MKclass::perfMeasures(predict_testing, truth = testing$case, namePos = 1)
should also work and computes 80 performance measures.
Best Matthias
Am 25.10.22 um 06:42 schrieb Jin Li:
Hi Greg,
This can be done by:
spm::pred.acc(testing$case, predict_testing)
It will return both sensitivity and
Hi Greg,
This can be done by:
spm::pred.acc(testing$case, predict_testing)
It will return both sensitivity and specificity, along with a few other
commonly used measures.
Hope this helps,
Jin
On Tue, Oct 25, 2022 at 6:01 AM Rui Barradas wrote:
> Às 16:50 de 24/10/2022, greg holly escreveu:
>
Thanks to all, who have helped greatly. I essentially followed Rui to do:
fmt_string<-paste0("\ntol = %.1e","\nreltol = %.1e","\nsteptol =
%.1e","\ngradtol = %.1e")
#msg<-sprintf(fmt_string,mycontrol$tol,mycontrol$reltol,mycontrol$steptol,mycontrol$gradtol)
#works
msg<-with(mycontrol,sp
Às 16:21 de 24/10/2022, Steven T. Yen escreveu:
Thanks to everyone. I read ? sprint and the following is best I came up
with. If there are ways to collapse the lines I'd be glad to know.
Otherwise, I will live with this. Thanks again.
cat(sprintf("\ntol = %e",mycontrol$tol),
sprintf("
Às 16:50 de 24/10/2022, greg holly escreveu:
Hi Michael,
I appreciate your writing. Here are what I have after;
predict_testing <- ifelse(predict > 0.5,1,0)
head(predict)
1 2 3 5 7 8
0.29006984 0.28370507 0.10761993 0.02204224 0.12873872
Andrew,
Thanks. I reviewed the code for "require" and saw:
"if (!character.only)
package <- as.character(substitute(package))"
#This helps me better understand what is going on. I am sharing this
here because I think it might help others understand.
as.character( substitute("this_pac_does_
THanks Michael for this.This is much appreciated. So, how can I estimate
the sensitivity and specificity after having the prediction on testing
data. Any thoughts?
Kind regards,
Greg
On Mon, Oct 24, 2022 at 12:10 PM Michael Dewey
wrote:
> So predict is a one-dimensional vector of predictions
In the first one, the argument is a character vector of length 1, so the
code works perfectly fine.
The second is a call, and when coerced to a character vector should look
like
c("[", "packages_i_want_to_use", "1")
You can try this yourself with quote(packages_i_want_to_use[1]) which
returns it
I agree that the documentation should be clarified. Moreover, my last example
shows that the class can be different even when no mode coercion is required. I
don't know enough about S3 & S4 to comment on your last point.
Regards,
Jorgen Harmse.
From: Bert Gunter
Date: Monday, 24October, 2022
So predict is a one-dimensional vector of predictions but you are
treating it as a two-dimensional matrix (presumably you think those are
the totals).
Michael
On 24/10/2022 16:50, greg holly wrote:
Hi Michael,
I appreciate your writing. Here are what I have after;
> predict_testing <- ifel
I wanted to follow up.
A more careful reading of the following:
"A vector of the same length and attributes (including dimensions and
"class") as test..."
So the above **refers only to a "class" attribute that appears among
the attributes of test and result**. Using my previous example, note
that:
Thanks!
# Please, can you help me understand why
require( 'base' ) # works, but
require( packages_i_want_to_use[1] ) # does not work?
# In require( 'base' ), what is the "first argument"?
On Mon, Oct 24, 2022 at 12:29 PM Andrew Simmons wrote:
>
> require(), similarly to library(), does not eval
Don't load base. It is already loaded.
On October 24, 2022 9:07:44 AM PDT, Kelly Thompson wrote:
># Below, when using require(), why do I get the error message "Error
>in if (!loaded) { : the condition has length > 1" ?
>
># This is my reproducible code:
>
>#create a vector with the names of the
Reread ?require more carefully. Especially for the 'package' argument.
Incidentally, you don't need to require base. It's always available.
-- Bert
On Mon, Oct 24, 2022 at 9:25 AM Kelly Thompson wrote:
>
> # Below, when using require(), why do I get the error message "Error
> in if (!loaded) {
You need to pass character.only = TRUE to require() whenever you
specify the package using a character variable.
I agree, the error message is confusing.
