Thanks so much, David!
I will try.
On Thu, Feb 9, 2012 at 3:04 PM, David Winsemius wrote:
>
> On Feb 9, 2012, at 5:01 PM, Changbin Du wrote:
>
> test<-c("20120111_181515_001_**CCL54D_A01_S02_APL932_PL11_DL_**
>> 20120111.CEL",
>> "20120111_181516_002_
test<-c("20120111_181515_001_CCL54D_A01_S02_APL932_PL11_DL_20120111.CEL",
"20120111_181516_002_CCL54D_A02_S08_APL932_PL11_DL_20120111.CEL")
> test
[1] "20120111_181515_001_CCL54D_A01_S02_APL932_PL11_DL_20120111.CEL"
[2] "20120111_181516_002_CCL54D_A02_S08_APL932_PL11_DL_20120111.CEL"
fields1<-str
C-7SARK 1 849467 10 10 10 10 10
> 10
> 6C-7SCGC 1 874609 10 10 10 10 10
> 10
> array07 array08 array09 array10
> 1 10 10 10 10
> 2 10 10 10 10
> 3 10 10
10 10 10 10
5 10 10 10 10
6 10 10 10 10
Appreciated your great helps!
On Thu, Sep 22, 2011 at 10:55 AM, Changbin Du wrote:
> Thanks so much, Michael!
>
>
> head(first)
> probe_name chr_id position array1
> 1C-7S
s? What
> about (untested):
>
> for (i in 1:2) {
> first <-cbind(first, result.fun[[i]])
> }
>
> you will then have to look at
> names(first)
> and change the last part of it to
> paste("array",2:3,sep="")
>
> Hope that works!
> JC
&
)
>
> first.out <- do.call("cbind", list(first, result.fun))
> print(first.out)
>
> which provides this output.
>
> x y z a b
> 1 1 6 11 1 4
> 2 2 7 12 2 5
> 3 3 8 13 3 6
>
> More generally, you really should read about how arguments and ass
ments for a function are.
>
> More generally, this sort of thing may be best handled in a list rather
> than an set of independent variables.
>
> Michael Weylandt
>
> On Thu, Sep 22, 2011 at 1:07 PM, Changbin Du wrote:
>
>> HI, Dear R community,
>>
>>
HI, Dear R community,
I am trying to created new variables and put into a data frame through a
loop.
My original data set:
head(first)
probe_name chr_id position array1
1C-7SARK 1 849467 10
2C-4WYLN 1 854278 10
3C-3BFNY 1 854471 10
4C-7ONNE
e to be
> *very* large before it mattered.
>
> -Don
>
>
> --
> Don MacQueen
>
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
>
>
>
>
>
> On 9/1/11 11:08 AM, "Changbin Du" wrote:
>
Thanks for the great helps from David, Jim and Liviu. It solved my problem.
Appreciated!
On Thu, Sep 1, 2011 at 11:01 AM, David Winsemius wrote:
>
> On Sep 1, 2011, at 1:53 PM, Changbin Du wrote:
>
> HI, Dear R community,
>>
>> I want to split a data frame by using
HI, Dear R community,
I want to split a data frame by using two variables: let and g
> x = data.frame(num =
c(10,11,12,43,23,14,52,52,12,23,21,23,32,31,24,45,56,56,76,45), let =
letters[1:5], g = 1:2)
> x
num let g
1 10 a 1
2 11 b 2
3 12 c 1
4 43 d 2
5 23 e 1
6 14 a 2
7
PS -- Is your data guaranteed to be an integer? If you have floating point
> data, it's good practice to use something more like
>
> abs(x - 10) > 1e-8
>
> rather than x != 0 in your code. If you need to use this formulation, just
> put it inside the any() statement.
