Dear R-helpers,
I have run a basket analysis in Rattle. I've used 'arules' package.
> crs$apriori <- apriori(crs$transactions, parameter = list(support=0.100,
confidence=0.100, minlen=2))
and
> str(crs$apriori)
Formal class 'rules' [package "arules"] with 4 slots
..@ lhs:Formal class 'ite
Hi
or
> lapply(foo, "[",1)==1
A B C
TRUE TRUE FALSE
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of arun
> Sent: Friday, June 27, 2014 5:16 AM
> To: ce; r-help@r-project.org
> Subject: Re: [R] Erro
You could use:
mapply(`==`, sapply(foo,`[`,1),1)
# A B C
# TRUE TRUE FALSE
A.K.
On Thursday, June 26, 2014 10:50 PM, ce wrote:
Dear all,
I have a list of arrays :
foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1))
> foo
$A
[1] 1 3
$B
[1] 1 2
$C
[1] 3 1
I want to use all foo$A
Hi,
May be you can use ?cut or ?findInterval for the range
dat1 <- read.table(text="100, 100, 200
250, 300, 350
100, 350, 100
400, 250, 300
200, 450, 200
150, 501, 300
150, 250, 300",sep=",",header=F)
sapply(dat1, findInterval, c(400,500))==1
# V1 V2 V3
#[1,] FALSE FALSE FALSE
#[2,]
Hi all,
I'm creating a set of stacked bar charts, where each bar is colored by
a category. The problem I'm having is that a given category is being
represented by different colors across plots (category A might be red
in one plot and blue in another). This is because the number of
categories in
On Jun 26, 2014, at 6:11 PM, C Lin wrote:
> Hi Duncan,
>
> Thanks for trying to help. Sorry for not being clear.
> The string I'd like to get is 'AARSD1'
> It can be followed or preceded by white space or // or nothing
>
> so, from test <- c('AARSD11','AARSD1-','AARSD1//','AARSD1
> //','//AARS
foo[[1:3]] asks for the 3rd member of the second component of the list
that is nested within the 1st component of the top level list. You
have no nested sublists, so this is clearly nonsense. You need to
re-read ?"[" or consult The Introduction to R or other online tutorial
to understand the seman
Dear all,
I have a list of arrays :
foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1))
> foo
$A
[1] 1 3
$B
[1] 1 2
$C
[1] 3 1
I want to use all foo$A , foo$B and foo$C in a test :
> foo$A[1] == 1
[1] TRUE
> foo[[1]][1] == 1
[1] TRUE
> foo[[1:3]][1] == 1
Error in foo[[1:3]] : recursive indexi
Duncan,
How embarrassing! Thanks.
-Ben
On 06/26/2014 10:25 PM, Duncan Murdoch wrote:
> On 27/06/2014, 4:08 AM, Benjamin Tyner wrote:
>> Hi
>>
>> I know that subset() is not intended for use in programming. However I
>> am still curious to learn why, in non-interactive mode, if I take away
>> the
On 27/06/2014, 4:08 AM, Benjamin Tyner wrote:
> Hi
>
> I know that subset() is not intended for use in programming. However I
> am still curious to learn why, in non-interactive mode, if I take away
> the quotes around 'bar'
>
>Rscript -e "foo <- list(bar = iris); head(subset(foo$'bar',
> Spe
Hi
I know that subset() is not intended for use in programming. However I
am still curious to learn why, in non-interactive mode, if I take away
the quotes around 'bar'
Rscript -e "foo <- list(bar = iris); head(subset(foo$'bar',
Species=='setosa'))"
Sepal.Length Sepal.Width Petal.Length P
expand works perfectly, thanks VERY much Ista!
-Seth
- Original Message -
From: Ista Zahn
To:
Cc:
Sent:Thu, 26 Jun 2014 14:35:19 -0400
Subject:Re: [R] decreasing blank space in ggplot2 geom_area
Hi Seth,
See the "expand" argument to ?discrete_scale
Best,
Ista
On Thu, Jun 26, 2014
Hi Duncan,
Thanks for trying to help. Sorry for not being clear.
