, no) We want to
do some machine learning with SVM to predict the yes outcome
My problem is that if I just use the as.factor function to convert, then
it reverses the levels.
--
x - c(no, no, no, yes, yes, no, no)
Hi,
I am trying to figure out a short way to access two values output from
the sort function.
x - c(3,4,3,6,78,3,1,2)
sort(x, index.return=T)
$x
[1] 1 2 3 3 3 4 6 78
$ix
[1] 7 8 1 3 6 2 4 5
It would be great to do something like this (doesn't work.):
c(y, indexes) - sort(x,
Steve,
Couldn't he also just use the decision.value property to see the
equivilent of t(x) %*% b for each row?
-N
On 7/9/10 7:11 PM, Steve Lianoglou wrote:
Hi,
On Fri, Jul 9, 2010 at 12:15 PM, manuel.martin
manuel.mar...@orleans.inra.fr wrote:
Dear all,
after having calibrated a svm
Thanks!
-N
On 7/9/10 2:20 AM, Noah Silverman wrote:
Hi,
I am trying to figure out a short way to access two values output from
the sort function.
x - c(3,4,3,6,78,3,1,2)
sort(x, index.return=T)
$x
[1] 1 2 3 3 3 4 6 78
$ix
[1] 7 8 1 3 6 2 4 5
It would be great
Hello,
I'm brand new to using R. (I've been using Rapid Miner, but would like
to move over to R since it gives me much more functionality.)
I'm trying to learn how to do a conditional logit model.
My data has one dependent variable, 2 independent variables and a
group variable.
example:
Hello,
I'm using the e1071 library for SVM functions.
I can quickly train an SVM with:
svm(formula = label ~ ., data = testdata)
That works well.
I want to tune the parameters, so I tried:
tune.svm(label ~ ., data=testdata[1:2000, ], gamma=10^(-6:3), cost=10^(1:2))
THIS FAILS WITH AN ERROR:
Hello,
I'm coming from RapidMiner, so some of the easy things there are a bit
difficult for me to find in R
How do I normalize data in a data frame. Ideally I want to scale the
values for each column in the range of (-1,1)
Thank You,
__
Hi,
I am testing out some things with the kernlab library.
The dataframe is 22,000 rows of 32 columns.
The command I execute is:
model - ksvm(label ~ ., data = traindata, type=C-svc, kernel =
rbfdot, class.weights= c(0 =1, 1 =3), kpar = automatic, C = 10,
cross = 3, prob.model = TRUE)
I
Hi,
Quick question.
I'm working on training an SVM.
I have a dataframe with about 50 columns. I want to train on 46 of them.
Is there a way to say All except columns 22,23,25 and 31?
It would be nice to not have to do +c1 +c2 +c3 +c4, etc for all 48 columns.
Thanks!
-N
Hi,
I'm not sure that would work for the formula format of an SVM function.
the idea is normally
svm(label ~ c1 + c2 +c3, data=mydata);
It doesn't work to say
svm(label ~ -c(22,23,24), data=mydata)
On 7/27/09 12:17 PM, Steve Lianoglou wrote:
Hi,
On Jul 27, 2009, at 3:01 PM, Noah
Hi,
I'm switch over from RapidMiner to R. (The learning curve is steep, but
there is so much more I can do with R and it runs much faster overall.)
In RapidMiner, I can tune a parameter of my svm in a nice cross
validation loop. The process will print out the progress as it goes.
So for a
Hi,
This should be an easy one, but I have some trouble formatting the data
right
I'm trying to replace the column of a subset of a dataframe with the
scaled data for that column of the subset
subset(rawdata, code== foo, select = a) - scale( subset(rawdata,
code== foo, select = a) )
It
That works perfectly.
Thanks!
-N
On 7/31/09 2:04 PM, Steve Lianoglou wrote:
Hi,
On Jul 31, 2009, at 4:13 PM, Noah Silverman wrote:
Hi,
This should be an easy one, but I have some trouble formatting the data
right
I'm trying to replace the column of a subset of a dataframe
Hello,
I'm trying to duplicate what's an easy process in RapidMiner.
In RM, we can simply use two operators:
subgroup iteration
attribute value selection (Can use a regex for the attrribute name.)
I can do this in R with a lot of code and manual steps. It would be
really nice to
Hi,
I am reading in a dataframe from a CSV file. It has 70 columns. I do
not have any kind of unique row id.
rawdata - read.table(r_work/train_data.csv, header=T, sep=,,
na.strings=0)
When training an svm, I keep getting an error
So, as an experiment, I wrote the data back out to a new
Jim,
The write.table was simply a diagnostic step.
My problem is that R is automatically adding row_names and then shifting
my column labels over. (The shifting creates a bunch of related problems.)
Thanks for the help.
