[R] Sum of columns of a data frame equal to NA when all the elements are NA

[Bert Gunter]

"I see: consistency with additive identity. "

Ummm, well:

> 1+NULL
numeric(0)

> sum(1,NULL)
[1] 1

Of course, there could well be something here I don't get, but that doesn't
look very consistent to me. However, as I said privately, so long as the
corner case behavior is documented, which it is, I don't care.

Cheers,
Bert



Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Wed, Mar 21, 2018 at 10:26 AM, Boris Steipe <boris.steipe at utoronto.ca>
wrote:

> I see: consistency with additive identity. That makes sense. Thanks.
>
> B.
>
>
> > On Mar 21, 2018, at 1:22 PM, peter dalgaard <pdalgd at gmail.com> wrote:
> >
> > No. The empty sum is zero. Adding it to another sum should not change
> it. Nothing audacious about that. This is consistent; other definitions
> just cause trouble.
> >
> > -pd
> >
> >> On 21 Mar 2018, at 18:05 , Boris Steipe <boris.steipe at utoronto.ca>
> wrote:
> >>
> >> Surely the result of summation of non-existent values is not defined,
> is it not? And since the NA values have been _removed_, there's nothing
> left to sum over. In fact, pretending the the result in that case is zero
> would appear audacious, no?
> >>
> >> Cheers,
> >> Boris
> >>
> >>
> >>
> >>
> >>
> >>> On Mar 21, 2018, at 12:58 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
> wrote:
> >>>
> >>> What do you mean by "should not"?
> >>>
> >>> NULL means "missing object" in R. The result of the sum function is
> always expected to be numeric... so NA_real or NA_integer could make sense
> as possible return values. But you cannot compute on NULL so no, that
> doesn't work.
> >>>
> >>> See the note under the "Value" section of ?sum as to why zero is
> returned when all inputs are removed.
> >>> --
> >>> Sent from my phone. Please excuse my brevity.
> >>>
> >>> On March 21, 2018 9:03:29 AM PDT, Boris Steipe <
> boris.steipe at utoronto.ca> wrote:
> >>>> Should not the result be NULL if you have removed the NA with
> >>>> na.rm=TRUE ?
> >>>>
> >>>> B.
> >>>>
> >>>>
> >>>>
> >>>>> On Mar 21, 2018, at 11:44 AM, Stefano Sofia
> >>>> <stefano.sofia at regione.marche.it> wrote:
> >>>>>
> >>>>> Dear list users,
> >>>>> let me ask you this trivial question. I worked on that for a long
> >>>> time, by now.
> >>>>> Suppose to have a data frame with NAs and to sum some columns with
> >>>> rowSums:
> >>>>>
> >>>>> df <- data.frame(A = runif(10), B = runif(10), C = rnorm(10))
> >>>>> df[1, ] <- NA
> >>>>> rowSums(df[ , which(names(df) %in% c("A","B"))], na.rm=T)
> >>>>>
> >>>>> If all the elements of the selected columns are NA, rowSums returns 0
> >>>> while I need NA.
> >>>>> Is there an easy and efficient way to use rowSums within a function
> >>>> like
> >>>>>
> >>>>> function(x) ifelse(all(is.na(x)), as.numeric(NA), rowSums...)?
> >>>>>
> >>>>> or an equivalent function?
> >>>>>
> >>>>> Thank you for your help
> >>>>> Stefano
> >>>>>
> >>>>>
> >>>>>
> >>>>>      (oo)
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> >>
> >> ______________________________________________
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> > --
> > Peter Dalgaard, Professor,
> > Center for Statistics, Copenhagen Business School
> > Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> > Phone: (+45)38153501
> > Office: A 4.23
> > Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com
> >
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>
> ______________________________________________
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