The string I'd like to get is 'AARSD1'
It can be followed or preceded by white space or // or nothing
so, from test <- c('AARSD11','AARSD1-','AARSD1//','AARSD1
//','//AARSD1','AARSD1');
I want to match only 'AARSD1//','AARSD1 //',
Hi
You only have a vector of length 5 and I am not quite sure of the string you
are testing
so try this
grep('[/]*\\[/]*',test)
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
-Original Message
As Greg has listed lattice
Here are ways in lattice
quick 1 panel
library(lattice)
densityplot(~ mu1+mu2+mu3+mu4)
dat = data.frame(mu = c(mu1,mu2,mu3,mu4), gp = rep(1:4,
sapply(list(mu1,mu2,mu3,mu4), length)) )
densityplot(~ mu|gp, data = dat)
densityplot(~ mu|gp, dat, pch = "|")
see
?xyplot
?
On Jun 26, 2014, at 12:59 AM, Ron Crump wrote:
> Hi Carol,
>> It might be a primitive question but I have a file of text and there is no
>> separator between character on each line and the strings on each line have
>> the same length. The format is like the following
>>
>> absfjdslf
>> jfdldsk
On Jun 26, 2014, at 3:03 AM, FV wrote:
> Hi
>
> I have a 2D array and I would like to compute finite differences in each
> dimension!
> Is there any function on R that computes that automatically.
> thanks in advance
>
It doesn't appear that there is a matrix method for diff but why not these:
Dear R users,
I need to match a string. It can be followed or preceded by whitespace or // or
nothing.
How do I code it in R?
For example:
test <- c('AARSD11','AARSD1-','AARSD1//','AARSD1 //','//AARSD1');
grep('AARSD1(\\s*//*)',test);
should return 3,4,5 and 6.
Thanks in advance for your help.
On Thu, 26 Jun 2014 10:42:54 AM Mangalani Peter Makananisa wrote:
> Hi,
>
> I am try to fix the data frame and I see that we only have two fields
> defined characters and numeric fields. How do I add date field on
the
> same data after using the as.Date & as.yearmon and be able to
manipulat
This probably doesn't count as simple, but you can convert your format, call it
degrees.minutes, to decimal degrees, do the math and then convert back. The
functions would be simple:
dm2dd <- function(x) trunc(x) + (x - trunc(x))/60*100
dd2dm <- function(x) trunc(x) + (x - trunc(x))*60/100
So y
Hi,
On Thu, Jun 26, 2014 at 12:17 PM, VINCENT DEAN BOYCE
wrote:
> Hello,
>
> Using R, I've loaded a .cvs file comprised of several hundred rows and 3
> columns of data. The data within maps the output of a triaxial
> accelerometer, a sensor which measures an object's acceleration along the
> x,y
You could define a simple function to detect whether a value is within a
given range. For example,
inrange <- function(vec, range) {
!is.na(vec) & vec >= range[1] & vec <= range[2]
}
x <- 1:30
inrange(x, c(5, 20))
If you wanted to apply this function to all three columns at once, you
could use
Rook apps can be run using the httpd built into R (great feature!), and can
be deployed using rApache server.
For those who prefer to deploy the apps under FastRWeb, below is the code
to do so:
-- in your web.R/myrookapp.R file, we ran the Rook app instance (stored in
variable app):
run <- f
Hello,
I have been using the QQ-Plot functions in R for a while now and I noticed
what I believe is a major inconsistency in the way R estimates and plots
the quantiles of a data set. To see this, let x be a vector of data (or
just simulate some data) say 15 points. Then type
qqnorm(x)
By defaul
Hello,
Using R, I've loaded a .cvs file comprised of several hundred rows and 3
columns of data. The data within maps the output of a triaxial
accelerometer, a sensor which measures an object's acceleration along the
x,y and z axes. The data for each respective column sequentially
oscillates, and
Hi all,
I posted this question on Cross Validated a while ago but have never gotten an
answer.
