If you wanted this for all values in x that are smaller, i'd use
x[x$a < y$a,] <- y
for just the smallest:
x[intersect(which(x$a < y$a),which.min(x$a)),] <- y
On 29.01.2013, at 22:11, Dimitri Liakhovitski wrote:
> Hello!
>
> I have a large data frame x:
> x<-data.frame(item=letters[1:5],a=1:
Professor of Anthropology
> Texas A&M University
> College Station, TX 77843-4352
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of Jessica Streicher
>> Sent: Wednesday, January 23
As an example:
chars<-c("A","A","B")
numbers<-as.numeric(as.factor(chars)) #make this numerical
plot(numbers,c(0.4,0.5,0.6),xaxt="n") #xaxt="n" says to not plot the x-axis
axis(side=1,at=numbers,labels=chars) #make the axis with labels
On 23.01.2013, at 10:16, Ng Wee Kiat Jeremy wrote:
> Dea
Or maybe
x<-matrix(test,nrow=10)
apply(x,2,mean)
On 23.01.2013, at 00:09, Wim Kreinen wrote:
> Hello,
>
> I have vector called test. And now I wish to measure the mean of the first
> 10 number, the second 10 numbers etc
> How does it work?
> Thanks Wim
>
>> dput (test)
> c(0, 0, 0, 0, 0, 0, 0
You're not giving people much to work with. I googled the error, and it seems
to come from the call to optim and has likely to do with bad starting
parameters.
That said, the documentation of fitdistr doesn't suggest it even supports
"dbeta", there is only a "beta" mentioned.
On 22.01.2013, at
Next time please provide sample data in a form we can easily read in (look at
?dput for example)
If i understand this right:
yourData<-read.table(header=T,text="
datedays rate
1996_01_02 155.74590
1996_01_02 505.67332
1996_01_02 785.60888
1996_01_0
Rather unspecific.
Basically you'd need a loop to create the sets, and a way to write them into a
file.
You did not specify the format of your data. You might be able to use write,
write.table or write.csv and the like.
You could also have a look at ?save which allows you to save any R object.
Sorry, but i don't get the problem at all.
Could you provide a bit of y by using dput()?
Can you provide an example of how you want the data to look like after the
transformation?
On 16.01.2013, at 16:42, Andrea Goijman wrote:
> Dear list,
>
> I'm working with a large data set, where I groupe
Maybe rle can help a little here
rle(b>1)
Run Length Encoding
lengths: int [1:5] 3 5 5 8 3
values : logi [1:5] FALSE TRUE FALSE TRUE FALSE
r<-rle(b>1)
r$lengths[r$values]
[1] 5 8
# started for the maximum but need to go home now, sorry. Will continue
tomorrow if noone else finishes it.
g
Sorry, you wanted rows, i wrote for columns
#rows would be:
test2<-apply(test[,-c(1:4)],1,function(x){table(t(x))})
#find single values in a row
sapply(test2,function(row){
allVars<-paste(names(row),collapse="")
u <- unique(strsplit(allVars,"")[[1]])
parts<-sapply(names(ro
If test is the structure, will
test2<-sapply(test[,-c(1:4)],function(x){table(t(x))})
to what you want?
On 09.01.2013, at 15:48, jim holtman wrote:
> forgot the data. this will count the characters; you can add logic
> with 'table' to count groups
>
>
> x <-
> structu
Well for one, if ever someone looks at your code, cvtest[[lambda.rule]] is much
easier to read and understand than
eval(parse(text=paste0("cvtest$",lambda.rule)))
On 27.12.2012, at 11:44, Heramb Gadgil wrote:
> I am not sure why "Never ever!"
>
> Can you please elaborate. What are the negative
A bit of data might be useful. Make a small example and post the data with
dput().
On 24.12.2012, at 20:21, jenna wrote:
> I am trying to get the means of each participants avg saccade amplitude as a
> function of the group they were in (designated by "shape_"), but there are
> missing values in
@David : In my mind it was quite complete enough.
@William: Thanks, didn't know you could do that with data frames, if i ever
have to do something similar again i might try this.