>
ch rows to
> keep
>
> Answer = M[rowsToKeep,] # keep only those rows
>
> Hope this helps,
>
> Michael
>
>
> On Mon, Aug 22, 2011 at 12:56 PM, Changbin Du wrote:
>
>> HI, Michael,
>>
>> What I want to do is remove all the rows, for which array1, arr
n(x){any (x != 10)})
> # apply the test function row-wise to get a logical vector of which rows to
> keep
>
> Answer = M[rowsToKeep,] # keep only those rows
>
> Hope this helps,
>
> Michael
>
>
> On Mon, Aug 22, 2011 at 12:56 PM, Changbin Du wrote:
>
>> H
op
> based on that test?
>
> Michael
>
> On Mon, Aug 22, 2011 at 12:35 PM, Changbin Du wrote:
>
>> Dear R community,
>>
>> I have a data set like the following:
>>
>> probe_name chr_id position array1 array2 array3 array4 array5 array6
>> array7
Dear R community,
I have a data set like the following:
probe_name chr_id position array1 array2 array3 array4 array5 array6
array7
1C-3 10 16566949 10 10 10 10 10 10
10
2C-3AAAB 17 33478940 10 10 10 10 10 10
10
3C-3AAAC
Nabble.com.
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self
"W.0.N" "D.1.H"
> > sub("\\.([0-9]+)\\.", "{\\1}", test)
> [1] "H{1}Y" "N{0}E" "G{1}N" "E{0}P" "W{2}G" "W{4}G" "W{3}W" "W{0}N" "D{1}H"
> >
>
>
> On
HI, Dear R-community,
I have one vector of the following string,
> test<-c("H.1.Y", "N.0.E", "G.1.N", "E.0.P", "W.2.G", "W.4.G", "W.3.W",
"W.0.N", "D.1.H")
> test
[1] "H.1.Y" "N.0.E" "G.1.N" "E.0.P" "W.2.G" "W.4.G" "W.3.W" "W.0.N" "D.1.H"
*I want to change them into*
"H{1}Y" "N{0}E" "G{1}N" "
#x27;sqldf' had non-zero exit status
When I try to remove the */house/homedirs/c/cdu/library/00LOCK'* by rm -r
00LOCK,
I got the following errors:
cdu@nuuk:~/library$ rm -r 00LOCK
*rm: cannot remove `00LOCK/RSQLite/libs/.nfs0001301e52e00004':
Device or resource busy*
I will try this first.
Thanks, Gabor!
On Wed, Jun 22, 2011 at 3:15 PM, Gabor Grothendieck wrote:
> On Wed, Jun 22, 2011 at 6:10 PM, Changbin Du wrote:
> > My R is 2.12.0.
> >
> >> R.version.string
> > [1] "R version 2.12.0 (2010-10-15)"
>
On Wed, Jun 22, 2011 at 3:04 PM, Gabor Grothendieck wrote:
> On Wed, Jun 22, 2011 at 6:01 PM, Changbin Du wrote:
> > I found the following errors:
> >
> >
> >> library(sqldf)
> > Loading required package: DBI
> > Loading required package: RSQLite
> &
quot;/house/homedirs/c/cdu/operon/gh5/hypo_re.dimer",
header=FALSE, sep="\t",sql="select varr from file", quote="", fill=T)
*Error: could not find function "read.csv.sql"*
On Wed, Jun 22, 2011 at 2:57 PM, Changbin Du wrote:
> Hi, Gabor,
>
&g
Hi, Gabor,
Thanks so much, I will try it and let you know the results.
Appreciated!
On Wed, Jun 22, 2011 at 2:54 PM, Gabor Grothendieck wrote:
> On Wed, Jun 22, 2011 at 5:45 PM, Changbin Du wrote:
> > HI, Dear R community,
> >
> > I have a large data set names dd
HI, Dear R community,
I have a large data set names dd.txt, the columns are: there are 2402
variables.
a1, b1, ..z1, a11, b11, ...z11, a111, b111, ..z111..
IF I dont know the relative position of the columns, but I know I need the
following variables:
var<-c(a1, c1,a11,b11,f111)
Can I use read.
list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>
--
Sincerel
atrix to create dummy variables:
>
> mydummies = model.matrix(~myfactor)[, -1]
>
> You'll get as many dummy variables as values you have in ack$ID -
> minus 1 (for the reference level).
