Displaying 20 results from an estimated 700 matches similar to: "Calculating the probability for a logistic regression"
2011 Jan 07
1
[RFC/PATCH] ssh: config directive to modify the local environment
This provides a mechanism to attach arbitrary configure options into the
ssh_config file and use them from the LocalCommand and ProxyCommand.
Examples:
# set FOO to foo
LocalEnvMod FOO = foo
# append bar to FOO with default separator ","
LocalEnvMod FOO += bar
# unset FOO
LocalEnvMod FOO =
# append foo to BAR with separator ":", if BAR is empty
2011 May 12
1
Saving misclassified records into dataframe within a loop
Greetings R world,
I know some version of the this question has been asked before, but i need
to save the output of a loop into a data frame to eventually be written to a
postgres data base with dbWriteTable. Some background. I have developed
classifications models to help identify problem accounts. The logic is this,
if the model classifies the record as including variable X and it turns out
2005 Sep 26
4
p-level in packages mgcv and gam
Hi,
I am fairly new to GAM and started using package mgcv. I like the
fact that optimal smoothing is automatically used (i.e. df are not
determined a priori but calculated by the gam procedure).
But the mgcv manual warns that p-level for the smooth can be
underestimated when df are estimated by the model. Most of the time
my p-levels are so small that even doubling them would not result
2011 Jan 26
2
Extracting the terms from an rpart object
Hello all,
I wish to extract the terms from an rpart object.
Specifically, I would like to be able to know what is the response variable
(so I could do some manipulation on it).
But in general, such a method for rpart will also need to handle a "." case
(see fit2)
Here are two simple examples:
fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
fit1$call
fit2 <-
2011 Apr 22
3
Parametrized object name in Save statement
Greetings All,
I am looking to write a parametrized Rscript that will accept a variable
name(that also is the name of the flat file), transform the data into a data
frame and preform various modeling on the structure and save the output and
plot of the model. In this example i am using a rpart decision tree. The
only problem i am having is integrating the parameter into the internal
object name
2007 Jun 15
2
model.frame: how does one use it?
Philipp Benner reported a Debian bug report against r-cran-rpart aka rpart.
In short, the issue has to do with how rpart evaluates a formula and
supporting arguments, in particular 'weights'.
A simple contrived example is
-----------------------------------------------------------------------------
library(rpart)
## using data from help(rpart), set up simple example
myformula <-
2002 Apr 29
2
RPart
I am using the rpart package and seem to have trouble with data sets that
have columns with no data. I look at the column data in R and all values are
NA. When this occurs, I get nothing back from the rpart function. Is there a
way to get the rpart package to ignore these columns, without knowing what
columns are empty? I have tried the na.action=na.omit and
na.action=na.exclude, but neither one
2003 May 24
1
...listable functions...
Hi R-helpers.
I have the following problem:
I would like to apply my function gain(df,X,A) to a
list of arguments.
df is a data frame
X,A are the varibales od data frame.
When I do
> gain(kyphosis,"Kyphosis",c("Start","Number"))
[1] "Start" "Number"
I get the following error...
Error in unique.default(x) : unique() applies only to
vectors
I
2006 Dec 27
1
Question about predict function
I am working with a non-parametic smoothing operation using a
Generalized Additive Model. It is a bivariate data set. I know how to
do the smooth, and out comes a nice smooth curve.
Now I want to find the value of the smoothed curve for several values
of x (the abscissa). This can be done (please correct me if I am
wrong) by using the predict.gam function. You feed the predict.gam
function a
2010 Dec 13
2
rpart.object help
Hi,
Suppose i have generated an object using the following :
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
And when i print fit, i get the following :
n= 81
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 81 17 absent (0.7901235 0.2098765)
2) Start>=8.5 62 6 absent (0.9032258 0.0967742)
4) Start>=14.5 29 0 absent (1.0000000
2012 Mar 04
1
rpart package, text function, and round of class counts
I run the following code:
library(rpart)
data(kyphosis)
fit <- rpart(Kyphosis ~ ., data=kyphosis)
plot(fit)
text(fit, use.n=TRUE)
The text labels represent the count of each class at the leaf node.