On 20.12.2012, at 22:39, David Winsemius wrote:
>
> On Dec 20, 2012, at 10:01 AM, Jessica Stre
w, sep="/") # doSomething()
> }
> }
>
> grid <- expand.grid(height, width, stringsAsFactors=FALSE)
> as.character(grid[1,])
> # [1] "high" "slim", not the [1] "1" "1" you get with stringsAsFactors=TRUE
> l[[ as.c
705 450 | Mobile. +44 (0) 7929 628 349
> mailto:ccampb...@mango-solutions.com | http://www.mango-solutions.com
> Mango Solutions
> 2 Methuen Park
> Chippenham
> Wiltshire
> SN14 OGB
> UK
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-b
Following problem:
Say you have a bunch of parameters and want to produce results for all
combinations of those:
height<-c("high","low")
width<-c("slim","wide")
then what i used to do was something like this:
l<-list()
for(h in height){
l[[h]]<-list()
for(w in width){
f<-factor(c(1,1,2,3))
n<-c(1,1,2,3)
df<-data.frame(f,n)
sapply(df,is.factor)
f n
TRUE FALSE
df[sapply(df,is.factor)]
f
1 1
2 1
3 2
4 3
df[sapply(df,is.numeric)]
n
1 1
2 1
3 2
4 3
something like that?
On 17.12.2012, at 11:02, Martin Spindler wrote:
> Dear R users,
>
> I have a dataf
dataset<-data.frame(id=c(1,1,2,3,3,3),time=c(3,5,1,2,4,6))
dataset
id time
1 13
2 15
3 21
4 32
5 34
6 36
ids<-unique(dataset$id)
for(id in ids){
+ dataset$time[dataset$id==id]<-c(0,diff(dataset$time[dataset$id==id]))
+ }
dataset
id time
1 10
2 1
You need to install the party package and then load the package before you can
use its functions:
install.packages("party")
library(party)
On 11.12.2012, at 16:56, Peterson, Kim - DNR wrote:
> Greetings! I'm trying to use function varimpAUC in the party package
> (party_1.0-3 released Septemb
Nevermind, found i can at least do it with pdf() and consorts, so i'll get it
right in my images, and thats the main point.
On 03.12.2012, at 18:06, Jessica Streicher wrote:
> I'd like to make plots that do not have the quadratic layout, like having a
> plot that is twice as wi
I'd like to make plots that do not have the quadratic layout, like having a
plot that is twice as wide as it is high (without distorting everything).
Sadly i wasn't able to find anything in par that does that.
Best would be with plot(), but i'd use the ggplot package as well if necessary.
thanks
This is not a reproducible example ;)
anyways, dcast has an attribute:
drop
should missing combinations dropped or kept?
does that not do what you want?
On 03.12.2012, at 16:07, michael.laviole...@dhhs.state.nh.us wrote:
>
> Hello--I'm doing a simple crosstab using dcast:
>
> rawfreq <- d
Thats not a very precise question. I'll try anyway..
- if you use c, you need to separate the values by commas
- i think you mean seq(1,100,0.1), otherwise x only has one value
- function sen is not defined
- If you call int(1), upper will be 1, not x[i]
- why are you making a function and calling
TRUE
mtest[!duplicated(mtest[,1:2]),]
id time value
[1,] 13 1
[2,] 12 3
[3,] 11 4
[4,] 21 5
[5,] 23 6
On 30.11.2012, at 17:15, Jessica Streicher wrote:
> Hello Bikek,
>
> please use dput() next time to provide the data, its easier to
Hello Bikek,
please use dput() next time to provide the data, its easier to use that.
also: looking at the data provided, how would you want to decide which value of
the non-unique times to retain? Just take the first one? They aren't all the
same.
On 30.11.2012, at 16:59, bibek sharma wrote:
Anyway, need to go home. If you really only want to plot the values of all
variables, and its for some reason not important what value belongs to what
exact variable:
plot(variances,ylim=range(variances,sdevs)) # black circles
points(sdevs,col=2,pch=3) #red crosses
On 20.11.2012, at 19:32, Elli
rly visible in just one plot with hundreds of
variables.
some kind of distribution maybe, like how likely some value of variance is?
do you have an example of what it should look like?
Whats the purpose of plotting it?