> Dimitri
>
> On Thu, Feb 24, 2011 at 2:30 PM, Changbin Du wrote:
> > Thanks to al
Thanks, David!
On Thu, Feb 24, 2011 at 11:30 AM, David Winsemius wrote:
>
> On Feb 24, 2011, at 1:23 PM, Changbin Du wrote:
>
> HI, Dear R community,
>>
>> I try to create 100 dummy variables like the following:
>>
>> ack$id_1 <- (ack$ID==1)*1
>>
> "Is the room still a room when its empty? Does the room,
> the thing itself have purpose? Or do we, what's the word... imbue it."
> - Jubal Early, Firefly
>
> r-help-boun...@r-project.org wrote on 02/24/2011 01:23:54 PM:
>
> > [image removed]
> &
HI, Dear R community,
I try to create 100 dummy variables like the following:
ack$id_1 <- (ack$ID==1)*1
ack$id_2 <- (ack$ID==2)*1
..
.
ack$id_100 <- (ack$ID==100)*1
I used the following codes:
for(i in 1:100){
ack$id_[i] <- (ack$ID==i)*1
}
But only one colum
_
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
ttp://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alternative HTML v
Thanks Jim1 I will split the data and rum again.
On Thu, Dec 16, 2010 at 2:57 PM, Jim Holtman wrote:
> I think that your object exceeds the limit of 2^31 elements.
>
> Sent from my iPad
>
> On Dec 16, 2010, at 17:44, Changbin Du wrote:
>
> > Dear R community,
> &g
er.nn))
*
Error in unlist(X, recursive = FALSE, use.names = FALSE) :
negative length vectors are not allowed*
Thanks so much!
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alter
HI, Dear R community,
I have used the following codes this morning, but this afternoon, I got the
following errors:
> x <- seq(0,10, by=1)
> y <- c(0.952, 0.947, 0.943, 0.941, 0.933, 0.932, 0.939, 0.932, 0.924,
0.918, 0.920) # missense
> z <- c(0.068, 0.082, 0.080, 0.099, 0.108, 0.107, 0.101, 0.1
0.02
> > identical(v_3,v)
> [1] TRUE
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
> > -Original Message-
> > From: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] On Behalf Of Changbin Du
> > Sent: Frida
Thanks Martin! I will try it and will let your guys know how it goes.
On Fri, Nov 5, 2010 at 9:42 AM, Martin Morgan wrote:
> On 11/05/2010 09:13 AM, Changbin Du wrote:
> > HI, Phil,
> >
> > I used the following codes and run it overnight for 15 hours, this
> morning,
&g
umeric(l)
+ for(i in 1:l)output[i] = sum(data >= data[i])
+ 100 * output / l
+ }
> result3<-cover_per_2(cover)
On Thu, Nov 4, 2010 at 10:37 AM, Changbin Du wrote:
> Thanks Phil, that is great! I WILL try this and let you know how it goes.
>
>
>
>
> On Thu, N
pector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
>
>
>
>
Thanks Martin, I will try this.
On Thu, Nov 4, 2010 at 10:06 AM, Martin Morgan wrote:
> On 11/04/2010 09:45 AM, Changbin Du wrote:
> > Thanks, Jim!
> >
> > This is not what I want, What I want is calculate the percentage of
> reads
> > bigger or equal to that re
!
On Thu, Nov 4, 2010 at 9:12 AM, Henrique Dallazuanna wrote:
> Try this:
>
> rev(100 * cumsum(matt$reads > 1) / length(matt$reads) )
>
> On Thu, Nov 4, 2010 at 1:46 PM, Changbin Du wrote:
>
>> HI, Dear R community,
>>
>> I have one data set like t
647
> 5 Contig79:5 17 50.0
> 6 Contig79:620 58.82353
> 7 Contig79:725 73.52941
> 8 Contig79:827 79.41176
> 9 Contig79:932 94.11765
> 10 Contig79:1033 97.05882
> 11 Contig79:1134 100.0
>
>
> On Thu, Nov 4, 2010 at 1
HI, Dear R community,
I have one data set like this, What I want to do is to calculate the
cumulative coverage. The following codes works for small data set (#rows =
100), but when feed the whole data set, it still running after 24 hours.
