cause of things like character
strings that can be evaluated as numeric but may not be evaluated correctly if
they are written in some way as in hexadecimal notation or scientific notation.
-Original Message-
From: R-help On Behalf Of Adrian Dusa
Sent: Tuesday, June 25, 2024 4:59 AM
To: Mar
Oh I see...
It's not that "-7" gets coerced to numeric, but 0 gets coerced to "0".
Of course...
On Tue, Jun 25, 2024 at 11:02 AM Martin Maechler
wrote:
> > Adrian Dusa
> > on Tue, 25 Jun 2024 10:56:07 +0300 writes:
>
> > Dear R fellows,
>
> >> From time to time, just when I t
Also notice that
> "+5" < 0
[1] TRUE
> as.numeric("+5") < 0
[1] FALSE
So the presumption is wrong. It is really about
> sort(c(0:3, "+", "-"))
[1] "-" "+" "0" "1" "2" "3"
-pd
> On 25 Jun 2024, at 10:02 , Martin Maechler wrote:
>
>> Adrian Dusa
>>on Tue, 25 Jun 2024 10:56:07 +03
> Adrian Dusa
> on Tue, 25 Jun 2024 10:56:07 +0300 writes:
> Dear R fellows,
>> From time to time, just when I thought I knew my R, I get
>> bitten by some
> small things that reminds one to constantly return to the
> basics.
> I knew for instance that "-1" <
Dear R fellows,
>From time to time, just when I thought I knew my R, I get bitten by some
small things that reminds one to constantly return to the basics.
I knew for instance that "-1" < 0 is TRUE, presumably because R first
coerces to numeric before comparing with 0.
But I did not expect that
Thank you Mark & Dunlap,
Will make changes to the variable as suggested. Thank you for your time &
assistance.
On Wed, Aug 31, 2016 at 9:16 PM, Mark Sharp wrote:
> Shivi,
>
> Looking at the help from ?WOE, ?WOETable, and ?IV, your Y vector in all
> cases is to be categorical and it is numeric.
Shivi,
Looking at the help from ?WOE, ?WOETable, and ?IV, your Y vector in all cases
is to be categorical and it is numeric.
Mark
R. Mark Sharp, Ph.D.
Director of Primate Records Database
Southwest National Primate Research Center
Texas Biomedical Research Institute
P.O. Box 760549
San Antonio
I see the printout (not an "error") that you describe on the 2nd example
you gave:
> t2 <- WOETable(X=SFDC1$case_age, Y=SFDC1$survey)
> print(t2)
[1] GOODS BADS TOTAL PCT_G PCT_B WOE IV
<0 rows> (or 0-length row.names)
The result of WOETable is perfectly legal - it is a data.frame with no rows
These are the packages i am using:
library(woe) #WEIGHT OF EVIDENCE
library(InformationValue) #INFORMATION VALUE
The syntax used is :
WOE(X=SFDC1$log_caseage, Y=SFDC1$survey)
WOETable(X=SFDC1$case_age, Y=SFDC1$survey)
IV(X=SFDC1$case_age, Y=SFDC1$survey)
On Wed, Aug 31, 2016 at 12:54 AM, Mark S
You need to show what R expressions you ran before running into this
problem, including calls to library() or require().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Aug 30, 2016 at 12:15 PM, Shivi Bhatia wrote:
> Hi Mark,
> What i understand, probably when i run the WOE & IV to check s
Shivi,
What package(s) are you using. What functions are you using. How are you
calling the functions. A reproducible sample has all of the actual code needed
to create a representative error. There are multiple packages you could be
using to look at weight of evidence and information value. Fo
Hi Mark,
What i understand, probably when i run the WOE & IV to check significant
variables that is where i get this error. Thanks for your assistance Mark
really appreciate i will look into some other measure on this.
On Wed, Aug 31, 2016 at 12:35 AM, Mark Sharp wrote:
> Shivi,
>
> It is likely
Shivi,
It is likely that William knows what you are trying to do because of his
considerable experience, but I am not able to figure it out from what you have
written. You have apparently sent the output from something like
dput(SFDC[1:50, ]), but I still do not know what you did to get the err
Hi William/ Mark,
I am using WOE & IV (weight of evidence) reduce the number of independent
vars.
I have read this data as a csv file.
reproducible example for your reference please:
structure(list(date = structure(c(6L, 6L, 6L, 6L, 6L, 6L, 14L,
14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 1
You did not say what operation gave you the error.