Unfortunately, the numbers are rounded and in scientific notation rather
than the exact number of examples sorted by that node in each class.
The plot is supposed to look like
2009 Dec 14
1
RPART - printing full splitting rule number on tree plot
Dear R-users
I am using RPART package to get regression trees. However having trouble getting the text function to put the full splitting rule number on the plot, instead to puts it in scientific notation. When a covariate has 1e4 or greater number of digits then the splitting rule number displayed on the plot is in scientific notation. But print.rpart displays the splitting rules in full.
2012 Apr 12
2
enableJIT(2) causes major slow-up in rpart
Hello,
Due to exploration of the JIT capabilities offered through the {compiler}
package, I came by the fact that using enableJIT(2) can *slow* the rpart
function (from the {rpart} package) by a magnitude of about 10 times.
Here is an example code to run:
library(rpart)
require(compiler)
enableJIT(0) # just making sure that JIT is off # We could also use
enableJIT(1) and it would be fine
fo
2011 Mar 31
2
fit.mult.impute() in Hmisc
I tried multiple imputation with aregImpute() and
fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1.
The warning message below suggests that summary(f) of
fit.mult.impute() would only use the last imputed data set.
Thus, the whole imputation process is ignored.
"Not using a Design fitting function; summary(fit)
will use standard errors, t, P from last imputation only.
Use
2009 Feb 03
5
Large file size while persisting rpart model to disk
I am using rpart to build a model for later predictions. To save the
prediction across restarts and share the data across nodes I have been
using "save" to persist the result of rpart to a file and "load" it
later. But the saved size was becoming unusually large (even with
binary, compressed mode). The size was also proportional to the amount
of data that was used to create the
2009 Feb 25
1
how to label the branches of a tree
Hi,
I am using rpart package to fit classification trees.
library(rpart)
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
plot(fit,uniform=T)
text(fit, use.n=TRUE)
But I am unable to label the branches (not the nodes) of the tree. Can somebody help me out in this?
Thank you,
Regards
Utkarsh Singhal | Amba Research
Ph +91 80 3980 8017 | Mob +91 99 0295 8815
2009 Sep 14
1
summary of rpart-Object in tktext window?
Hi,
is it possible to put a summary of an rpart-Object into a tktext-window?
Here is what I'm trying to do:
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
tt <- tktoplevel()
tex <- tktext(tt)
tkpack(tex)
tkinsert(tex, "end", summary(fit))
But since the summary of an object is a list, I always get back the
following error-message:
cannot handle object of
2010 Mar 07
1
Is there an equivalence of lm's “anova” for an rpart object ?
Simple example:
# Classification Tree with rpart
library(rpart)
# grow tree
fit <- rpart(Kyphosis ~ Age + Number + Start,
method="class", data=kyphosis)
Now I would like to know how can I measure the "importance" of each of my
three explanatory variables (Age, Number, Start) in the model?
If this was a regression model, I could have looked at p values from the
2010 May 03
1
rpart, cross-validation errors question
I ran this code (several times) from the Quick-R web page (
http://www.statmethods.net/advstats/cart.html) but my cross-validation
errors increase instead of decrease (same thing happens with an unrelated
data set).
Why does this happen?
Am I doing something wrong?
# Classification Tree with rpart
library(rpart)
# grow tree
fit <- rpart(Kyphosis ~ Age + Number + Start,
2011 Aug 08
1
Classification trees problem.
Hello Everyone,
I'm doing a Classification trees with categorical explanatory variables using library rpart and I would like to do a prediction for some data imputs. I don't know where's a function or how can I do it?. Is there someone can help ?? ¿. Here's the code that I'm using.
library(rpart)
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
plot(fit)