Hi,
I want to import data from teradata.Data set is very large in size , it has
around 1 million rows and 40 columns.
Could anyone tell me an efficient and fast way to import data sets with this
size.
Thanks.
Sumit
[[alternative HTML version deleted]]
Dear users,
How to get a symmetric square root of a positive definite matrix? I have
tried using spectral decomposition, but some eigen values come out to be
complex. Is there any function in R that can give the symmetric square root
of a pd matrix?
--
Arindam Fadikar
M.Stat
Indian Statistical I
Peter,
Thank you. I'll be sure to include a as.character.
Greg
On 4/7/10 11:40 AM, Peter Ehlers wrote:
On 2010-04-07 11:40, Greg Hirson wrote:
Dear list,
I have come across this issue:
combn(letters[1:5], 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "a" "a" "a" "a" "a" "a" "
On 20 February 2010 11:22, Sam Albers wrote:
> #I've tried to alter the panel function according to the thread to produce
> vertical error bars in my barcharts
>
> prepanel.ci <- function(x, y, ly, uy, subscripts, ...) {
>
> y <- as.numeric(y)
> ly <- as.numeric(ly[subscripts])
>
Thanks Dennis and Daniel for your reply!
On Wed, Apr 7, 2010 at 12:24 PM, Daniel Malter wrote:
>
> Assume you have suitable "week" indicator and your data is stored in a
> data.frame named "data." Then you get the weekly averages by
>
> #simulate data
> week <- rep(1:52,each=7)
> x <- week+rnorm
On 8 April 2010 06:03, Ivan Gregoretti wrote:
> Hi Sam and everybody,
>
> Can you educate me a little bit on the use of barchart?
>
> A command like this
>
> barchart(~table(someProperty), data=A)
>
> produces a barchart with horizontal bars.
That should be
barchart(table(someProperty), data=A)
Hullo,
I am trying to use the leaps package, & keep getting the following error:
Error in leaps.setup(x, y, wt = wt, nbest = nbest, nvmax = NCOL(x) + int, :
y and x different lengths
My data set is attached below.
I am rather new to R, & would appreciate any help that could be given.
Thanks
Thank you very much for the help.
I installed data.table package, but I keep getting the following warnings:
> setkey(DT,id,date)
Warning messages:
1: In `[.data.table`(deref(x), o) :
This R session is < 2.4.0. Please upgrade to 2.4.0+.
I'm using R 2.10, but why I keep getting warnings on upgr
Thank you for your reply. The WoodEnergy example helped a lot. I
understand now that it is inappropriate to make all pairwise comparisons
with an interaction present and better to make comparisons between levels of
one factor within a constant level of the second factor. As I understand it,
the s
On Wed, 7 Apr 2010, Jun Shen wrote:
Dear all,
I understand identical (a,b) will tell me if a and b are exactly the same or
not. But what if they are different, is there anyway to tell which
element(s) are different? Thanks.
which( a != b, arr.ind = TRUE)
HTH,
Chuck
Jun
[[alternat
Hi Ivan,
Can you educate me a little bit on the use of barchart?
>
Unfortunately no... For this post I eventually used the barplot2() in the
gplots packages. I got bogged down trying to do it in lattice so I looked
for an alternative. It was quite straight forward which was nice and I was
able to
On Apr 7, 2010, at 9:55 PM, Erik Iverson wrote:
Jun Shen wrote:
Hi, David,
Thanks for the reply. However str() doesn't tell me exactly which
element is
different. I expect to see a is identical to b. But if there is
some minor
difference (usually by human mistake), I want to know which el
Hi,
My problem maybe a little bit complicated, so forgive me if the following
words are too much.
#date set
a0<-matrix(c(1.1,1.3,1.1,1.3,1.3, 2.0,1.8,2.0,1.8,1.8,
"12/01/2008","05/20/2007","12/06/2008","05/10/2007","05/06/2007",
"N","N","A","C","A", 1,2,3,4,5),ncol=5,byrow=FALSE)
a0<-data.frame(
Jun Shen wrote:
Hi, David,
Thanks for the reply. However str() doesn't tell me exactly which element is
different. I expect to see a is identical to b. But if there is some minor
difference (usually by human mistake), I want to know which element
(numerical or character) is different.
So you'
Hi, David,
Thanks for the reply. However str() doesn't tell me exactly which element is
different. I expect to see a is identical to b. But if there is some minor
difference (usually by human mistake), I want to know which element
(numerical or character) is different.