I can get that message (which is not an "error") if I print
an illegally constructed data.frame, one without the
row.names attribute.
> illegalDF <- structure(class="data.frame", list(ColumnA = 1:3))
> illegalDF
[1] ColumnA
<0 rows> (or 0-length
Shivi,
Can you show the code that throws the error?
<0 rows> (or 0-length row.names)
Of course as always a reproducible sample would be great. Perhaps you can make
a small subset of the data and use dput() to provide a defined object.
R. Mark Sharp, Ph.D.
Director of Primate Records Database
So
Hi Mark,
It gives me num [1:5083]. I have used head also to see first 10 obs:
head(SFDC$case_age,10)
[1] 24.84 0.05 13.38 0.15 11.11 4.16 8.13 0.07 3.61 0.00
Thanks.
On Tue, Aug 30, 2016 at 10:05 PM, Mark Sharp wrote:
> What do you get from
> str(SFDC$case_age)
>
> Mark
>
> R. Mark Sh
What do you get from
str(SFDC$case_age)
Mark
R. Mark Sharp, Ph.D.
Director of Primate Records Database
Southwest National Primate Research Center
Texas Biomedical Research Institute
P.O. Box 760549
San Antonio, TX 78245-0549
Telephone: (210)258-9476
e-mail: msh...@txbiomed.org
> On Aug 30,
I know this question has been asked zillion times but even after consulting
Stack Overflow & other forum cant figure out the reason.
I have one var in my data-set names case age. This variable is numeric as:
class(SFDC$case_age)
*numeric*
however it throws this error:
<0 rows> (or 0-length row
On May 26, 2014, at 9:42 AM, Prof Brian Ripley wrote:
> On 26/05/2014 13:16, ritwi...@isical.ac.in wrote:
>> Dear R helpers,
>>
>>
>> today I found something interesting in R. 0^0 gives value 1 in R. But it
>> is undefined in mathematics. During debugging a R c
evity.
On May 26, 2014 5:16:21 AM PDT, ritwi...@isical.ac.in wrote:
>Dear R helpers,
>
>
>today I found something interesting in R. 0^0 gives value 1 in R. But
>it
>is undefined in mathematics. During debugging a R code, I found it and
>it
>effects my program severely
On 26/05/2014 13:16, ritwi...@isical.ac.in wrote:
Dear R helpers,
today I found something interesting in R. 0^0 gives value 1 in R. But it
is undefined in mathematics. During debugging a R code, I found it and it
effects my program severely. So my question is why it is defined 1 in R?
Is there
Dear R helpers,
today I found something interesting in R. 0^0 gives value 1 in R. But it
is undefined in mathematics. During debugging a R code, I found it and it
effects my program severely. So my question is why it is defined 1 in R?
Is there any particular reason or its a bug in the R
Sam:
Yes. Good point. (which is why my "??" was necessary).
-- Bert
On Mon, Mar 18, 2013 at 10:05 AM, Sam Steingold wrote:
> > * Bert Gunter [2013-03-17 20:30:56 -0700]:
> >
> > I also think it fair to say that all (??) languages have these sorts
> > of malapropisms due to operator precedenc
> * Bert Gunter [2013-03-17 20:30:56 -0700]:
>
> I also think it fair to say that all (??) languages have these sorts
> of malapropisms due to operator precedence.
Except for those languages which do _not_ have "operator precedence".
Like, e.g., Lisp.
--
Sam Steingold (http://sds.podval.org/) o
Ben Bolker gmail.com> writes:
> Maybe FAQ 7.31 was referred to not for its direct relevance but as
> a measure of the "old-hand-ness" of the people who will get the joke.
!1i|!0
Chuck
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
Bert Gunter gene.com> writes:
>
> But this has nothing to do with 7.31 and everything to do with operator
> precedence and automatic casting from integers to logical and vice-versa.
>
> I also think it fair to say that all (??) languages have these sorts of
> malapropisms due to operator prece
But this has nothing to do with 7.31 and everything to do with operator
precedence and automatic casting from integers to logical and vice-versa.
I also think it fair to say that all (??) languages have these sorts of
malapropisms due to operator precedence.
-- Bert
On Sun, Mar 17, 2013 at 7:17
Chuck,
What an absolutely wonderful R Infernoism.
Pat
On 18/03/2013 02:17, Charles Berry wrote:
Hi all,
The subject line is TRUE.