On 20.11.2012, at 19:22, Jessica Streicher wrote:
> At least for me this is
At least for me this is not enough information to go on Elli.
What exactly is the problem? Can you post example data?
On 20.11.2012, at 18:39, Elli wrote:
> How I can plot the standard deviation and variance separately or is there a
> graph where I can show the two?
>
>
>
> --
> View this mes
wrong with the asumption, that if i get those
probabilities, that 0.5 is the threshold to get "the default" result?
On 16.11.2012, at 16:32, Jessica Streicher wrote:
> Hi again!
>
> This might be more of a statistical question, but anyway:
> If i train several support vector m
Fixed it by using e1071 instead..
On 15.11.2012, at 17:43, Jessica Streicher wrote:
> Guess it has something t do with the values in prob.model. The classifiers
> with bad predicitons have positive values and the ones with good predictions
> have negative values.
>
> On 15.11
Hi again!
This might be more of a statistical question, but anyway:
If i train several support vector machines with different degrees of
polynomials, and as result, get that higher degrees not only have a higher test
error, but also a higher in-sample error, why is that?
I would assume i shoul
Guess it has something t do with the values in prob.model. The classifiers with
bad predicitons have positive values and the ones with good predictions have
negative values.
On 15.11.2012, at 17:18, Jessica Streicher wrote:
> I'll try .. lets see
>
> dput(dframe)
> struc
ways occur however, you may have to run it a few times.
On 15.11.2012, at 16:47, Sarah Goslee wrote:
> Hi Jessica,
>
> Talking may help you, but it's kind of a waste of time for the
> thousands of readers of R-help unless you provide a reproducible
> example and full contex
Now i let it run for one specific set and got the same bad result, then i
deactivated the probabilities and got a good result, then i activated the
probabilities again and got a good result .. huh???
On 15.11.2012, at 15:32, Jessica Streicher wrote:
> Its not scaling.. so..
>
> I g
with
pred.pca<-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs]
dframe<-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2)));
results[[i]]$classifier<-ksvm(class~.,data=dframe,scaled=T,kernel="polydot",type="C-svc",
C=C,kpar=list(degree=degree,scale=scale,offset=
Its not scaling.. so..
I guess i'll stay severely frustrated, and yes i know this is probably not
enough information for anyone to help.
Still, talking helps ;)
On 15.11.2012, at 15:15, Jessica Streicher wrote:
> with
>
> pred.pca<-predict(splits[[i]]$pca,trainingData
If i understand this correctly, what you want to use is a vector or a list as a
parameter to that function.
getdata<-function(letterVec){
...
query<-paste("SELECT *FROM myfile1 where MODEL in",
sqlListFromVector(letterVec))
...
}
sqlListFromVector<-function(v){
Not that i really understand what that shall do, but..
why do you define 2 functions that do exactly the same?
why do you want to make a matrix of functions instead of a matrix with the
results of functions?
On 06.11.2012, at 05:11, Aimee Kopolow wrote:
> Hi all,
>
>
> I have a matrix simulat
I'll second the full path option.
Just create a variable path_to_data and concatenate it with the names of any
files you need in that directory.
You could also try to use relative paths if the data is "nearby" and is likely
to not change that position.
e.g. if the data was in '/home/r/Documents
It depends a bit on your setup, but you could have a look at
?memory.size
or
?memory.limit
and see if any of that helps
Jessi
On 05.11.2012, at 12:49, Haris Rhrlp wrote:
> Dear R users,
>
> I have this problem with memory i guess
>
> AA<-array(rep(0,96096000),dim=c(16,5,3003,400))
> Error
Actually forget that loop, that was old code. Now remember i had to actually
split the one query down into several queries to get everything.
On 01.11.2012, at 18:38, Jessica Streicher wrote:
> I didn't have the same exception yet, but please make sure you have a stable
> connec
I didn't have the same exception yet, but please make sure you have a stable
connection (avoid wlan).
I personally ran out of memory very often and had to make several queries,
combining the results in R instead.
This is probably not the best thing to do, but worked for me to fetch large
amount
I really don't understand what you want to achieve and how your data presents
itself.
You have several combinations of trial/subject which are unique?
There is a variable for each such combination that you want to test against
(tt)?