Jun
On Wed, Apr 7, 2010 at
On Apr 7, 2010, at 9:31 PM, Jun Shen wrote:
Dear all,
I understand identical (a,b) will tell me if a and b are exactly the
same or
not. But what if they are different, is there anyway to tell which
element(s) are different? Thanks.
You could try:
str(a); str(b)
David Winsemius, MD
Wes
See the bottom of the R-help page:
https://stat.ethz.ch/mailman/listinfo/r-help
You can unsubscribe or change your mode to digest. (Just like any
other mailing list.)
--
David
On Apr 7, 2010, at 8:15 PM, Philip Wong wrote:
hello people,
I didn't know where should post this thread, I aske
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Philip Wong
> Sent: Wednesday, April 07, 2010 5:15 PM
> To: r-help@r-project.org
> Subject: [R] how can I stop getting email from every comment posted in
> this forum?
>
>
> hell
Dear all,
I understand identical (a,b) will tell me if a and b are exactly the same or
not. But what if they are different, is there anyway to tell which
element(s) are different? Thanks.
Jun
[[alternative HTML version deleted]]
__
R-help@r-pr
> url <- "http://dl.dropbox.com/u/41902/22240.csv";
>
> read.csv(url)[, 1]
[1] "oppose" NA"oppose" "support"
> read.csv(url, header = F)[, 1]
[1] "url"
[2] "http://maplight.org/us-congress/bill/109-hr-5825/387248";
[3] "http://maplight.org/us-congress/bill/110-hr-3546/378743";
[4] "http:/
On Apr 7, 2010, at 8:58 PM, array chip wrote:
With the help document, i finally find a set of values of for x=,y=
and z= in "screen" argument that gives me the correct rotation of
the plot. But now it plots x and y axis (tick marks and labels)
along the top of the plot. Is there one way to
df <- data.frame(a = "a\"b")
write.table(df, "test.csv", sep = ",", row = F)
Is there any to load test.csv into R correctly? I've tried the following:
> read.table("test.csv", sep = ",")
[1] V1
<0 rows> (or 0-length row.names)
Warning message:
In read.table("test.csv", sep = ",") :
incomplete
With the help document, i finally find a set of values of for x=,y= and z= in
"screen" argument that gives me the correct rotation of the plot. But now it
plots x and y axis (tick marks and labels) along the top of the plot. Is there
one way to plot x and y axis on the bottom of the plot?
Thank
hello people,
I didn't know where should post this thread, I asked Hugo the nabble
administrator and his response was I should be able to unsubscribed from
getting an email from every comment or response posted here. Yet I couldn't
find the method nor the button to click to do so. I even change
On 7 April 2010 16:27, Steve Lianoglou wrote:
> On Wed, Apr 7, 2010 at 7:16 PM, Robin Evans wrote:
>> On 7 April 2010 16:12, Steve Lianoglou
>> wrote:
>>> Hi,
>>>
>>> On Wed, Apr 7, 2010 at 5:44 PM, Robin Evans
>>> wrote:
Dear all,
I'm wondering if there exist vectorized forms
Cool.
Using [[var]] instead of $var to extract the expected variable is really
what i want.
Thanks. Guys.
2010/4/7 Steve Lianoglou
> Hi,
>
> On Wed, Apr 7, 2010 at 6:27 PM, rusers.sh wrote:
> > Hi all,
> > My problem may be related with the list manipulations. See below.
> > #example data
> >
On 8 April 2010 03:34, Luigi Ponti wrote:
> Thanks for the hint, Felix: the following code makes it but (don't know why)
> the median dots disappear.
>
> bwplot(count ~ spray, data = InsectSprays,
> groups = spray,
> panel = panel.superpose,
> panel.groups = function(..., bo
Thank you for the pointer to: help("BinaryTree-class").
