It helps. But it is overly sophisticated.
I have already downloaded and used the Excel file containing the validated
stuff.
Since there are R commands to download gzip as well as FASTA files, I wonder
whether it is possible to
automatically download the Excel file from
http://mirecords.umn.edu
Hi Malcolm,
You need to tell geom_boxplot not to use stat_boxplot:
geom_boxplot(aes(lower=y_q1, upper=y_q3, middle=y_med, ymin=y_min,
ymax=y_max), stat = "identity")
Hadley
On Mon, Jul 6, 2009 at 6:55 AM, Malcolm Ryan wrote:
> Is there anyway in ggplot2 to set the aesthetics for a geom_boxplot
>
Is there anyway in ggplot2 to set the aesthetics for a geom_boxplot
directly, rather than having them computed by an implicit stat_boxplot?
If I try:
ggplot(data = t, aes(x = factor(x))) + geom_boxplot(coef=NULL,
aes(lower=y_q1, upper=y_q3, middle=y_med, ymin=y_min, ymax=y_max))
I get the
You can also derive what you need using the rle()
function, though the which() solution may be
easier.
-Don
At 4:20 PM -0300 7/5/09, Henrique Dallazuanna wrote:
Content-Type: text/plain
Content-Disposition: inline
Content-length: 944
Try this:
which(c(0, diff(x)) != 0)
On Sun, Jul 5, 2009
At 10:42 AM -0700 7/5/09, Mark Knecht wrote:
2009/7/5 Uwe Ligges :
<- a lot of other conversation omitted, to focus on the following>
Currently my data is one experiment per row, but that's wasting space
as most experiments only take 20% of the row and 80% of the row is
filled with 0's. I
Hi,
Thank you for using Rdonlp2.
2009/7/4 Lars Bishop :
> Error in tryCatchList(expr, classes, parentenv, handlers) :
> SET_VECTOR_ELT() can only be applied to a 'list', not a 'character'
I guess you are using an older version of Rdonlp2 that is incompatible
to R-2.9.0. I have fixed the "SET_V
On Sun, Jul 5, 2009 at 1:44 PM, hadley wickham wrote:
>> I think the root cause of a number of my coding problems in R right
>> now is my lack of skills in reading and grabbing portions of the data
>> out of arrays. I'm new at this. (And not a programmer) I need to find
>> some good examples to r
Try,
> library("fGarch")
> fitmodel = garchFit(formula=~ar(1)+~garch(1,1), data=garat)
> fitmo...@fit$ics #is a vector with Information Criterion Statistics
> fitmo...@fit$ics[1] #is the AIC
Murilo Eiji Doi
Ben Bolker wrote:
>
> Vasileios Ismyrlis gmail.com> writes:
>
>>
>> library(
Try,
> library("fGarch")
> fitmodel = garchFit(formula=~ar(1)+~garch(1,1), data=garat)
> fitmo...@fit$ics #is a vector with Information Criterion Statistics
> fitmo...@fit$ics[1] #is the AIC
Ben Bolker wrote:
>
> Vasileios Ismyrlis gmail.com> writes:
>
>>
>> library("fGarch")
>> fit = garch
> I think the root cause of a number of my coding problems in R right
> now is my lack of skills in reading and grabbing portions of the data
> out of arrays. I'm new at this. (And not a programmer) I need to find
> some good examples to read and test on that subject. If I could locate
> which co
We’re very pleased to announce another update to
http://crantastic.org/, the community site for R packages. Notable new
features include:
* More package relationships. We now have enhances, imports, and
reverse relationships.
* Easier rating: just log-in, and click the number of stars you
On Sun, Jul 5, 2009 at 1:00 PM, Henrique Dallazuanna wrote:
> Yes,
>
> First, select only columns C1 to C6, then look for values greater than 0,
> after use this to select the columns in original subset.
>
> On Sun, Jul 5, 2009 at 4:48 PM, Mark Knecht wrote:
>>
>> On Sun, Jul 5, 2009 at 12:30 PM,
Fala ai cara!
Era o RMySQL mesmo q eu precisava, vlw
Bernardo Rangel tura wrote:
>
> On Sat, 2009-07-04 at 22:24 -0700, JoK LoQ wrote:
>> I'm dealing with lots of columns and conditions, wats t best way to deal
>> with
>> that?
>> How do I work with SQL on R? the manual is quite confuse talki
On Sun, 05-Jul-2009 at 12:10PM -0700, megh wrote:
|>
|> Suppose, I have following
|>
|> vec <- vector("list", length=3)
|> for (i in 1:3) vec[[i]] <- matrix(rnorm((i+3)*2), (i+3))
|> vec
|>
do.call("rbind", vec)
|>
|> Now I want to change the type of "vec" from list to a matrix with (4+5
Yes,
First, select only columns C1 to C6, then look for values greater than 0,
after use this to select the columns in original subset.
