gt;> On Behalf
Of Spencer Brackett
Sent: February 14, 2019 12:32 AM
To: R-help mailto:r-help@r-project.org>>; Sarah Goslee
mailto:sarah.gos...@gmail.com>>; Caitlin Gibbons
mailto:bioprogram...@gmail.com>>; Jeff Newmiller
mailto:jdnew...@dcn.davis.ca.us>>
Subject: [R] R
] 2 7 2NA
> [3,] 3 8 3NA
> [4,] 4 9 4NA
> [5,] 5 10 5NA
>
> -Original Message-
> From: R-help On Behalf Of Spencer Brackett
> Sent: February 14, 2019 12:32 AM
> To: R-help ; Sarah Goslee ;
> Caitlin Gibbons ; Jeff Newmiller <
> jdnew...@dcn.da
NA
[4,] 4 9 4NA
[5,] 5 10 5NA
-Original Message-
From: R-help On Behalf Of Spencer Brackett
Sent: February 14, 2019 12:32 AM
To: R-help ; Sarah Goslee ;
Caitlin Gibbons ; Jeff Newmiller
Subject: [R] R Data
Hello everyone,
The following is a portion of coding that a coll
Hello everyone,
The following is a portion of coding that a colleague sent. Given my lack
of experience in R, I am not quite sure what the significance of the
following arguments. Could anyone help me translate? For context, I am
aware of the downloading portion of the script... library(data.table
Thanks Bert, worked nicely. Yes, genotypes with only one ID will be
eliminated before partitioning the data.
Best regards
Ahmed Attia
On Mon, Aug 27, 2018 at 8:09 PM, Bert Gunter wrote:
> Just partition the unique stand_ID's and select on them using %in% , say:
>
> id <- unique(dataGenoty
Sorry, my bad -- careless reading: you need to do the partitioning within
genotype.
Something like:
by(dataGenotype, dataGenotype$Genotype, function(x){
u <- unique(x$standID)
tst <- x$x2 %in% sample(u, floor(length(u)/2))
list(test = x[tst,], train = x[!tst,]
})
This will give a
And yes, I ignored Genotype, but for the example data none of the stand_ID
values are present in more than one Genotype, so it doesn't matter. If that's
not true in general, then constructing the grp variable is a little more
complex, but the principle is the same.
--
Don MacQueen
Lawrence Live
You could start with split()
grp <- rep('', nrow(mydata) )
grp[mydata$stand_ID %in% c(7,9,67)] <- 'A-training'
grp[mydata$stand_ID %in% c(3,18,20,21,32)] <- 'B-testing'
split(mydata, grp)
or perhaps
grp <- ifelse( mydata$stand_ID %in% c(7,9,67) , 'A-training', 'B-testing' )
split(mydata, grp)
Just partition the unique stand_ID's and select on them using %in% , say:
id <- unique(dataGenotype$stand_ID)
tst <- sample(id, floor(length(id)/2))
wh <- dataGenotype$stand_ID %in% tst ## logical vector
test<- dataGenotype[wh,]
train <- dataGenotype[!wh,]
There are a million variations on this t
I would like to partition the following dataset (dataGenotype) based
on two variables; Genotype and stand_ID, for example, for Genotype
H13: stand_ID number 7 may go to training and stand_ID number 18 and
21 may go to testing.
Genotypestand_IDInventory_date stemC mheight
H13
To slightly correct what's been said: In general
lists are linear objects, but a list can have
dimension.
An example is in Circle 8.1.8 of 'The R Inferno'.
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
Pat
On 16/08/2012 21:50, Schumacher, Jay S wrote:
are these correct/accurate/sensib
On Thu, Aug 16, 2012 at 5:44 PM, MacQueen, Don wrote:
> Whereas for a matrix or data frame, one must supply *two* index values
> (even if one of them may be omitted)
> mydf[ 1 , 3 ]
> mydf[ , 5 ]
>
> mymat[ 2:5 , ]
> mymat[ 3 , 4:6 ]
> are valid statements.
>
Not quite:
x <- matr
I don't disagree with Michael, but I would add that to me it also depends.