It probably isn't the prettiest solution, but I was able to extract what
I needed by doing something like the following:
nodes = where(fit)
freq = treeresponse(fit)
for(i in 1:100)
{
if(nodes[i]==4
On Wed, Apr 7, 2010 at 7:16 PM, Robin Evans wrote:
> On 7 April 2010 16:12, Steve Lianoglou wrote:
>> Hi,
>>
>> On Wed, Apr 7, 2010 at 5:44 PM, Robin Evans
>> wrote:
>>> Dear all,
>>>
>>> I'm wondering if there exist vectorized forms of 'isTRUE()',
>>> 'identical()' and 'all.equal()'. My probl
On 7 April 2010 16:12, Steve Lianoglou wrote:
> Hi,
>
> On Wed, Apr 7, 2010 at 5:44 PM, Robin Evans wrote:
>> Dear all,
>>
>> I'm wondering if there exist vectorized forms of 'isTRUE()',
>> 'identical()' and 'all.equal()'. My problem is that I wish to test if
>> each element of a vector is equal
Hi,
On Wed, Apr 7, 2010 at 5:44 PM, Robin Evans wrote:
> Dear all,
>
> I'm wondering if there exist vectorized forms of 'isTRUE()',
> 'identical()' and 'all.equal()'. My problem is that I wish to test if
> each element of a vector is equal to a particular value (or
> numerically close), whilst d
Hi,
On Wed, Apr 7, 2010 at 6:27 PM, rusers.sh wrote:
> Hi all,
> My problem may be related with the list manipulations. See below.
> #example data
> a<-list()
> a[[1]]<-data.frame(matrix(c(1:4),ncol=2));
> a[[2]]<-data.frame(matrix(c(5:8),ncol=2))
> I can use a[[1]]$X1,a[[1]]$X2,a[[2]]$X1,a[[
Last point ...
On Wed, Apr 7, 2010 at 6:39 PM, Stuart Andrews wrote:
>
> Ahhh, I see my error, thanks to Steve and others who mailed me off list.
>
> Perhaps reading a little too quickly, I mis-interpreted the help for ddply,
> in particular, the second argument:
>
>> ".variables: variables
I'm a big fan of both Weka and R (quite new at R :) ), and jumped at the chance
to use them together. Unfortunately, I'm running into what is probably a dumb
error when trying to view info about my model. A Google search turned up 0 hits
for the actual error I got (last line), but you all are s
Ahhh, I see my error, thanks to Steve and others who mailed me off list.
Perhaps reading a little too quickly, I mis-interpreted the help for
ddply, in particular, the second argument:
>".variables: variables to split data frame by, as quoted variables, a
>formula or characte
Hi all,
My problem may be related with the list manipulations. See below.
#example data
a<-list()
a[[1]]<-data.frame(matrix(c(1:4),ncol=2));
a[[2]]<-data.frame(matrix(c(5:8),ncol=2))
I can use a[[1]]$X1,a[[1]]$X2,a[[2]]$X1,a[[2]]$X2 to use the corresponding
variables. But what i need to do is
Howdy,
I'm no plyr master, but here's my 2 cents ...
On Wed, Apr 7, 2010 at 5:15 PM, Stuart Andrews wrote:
> Hi,
>
> I am confused by results from:
>
>> ddply(aa, names(aa), colwise(sum))
>
> I thought ddply was just calling colwise(sum)() with each column. However
> ddply() returns a 13 x 5 r
Dear all,
I'm wondering if there exist vectorized forms of 'isTRUE()',
'identical()' and 'all.equal()'. My problem is that I wish to test if
each element of a vector is equal to a particular value (or
numerically close), whilst dealing carefully with NAs and so on.
However, using sapply() with id
Hi,
I am confused by results from:
> ddply(aa, names(aa), colwise(sum))
I thought ddply was just calling colwise(sum)() with each column.
However ddply() returns a 13 x 5 result !!
The general result I expected is similar to that of apply() , or
using colwise(sum)() alone. Shouldn
The following grob might be a starting point. I couldn't find a clean
way of orienting the linking arcs though...
Best,
baptiste
library(grid)
paragraph <- "Lorem ipsum dolor sit amet, consectetur adipiscing elit.
Praesent adipiscing lobortis placerat. Nunc vel arcu mauris. Aliquam
erat volutpa
Hi All,
I am currently trying to transform my package to s4 methods, I don't know if
there is any package or easy way like Roxygen to document the S4 methods. I
use emacs+ess+roxygen to document simple function and package before, I
don't know if Roxygen support S4 method too.
Any suggestions for
On Apr 7, 2010, at 3:50 PM, sumit gupta wrote:
> I imported data from teradata to R data frames.Now i want to export this
> data to .xlsx file ( not .xls since it doesn't support large data set).
>
> I have tried dataframes2xls package but it needs python/perl which I don't
> have on my computer.
Thanks a lot, Tom.
In fact, I am likely to have no NAs at all - only zeros!