In addition you have data on each such combination (several set
please provide a bit of the dataframe in the future using the dput() function
> x<-data.frame(height=1:10,
> color=sample(c("blue","green","brown"),10,replace=T))
> x
height color
1 1 blue
2 2 green
3 3 blue
4 4 brown
5 5 blue
6 6 brown
7 7 green
8
A look at the tutorial might help here, but anyway:
Say you have that dataframe down there with the name myData (you should use
dput() to give us the data btw),
then you can subset that by using myData[rows,columns], where left of the comma
you define which rows you want, and right of the comma
s <- seq(0.8, 1.2, by = 0.1)
as.list(s)
to get the first type of list from the sequence.
On 30.10.2012, at 16:51, ONKELINX, Thierry wrote:
> You first example is a list of 5 items, each item is a number
> The second example is a list with one item: a vector with 5 elements.
>
> You'll need c()
a,
>
> Thanks for inform this and very sorry for inconvenience, Here I have
> attached two Files
> 1. crash.png- For Issue with Amazon Instance
> 2. localmachine_error.bmp - for Issue with local machine
>
> Thanks
>
> On Mon, Oct 22, 2012 at 1:42 PM, Jessica Streich
Hello Vignesh, we did not get any attachments, maybe you could upload them
somewhere?
On 19.10.2012, at 09:46, Vignesh Prajapati wrote:
> As I found the memory problem with local machine/micro instance(amazon) for
> building SVM model in R on large dataset(2,01,478 rows with 11 variables),
> the
riment so its only around once.
I'll take a look at the lazy-load stuff too, thanks.
On 10.10.2012, at 19:47, J Toll wrote:
> On Tue, Oct 9, 2012 at 10:35 AM, Jessica Streicher
> wrote:
>> Can i somehow append objects to an .Rdata file?
>>
>> I didn't see an o
- Why are you making a matrix for lookzone if it is ever only one number?
- changing a variable used for iteration is bad practice (input in this case)
though you aren't changing the rows i guess.
- There are too many x and too few y in those ifs.
Plus this is shorter:
test<-apply(input,
Actually i just saw that i copied the wrong code
apply(m,2,function(x){median(c(x,m[,5]))})
it should be.
Sorry if this confused anyone.
On 15.10.2012, at 15:32, eliza botto wrote:
>
> Dear Arun and Jessica,
> thnx for your reply. it realy worked out.
> regards
> eliza
>
>
>> Date: Mon, 15
m<-matrix(1:10,ncol=5)
> m
[,1] [,2] [,3] [,4] [,5]
[1,]13579
[2,]2468 10
apply(m,2,function(x){median(x,m[,5])})
[1] 5.5 6.5 7.5 8.5 9.5
Though i am not 100% sure thats what you mean. But the median of each row
between two columns kind of doesn't make
This is pretty confusing writeup, but if you have iterations, then yes, you
need a loop or a recursion.
A loop is probably easier, take a look at
?"for"
or
?apply
for this.
there are several derivates of apply
for example, say your outputs are what the code below gives you for each file,
From: 발송실패알림
Subject:[발송실패 안내]
envy721c@naver.cobSDsnLzroZwg66mU7J287J20IOyghOyGoeuQmOyngCDrqrvtlojsig==teuLiOuLpC4=
The only plain english in the message is that the mail was denied by the
receiver
Anyone else getting this?
__
R-he
You could set xlim and slim when using plot()
plot(vector,xlab="Period",ylab="Values",xlim=range(0,length(vector)+1),ylim=range(vector,est_vector,forecast))
i think - you forgot to provide data for the vectors :)
On 10.10.2012, at 11:31, piranha piranha wrote:
> Hello,
>
> i have been doing br
Can i somehow append objects to an .Rdata file?
I didn't see an option for it in the save() method.
dump() won't work since i have s4 objects in there.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read th
ying
> Research Engineer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
&g
---
> Sent from my phone. Please excuse my brevity.
>
> Jessica Streicher wrote:
>
>> I'm making tables for prediction results of classifiers (2 classes)
>> that show the usual numbers, true positives, false positives, etc
>>
>> I used the c
I'm making tables for prediction results of classifiers (2 classes) that show
the usual numbers, true positives, false positives, etc
I used the command
table(predictedLabels,realLabels)
to make those.
