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
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
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
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
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.
>
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
>
>
>
> 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
Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
> On Wed, 2 Jun 2010, Changbin Du wrote:
>
> dim(tmp)
>>>
>> [1] 576 12
>>
>
_
> 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.
>
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
> "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]
> &
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
>>
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
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
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
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
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
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
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.
>
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
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
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:
>
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.
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
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
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
> &
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)"
>
#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*
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" "
"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
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
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
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,
>>
>>
)
>
> 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
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
&
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
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
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
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_
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