Dimitri
On Wed, Apr 7, 2010 at 4:54 PM, Tom Short wrote:
> Another way that Matthew Dowle showed me for this type of problem is
> to reshape frame to a long format. It makes it easier to manipulate
> and can be faster.
>
On Wed, 7 Apr 2010, Steven J. Clark wrote:
Hi,
I am new to R and am using the ctree() function to do customer
segmentation. I am using the following code to generate the tree:
treedata$Response<-factor(treedata$Conversion)
fit<-ctree(Response ~
.,controls=ctree_control(mincriterion=0.99,maxde
Another way that Matthew Dowle showed me for this type of problem is
to reshape frame to a long format. It makes it easier to manipulate
and can be faster.
> longdt <- with(frame, data.table(group = unlist(rep(group, each=7)), x =
> c(a,b,c,d,e,f,g)))
>
> system.time(new.frame4 <- longdt[, x/mean
You can also read the help page for the conflicts() function. And try
a commend like
find('acid')
(or any of the objects listed as being masked)
to find out where the two or more objects with the same name are located.
Oh, and looking at ?search would be good also.
-Don
At 4:31 PM -0400 4/7
I imported data from teradata to R data frames.Now i want to export this
data to .xlsx file ( not .xls since it doesn't support large data set).
I have tried dataframes2xls package but it needs python/perl which I don't
have on my computer.
Could any one help me resolve this issue.
Thanks
Sumit
Dear all,
Can someone help me with a C++ source code that underlies MANOVA
(Multivariate Analysis of Variance) in R
Kind regards,
Jes
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listi
Hi,
On Wed, Apr 7, 2010 at 4:24 PM, Changbin Du wrote:
> Hi, r-community,
>
> This morning, I MET the following problem several times when I try to attach
> the data set.
>
> When I closed the current console and reopen the R console, the problem
> disappear. BUt with the time passed on, the prob
I found the following message maybe help. And I will try it.
Hi there,
I have just found that the ``attach'' function
can get you into trouble when called many times.
For example, you have a simulation routine called ``f()'',
in which you used ``attach'' and no corresponding ``detach''.
Then y
Hi, r-community,
This morning, I MET the following problem several times when I try to attach
the data set.
When I closed the current console and reopen the R console, the problem
disappear. BUt with the time passed on, the problem occurs again.
Can anyone help me with this?
> attach(total)
Yes, Tom's solution is indeed the fastest!
On my PC it took .17-.22 seconds while using ave() took .23-.27 seconds.
And of course - the last two methods I mentioned took 1.3 SECONDS, not
MINUTES (it was a typo).
All that is left to me is to understand what .SD stands for.
:-)
Dimitri
On Wed, Apr
Leave it up to Tom to solve things wickedly fast :)
Just as an fyi Dimitri, Tom is one of the developers of data.table.
-Rob
On Wed, Apr 7, 2010 at 2:51 PM, Dimitri Liakhovitski wrote:
> Wow, thank you, Tom!
>
> On Wed, Apr 7, 2010 at 3:46 PM, Tom Short wrote:
> > Here's how I would have done
Hi Sam and everybody,
Can you educate me a little bit on the use of barchart?
A command like this
barchart(~table(someProperty), data=A)
produces a barchart with horizontal bars.
I want to produce that graph but with VERTICAL bars.
clearly, horizontal=FALSE does not work.
Thank you,
Ivan
Hi,
On Wed, Apr 7, 2010 at 3:15 PM, Cable, Samuel B Civ USAF AFMC
AFRL/RVBXI wrote:
>
> OK, my apologies. I am sure this is a question that has been answered
> before. But I have looked all over the web and can't find an answer for
> it. I promise, wasting your time and bandwidth is my last re
Wow, thank you, Tom!
On Wed, Apr 7, 2010 at 3:46 PM, Tom Short wrote:
> Here's how I would have done the data.table method. It's a bit faster
> than the ave approach on my machine:
>
>> # install.packages("data.table",repos="http://R-Forge.R-project.org";)
>> library(data.table)
>>
>> f3 <- funct
Here's how I would have done the data.table method. It's a bit faster
than the ave approach on my machine:
> # install.packages("data.table",repos="http://R-Forge.R-project.org";)
> library(data.table)
>
> f3 <- function(frame) {
+ frame <- as.data.table(frame)
+ frame[, lapply(.SD[,2:ncol(.SD
Thanks so much. Solved.