On Sun, Jul 5, 2009 at 4:48 PM, Mark Knecht wrote:
> On Sun, Jul 5, 2009 at 12:30 PM, Henrique Dallazuanna
> wrote:
> > Try this:
> >
> > subset(test[3,], sele
On Sun, Jul 5, 2009 at 12:30 PM, Henrique Dallazuanna wrote:
> Try this:
>
> subset(test[3,], select=C1:C6)[,subset(test[3,], select = C1:C6) > 0]
>
> subset(test[6,], select=C1:C6)[,subset(test[6,], select = C1:C6) > 0]
>
>
I must admit I like this one. Pleasing to look at. It seems
approachable.
mau...@alice.it wrote:
> I tried to apply the scheme you suggested to open the web page on
> "http://mirecords.umn.edu/miRecords/index.php"; and got the followiing:
>
>> result <- postForm("http://mirecords.umn.edu/miRecords/index.php";,
> + searchType="miRNA", species="Homo sapiens",
> + searchB
Try this:
subset(test[3,], select=C1:C6)[,subset(test[3,], select = C1:C6) > 0]
subset(test[6,], select=C1:C6)[,subset(test[6,], select = C1:C6) > 0]
On Sun, Jul 5, 2009 at 1:19 PM, Mark Knecht wrote:
> On Sun, Jul 5, 2009 at 8:18 AM, David Winsemius
> wrote:
> >
> > On Jul 5, 2009, at 10:50
Try this:
which(c(0, diff(x)) != 0)
On Sun, Jul 5, 2009 at 4:03 PM, Arun Kumar Saha
wrote:
> Hi all,
>
> say, I have following vector :
>
> x <- c(rep(5, 5), rep(3,4), rep(5,10))
>
> Now I want to get the index numbers where elements of that vector changes
> i.e. in above example I want to get
Try:
do.call(rbind, vec)
On Sun, Jul 5, 2009 at 4:10 PM, megh wrote:
>
> Suppose, I have following
>
> vec <- vector("list", length=3)
> for (i in 1:3) vec[[i]] <- matrix(rnorm((i+3)*2), (i+3))
> vec
>
>
> Now I want to change the type of "vec" from list to a matrix with (4+5+6)
> rows and 2 co
I tried to apply the scheme you suggested to open the web page on
"http://mirecords.umn.edu/miRecords/index.php"; and got the followiing:
> result <- postForm("http://mirecords.umn.edu/miRecords/index.php";,
+ searchType="miRNA", species="Homo sapiens",
+ searchBox="hsa-let-7a", submitButton="Se
Suppose, I have following
vec <- vector("list", length=3)
for (i in 1:3) vec[[i]] <- matrix(rnorm((i+3)*2), (i+3))
vec
Now I want to change the type of "vec" from list to a matrix with (4+5+6)
rows and 2 columns. How can I do that?
Thanks and regards,
--
View this message in context:
http:/
Hi all,
say, I have following vector :
x <- c(rep(5, 5), rep(3,4), rep(5,10))
Now I want to get the index numbers where elements of that vector changes
i.e. in above example I want to get a vector with elements : 6, 10. Because
at that indices, element of original vector changes value.
Is there
On Jul 5, 2009, at 2:43 PM, Martin Morgan wrote:
Hi ZRL,
zrl wrote:
Dear list,
I have a question regarding how to allocate or initialize the
proper type of
the arrays or lists.
I have a for loop, each iteration, I want it to assign a subset of
a data
frame to an element of an array or
David Winsemius wrote:
On Jul 5, 2009, at 1:19 PM, Uwe Ligges wrote:
snippedpreample
test <- data.frame(A=1:10, B=100, C1=runif(10), C2=runif(10),
C3=runif(10), C4=runif(10), C5=runif(10), C6=runif(10))
test<-round(test,2)
#Make array ragged
test$C3[2]<-0;test$C4[2]<-0;test$C5[2]<-0;test$C
Hi ZRL,
zrl wrote:
> Dear list,
>
> I have a question regarding how to allocate or initialize the proper type of
> the arrays or lists.
>
> I have a for loop, each iteration, I want it to assign a subset of a data
> frame to an element of an array or list.
> However, I am wondering how I should
On Jul 5, 2009, at 1:19 PM, Uwe Ligges wrote:
snippedpreample
test <- data.frame(A=1:10, B=100, C1=runif(10), C2=runif(10),
C3=runif(10), C4=runif(10), C5=runif(10), C6=runif(10))
test<-round(test,2)
#Make array ragged
test$C3[2]<-0;test$C4[2]<-0;test$C5[2]<-0;test$C6[2]<-0
test$C4[3]<-0;test
Try
mylist <- NULL
and then just assign the subsets to each element of the list
mylist[[1]] <- data1[,1]
as an example
--- On Sun, 7/5/09, zrl wrote:
> From: zrl
> Subject: [R] allocation/initialization of arrays/lists
> To: r-h...@stat.math.ethz.ch
> Received: Sunday, July 5, 2009, 1:58 P
Dear list,
I have a question regarding how to allocate or initialize the proper type of
the arrays or lists.
