Hi,
I have been using some code for multivariate random forests. The output
from this code is a list object with all the same values as from
randomForest, but the model object is, of course, not of the class
randomForest. So, I was hoping to modify the code for predict.randomForest
to work for
On 10.04.2012 23:03, Katharine Miller wrote:
Hi,
I have been using some code for multivariate random forests. The output
from this code is a list object with all the same values as from
randomForest, but the model object is, of course, not of the class
randomForest. So, I was hoping to
On 10.04.2012 20:24, Josh O'Brien wrote:
I am attempting to build a customized R installer on Windows, using the Inno
Setup installer.
I am following the instructions in Section 3.1.8 of the R Installation and
Administration Manual (Building the Inno Setup installer), which includes
the
hello everyone,
i wanted help to build a* user defined function in R for multiple sequence
alignment*.
i tried using pairwisealignment () for the multiple sequences, but the
results seem to be wrong.
i have 10 sequences which i have to align ( by building profiles).
is there a function to build a
On Apr 10, 2012, at 22:03 , nerak13 wrote:
Hi,
I've got the following data:
x-c(1,3,5,7)
y-c(37.98,11.68,3.65,3.93)
penetrationks28-dataframe(x=x,y=y)
now I need to fit a non linear function so I did:
fit - nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start =
Sorry this is so late:
But you could try a nerge (from the 'caroline' package)
nerge(list(a,b,c))
Just have to make sure that the rows for each dataframe are renamed with
the date columns.
On 1/30/2012 11:44 PM, Massimo Bressan wrote:
thanks don
I have here enough to study for a while
Friends
I am extracting sub-sets of the rows of a matrix. Generally the result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are 0, or more than 1 rows returned the result is a
On Apr 10, 2012, at 7:33 PM, Worik R wrote:
Friends
I am extracting sub-sets of the rows of a matrix. Generally the
result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there
Thank you.
That was exactly what I need.
Looking at '?[' I see...
drop: For matrices and arrays. If TRUE the result is coerced to
the lowest possible dimension (see the examples). This only
works for extracting elements, not for the replacement. See
drop
On Apr 10, 2012, at 8:01 PM, Worik R wrote:
Thank you.
That was exactly what I need.
Looking at '?[' I see...
drop: For matrices and arrays. If TRUE the result is coerced to
the lowest possible dimension (see the examples). This only
works for extracting
Many thanks for your help.
I mistakenly deleted my original message (not even knowing that was
possible). Apologies for that.
For future reference, when a section of an R manual (like the bit from the
'Installing R Under Windows Section' of the R-admin manual) no longer
applies to the current
The Milwaukee Chapter of the ASA (MILWASA) in cooperation
with The Medical College of Wisconsin,
Marquette University,
The Children's Research Institute,
The Clinical and Translational Science Institute (CTSI)
and Quantitative Health Sciences
are proud to announce
R Programming Workshops with
On Apr 10, 2012, at 7:57 PM, Josh O'Brien wrote:
Many thanks for your help.
I mistakenly deleted my original message (not even knowing that was
possible). Apologies for that.
Heh. I suspect it disappeared from your mail-client but not from the
Archive (or for that matter the multiple
Dear all, I get an error using the return function. The following is a
simpler version.
for (j in 1:10)
{
samples = 5*j
return(samples)
}
Error: no function to return from, jumping to top level
Similar warning happens with if statement.
Why do I get an error? print() works fine. I
hi, I'm doing some data on least square curve fitting. What I like to have is
to compare the scatter plot whilst having the fitting curve on the same
coordinates. Any suggestting command besides plot(x,y). TaweeMac OSX 10.7.3
[[alternative
Hello,
I am writing codes for time series computation but encountering some
problems
Given the quarterly data from 1983Q1 to 1984Q2
PI1-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485,
-1.190061246, -0.553031799, 0.686874720, 0.953911035),
start=c(1983,1), frequency=4)
PI1
The error message should make it pretty clear -- you aren't inside a
function so you can't return() a value which is, by definition, what
functions (and only functions) do.** Not sure there's a great
reference for this though beyond the error message
Incidentally, what are you trying to do
This is the same malformatted message you posted on R-SIG-Mac even
after David specifically asked for clarification not to reward bad
behavior, but perhaps this will enlighten:
# Minimal reproducible data!
x - runif(15, 0, 5)
y - 3*x - 2 + runif(15)
dat - data.frame(x = x, y = y)
rm(list =
Two ways around this:
I = Easy) Just use zoo/xts objects. ts objects a real pain in the
proverbial donkey because of things like this.
Something like:
library(xts)
PI1.yq - as.xts(PI1) # Specialty class for quarterly data (or regular
zoo works)
lag(PI1.yq)
II = Hard) lag on a ts actually
Dear Michael (and Davis), Your answer is not what I want to know. My question
is to find any command to plot the data I got from the field; such as a set of
(x,y) data ( I actually have these data) and together withe the derived ones .
I brought these data to plot on x-y plane, getting a
Thanks.
So, I want the function to return results from the if statement.
bob - function(var1, func)
{
#func: a simple function
num1 - var1
num2 - func(var1)
if(ruinf(1)num1)
{
return(num1)
}else{
return(num2)
}
}
And then, run 20 iterations of the function.
You were told before this isn't a Mac question so please don't cc R-SIG-Mac.
I'm not sure what this bit of your reply means My question is to find
any command to plot the data I got from the field; but your reply
later suggests that your problem is that you are overriding previous
plots on a
Hi all,
I wish to merge 24 .csv files, each having a common identifier-column
(Name) and do two things:
1. Retrieve the common one's. [Analogy: while merging 2-dataframes, similar
to using: merge ( ,by=Name, all=FALSE) ]
2. Retrieve all, i.e., the union of the rows of 24 files. [again,
Hello,
I wish to censor 10% of my sample units of 50 from a Weibull distribution.
Below is the code for it.
I will need to know whether what i have done is correct and if not, can i have
any suggestion to improve it?
Thank you
p=2;b=120
n=50
r=45
t-rweibull(r,shape=p,scale=b)
You don't need to return() from the for loop -- just put your outputs
in a variable:
set.seed(1) # For reproducibility
x - numeric(20)
for(i in 1:20) x[i] - bob(0.5, sqrt)
or (more elegant but basically the same thing)
set.seed(1)
x1 - replicate(20, bob(0.5, sqrt)) # Same calculation done 20x
Your problem is that you can't merge the file names, but you need to
load them into R and merge the resulting objects.
This should be straightforward enough to do:
file_list - list.files()
list_of_files - lapply(file_list, read.csv) # Read in each file
merge_all(list_of_files, by = Name)
On Apr 10, 2012, at 11:48 PM, Christopher Kelvin wrote:
Hello,
I wish to censor 10% of my sample units of 50 from a Weibull
distribution. Below is the code for it.
I will need to know whether what i have done is correct and if not,
can i have any suggestion to improve it?
Thank you
On Apr 11, 2012, at 12:17 AM, R. Michael Weylandt wrote:
Your problem is that you can't merge the file names, but you need to
load them into R and merge the resulting objects.
This should be straightforward enough to do:
file_list - list.files()
list_of_files - lapply(file_list, read.csv) #
Thanks to David and Michael
Michael:
That works, however with a glitch. Each of my 24 files has got two columns:
Name, and Rank/score.
file_list - list.files()
list_of_files - lapply(file_list, read.csv) # Read in each file
# I can see the 2-columns at this stage. However, the following line:
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