Joh
hadley wickham wrote:
> Because of the way you've constructed the plot with qplot, you need to
> use:
>
> myPlot + geom_point(
> data=medians,
> aes(x=med,shape=cut, y=0),
> size=2.5,
> )
>
> Hadley
>
> On Wed, Apr 7, 2010 at 5:11 AM, Johannes Graumann
> wrote:
Thank you David. The document is very helpful.
I had a typo in my 2nd example plot, one of the "z" is "x", "z" is the vertical
one.
Thanks
John
--- On Wed, 4/7/10, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] 3-D response surface using wireframe()
> To: "array chip"
>
lm1 <- lm(y~x, data=mydata, subset=mydata$patchchoice==1)
lm2 <- lm(y~x, data=mydata, subset=mydata$patchchoice==2)
abline(lm1, col="Blue")
abline(lm2, col="Red" )
Should do it. Not likely the best way but a way.
Andy.
andydol...@gmail.com
On 7 April 2010 19:17, Samantha Reynolds wrote:
OK, my apologies. I am sure this is a question that has been answered
before. But I have looked all over the web and can't find an answer for
it. I promise, wasting your time and bandwidth is my last resort.
So here goes:
I have an ASCII file formatted like so:
Label 1.1
Time 1
La
Dear Vlatka,
It's impossible to know what the problem is without knowing something about
your data, which you didn't tell us either in this message or your
subsequent one.
The recode command should work:
> (x <- c(rep(0, 5), sample(1:30, 5, replace=TRUE)))
[1] 0 0 0 0 0 17 27 19 19 2
> r
You might want to investigate the
'data.table' package.
On 07/04/2010 16:15, bo wrote:
Hi all,
I have a huge dataset(5000k observations), which contains the daily sales
for each company.
If I want to find out the a company with unique company id number, which
function I should use that is more
Hello
I am trying to plot two lines of best fit on a scatter plot.
I'm plotting number of visits (y) against personality (x), half the
plots are blue for patchchoice one and the other plots are red for
patchchoice two (these options come from one factor, see below) . So i
need a line for
You might try ?system.time
Hi all,
I have a huge dataset(5000k observations), which contains the daily sales
for each company.
If I want to find out the a company with unique company id number, which
function I should use that is more efficient, match or "=="? For example,
use<-dataset[!is.na
Dear all,
I'm trying to use the fold function as described here:
http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-cox-regression.pdf
Page9
It does say that you can use this when you have more than one time varying
covariate: in the description of the argument cov it says:
"cov: A ve
Ramon,
for me this works
> setwd("d:/tmp")
> ffd <- as.ffdf(d, col_args=list(pattern = paste(getwd(), "/fftmp", sep = "")))
> filename(ffd)
$x
[1] "d:/tmp/fftmp35c34861.ff"
$y
[1] "d:/tmp/fftmp5be946bb.ff"
$z
[1] "d:/tmp/fftmp26c49ce.ff"
Jens
-Ursprüngliche Nachricht-
Von: Ramon Diaz-
Hi all,
I have a huge dataset(5000k observations), which contains the daily sales
for each company.
If I want to find out the a company with unique company id number, which
function I should use that is more efficient, match or "=="? For example,
use<-dataset[!is.na(match(dataset$companyID, 123
On 2010-04-07 11:40, Greg Hirson wrote:
Dear list,
I have come across this issue:
combn(letters[1:5], 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "a" "a" "a" "a" "a" "a" "b" "b" "b" "c"
[2,] "b" "b" "b" "c" "c" "d" "c" "c" "d" "d"
[3,] "c" "d" "e" "d" "e" "e" "d" "e" "e" "e"
c
Does anyone know of an R package that does decompounding for German?
For those not familiar with the term, decompounding is the process of
breaking down a word that has been created by stringing several words
together into the component words, e.g., Schulhaus -> Schule + Haus.
Thanks,
Ric
Hello,
Recently I installed R version 2.10 from 2.9.2 and after the update
some the commands like write.table is not functioning properly. Does
anyone has similar issues. Am I missing something ?
Say for example I generated a simulation data (simd1) and used this
syntax to store it
writ
On Apr 7, 2010, at 1:50 PM, Vlatka Matkovic Puljic wrote:
atomic [1:1578] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, "levels")= chr "0=0,1:33=1"
Can you go back to where you got this data and start over? You have
made something that looks quite strange. It appears to be neither a
vector of mode a
Linda,
Thanks for the example.