I have a for loop, each iteration, I want it to assign a subset of a data
frame to an element of an array or list.
However, I am wondering how I should initialize or allocate the empty arr
2009/7/5 Uwe Ligges :
>
>
> David Winsemius wrote:
>>
>> On Jul 5, 2009, at 12:19 PM, Mark Knecht wrote:
>>
>>> On Sun, Jul 5, 2009 at 8:18 AM, David Winsemius
>>> wrote:
On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
>
>
> David Winsemius wrote:
>>
>> So if your
David Winsemius wrote:
On Jul 5, 2009, at 12:19 PM, Mark Knecht wrote:
On Sun, Jul 5, 2009 at 8:18 AM, David
Winsemius wrote:
On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
David Winsemius wrote:
So if your values are calculated from other values then consider using
all.equal()
And r
On Jul 5, 2009, at 12:19 PM, Mark Knecht wrote:
On Sun, Jul 5, 2009 at 8:18 AM, David
Winsemius wrote:
On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
David Winsemius wrote:
So if your values are calculated from other values then consider
using
all.equal()
And repeated applications o
On Sun, Jul 5, 2009 at 8:18 AM, David Winsemius wrote:
>
> On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
>
>>
>>
>> David Winsemius wrote:
>>>
>>> So if your values are calculated from other values then consider using
>>> all.equal()
>>> And repeated applications of the testing criteria process ar
On Sun, Jul 5, 2009 at 3:40 PM, hadley wickham wrote:
>> 2) Related to the above, how do I tell what packages are currently
>> loaded at any given time so that I don't waste time loading things
>> that are already loaded? search() tells me what's available, but
>> what's loaded? The best I can find
Thanks very much Gabor
Bill
On Sun, Jul 5, 2009 at 4:44 PM, Gabor
Grothendieck wrote:
> In terms of x this is:
>
> k <- 3
> x <- 1:5
>
> n <- length(x)
> matrix(x, 2*n-1, k)[1:n, ]
>
> On Sun, Jul 5, 2009 at 10:01 AM, Gabor
> Grothendieck wrote:
>> Try this (ignore the warning):
>>
>> k <- 3
>> m
In terms of x this is:
k <- 3
x <- 1:5
n <- length(x)
matrix(x, 2*n-1, k)[1:n, ]
On Sun, Jul 5, 2009 at 10:01 AM, Gabor
Grothendieck wrote:
> Try this (ignore the warning):
>
> k <- 3
> matrix(1:5, 9, k)[1:5, ]
>
>
> On Sun, Jul 5, 2009 at 8:54 AM, William
> Simpson wrote:
>> I want a function t
On Jul 5, 2009, at 10:50 AM, Uwe Ligges wrote:
David Winsemius wrote:
So if your values are calculated from other values then consider
using all.equal()
And repeated applications of the testing criteria process are
effective:
test[3,][which(names(test)=="C1"):(which(test[3,] == 0.0)-1)
On Sun, Jul 5, 2009 at 7:35 AM, David Winsemius wrote:
>
> On Jul 5, 2009, at 9:53 AM, Mark Knecht wrote:
>
>> On Sat, Jul 4, 2009 at 5:22 PM, jim holtman wrote:
>>>
>>> See if this example helps; show how to either plot the row or columns
>>> of a data frame:
>>>
test <- data.frame(C1=runif(1
David Winsemius wrote:
On Jul 5, 2009, at 9:53 AM, Mark Knecht wrote:
On Sat, Jul 4, 2009 at 5:22 PM, jim holtman wrote:
See if this example helps; show how to either plot the row or columns
of a data frame:
test <- data.frame(C1=runif(10), C2=runif(10), C3=runif(10))
test
C1
Also make sure to check roxygen (from roxygen.org) - it makes package
documentation much much easier. Ironically, the documentation for
roxygen currently leaves something to be desired but I think Peter and
Manuel are working on it.
Hadley
On Sat, Jul 4, 2009 at 3:59 PM, Jason Rupert wrote:
>
>
Try this:
foo <- function(x, ncols){
tail(embed(rep(x, ncols), ncols), length(x))
}
foo(x, 3)
On Sun, Jul 5, 2009 at 9:54 AM, William Simpson wrote:
> I want a function that takes an input vector, the number of columns
> and returns a matrix as follows.