I just had a case though ,where one of the label vectors had only one class in
it. This wil
> x<-c(NA , 1 ,NA, 1 , 1 , 1 , 1 , 1 ,1 ,NA , 1)
> embed(x,3)
[,1] [,2] [,3]
[1,] NA1 NA
[2,]1 NA1
[3,]11 NA
[4,]111
[5,]111
[6,]111
[7,]111
[8,] NA11
[9,]1 NA1
> which(rowSums(embed(x,
> paste("Trial and",x[1],"sheet")
[1] "Trial and a sheet"
> cat("Trial and",x[1],"sheet")
Trial and a sheet
On 18.09.2012, at 11:40, Sri krishna Devarayalu Balanagu wrote:
> Suppose I want the output as "Trial and a sheet" without quotes
> x=c("a", "b", "c")
> print("Trial and x[1] sheet")
>
>
Now i know why using else never worked for me - really, i'd consider this more
of a bug than a feature of the language ;)
On 17.09.2012, at 14:35, R. Michael Weylandt wrote:
> And now it is easy to see why we should all the one true brace style ;-)
>
> Michael
>
> On Mon, Sep 17, 2012 at 12:29
Since noone but you knows what you are interested in - unlikely.
You could try to set up your mailclient with rules, deleting mails coming from
the list that do not contain certain keywords.
On 05.09.2012, at 12:00, Marcus Tullius wrote:
> Hello there,
>
> is there a way I can erase my name fr
I'm not quite sure what you expect as a result there..
On 28.08.2012, at 12:34, Akkara, Antony (GE Energy, Non-GE) wrote:
> Ok.
> Here I have a variable called "Variable_1".
>
> Variable_1 <- "MyDataFrame"
>
> Here I want to create another variable, by assigning the value of
> "Variable_1" .
>
This?
> x<-"y"
> assign(x,1)
> y
[1] 1
On 28.08.2012, at 10:16, Bert Gunter wrote:
> Post a small reproducible example to clarify what you want to do.
>
> -- Bert
>
> On Mon, Aug 27, 2012 at 11:48 PM, Rantony wrote:
>> Hi,
>>
>> Here i have a variable named "MyVariable" containing particula
filled.contour(Nooshin, main="Permeability",color.palette=rainbow)
filled.contour(Nooshin, main="Permeability",color.palette=terrain.colors)
filled.contour(Nooshin, main="Permeability",color.palette=topo.colors)
not quite sure what to do with the heat map, but it has a "col" attribute.
On 21.08.2
Dear Nooshin,
most attachments will not make it into the list. Please upload it somewhere and
link to it instead.
alternatively you could post reproducible code to generate your plot so far.
On 20.08.2012, at 15:53, nooshin bahar wrote:
> Hello guys,
>
> I want to increase contrast of colors i
ow to plot points and
lines, whats there more to explain?
You can use the commands ?plot ? points ?lines and so on to find help on it.
On 17.08.2012, at 13:18, Jessica Streicher wrote:
> Above mentioned formula is wrong - maybe a typo
> http://en.wikipedia.org/wiki/Receiver_operating_cha
Above mentioned formula is wrong - maybe a typo
http://en.wikipedia.org/wiki/Receiver_operating_characteristic
The false positive rate is the rate of false positives, meaning how many of the
total negatives (all in reality negatives(N), that is, all negatives falsely
classified as positives(fp)
Pretty much every time i reply to a post in this list i get a bounce saying the
above, + waiting for approval.
I first thought it was because i didn't send in plain, but now they are and it
hasn't become any better. Any other ideas? Is this normal?
__
than
in threshold 1.
It needs to be the same, so what the hell have you DONE? ^^
greetings
Jessi
On 16.08.2012, at 17:49, Jessica Streicher wrote:
> To clarify:
>
> Is TN = 0 or do you not know TN (N)?
>
> On 16.08.2012, at 11:51, vjyns wrote:
>
>> Hi,
>>
To clarify:
Is TN = 0 or do you not know TN (N)?