I did this to make it more reproducible:
set.seed(1)
X<-matrix(rnorm(50*100),nrow=50)
y<-rnorm(50*1)
dimnames(X)
colnames(X) <- paste("V", 1:nrow(X))
# Applying caret package
set.seed(2)
con<-trainControl(method="cv",number=10)
data<-NULL
d
On Apr 7, 2010, at 1:31 PM, Vlatka Matkovic Puljic wrote:
Hi,
I have numerical variable that I want to recode into categories '0'
and '1
and more' and do analysis with that data.
I have tried various of possibilities to do so, but I am sucked and
nothing
is working.
recode(Q12, "0='A';1
atomic [1:1578] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, "levels")= chr "0=0,1:33=1"
2010/4/7 David Winsemius
>
> On Apr 7, 2010, at 1:31 PM, Vlatka Matkovic Puljic wrote:
>
> Hi,
>>
>> I have numerical variable that I want to recode into categories '0' and '1
>> and more' and do analysis with that d
On 07/04/2010 02:10, Felix Andrews wrote:
> On 6 April 2010 20:14, Luigi Ponti wrote:
>
>> On 04/04/2010 06:21, Deepayan Sarkar wrote:
>>
>>> On Thu, Apr 1, 2010 at 4:10 AM, Luigi Ponti wrote:
>>>
>>>
Hello,
I am trying to give different colors to boxes in a violin p
Hi,
I have numerical variable that I want to recode into categories '0' and '1
and more' and do analysis with that data.
I have tried various of possibilities to do so, but I am sucked and nothing
is working.
recode(Q12, "0='A';1:30='B'")
cut(Q12, breaks=c(0,1,30), lables=c('0', '1 and more'))
ca
Dear list,
I have come across this issue:
combn(letters[1:5], 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "a" "a" "a" "a" "a" "a" "b" "b" "b" "c"
[2,] "b" "b" "b" "c" "c" "d" "c" "c" "d" "d"
[3,] "c" "d" "e" "d" "e" "e" "d" "e" "e" "e"
combn(facto
Hello
I am trying to plot two lines of best fit on a scatter plot.
I'm plotting number of visits (y) against personality (x), half the
plots are blue for patchchoice one and the other plots are red for
patchchoice two (these options come from one factor, see below) . So i
need a line for
On Wed, 7 Apr 2010, Ivan Calandra wrote:
Hi,
I'm not sure what your goal is, but why don't you use a list?
You can store each new dataframe name in a new element of the list.
At the end, if each element has the same length, you can use as.data.frame()
or something similar.
Using a list would
HI, Stephan,
Thanks so much! IT works, I appreciated very much!
On Wed, Apr 7, 2010 at 10:06 AM, Stephan Kolassa wrote:
> Hi,
>
> first get the densities without plotting the histogram:
>
> foo <- hist(x, plot=FALSE)
>
> then plot the histogram and feed the rounded densities, converted to
> ch
I have some output from a survey shown below and would like to insert
total rows for each category to then be able to use it as a data set for
further analysis. Is it possible to do this in R. In this example I
would like to insert a row between rows 5 and 6, 10 and 11, 15 and 16,
20 and 21 and p
Hi,
first get the densities without plotting the histogram:
foo <- hist(x, plot=FALSE)
then plot the histogram and feed the rounded densities, converted to
character, to the labels argument (instead of just labels=TRUE):
hist(x, freq=F, xlab='',ylab="Percent of Total", col="skyblue",
labels
HI, Dear R-community:
I have the following codes to plot the conditional histogram, is a way to
label the bars by the percentage values in the conditional histogram?
h<- sample(1:14, 319, rep=T)
c<- sample(1:14, 608, rep=T)
n<- sample(1:14, 1140, rep=T)
vt<-c(h, c, n)
ta<-rep(c("h", "c", "n"), c
Dear All,
I am using package ff. In version 2.1-1 it was possible to use
"pattern" with as.ffdf.data.frame:
d <- data.frame(x=1:26, y=letters, z=Sys.time()+1:26)
as.ffdf(d, pattern = paste(getwd(), "/fftmp", sep = ""))
With the latest version, the last command crashes. I wonder if the new
behavi
On 07/04/2010 11:19 AM, jgar...@ija.csic.es wrote:
Hi Duncan,
I must be misunderstanding something. It is not that I want them built at
installation time, but when I try to display the html help for any
function the html page is not built; e.g. cliking on the link to get() in
the base library, th
x<- sample(1:14, 319, rep=T)
hist(x, freq=F, xlab='',ylab="Percent of Total", col="skyblue",
labels=TRUE, right=FALSE,main="Position of Hypothetical Protein")
Is there is way to round the labels to 2 decimal digits, for example, 0.088
is changed to 0.09.