Can someone give some suggestions for long vector?
id
gt; On 11/1/2010 2:24 AM, Changbin Du wrote:
> > Thanks Joshua! Yes, i is not going up sequentially by 1, as i here is
> the
> > raw number of reads for each DNA base. Thanks so much for the great help!
> >
> >
> > On Sun, Oct 31, 2010 at 6:03 PM, Joshua Wiley >w
gt; for (i in min(cover):max(cover)) {
> output[j] <- 100*sum(ifelse(cover >= i, 1, 0))/length(cover)
>j <- j + 1
> }
> return(output)
> }
>
> Josh
>
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
min(data):max(data)) {
x<-(100*sum(ifelse(data >= i, 1, 0))/length(data))
output<-c(output, x)
}
return(output)
}
result<-cover_per(test)
On Sun, Oct 31, 2010 at 5:46 PM, David Winsemius wrote:
>
> On Oct 31,
HI, Dear R community,
I have the following codes to calculate the commulative coverage. I want to
save the output in a vector, How to do this?
test<-seq(10, 342, by=2)
#cover is a vector
cover_per<-function (cover) {
for (i in min(cover):max(cover)) {print(100*sum(ifelse(cover >= i, 1,
0))/lengt
Sorry, I was a little numb at that time.
On Thu, Oct 28, 2010 at 10:45 AM, David Winsemius wrote:
> In a sense you deserve what you have asked for. You have asked thousands of
> people to send you a copy when you could have instead searched the archives
> yourself and gotten a much quicker ans
Hi, Dear Community,
Several days ago, I received one email about the online svm course in R, I
try to find it. Can someone forward the information to me.
Thanks!
--
Sincerely,
Changbin
--
[[alternative HTML version deleted]]
__
R-help@r-pro
red", yaxt="n", lty=3, xlab="",
> ylab="", ylim = c(min(c(y, z)), max(c(y, z
>
> # add x vs. fp
> lines(x, z, type="b", pch=22, col="blue", lty=2)
>
>
> Cheers,
>
> Josh
>
> On Mon, Oct 25, 2010 at 2:38 PM,
HI, Dear R community,
I am using the following codes to plot, however, the lines code works. But
the line was not drawn on the previous plot and did not shown up.
How comes?
# specify the data for missense simulation
x <- seq(0,10, by=1)
y <- c(0.952, 0.947, 0.943, 0.941, 0.933, 0.932, 0.939, 0
I think you should use 10 fold cross validation to judge your performance on
the validation parts. What you did will be overfitted for sure, you test on
the same training set used for your model buliding.
On Sat, Oct 23, 2010 at 6:39 AM, mxkuhn wrote:
> I think the issue is that you really can'
Thanks, Claudia!
On Tue, Oct 12, 2010 at 9:54 AM, Claudia Beleites wrote:
> I'm not sure how much fun it is to fit > 7000 weights with 1800 samples,
> but you can tell nnet to allow more weights with MaxNWts, see ?nnet
>
>
>
> On 10/12/2010 06:45 PM, Changbin Du wrote:
HI, Dear R community,
My data set has 2409 variables, the last one is response variable. I have
used the nnet after feature selection and works. But this time, I am using
nnet to fit a model without feature selection. I got the following error
information:
> dim(train)
[1] 1827 2409
nnet.fit<
quot;
>
> Hope this helps.
>- Phil Spector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
>
Dear community,
I have one file named ca_boost_feature.txt,
Feature selection (Boosting:0.0025,5)!
H.2.C C.1.D C.3.R E.0.N C.2.S C.0.G H.3.G
log file: ep
If I want to use the second line of this file, how to read it into R?
varr<-read.table("/home/cdu/operon/carbonic/ca_boost_feature.txt", sep
Thanks so much, David!