-Noah
On 8/2/09 2:22 PM, jim holtman wrote:
try 'row.names=FALSE' in
The column names have to obfuscated, but here are 10 rows of the data.
label c0 c1 c2 c3 c4 c5 c6 c7 c8
c9 c10 c11 c12 c13
c14 c15 c16 c17 c18 c19 c20 c21 c22 c23
c24 c25 c26 c27
Somehow, my data is still getting mangled.
Running the SVM gives me the following error:
names attribute[1994] must me the same length as the vector[1950]
Any thoughts?
-N
On 8/2/09 2:35 PM, (Ted Harding) wrote:
On 02-Aug-09 21:10:12, Noah Silverman wrote:
Hi,
I am reading
command.
But, issuing the same 0 substitution AFTER the scale command makes
everything work again.
rawdata[is.na(rawdata)] - 0
VERY strange behavior.
-N
On 8/2/09 3:57 PM, J Dougherty wrote:
On Sunday 02 August 2009 02:34:43 pm Noah Silverman wrote:
The column names have to obfuscated
. Additionally, R seems MUCH MUCH faster.)
I'm open to ideas.
Thanks!
-N
On 8/2/09 4:14 PM, David Winsemius wrote:
On Aug 2, 2009, at 7:02 PM, Noah Silverman wrote:
Hi,
It seems as if the problem was caused by an odd quirk of the scale
function.
Some of my data have NA entries.
So
Just tried your suggestion.
rawdata[is.na(rawdata), ] - 0
It FAILS with the following error:
Error in `[-.data.frame`(`*tmp*`, is.na(rawdata), , value = 0) :
non-existent rows not allowed
__
R-help@r-project.org mailing list
, at 7:02 PM, Noah Silverman wrote:
Hi,
It seems as if the problem was caused by an odd quirk of the scale
function.
Some of my data have NA entries.
So, I substitute 0 for any NA with:
rawdata[is.na(rawdata)] - 0
Perhaps this would have done what you intended:
rawdata[is.na(rawdata), ] - 0
Hi,
More questions in my ongoing quest to convert from RapidMiner to R.
One thing has become VERY CLEAR: None of the issues I'm asking about
here are addressed in RapidMiner. How it handles misisng values,
scaling, etc. is hidden within the black box. Using R is forcing me
to take a much
Hi,
Thanks for the continued support.
I've been working on this all night, and have learned some things:
1) Since I'm really committed to using an SVM, I need to skip the
examples with missing data. I have a training set of approximately
22,000 examples of which about 500 have missing
Hello,
I'm using the e1071 package for training an SVM. It seems to be working
well.
This question has two parts:
1) Once I've trained an SVM model, I want to USE it within R at a later
date to predict various new data. I see the write.svm command, but
don't know how to LOAD the model back
with data frame of:
label, v1, v2, v3
After svm prediction, ending up with data frame of:
label, v1, v2, v3, prediction, probability
Thanks again!
-N
On 8/3/09 8:15 PM, Steve Lianoglou wrote:
Hi,
On Aug 3, 2009, at 10:55 PM, Noah Silverman wrote:
Hello,
I'm using the e1071 package for training
Hello,
I've come across a strange error...
Here is what happens:
model - svm(traindata,trainlabels, type=C-classification,
kernel=radial, cost=10, class.weights=c(win=3,lose=1),
scale=FALSE, probability = TRUE)
predictions - predict(model, traindata)
pred - prediction(predictions,
Good point. I'm not sure how I missed that.
This does lead to an additional question:
Is the probability of the true label the best prediction to feed to
the ROCR package, or is it better to use the decision.value
Anybody have any experience on this one?
Thanks!
-N
On 8/4/09 3:28 AM,
I hadn't thought of that. I'll run some tests...
-N
On 8/4/09 11:49 AM, Tobias Sing wrote:
Is the probability of the true label the best prediction to feed to
the ROCR package, or is it better to use the decision.value
Since AFAIK they are related by a monotonous transformation, both
Hi,
Trying to setup a logistic regression model. (Something new to me. I
usually use SVM.)
The person explaining the concept explained to me that I can include a
group variable so that the probabilities predicted by the model will
be per group
Does this make sense to anyone? If so, how
Thanks David,
But HOW do I indicate the grouping variable in the formula?
Thanks!
-N
On 8/4/09 3:37 PM, David Winsemius wrote:
On Aug 4, 2009, at 6:33 PM, Noah Silverman wrote:
Hi,
Trying to setup a logistic regression model. (Something new to me. I
usually use SVM.)
The person
be:
lrm( label ~ v1 + v2, group_by(group)
-N
On 8/4/09 3:41 PM, David Winsemius wrote:
On Aug 4, 2009, at 6:38 PM, Noah Silverman wrote:
Thanks David,
But HOW do I indicate the grouping variable in the formula?