Thanks!
--
Sincerely,
Changbin
--
On Wed, 7 Apr 2010, ONKELINX, Thierry wrote:
Dear all,
We are analysing some survey data and we are not sure if we are using
the correct syntax for our design.
The population of interest is a set of 4416 polygons with different
sizes ranging from 0.003 to 45.6 ha, 7460 ha in total. Each polygo
I would like to thank once more everyone who helped me with this question.
I compared the speed for different approaches. Below are the results
of my comparisons - in case anyone is interested:
### Building an EXAMPLE FRAME with N rows - with groups and a lot of NAs:
N<-10
set.seed(1234)
frame
Hi,
I am new to R and am using the ctree() function to do customer
segmentation. I am using the following code to generate the tree:
treedata$Response<-factor(treedata$Conversion)
fit<-ctree(Response ~
.,controls=ctree_control(mincriterion=0.99,maxdepth=4),data=treedata)
plot(fit)
print(fit)
On Wed, 7 Apr 2010, cheba meier wrote:
Dear Thomas,
Thank you very much for your answer! Can I use your function in my analysis?
Sure.
The wilcox.test() tests the differences in average ranks (H0: F(X)=F(Y)),
why then do people in many studies compare t.test() with wilcox.test()?
The Wil
Hi,
I'm not sure what your goal is, but why don't you use a list?
You can store each new dataframe name in a new element of the list.
At the end, if each element has the same length, you can use
as.data.frame() or something similar.
Since you didn't provide any sample data, I'm not sure if tha
Assume you have suitable "week" indicator and your data is stored in a
data.frame named "data." Then you get the weekly averages by
#simulate data
week <- rep(1:52,each=7)
x <- week+rnorm(52*7,)
y <- rev(week)+rnorm(52*7)
#create data frame
data=data.frame(week,x,y)
#get weekly averages in a l
Hi,
I'd greatly appreciate any insight into the problem in my previous
message below: Please note that I would like to loop through the
following steps for a list of data frames whose names are dynamically
assigned at run time, so I have a character string variable, called
"variable.df", which stor
Because of the way you've constructed the plot with qplot, you need to use:
myPlot + geom_point(
data=medians,
aes(x=med,shape=cut, y=0),
size=2.5,
)
Hadley
On Wed, Apr 7, 2010 at 5:11 AM, Johannes Graumann
wrote:
> Hi,
>
> Please consider the example below. How can I manage to overlay the p
On Apr 7, 2010, at 11:41 AM, Peter Ehlers wrote:
On 2010-04-07 8:37, Dimitri Liakhovitski wrote:
Dear R-ers!
# I have a data frame with one factor and 2 numeric variables:
x<-data.frame(group=c("b","b","d","d","e","e"),a=c(1,NA,
10,20,100,200),b=c(5,15,20,NA,10,30))
x
# I want to divide ea
On 2010-04-07 8:37, Dimitri Liakhovitski wrote:
Dear R-ers!
# I have a data frame with one factor and 2 numeric variables:
x<-data.frame(group=c("b","b","d","d","e","e"),a=c(1,NA,10,20,100,200),b=c(5,15,20,NA,10,30))
x
# I want to divide each value of each variable by its group mean -
using ply
Hi All,
I have a time series data with two continuous variables (say Var1 and Var2)
for 4 years (***not continuous, do have some breaks because of missing
data***). Something like this:
Date Var1 Var2
12/01/2004 7 0
12/01/2004 0 0
12/01/2004 0 7
12/01/2004 7 0
12/01/2004 0 7
12/01/2004 0 7
12/02/
Hi Duncan,
I must be misunderstanding something. It is not that I want them built at
installation time, but when I try to display the html help for any
function the html page is not built; e.g. cliking on the link to get() in
the base library, the message appears:
An error occurred while loading
f
Hi,
Sorry about my last question. I've found the solution in the
documentation, and yes; there is a flag in ./configure I did not see.
Thanks
Javier
---
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