Thank you. Will take look.
Sent from my iPhone
> On Aug 27, 2018, at 1:53 PM, John Kane wrote:
>
> A simple google would have let you here SAS Enterprise Guide Implemented at
> MDACC.
> It looks like you have a SAS transport file . Check out the SASxport package.
> It may do what you want.
>
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
Thanks Bert, I'll reference this repository to determine the firewall rules
we will need.
--
Tully Holmes
Business Applications Analyst
State of Wyoming
Wyoming Community College Commission
2300 Capitol Ave., 5th Floor, Suite B
Cheyenne, WY 82002
307-777-6832
On Mon, Aug 27, 2018 at 2:3
A simple google would have let you here SAS Enterprise Guide Implemented at
MDACC.It looks like you have a SAS transport file . Check out the SASxport
package. It may do what you want.
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|
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SAS Enterprise Guide Implemented at MDACC
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On Thursday, August 23, 2018, 4:39:45
A quick guess it that your version of R is outdated.sessionInfo()
R version 3.5.0
package ‘taskscheduleR’ was built under R version 3.5.1
I don't know if that is the source of the error but I'd suggest updating to
3.5.1 as a first step.
On Friday, August 24, 2018, 9:12:36 a.m. EDT, Christofer
The main CRAN repository is at:
https://cran.r-project.org/
A full list of repositories can be found under the "Mirrors" link there.
Cheers,
Bert
On Mon, Aug 27, 2018 at 1:19 PM Tully Holmes wrote:
> Good afternoon,
>
> I'm trying to install a package with the "install.packages" command in
>
Good afternoon,
I'm trying to install a package with the "install.packages" command in
RGUI, and get the following error message:
> install.packages ("tidyverse")
Warning: unable to access index for repository
https://mran.microsoft.com/snapshot/2017-05-01/src/contrib:
cannot open URL ''
Warni
Hi Ivan,
Unfortunately I cannot answer your question.
However, I do have quite a bit of experience using R's reference classes
and you might want to consider the more recent R6 package.
It provides R6 classes which have advantages over
reference classes. See for example:
1. Hadley Wickham on R6 (
Hi!
I'm trying to create a persistent memoising class with a destructor and
an option to evaluate cache misses in parallel. I want to lock all
its fields because it doesn't make sense to change them after the
filename, the environment object and the function are set in the
object. (I'm not sure wh
Thanks you Petr. This worked.
Regards, Shivi
Sent from Yahoo Mail for iPhone
On Monday, August 27, 2018, 14:31, PIKAL Petr wrote:
Hi
the output seems to me rather weird. Unles you have NaNs in input data frame
you should not get NaN as a result.
Anyway, your aggregate will give you NA or
Hi
the output seems to me rather weird. Unles you have NaNs in input data frame
you should not get NaN as a result.
Anyway, your aggregate will give you NA or NaN even when there is only one NA
or NaN in your input data frame. So I suggest to use
sentiments_per_Category <- aggregate(relative_s
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