>
> x<- 1:5
>
> foo(x, nc=3)
>
> 1 5
> 2) Related to the above, how do I tell what packages are currently
> loaded at any given time so that I don't waste time loading things
> that are already loaded? search() tells me what's available, but
> what's loaded? The best I can find so far goes like this:
Loading something a second time t
On Jul 5, 2009, at 9:53 AM, Mark Knecht wrote:
On Sat, Jul 4, 2009 at 5:22 PM, jim holtman wrote:
See if this example helps; show how to either plot the row or columns
of a data frame:
test <- data.frame(C1=runif(10), C2=runif(10), C3=runif(10))
test
C1C2C3
1 0.91
Try this (ignore the warning):
k <- 3
matrix(1:5, 9, k)[1:5, ]
On Sun, Jul 5, 2009 at 8:54 AM, William
Simpson wrote:
> I want a function that takes an input vector, the number of columns
> and returns a matrix as follows.
>
> x<- 1:5
>
> foo(x, nc=3)
>
> 1 5 4
> 2 1 5
> 3 2 1
> 4 3 2
> 5 4 3
>
On Sun, Jul 5, 2009 at 12:00 AM, Mark Wardle wrote:
> Hi. Essentially your data is currently in "wide" format, with repeated
> measures in different columns. For most analysis and in particular for
> graphing, it is frequently helpful to reshape your data into a "long"
> format, with one row per da
On Sat, Jul 4, 2009 at 5:22 PM, jim holtman wrote:
> See if this example helps; show how to either plot the row or columns
> of a data frame:
>
>> test <- data.frame(C1=runif(10), C2=runif(10), C3=runif(10))
>> test
> C1 C2 C3
> 1 0.91287592 0.3390729 0.4346595
> 2 0.29360
On Jul 5, 2009, at 8:54 AM, William Simpson wrote:
I want a function that takes an input vector, the number of columns
and returns a matrix as follows.
x<- 1:5
foo(x, nc=3)
1 5 4
2 1 5
3 2 1
4 3 2
5 4 3
See if this gives you any ideas:
sapply(1:3, function(z) { ((x - z) %% 5) +1 } )
Da
On Jul 4, 2009, at 9:22 PM, nyk wrote:
I have a data matrix containing quite a lot of missing values (NA).
I know
how to remove all column or rows containing NA values, but is there
a some
standard method for removing not all NA containing rows/column, but
only
those which have signific
I want a function that takes an input vector, the number of columns
and returns a matrix as follows.
x<- 1:5
foo(x, nc=3)
1 5 4
2 1 5
3 2 1
4 3 2
5 4 3
Thanks again for any help.
Bill
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Thanks everyone for your help!
Bill
__
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, reproducib
On Sat, 2009-07-04 at 22:24 -0700, JoK LoQ wrote:
> I'm dealing with lots of columns and conditions, wats t best way to deal with
> that?
> How do I work with SQL on R? the manual is quite confuse talking about that.
> Do I need a package?
I don't understand your question, but
if you think use S
I'm dealing with lots of columns and conditions, wats t best way to deal with
that?
How do I work with SQL on R? the manual is quite confuse talking about that.
Do I need a package?
--
View this message in context:
http://www.nabble.com/SQL-on-R-tp24340314p24340314.html
Sent from the R help mail
Ben
that subset(..) works, and it solves all my problems. thnks!
Ben Bolker wrote:
>
>
>
> JoK LoQ wrote:
>>
>> Just a quickly beginner's question.
>>
>> I wanna find the mean only from the values from a column related to
>> specific values from another one. Like, theres a 'region' colu
I have a data matrix containing quite a lot of missing values (NA). I know
how to remove all column or rows containing NA values, but is there a some
standard method for removing not all NA containing rows/column, but only
those which have significantly more NAs than others?
--
View this message
2009/7/2 Chris Anderson :
> Is there a way to export multiple datasets into multiple excel spreadsheets
> using xlsReadWritePro or with xlsReadWrite?
I have answered the question in your other thread. See:
http://thread.gmane.org/gmane.comp.lang.r.general/154614/focus=154752
Cheers,
Hans-Peter
> what happened to the xlsReadWrite package?
[snip]
> Are there any plans for it be available again?
Yes. (I do have a halfway finished version but then had much work on
my regular job, my 'free-programming-time' got eaten by a e-texteditor
plugin (http://github.com/swissr/rmate.tmbundle/tree/mast
Hi David,
It looks like we will be in minority this time around.
See you there,
Tal
On Wed, Jul 1, 2009 at 8:04 PM, David M Smith <
da...@revolution-computing.com> wrote:
> I'm going to both UseR! (in Rennes) and DSC (in Copenhagen), and will
> be blogging about the talks and other interest
Hi. Essentially your data is currently in "wide" format, with repeated
measures in different columns. For most analysis and in particular for
graphing, it is frequently helpful to reshape your data into a "long"
format, with one row per data value and additional variables to list
experiment or subj
58 matches
Mail list logo