If one thinks in terms of subsetting an object (for objects that can be
subsetted)
To subset a vector, one supplies *one* value for the index:
myvector[3]
myvector[ 2:5 ]
are valid statements.
Similarly for a list
m
On Aug 16, 2012, at 1:50 PM, Schumacher, Jay S wrote:
are these correct/accurate/sensible statements:
a vector is a one dimensional object.
a matrix is a two dimensional object.
a list is a one dimensional object.
i'm working from this web page:
http://www.agr.kuleuven.ac.be/vakken/s
yes, thank you, conceptual model (rather than formal dimension attribute)
is where i'm coming from at this point.
It would be helpful to distinguish between a formal dimension attribute,
and a (personal) conceptual model of whether or not any particu
On Thu, Aug 16, 2012 at 4:50 PM, Schumacher, Jay S wrote:
>
> are these correct/accurate/sensible statements:
>
> a vector is a one dimensional object.
> a matrix is a two dimensional object.
>
> a list is a one dimensional object.
>
> i'm working from this web page:
> http://www.agr.kul
are these correct/accurate/sensible statements:
a vector is a one dimensional object.
a matrix is a two dimensional object.
a list is a one dimensional object.
i'm working from this web page:
http://www.agr.kuleuven.ac.be/vakken/statisticsbyR/someDataStructures.htm
It would be helpful to distinguish between a formal dimension attribute,
and a (personal) conceptual model of whether or not any particular R
object, or type of object, has dimension. Mention of data frames having
dimension can be found in the help page for the dim() function.
> foo <- 1:10
> is.
On Aug 16, 2012, at 11:49 AM, Schumacher, Jay S wrote:
hi,
i'm trying to understand r data structures. i see that vectors,
matrix, factors and arrays have a "dimension."
there seems to be no mention of dimensionality anywhere for lists
or dataframes. can i consider lists and frames to
Hi,
On Thu, Aug 16, 2012 at 2:49 PM, Schumacher, Jay S wrote:
>
>
> hi,
> i'm trying to understand r data structures. i see that vectors, matrix,
> factors and arrays have a "dimension."
Out of curiosity, where do you "see" that vectors and factors have a
dimension? I mean -- I guess they're
hi,
i'm trying to understand r data structures. i see that vectors, matrix,
factors and arrays have a "dimension."
there seems to be no mention of dimensionality anywhere for lists or
dataframes. can i consider lists and frames to be of fixed dimension 2?
thanks,
jay s
_
I should note that the form is compiling in C# but upon clicking the button
nothing happens to the dataGridView.
--
View this message in context:
http://r.789695.n4.nabble.com/R-Data-Frame-to-C-DataGridView-R-NET-tp3929413p3929415.html
Sent from the R help mailing list archive at Nabble.com.
___
This is more of a C# question, but it has to do with the conversion of a data
frame using R.NET into an appropriate format to display on a C#
DataGridView.
My code is as follows (focusing on the button_Click)
/private void button_displayDF_Click(object sender, EventArgs e)
{
On Thu, Feb 3, 2011 at 10:05 PM, Peter Ehlers wrote:
> On 2011-02-03 14:09, Dennis Murphy wrote:
>>
>> Hi:
>>
>> This also works, except that the result is of class 'table':
>>
>> xtabs(COUNTS ~ USER_ID + SITE, data = RAW)
>> SITE
>> USER_ID SITE1 SITE2 SITE3
>> 1 10 13 22
>>
On 2011-02-03 14:09, Dennis Murphy wrote:
Hi:
This also works, except that the result is of class 'table':
xtabs(COUNTS ~ USER_ID + SITE, data = RAW)
SITE
USER_ID SITE1 SITE2 SITE3
1101322
2101212
31344 0
4 0 09
Hi:
This also works, except that the result is of class 'table':
xtabs(COUNTS ~ USER_ID + SITE, data = RAW)
SITE
USER_ID SITE1 SITE2 SITE3
1101322
2101212
31344 0
4 0 099
If you need a data frame, then the earlier res
Hi,
On Thu, Feb 3, 2011 at 2:41 PM, Mike Schumacher
wrote:
> Hello,
>
> I'd like to transpose data to create an analysis-friendly dataframe. See
> below for an example, I was unable to use t(x) and I couldn't find a
> function with options like PROC TRANSPOSE in SAS.