I am trying to see if a discrete distribution will fit a family of zero
inflated distributions. I used fitdist from fitdistrplus and several gamlss
ZIP functions. Here is a descriptive stat of
Is there a simple way to subtract base-60 (coordinate system) values?
A trivial example:
I have a Lat of 44.1 degrees and I want to subtract 0.2 degrees from it.
Therefore the answer should be 43.5 degrees (base 60).
I can do a change to character; stringsplit ; change back - deal with the
leadi
Does this do what you want?
d1 <- density(mu1)
d2 <- density(mu2)
d3 <- density(mu3)
d4 <- density(mu4)
matplot( cbind( d1$x, d2$x, d3$x, d4$x ), cbind( d1$y, d2$y, d3$y,
d4$y ), type='l')
Or in a more expandable way:
mus <- mget( ls(pat='^mu') )
ds <- lapply( mus, density )
xs <- sapply( ds, `
You can do it, but mu1 has a much smaller variance and mu3 and mu4 are almost
identical so they overplot.
> xy1 <- density(mu1)
> xy2 <- density(mu2)
> xy3 <- density(mu3)
> xy4 <- density(mu4)
> Density <- cbind(xy1$y, xy2$y, xy3$y, xy4$y)
> x <- cbind(xy1$x, xy2$x, xy3$x, xy4$x)
> matplot(x, De
Hi Seth,
See the "expand" argument to ?discrete_scale
Best,
Ista
On Thu, Jun 26, 2014 at 7:01 AM, wrote:
> I wish to shrink the automatically inserted blank space at either end
> of the x axis of my area plot, so that the colorful graphic in the
> center takes up more of the available space. W
For your 2nd question (which also answers your first question) I use
the permn function in the combinat package, this function is nice that
in addition to generating all the permutations it will also,
optionally, run a function on each permutation for you:
> t(simplify2array( permn( c("A","B","C")
On 26/06/2014 16:57, Greg Snow wrote:
Ailan, Most of the readers of this list speak English and you may
have an easier time getting an answer if you translate to English
(though there may be some native Portuguese speakers on the list that
can help you directly). This is a text only list so ple
I wish to shrink the automatically inserted blank space at either end
of the x axis of my area plot, so that the colorful graphic in the
center takes up more of the available space. When I use the
scale_x_discrete(limits...) command to expand the displayed area, the
graphic shrinks away from the x-
Hi,
I am try to fix the data frame and I see that we only have two fields defined
characters and numeric fields. How do I add date field on the same data
after using the as.Date & as.yearmon and be able to manipulate the data on
the date field(s)?
Please advise
Kind regards
peter
Ple
Bom dia!
Estou tentando rodar o pacote rjava do R e não funciona. Muito do que li na
internet sugeriu atualização do Java. Entretanto, não estou conseguindo
executar o processo, sempre é interrompido. Segue abaixo o erro que aparece no
R.
Error : .onLoad falhou em loadNamespace() para 'rJava'
Ailan, Most of the readers of this list speak English and you may
have an easier time getting an answer if you translate to English
(though there may be some native Portuguese speakers on the list that
can help you directly). This is a text only list so please in the
future post in plain text, no
Hi
I have a 2D array and I would like to compute finite differences in each
dimension!