Today I accidentally typed rnorm(!0).
My old eyes took a minute to focus clearly enough to see what I really typed and
why I got '!0' random numbers instead of
Hi all,
The subject line is TRUE.
Today I accidentally typed rnorm(!0).
My old eyes took a minute to focus clearly enough to see what I really typed and
why I got '!0' random numbers instead of '10' random normal numbers.
If the subject line is disturbing, be assured that this is TRUE:
!0^2
I am not entirely sure what is going on. I am sure that I am missing
something. maptools won't detach... Thanks for all of your help.
R 2.11.1
OS Ubuntu 10.04
R --vanilla
install.packages("sp", dep=TRUE)
Loading required package: reshape
Loading required package: plyr
Loading required packag
On 05-Sep-09 10:00:26, Markku Karhunen wrote:
>> On 04-Sep-09 10:45:27, Markku Karhunen wrote:
>>> True. Should have read ?diag.
>>>
>>> However, this provokes a more general question: Is there some way I
>>> can declare some scalar and _all its functions_ as matrices?
>>>
>>> For instance, I would
On 04-Sep-09 10:45:27, Markku Karhunen wrote:
True. Should have read ?diag.
However, this provokes a more general question: Is there some way I
can declare some scalar and _all its functions_ as matrices?
For instance, I would like to
A = as.matrix(0.98)
B = function(A)
C = diag(sqrt(B))
s
On 04-Sep-09 10:45:27, Markku Karhunen wrote:
> True. Should have read ?diag.
>
> However, this provokes a more general question: Is there some way I
> can declare some scalar and _all its functions_ as matrices?
>
> For instance, I would like to
>
> A = as.matrix(0.98)
> B = function(A)
> C =
True. Should have read ?diag.
However, this provokes a more general question: Is there some way I
can declare some scalar and _all its functions_ as matrices?
For instance, I would like to
A = as.matrix(0.98)
B = function(A)
C = diag(sqrt(B))
so that all scalars are explicitly [1,1] matric
Duncan Murdoch wrote:
baptiste auguie wrote:
it's documented as "unexpected"
?diag
Note
Using diag(x) can have unexpected effects if x is a vector that could be of
length one. Use diag(x, nrow = length(x)) for consistent behaviour.
And the result follows from this part,
else if (leng
baptiste auguie wrote:
it's documented as "unexpected"
?diag
Note
Using diag(x) can have unexpected effects if x is a vector that could be of
length one. Use diag(x, nrow = length(x)) for consistent behaviour.
And the result follows from this part,
else if (length(x) == 1L && nargs() ==
it's documented as "unexpected"
?diag
Note
Using diag(x) can have unexpected effects if x is a vector that could be of
length one. Use diag(x, nrow = length(x)) for consistent behaviour.
And the result follows from this part,
else if (length(x) == 1L && nargs() == 1L) {
n <- as.int
Markku Karhunen wrote:
Hi,
Does anybody know, what is going on here?
diag( x ) produces a round(x) x round(x) identity matrix when x is
length 1. (This is the third case listed on the man page ?diag). See
the note there about a safer form if you wanted a matrix with x on the
diagonal.
On Fri, Sep 4, 2009 at 11:25 AM, Markku
Karhunen wrote:
> Hi,
>
> Does anybody know, what is going on here?
>
>> diag(sqrt(1))
>
> [,1]
> [1,] 1
>>
>> diag(sqrt(0.))
>
> <0 x 0 matrix>
>>
>> sqrt(1)
>
> [1] 1
>>
>> sqrt(0.)
>
> [1] 0.5773214
>
Read the help for diag yet?
'diag
On 09/04/2009 12:25 PM, Markku Karhunen wrote:
Hi,
Does anybody know, what is going on here?
diag(sqrt(1))
[,1]
[1,] 1
diag(sqrt(0.))
<0 x 0 matrix>
sqrt(1)
[1] 1
sqrt(0.)
[1] 0.5773214
BR, Markku Karhunen
researcher
University of Helsinki
Try this instead;
> diag( sqrt(.33
Hi,
Does anybody know, what is going on here?
diag(sqrt(1))
[,1]
[1,]1
diag(sqrt(0.))
<0 x 0 matrix>
sqrt(1)
[1] 1
sqrt(0.)
[1] 0.5773214
BR, Markku Karhunen
researcher
University of Helsinki
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