>
> The ideal solution handl
Hi Mike
reshape will be your friend.
reshape(RAW,direction="wide",timevar="SITE",idvar="USER_ID")
there is also the 'reshape'-package, which can do some more
sophisticated transformations.
hth.
Am 03.02.2011 20:41, schrieb Mike Schumacher:
> Hello,
>
> I'd like to transpose data to create an
Hello,
I'd like to transpose data to create an analysis-friendly dataframe. See
below for an example, I was unable to use t(x) and I couldn't find a
function with options like PROC TRANSPOSE in SAS.
The ideal solution handles variable quantities of SITE - but beggars can't
be choosers. :-)
Tha
If I understand your problem correctly, I think you need to be doing:
summary(data.name)
The functions read.dta and read.spss both return things (data frames,
if you use the to.data.frame=T argument with read.spss). So whatever
variable you set is what you should be doing a summary on. In this
Hi Alla,
The error message suggests that "kelleya" is a function. What happens
when you type: str(kelleya) ?
Josh
On Sat, Oct 2, 2010 at 7:01 PM, Alla Manukyan wrote:
> Dear Sir/Madam,
> I have just installed R for Windows. I am trying to open a stata file and I
> have
> a problem. I have
Dear Sir/Madam,
I have just installed R for Windows. I am trying to open a stata file and I
have
a problem. I have used the following commands:
> install.packages("foreign")
> library(foreign)
> data.name <- read.dta(file.choose())
# then I choose a kelleya.dta file and click on open and
It works perfectly !!!
Thank you so much !!
Angelo
-Messaggio originale-
Da: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Inviato: giovedì 6 maggio 2010 12.05
A: LINARDI ANGELO
Cc: r-help@r-project.org
Oggetto: Re: [R] Data frame "pivoting"
Hi Angelo,
try
x <- structure(li
The other respondants here have missed the point - write.table() as
invoked by the OP does not produce a csv file. The default separator is
" " (a space) in write.table(), so no wonder Excel cocked it up as it
assumes that a csv file is one that has elements separated by a comma
",". We don't need
>
> it doesn't seem indicate any installation instruction.
> http://cran.r-project.org/web/packages/xlsReadWrite/index.html
As Wensui Liu already mentioned, installation is the same as with other
packages: In the RGui menu select Packages->Install Packages...->(Select a
CRAN mirror)->(Select the
i think you simply install it in the way you install other R packages.
On Sun, Mar 2, 2008 at 10:40 PM, Christophe Lo <[EMAIL PROTECTED]> wrote:
> thanks for your response. How do i install it? I try looking at the manual
> it doesn't seem indicate any installation instruction. I also download a
>
thanks for your response. How do i install it? I try looking at the manual
it doesn't seem indicate any installation instruction. I also download a
windows version but it doesn't have an exe file.
http://cran.r-project.org/web/packages/xlsReadWrite/index.html
Newbie,
Kei
On 3/3/08, Wensui Liu <
hi,
did you try write.xls in xlsReadWrite package?
On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 <[EMAIL PROTECTED]> wrote:
>
> Here is my R Code
>
> x<-1:2
> y<-2:141
> data.matrix<-data.matrix(data[,y])#create data.matrix
> variableprobe<-apply(data.matrix[x,],1,var)
> variableprobe #outp
If you are asking how to convert to multiple columns in Excel, look at
the "text to column" option in I think the data tab.
On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 <[EMAIL PROTECTED]> wrote:
>
> Here is my R Code
>
> x<-1:2
> y<-2:141
> data.matrix<-data.matrix(data[,y])#create data.matrix
>
Here is my R Code
x<-1:2
y<-2:141
data.matrix<-data.matrix(data[,y])#create data.matrix
variableprobe<-apply(data.matrix[x,],1,var)
variableprobe #output variance across probesets
hist(variableprobe) #displaying histogram of variableprobe
write.table(cbind(data[1],
Variance=apply(data[,y],1,
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