R help - Feb 2005 - 2 small problems: integer division and the nature of NA

Hi,

I'm wondering why

48 %/% 2 gives 24
but
4.8 %/% 0.2 gives 23...
I'm not trying to round up here, but to find out how many times 
something fits into something else, and the answer should have been the 
same for both examples, no?

On a different topic, I like the behavior of NAs better in R than in 
SAS (at least they are not considered the smallest value for a 
variable), but at the same time I am surprised that the sum of NAs is 0 
instead of NA.

The sum of a vector having at least one NA but also valid data gives NA 
if we do not specify na.rm=T. But with na.rm=T, we are telling sum to 
give the sum of valid data, ignoring NAs that do not tell us anything 
about the value of a variable. I found out while getting the sum of 
small subsets of my data (such as when subsetting by several 
variables), sometimes a "cell" only contained NAs for my response 
variable. I would have expected the sum to be NA in such cases, as I do 
not have a single data point telling me the value of my response here. 
But R tells me the sum was zero in that cell! Was this behavior 
considered "desirable" when sum was built? If not, any hope it will be
fixed?

Sincerely,

Denis Chabot

Denis Chabot wrote:
> Hi,
> 
> I'm wondering why
> 
> 48 %/% 2 gives 24
> but
> 4.8 %/% 0.2 gives 23...
> I'm not trying to round up here, but to find out how many times 
> something fits into something else, and the answer should have been the 
> same for both examples, no?
No. Not from the perspective of a digital computer who cannot represent 
all real numbers exactly (well, only a very small subset, since we are 
using floating point arithmetics) ...

> On a different topic, I like the behavior of NAs better in R than in SAS 
> (at least they are not considered the smallest value for a variable), 
> but at the same time I am surprised that the sum of NAs is 0 instead of NA.
It *is* NA:
   sum(c(NA, NA)) # [1] NA
   sum(c(NA, 1))  # [1] NA
> The sum of a vector having at least one NA but also valid data gives NA 
> if we do not specify na.rm=T. But with na.rm=T, we are telling sum to 
> give the sum of valid data, ignoring NAs that do not tell us anything 
> about the value of a variable. I found out while getting the sum of 
> small subsets of my data (such as when subsetting by several variables), 
> sometimes a "cell" only contained NAs for my response variable. I
would
> have expected the sum to be NA in such cases, as I do not have a single 
> data point telling me the value of my response here. But R tells me the 
> sum was zero in that cell! Was this behavior considered
"desirable" when
> sum was built? If not, any hope it will be fixed?
???? I don't get your point!
If you *remove* NAs as in
   sum(c(NA, NA), na.rm=TRUE) # [1] 0
   sum(c(NA, 1), na.rm=TRUE)  # [1] 1
you are summing up not that much.... so what do you expect in the cases 
above?

Please read the docs on NA handling.

Uwe Ligges


> Sincerely,
> 
> Denis Chabot
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html

Denis Chabot <chabotd at globetrotter.net> writes:
> Hi,
> 
> I'm wondering why
> 
> 48 %/% 2 gives 24
> but
> 4.8 %/% 0.2 gives 23...
> I'm not trying to round up here, but to find out how many times
> something fits into something else, and the answer should have been
> the same for both examples, no?
Well, you can't trust floating point numbers to give you an exact
result:
> 4.8 / 0.2 - 24
[1] -3.552714e-15

and even
> (48/10) / (2/10) - 24
[1] -3.552714e-15

the basic issue being that tenths are not exactly representable in
binary floating point. I think very few people even expected you to
use integer division on non-integers, but I note that the claim on the
help page actually holds:
> 0.2 * 4.8 %/% 0.2  + 4.8 %% 0.2 == 4.8
[1] TRUE

 
> On a different topic, I like the behavior of NAs better in R than in
> SAS (at least they are not considered the smallest value for a
> variable), but at the same time I am surprised that the sum of NAs is
> 0 instead of NA.
> 
> The sum of a vector having at least one NA but also valid data gives
> NA if we do not specify na.rm=T. But with na.rm=T, we are telling sum
> to give the sum of valid data, ignoring NAs that do not tell us
> anything about the value of a variable. I found out while getting the
> sum of small subsets of my data (such as when subsetting by several
> variables), sometimes a "cell" only contained NAs for my response
> variable. I would have expected the sum to be NA in such cases, as I
> do not have a single data point telling me the value of my response
> here. But R tells me the sum was zero in that cell! Was this behavior
> considered "desirable" when sum was built? If not, any hope it
will be
> fixed?
Yes it was, and no there isn't. In math, the sum over an empty index
set is zero, which has some nice consistency properties (the sum over
a disjoint union of sets is the sum of the sums over each set, for
instance. 