The following codes works!
result.fun <- lapply(1:2, function(i) cv.fold(i, 3, 0.3))
On Tue, Sep 7, 2010 at 3:35 PM, David Winsemius wrote:
>
> On Sep 7, 2010, at 5:43 PM, Changbin Du wrote:
>
> cv.fold<-function(i, size=3, rang=0.3){
>>
cv.fold<-function(i, size=3, rang=0.3){
cat('Fold ', i, '\n')
out.fold.c <-((i-1)*c.each.part +1):(i*c.each.part)
out.fold.n <-((i-1)*n.each.part +1):(i*n.each.part)
train.cv <- n.cc[-out.fold.c, c(2:2401, 2417)]
train.nv <- n.nn[-out.fold.n, c(2:2401, 2417)]
Thanks all so much for your help! I went out for 2 days vacation and could
not reply your guys email. Yes, the CV=False works.
Thanks again!
On Sun, Jul 4, 2010 at 2:47 AM, Peter Ehlers wrote:
> On 2010-07-03 21:33, Changbin Du wrote:
>
>> HI, Dear community,
>>
>&g
HI, Dear community,
I am using the linear discriminant analysis to build model and make new
predictions:
> dim(train) #training data
[1] 1272 22
> dim(valid) # validation data
[1] 140 22
lda.fit <- lda(out ~ ., data=train, na.action="na.omit", CV=TRUE) # model
fitting of linear discriminan
Hi, Dear Community,
My data set logit.pred contains 2 columns and 1400 rows. When I want to use
the first column, it is very strange. Where the $ come out? Thanks so much!
> dim(logit.pred)
[1] 14002
> head(logit.pred)
tree.pred valid.out
754 0.6550606 1
1080 0.6353524
HI, Dear R community,
I am using the xtable to create the table, but how can I see the table?
The following is the codes I used:
> data(tli)
> tli.table <- xtable(tli[1:10, ])
> digits(tli.table)[c(2, 6)] <- 0
> print(tli.table, floating = FALSE)
% latex table generated in R 2.11.0 by xtable
posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alternative HTML version deleted]]
_
HI, Dear community,
How to distinguish bi-mode distribution from mono-mode distribution? I have
only the histograms of 3500 data set.
Thanks!
--
Sincerely,
Changbin
--
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
HI, GUYS,
I used the following codes to run SVM and get prediction on new data set hh.
dim(all_h)
[1] 2034 24
dim(hh)# it contains all the variables besides the variables in all_h
data set.
[1] 640 415
require(e1071)
svm.tune<-tune(svm, as.factor(out) ~ ., data=all_h,
ranges=list(gamma
HI, Guys,
I am trying to use the AdaBoosting. M.1 algorithm to integrate three models.
I found the sum of weights for each model is not equal to one.
How to deal with this?
Thanks, any response or suggestions are appreciated!
--
Sincerely,
Changbin
--
[[alternative HTML version del
> nnet.fit<-nnet(as.factor(out) ~ ., data=all_h, size=5, rang=0.3,
decay=5e-4, maxit=500) # model fitting
> summary(nnet.fit)
a 23-5-1 network with 126 weights
options were - entropy fitting decay=5e-04
HI, Guys,
I can not find the manual to describe how the model is built, is there a
more
HI, Dear R community,
I am using the nnet to fit a neural network model to do classification on
binary target variable (0, 1). I am using the following codes:
nnet.fit<-nnet(as.factor(out) ~ ., data=train, size=5, rang=0.3,
decay=5e-4, maxit=500)
I want to know what is the activation function f
HI, Dear R community,
I am using the nnet to fit a neural network model to do classification on
binary target variable (0, 1). I am using the following codes:
nnet.fit<-nnet(as.factor(out) ~ ., data=train, size=5, rang=0.3,
decay=5e-4, maxit=500)
I want to know what is the activation function f
at least once? IF SOME obs are not selected, how to
calculate the training error?
Thanks?