On 16.08.2012, at 11:51, vjyns wrote:
> Hi,
>
> I want to plot ROC curve for my detection algorithm which detects
> features in different images at two different thresholds.
>
> 6 different images used and obtained tp, fp and fn. No tn in my c
set.seed(2)
a=matrix(rnorm(400),ncol=100)
#use a list
aList<-list()
for(i in 1:dim(a)[2]){
aList[[i]]<-matrix(a[,i],ncol=2)
}
#get lots of variables
for(i in 1:dim(a)[2]){
assign(paste("a",i,sep=""),matrix(a[,i],ncol=2))
}
On 16.08.2012, at 08:07, bantex wrote:
> Hi all,
>
> I
Also, since there apparently were warnings, how about adding them?
On 13.08.2012, at 16:34, peter dalgaard wrote:
>
> On Aug 13, 2012, at 15:38 , Jennifer Kaiser wrote:
>
>>> Model: poisson, link: log
>>>
>>> Response: sb_ek_ber
>>>
>>> Terms added sequentially (first to last)
>>>
>>>
>>>
Tempting a use of let me google that for you..
Anyway, theres a package called Imputation. I myself used the zoo package.
There are probably lots of others since its a real common problem.
They usually fill in places in you data that are designated as NA.
I do not completely understand what yo
> x=c(5, 8, 28, 29, 30)
> y=c(5, 8, 28, 29, 31)
> x==y
[1] TRUE TRUE TRUE TRUE FALSE
> which(x==y)
[1] 1 2 3 4
> which(x!=y)
[1] 5
> x[x!=y]
[1] 30
> y[x!=y]
[1] 31
RTFM! - just kidding ;)
also, this might not work for floating point numbers, theres a section in the
documentation/fa
If its just a few tables, can't you just copy the text and color it in
PowerPoint?
If you'd like Latex output i could help maybe, but i'm not sure what you expect.
On 08.08.2012, at 14:19, namit wrote:
> Hi ,
>
>
> I have a data frame XYZ. i want to color the cells where ever we find "G" it
You should give us the data is what you should do :)
Aside from that: you can only make probability predictions if you activated it
when making the model.
On 07.08.2012, at 17:23, Camomille wrote:
> Hi, I have some difficulties in interpreting the prediction of a svm model
> using the package
?merge and ?unique might help
however: why is calee_id a floating point number? Ids are usually stuff thats
close to a factor, integers, strings and the like, you know stuff that has a
value that isn't dependant on precision. Floating points might just complicate
things..
On 03.08.2012, at 12
Actually this might be shorter for working with indices:
v1[names(v1)%in%names(v2)]+v2[names(v2)%in%names(v1)]
On 03.08.2012, at 15:41, Petr PIKAL wrote:
> Hi
>
> Your description is quite long but almost uninformative about what you
> really want.
>
> You do not say which values you want to
Is the Index of what you cut off or add unimportant or do you have indices on
which to compare them?
If you just want to have the same length, not depending on any indices:
v1<-1:9
v2<-1:10
minlength<-min(length(v1),length(v2))
v1[1:minlength]+v2[1:minlength]
If indices are important:
names(v
; I dotn want to take the pain of writing names of 40 companies
> rather than this i want to remove the date for 9 companies which i dont need
>
> On Thu, Aug 2, 2012 at 3:00 PM, Jessica Streicher
> wrote:
> data[data$Name %in% names, ]
>
> should work then. You do know
help? You really aren't explaining your problem well^^
On 02.08.2012, at 11:18, Akhil dua wrote:
> sry its not firms
> its banks
>
> On Thu, Aug 2, 2012 at 2:42 PM, Jessica Streicher
> wrote:
>
> On 02.08.2012, at 10:44, Akhil dua wrote:
>
> > Hi everyone
&g
On 02.08.2012, at 10:44, Akhil dua wrote:
> Hi everyone
>
> I have banking data set in long format with 4 columns.One of these columns
> is bank name which consist of 49 banks and I want the data for only 40
> banks out of these 49
> so can anyone help me on how to get this 40 banks data
>
> My
I assume "result" means the invalid variable.
works with sample data frame i made, so please provide an example of cdb_frame.