-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907

It's the difference between integers and reals:  48 and 24 are 
integers;  4.8 and 0.2 are floating point numbers.  Consider: 

 > (4.8+.Machine$double.eps) %/% (0.2-.Machine$double.eps)
[1] 24
 > (4.8-.Machine$double.eps) %/% (0.2+.Machine$double.eps)
[1] 23
 >
      Does this help?  spencer graves

Denis Chabot wrote:
> Hi,
>
> I'm wondering why
>
> 48 %/% 2 gives 24
> but
> 4.8 %/% 0.2 gives 23...
> I'm not trying to round up here, but to find out how many times 
> something fits into something else, and the answer should have been 
> the same for both examples, no?
>
> On a different topic, I like the behavior of NAs better in R than in 
> SAS (at least they are not considered the smallest value for a 
> variable), but at the same time I am surprised that the sum of NAs is 
> 0 instead of NA.
>
> The sum of a vector having at least one NA but also valid data gives 
> NA if we do not specify na.rm=T. But with na.rm=T, we are telling sum 
> to give the sum of valid data, ignoring NAs that do not tell us 
> anything about the value of a variable. I found out while getting the 
> sum of small subsets of my data (such as when subsetting by several 
> variables), sometimes a "cell" only contained NAs for my response
> variable. I would have expected the sum to be NA in such cases, as I 
> do not have a single data point telling me the value of my response 
> here. But R tells me the sum was zero in that cell! Was this behavior 
> considered "desirable" when sum was built? If not, any hope it
will be
> fixed?
>
> Sincerely,
>
> Denis Chabot
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html

It's convention in mathematics that the empty sum is 0. You can think of
this as a generalization of 0*x = 0. 

Reid Huntsinger

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Denis Chabot
Sent: Friday, February 04, 2005 11:01 AM
To: r-help at stat.math.ethz.ch
Subject: [R] 2 small problems: integer division and the nature of NA


Hi,

I'm wondering why

48 %/% 2 gives 24
but
4.8 %/% 0.2 gives 23...
I'm not trying to round up here, but to find out how many times 
something fits into something else, and the answer should have been the 
same for both examples, no?

On a different topic, I like the behavior of NAs better in R than in 
SAS (at least they are not considered the smallest value for a 
variable), but at the same time I am surprised that the sum of NAs is 0 
instead of NA.

The sum of a vector having at least one NA but also valid data gives NA 
if we do not specify na.rm=T. But with na.rm=T, we are telling sum to 
give the sum of valid data, ignoring NAs that do not tell us anything 
about the value of a variable. I found out while getting the sum of 
small subsets of my data (such as when subsetting by several 
variables), sometimes a "cell" only contained NAs for my response 
variable. I would have expected the sum to be NA in such cases, as I do 
not have a single data point telling me the value of my response here. 
But R tells me the sum was zero in that cell! Was this behavior 
considered "desirable" when sum was built? If not, any hope it will be
fixed?

Sincerely,

Denis Chabot

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html

Denis Chabot <chabotd <at> globetrotter.net> writes:
: The sum of a vector having at least one NA but also valid data gives NA 
: if we do not specify na.rm=T. But with na.rm=T, we are telling sum to 
: give the sum of valid data, ignoring NAs that do not tell us anything 
: about the value of a variable. I found out while getting the sum of 
: small subsets of my data (such as when subsetting by several 
: variables), sometimes a "cell" only contained NAs for my response 
: variable. I would have expected the sum to be NA in such cases, as I do 
: not have a single data point telling me the value of my response here. 
: But R tells me the sum was zero in that cell! Was this behavior 
: considered "desirable" when sum was built? If not, any hope it will
be
: fixed?

Think of it this way: If u and v are index vectors then its desirable that

	sum(x[u]) + sum(x[v]) == sum(x[c(u,v)])

hold for zero length index vectors too in which case
sum(numeric()) should be zero, not NA.

If you want a short expression that gives NA for zero length x try this:

        sum(x) + if (length(x)) 0 else NA

or define your own function, sum0, say.