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alternative HTML version deleted
Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
> On Wed, 2 Jun 2010, Changbin Du wrote:
>
> dim(tmp)
>>>
>> [1] 576 12
>>
>
> dim(tmp)
[1] 576 12
> tmp
1 10 11 12 13 2 3 4 5 6 7 9
10 0 0 0 0 0 0 12 0 0 0 0
20 0 0 0 0 0 0 15 0 0 0 0
30 0 0 0 0 0 0 11 0 0 0 0
40 0 0 0 0 0 0 11 0 0 0 0
50 0 0 0 0 0 0 12 0 0 0 0
60 0 0 0
as.factor(temp.predict))],
> main="Mapping plot",labels=label)
>
> It does not calculate the majority vote itself, it just assigns a label to
> the category based on the predicted labels. Which is equivalent in this
> case.
>
> Cheers
> Joris
>
>
>
ar Changbin,
>
> Please provide a self-contained, minimal example, meaning the whole code
> should run and create the plot as it is now, without having to load your
> dataset (which we don't have). Otherwise it's impossible to see what's going
> on and help you.
>
Thanks to all!
Yes, LETTERS[outcome.predict] works!
I appreciated your guys help!
On Tue, Jun 1, 2010 at 1:53 PM, Henrique Dallazuanna wrote:
> use "[" not "("
>
>
> On Tue, Jun 1, 2010 at 5:50 PM, Changbin Du wrote:
>
>> It does no
label<-intToChar(as.integer(outcome.predict))
?outcome.label<-LETTERS(outcome.predict)
plot(final.xyf, type="property", property=outcome.predict,
labels=outcome.label, palette.name =rainbow, main="Supervised:Prediction ")
Thanks so much!
On Tue, Jun 1, 2010 at 1:50 PM, Eri
It does not work.
> outcome.label<-LETTERS(outcome.predict)
Error: could not find function "LETTERS"
On Tue, Jun 1, 2010 at 1:48 PM, Henrique Dallazuanna wrote:
> Try this:
>
> LETTERS[outcome.predict]
>
> On Tue, Jun 1, 2010 at 5:43 PM, Changbin Du wr
Hi, Dear R- community,
I am use the intToChar function to convert the integers to letters. But the
output is mess. Can you guys give some suggestions? Thanks!
> outcome.predict
[1] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 7 4 4 4 4 4 4
4 4
[26] 4 4 4 4 4 4 4 4 4 4 4
ent to the global environment rather than the
> function's environment. In general this seems risky though as your
> function could be overwriting data in your main workspace without you
> knowing it.
>
> HTH,
>
> Josh
>
>
>
> On Wed, May 26, 2010 at 9:26 AM, Changb
valid.out 987
Fold 8
Dim of tree.pred 1128 2 length of valid.out 1128
Fold 9
Dim of tree.pred 1269 2 length of valid.out 1269
Fold 10
Dim of tree.pred 1410 2 length of valid.out 1410
Minsplit 5 Minbucket 5
10-cross validation is done!
if use return, it will print on the screen, you still can no
t.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnu
> your search (the first place I'll look for is the line where R stop reading.
> See if any thing strange there.)
>
> Also, changing "read.table" to "read.delim" often works.
>
> ...Tao
>
>
>
>
>
> - Original Message
> > From
ggest). I encounter this all the time. So try to be very thorough about
> your search (the first place I'll look for is the line where R stop reading.
> See if any thing strange there.)
>
> Also, changing "read.table" to "read.delim" often works.
>
&
o
>
> tail(gene)_name, 2)
>
> Come on, man, show some initiative.
>
> On May 25, 2010, at 12:12 PM, Changbin Du wrote:
>
> 644727344ABC-2 type transporterABC-2 type transporter
> 644727345conserved hypothetical proteinconserved hypothetical
> protein
&g
c...@nuuk:~/operon$ grep '^#' id_name_gh5.txt
c...@nuuk:~/operon$
no lines starts with #
On Tue, May 25, 2010 at 9:11 AM, Barry Rowlingson <
b.rowling...@lancaster.ac.uk> wrote:
> On Tue, May 25, 2010 at 4:42 PM, Changbin Du wrote:
> > HI, Dear R community,
> >
:42 AM, Changbin Du wrote:
>
> HI, Dear R community,
>>
>> My original file has 1932 lines, but when I read into R, it changed to
>> 1068
>> lines, how comes?