You can use dput(cdb_frame) for that, or dput(head(cdb_frame)) if you have a
lot of data.
For example it could be that $Publication is a string, not numeric.
On 01.08.2
rmind.
On 31.07.2012, at 16:25, Jessica Streicher wrote:
> Hi!
>
> The kernlab function kpca() mentions that new observations can be transformed
> by using predict. Theres also an example in the documentation, but as you can
> see i am getting an error there (As i do with my own data). I&
Hi!
The kernlab function kpca() mentions that new observations can be transformed
by using predict. Theres also an example in the documentation, but as you can
see i am getting an error there (As i do with my own data). I'm not sure whats
wrong at the moment. I haven't any predict functions wri
You could have googled "R solve equation", and would have easily reached
http://r.789695.n4.nabble.com/R-solve-equation-td882585.html
and would have in there found something like this:
> z <- matrix(c(-8.313,-1,1), ncol=1)
> polyroot(z)
[1] 3.42626+0i -2.42626+0i
> p<-polyroot(z)
> p^2-p
[1] 8.
)
For example will let you view a bit of that separating plane. I'm not getting
creative on how to find the best values there though..
On 27.07.2012, at 10:21, Jessica Streicher wrote:
> Now i'm scratching my head as well, thought it might have to do with scaling
> at first, s
Now i'm scratching my head as well, thought it might have to do with scaling at
first, so i turned it off, and also tried scaling the data for the plot
instead, but to no avail, it just switches the color, but doesn't show the
correct contours.
And it is at least predicting the stuff right, so
example
but a wild guess: if your class information is numeric the default is to do
eps-regression, not classification. Use factors or specify the type you want to
use.
?svm
might help there.
On 25.07.2012, at 15:31, Meffy wrote:
> Hello users!
> I'm calculating a simple model using svm(
Well, what do you want to control there?
Need a subset? Need an ordering?
On 20.07.2012, at 15:00, Carla Moreira wrote:
> Yes, I do.
>
> But I need to control how the permutations are done.
>
> Thank you.
>
> 2012/7/20 Jessica Streicher
> You mean execut
You mean executing the function for all combinations of values?
For example, if you have a<-b<-c<-1:2
you would get back the values of
myfunc(1,1,1)
myfunc(1,1,2)
myfunc(1,2,1)
myfunc(1,2,2)
myfunc(2,1,1)
myfunc(2,1,2)
myfunc(2,2,1)
myfunc(2,2,2)
?
On 20.07.2012, at 13:05, carla moreira wrote:
Copied the wrong lines, sry
l<-rep(list(1:5),20); boxplot(l,names=sample(1:20,20))
of course.
thanks for the answer
.
On 19.07.2012, at 16:17, Peter Ehlers wrote:
> On 2012-07-19 06:58, Jessica Streicher wrote:
>> When i make Boxplots with a lot of boxes, the names of them get
When i make Boxplots with a lot of boxes, the names of them get only written
down every second "column".
Since they aren't in any way ordered, you don't see anymore to what they belong.
example:
l<-rep(list(1:5),20); boxplot(l,names=sample(20,1:20))
Is there a way to show them all, or do i have
apply(mydataframe,2,function(x){ print(x);is.factor(x)})
[1] "1" "2" "3" "4"
[1] "16.99" "10.34" "21.01" "23.68"
[1] "1.01" "1.66" "3.50" "3.31"
[1] "Male" "Male" "Male" "Female"
X total_billtipsex
FALSE FALSE FALSE FALSE
> sapply(mydataframe,fu
Yepp, that makes much more sense, look at Eiks post for an answer.
I'm terribly sorry, i confused Rd with RData.
On 17.07.2012, at 15:18, Jessica Streicher wrote:
> Oh, i think i get it now, you're talking about formatting a table in a
> documentation file?
>
> O
Oh, i think i get it now, you're talking about formatting a table in a
documentation file?
On 17.07.2012, at 14:45, purushothaman wrote:
> Hi,
>
> i need to create table like this
>
> -
Try again please,
so you have a table/dataframe or some such with 4(?) columns:
module_name, class_name, function_name and function_description
I guess under the second line shall be an example of data in the table, but i
cannot see what belongs to what column.
If you have an example table alre
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