Is there any function on R that computes that automatically.
thanks in advance
Filipa
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
Hi Carol,
It might be a primitive question but I have a file of text and there is no
separator between character on each line and the strings on each line have the
same length. The format is like the following
absfjdslf
jfdldskjff
jfsldfjslk
When I read the file with read.table("myfile",colCl
Hello,
I am modelling in glmmADMB count data (I´m using a negative binomial
distribution to avoid possitive overdispersion) with four fixed and one
random effect. I´m also using MuMIn package to calculate the AICc and also
to model averaging using the function dredge. What I do not understand is
kindly guide me on how i can plot the following data on the same graph using
the kernel density. i will like to use as to compare performance
mu1<-c(500.0035, 501.2213, 500.7532, 500.2622, 500.3391, 500.1618, 499.9511,
500.1843, 499.8945, 499.8467)
mu2<-c(498.9623, 504.7938, 506.8957, 495.6634,
I am very new to R. I am using R package "pscl" for hurdle regression (Binary
with "cloglog" link and Poisson with "log" link). My problem is to model road
crash frequencies at given locations as a function of road geometry and surface
condition. All of my dependent variable are covariates. I wa
On Wed, 25 Jun 2014 14:16:08 -0700 (PDT)
Jeff Newmiller wrote:
> The brokenness of your perm.broken function arises from the attempted
> use of sapply to bind matrices together, which is not something
> sapply does.
>
> perm.fixed <- function( x ) {
>if ( length( x ) == 1 ) return( matrix( x
Bonjour,
Je suis absent jusqu'au 4 août prochain.
En cas d'urgence, merci de contacter Francis Meunier, directeur-adjoint :
f.meunier - at - conservatoirepicardie.org
J. Boutet
> From: r-help-requ...@r-project.org
> Subject: R-help Digest, Vol 136, Issue 26
> Date: Thu, 26 Jun 2014 12:0
On 26/06/14 21:21, carol white wrote:
Hi,
with read.fwf, it works.
But I still don't understand why it doesn't work with read.table
since the sep by default is "", which is the case and in one trial, I
used read.table("myfile",colClasses = "character",
stringsAsFactors=FALSE, and stil didn't
Carol,
while sep="" is the default, it really means 'whitespace', see the
documentation of 'sep'.
Göran Broström
On 2014-06-26 11:21, carol white wrote:
Hi,
with read.fwf, it works.
But I still don't understand why it doesn't work with read.table
since the sep by default is "", which is t
On 26/06/14 20:47, Frede Aakmann Tøgersen wrote:
Hi
Actually I had to read the man before answering Carol. Here it goes:
stringsAsFactors: logical: should character vectors be converted to
factors? Note that this is overridden by 'as.is' and
'colClasses', both of which al
Hi,
with read.fwf, it works.
But I still don't understand why it doesn't work with read.table since the sep
by default is "", which is the case and in one trial, I used
read.table("myfile",colClasses = "character", stringsAsFactors=FALSE, and stil
didn't work but it should have.
Regards,
Hi
Actually I had to read the man before answering Carol. Here it goes:
stringsAsFactors: logical: should character vectors be converted to
factors? Note that this is overridden by 'as.is' and
'colClasses', both of which allow finer control.
So setting colClasses should work
Hello,
Try using option stringsAsFactors = FALSE.
Hope this helps,
Rui Barradas
On 26/06/2014 09:32, carol white wrote:
It might be a primitive question but I have a file of text and there is no
separator between character on each line and the strings on each line have the
same length. The
On 26/06/14 19:32, carol white wrote:
It might be a primitive question
All questions are primitive; some questions are more primitive than others.
but I have a file of text and there
is no separator between character on each line and the strings on
each line have the same length. The format
Hi Carol
I cannot reproduce what you're seeing.
> tmp <- read.table(text = "absfjdslf
+ jfdldskjff
+ jfsldfjslk")
> str(tmp)
'data.frame': 3 obs. of 1 variable:
$ V1: Factor w/ 3 levels "absfjdslf","jfdldskjff",..: 1 2 3
> tmp <- read.table(text = "absfjdslf
+ jfdldskjff
+ jfsldfjslk",
+ co
It might be a primitive question but I have a file of text and there is no
separator between character on each line and the strings on each line have the
same length. The format is like the following
absfjdslf
jfdldskjff
jfsldfjslk
When I read the file with read.table("myfile",colClasses = "cha
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