>>
>
> We are being asked to investigate this quest, how?
>
> Have you looked at the last line to see
ount.fields("id_name_gh5.txt"))
> Regards Mohamed
>
>
>
>
> Changbin Du a écrit :
>
> HI, Dear R community,
>>
>> My original file has 1932 lines, but when I read into R, it changed to
>> 1068
>> lines, how comes?
>>
>>
>>
F, fill=T)
> dim(gene_name)
[1] 10683
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alternative HTML version deleted]]
__
R-help@r-p
HI, Dear R community,
I want to know how to select the optimal decision threshold from the ROC
curve? At what threshold will give the highest accuracy?
Thanks!
--
Sincerely,
Changbin
--
[[alternative HTML version deleted]]
__
R-help@r-projec
- Phil Spector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
> On Wed, 19 May
plot(svm.auc, col=2, main="ROC curves comparing classification performance\n
of six machine learning models")
legend(0.5, 0.6, c(ns, nb, nr, nt, nl,ne), 2:6, 9) # Draw a legend.
plot(bo.auc, col=3, add=T) # add=TRUE draws on the existing chart
plot(rf.auc, col=4, add=T)
plot(tree.auc, col=5, add=T
Thanks, David!
Yes, I found it just as you said. It works now after change to numeric.
On Tue, May 18, 2010 at 1:53 PM, David Winsemius wrote:
>
> On May 18, 2010, at 4:32 PM, Changbin Du wrote:
>
> head(en.id.pr)
>>>
>>valid.gene_id b.pred rf.pred svm.pred
&
gave me errors.
CAN someone help me with this?
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856
[[alternative HTML version deleted]]
__
R-help@r-proje
is this random decision tree, I dont know is there any package can run it.
If you know, please let me know.
On Fri, May 14, 2010 at 10:23 AM, Shi, Tao wrote:
> Hi list,
>
> Is there a way in "rpart" to force the variables only used once when doing
> the splits?
>
> This is how the question cam
Thanks so much, David!
On Wed, May 12, 2010 at 2:52 PM, David Winsemius wrote:
>
> On May 12, 2010, at 5:31 PM, Changbin Du wrote:
>
> fit.dimer <- rpart(as.factor(out) ~ ., method="class", data=p_df)
>>>
>>> fit.dimer$frame[, "var"]
>&
> fit.dimer <- rpart(as.factor(out) ~ ., method="class", data=p_df)
>
> fit.dimer$frame[, "var"]
[1] NE WC TA WG WD WW WC
[11]CT FC YG QT
[21] NW DP DY SK
[31]
401 Levels: AA AC AD AE AF AG AH AI AK AL AM AN AP AQ AR AS AT A
HI, Dear R community,
How to extract the variables actually used in tree construction? I want to
extract these variables and combine other variable as my features in next
step model building.
> printcp(fit.dimer)
Classification tree:
rpart(formula = outcome ~ ., data = p_df, method = "class")
V
Thanks, Steve and David!
svm.fit<-svm(as.factor(out) ~ ., data=all_h, method="C-classification",
kernel="radial", cost=bestc, gamma=bestg, cross=10, probability=TRUE)
It works this time!
On Wed, May 5, 2010 at 6:24 PM, Changbin Du wrote:
> Thanks Steve!
>
>
Thanks Steve!
I will try and let you know how it comes.
On Wed, May 5, 2010 at 6:07 PM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> Hi Changbin,
>
> On Wed, May 5, 2010 at 6:46 PM, Changbin Du wrote:
> > svm.fit<-svm(as.factor(out) ~ ., data=all_h,
variable. I trained the model svm.fit
in training data. And want to predict the out in the new data set hh.
WHy the probabilities are both 0 in 1 and 0 class?
--
Sincerely,
Changbin
--
Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927
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