/Henrik
On Mon, Oct 24, 2022 at 9:26 AM Kelly Thompson wrote:
>
> # Below, when using require(), why do I get the error message "Error
> in i
require(), similarly to library(), does not evaluate its first argument
UNLESS you add character.only = TRUE
require( packages_i_want_to_use[1], character.only = TRUE)
On Mon, Oct 24, 2022, 12:26 Kelly Thompson wrote:
> # Below, when using require(), why do I get the error message "Error
> in
В Mon, 24 Oct 2022 12:07:44 -0400
Kelly Thompson пишет:
> require( packages_i_want_to_use[1] )
> #Error in if (!loaded) { : the condition has length > 1
This seems to be a bug in require(). In addition to understanding
character strings as arguments, require() can load packages named by
unquoted
"...but 'same length and attributes (including dimensions and
‘"class"’) as ‘test’' looks wrong. The output seems to be `logical` or
something related to the classes of `yes` & `no`."
The documentation in fact says:
"A vector of the same length and attributes (including dimensions and
"class") as
# Below, when using require(), why do I get the error message "Error
in if (!loaded) { : the condition has length > 1" ?
# This is my reproducible code:
#create a vector with the names of the packages I want to use
packages_i_want_to_use <- c('base', 'this_pac_does_not_exist')
# Here I get error
Rather hard to know without seeing what output you expected and what
error message you got if any but did you mean to summarise your variable
predict before doing anything with it?
Michael
On 24/10/2022 16:17, greg holly wrote:
Hi all R-Help ,
After partitioning my data to testing and traini
Hi Michael,
I appreciate your writing. Here are what I have after;
> predict_testing <- ifelse(predict > 0.5,1,0)
>
> head(predict)
1 2 3 5 7 8
0.29006984 0.28370507 0.10761993 0.02204224 0.12873872 0.08127920
>
> # Sensitivity and Specificity
"collapse the lines" means ??
If you mean that you want to control the precision (# of decimals
places to show) then that is exactly what sprintf does. ?sprintf tells
you how. If you mean something else, please specify more clearly -- or
await a reply from someone with greater insight than I.
--
There were several interesting points about `ifelse`. The usual behaviour seems
to be that all three inputs are evaluated, and the entries of `yes`
corresponding to `TRUE` in `test` are combined with the entries of `no`
corresponding to `FALSE` in `test`. Moreover, `yes` & `no` seem to be recycl
Thanks to everyone. I read ? sprint and the following is best I came up
with. If there are ways to collapse the lines I'd be glad to know.
Otherwise, I will live with this. Thanks again.
cat(sprintf("\ntol = %e",mycontrol$tol),
sprintf("\nreltol = %e",mycontrol$reltol),
sprintf("\n
Hi all R-Help ,
After partitioning my data to testing and training (please see below), I
need to estimate the Sensitivity and Specificity. I failed. It would be
appropriate to get your help.
Best regards,
Greg
inTrain <- createDataPartition(y=data$case,
p=0.7,
Hello,
There's also ?message.
msg <- sprintf("(tol,reltol,steptol,gradtol): %E %E %E %E",
mycontrol$tol,mycontrol$reltol,mycontrol$steptol,mycontrol$gradtol)
message(msg)
Hope this helps,
Rui Barradas
Às 14:25 de 24/10/2022, Steven T. Yen escreveu:
Thank, Boris and Ivan.
The simple comma
On 10/24/22 7:39 AM, Steven T. Yen wrote:
I have a "list" containing four elements, as shown below:
> t(mycontrol)
tol reltol steptol gradtol
[1,] 0 0 1e-08 1e-12
Printing this in a main program causes no problem (as shown above).
But, using the command t(mycontrol) the line
Thank, Boris and Ivan.
The simple command suggested by Ivan ( print(t(mycontrol)) ) worked. I
went along with Boris' suggestion and do/get the following:
cat(sprintf("(tol,reltol,steptol,gradtol): %E %E %E %E",mycontrol$tol,
mycontrol$reltol,mycontrol$steptol,mycontrol$gradtol))
(tol,reltol,s
??? t() is the transpose function. It just happens to return your list
unchanged. The return value is then printed to console if it is not assigned,
or returned invisibly. Transposing your list is probably not what you wanted to
do.
Returned values do not get printed from within a loop or from
В Mon, 24 Oct 2022 20:39:33 +0800
"Steven T. Yen" пишет:
> Printing this in a main program causes no problem (as shown above).
> But, using the command t(mycontrol) the line gets ignored.
t() doesn't print, it returns a value. In R, there's auto-printing in
the toplevel context (see ?withAutopri
I have a "list" containing four elements, as shown below:
> t(mycontrol)
tol reltol steptol gradtol
[1,] 0 0 1e-08 1e-12
Printing this in a main program causes no problem (as shown above).
But, using the command t(mycontrol) the line gets ignored. Any idea? Thanks.
Steven Yen
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