Hard to tell. You have told us absolutely nothing about the problem
Hmmm.. I'll try that.
I recall reading somewhere that the group variable had to be indicated
in a special way.
-N
On 8/4/09 3:49 PM, David Winsemius wrote:
On Aug 4, 2009, at 6:45 PM, Noah Silverman wrote:
I guess I didn't explain it well enough.
I have a number of training examples
Thanks David,
My apologies for the HTML e-mail. Its the default of my desktop client.
-N
On 8/4/09 4:03 PM, David Winsemius wrote:
On Aug 4, 2009, at 6:52 PM, Noah Silverman wrote:
Hmmm.. I'll try that.
I recall reading somewhere that the group variable had to be
indicated in a special
LOL,
I'll happily support that. Can we possibly take my name off of it? :)
-N
On 8/4/09 4:31 PM, Rolf Turner wrote:
On 5/08/2009, at 11:10 AM, Greg Snow wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Noah
Hi,
Time for another of my newbie questions.
Is it possible to build up a data.frame row by row as I go
I'm going to be running a bunch of experiments (many in a loop) to test
different things. I'm using AUC as my main performance measure.
My thought was to add a row to a data.frame for
Hi,
Nice, but I need a few columns for the data. Don't know how to do this
with the method you suggest.
-N
On 8/4/09 4:57 PM, Remko Duursma wrote:
Hi Noah,
there are a few ways to do this. Easiest is to keep adding an element
to a list, and then make it into a dataframe at the end, like
I've completed an experiment and want to summarize the results.
There are two things I like to create.
1) A simple count of things from the data.frame with predictions
1a) Number of predictions with probability greater than x
1b) Number of predictions with probability greater than x
Hello,
I asked this as part of a previous message, but never really figured out
a usable solution. So this is a second attempt.
I have an process containing an SVM. The end result is the probability
that the class is true. That result is added back to the original data.
So I wind up
/how.
-N
On 8/5/09 11:22 AM, Steve Lianoglou wrote:
Hi,
On Aug 5, 2009, at 2:11 PM, Noah Silverman wrote:
Hello,
I asked this as part of a previous message, but never really figured
out a usable solution. So this is a second attempt.
I have an process containing an SVM. The end result
In my continuing quest to generate some summary data, I've come across
some useful suggestions in pasts posts.
The apply operation returns an error, and I can't figure out why.
Can someone help me fix this?
testlogdata - cbind(testlogdata, range_group=cut(testlogdata$lrm_score,
breaks=c(.9,
Hello,
I have a bit of a tricky puzzle with trying to implement a logit model
as described in a paper.
The particular paper is on horseracing and they explain a model that is
a logit trained per race, yet somehow the coefficients are combined
across all the training races to come up with a
Posted about this earlier. Didn't receive any response
But, some further research leads me to believe that MAYBE a GLMM or a
GEE function will do what I need.
Hello,
I have a bit of a tricky puzzle with trying to implement a logit model
as described in a paper.
The particular paper is on
rkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Urbana, IL 61801
On Aug 6, 2009, at 5:00 PM, Noah Silverman wrote:
Posted about this earlier. Didn't receive any response
But, some further
Yes,
I already have solid code to estimate the probabilities and gather the
public estimates.
What I'm stuck on is how to train the race-wise logit and then somehow
combine them to come up with a final set of coefficients.
I could just train a glm on the whole data set, but would be losing
Hello,
I've come up with some challenges with my process that are a bit too
complicated for the mailing list.
Is there anyone out there, preferably a real statistician, who is
willing to consult with me via phone/email for a few hours. I'm happy
to pay you for your time.
Thanks,
-Noah
Hi,
I'm training an SVM (C-classification from e1071 library)
Some of the variables in my data set are nominal. Is there some
easy/automatic way to convert them to numerical representations?
Thanks,
-N
__
R-help@r-project.org mailing list
newvariable=as.numeric(variablename). This converts your
factors into numeric variables, but not always with the desired
result. So
make sure that you check whether newvariable gives you what you want.
Otherwise recoding by hand is indicated.
Best,
Daniel
Noah Silverman-3 wrote:
Hi,
I'm
Hi,
The answers to my previous question about nominal variables has lead me
to a more important question.
What is the best practice way to feed nominal variable to an SVM.
For example:
color = (red, blue, green)
I could translate that into an index so I wind up with
color= (1,2,3)
But my
Lianoglou wrote:
Hi,
On Aug 12, 2009, at 2:53 PM, Noah Silverman wrote:
Hi,
The answers to my previous question about nominal variables has lead
me to a more important question.
What is the best practice way to feed nominal variable to an SVM.
For example:
color = (red, blue, green)
I could
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