R help - May 2002 - Why is.integer() doesn't work with single values?

On Wed, 29 May 2002, Sven Garbade wrote:
> Hi all,
>
> I don't understand the behavior of is.integer():
>
> > x <- integer()
> > is.integer(x)
> [1] TRUE
> > x <- 10
> > is.integer(x)
> [1] FALSE
> > x <- 1:10
> > is.integer(x)
> [1] TRUE
>
> Why is.interger() returns FALSE if x has only one element? And how can
> someone check if x is an integer but contains only one value? (R 1.5.0
> on Linux i386)
Your guesses are wrong:
> is.integer(as.integer(10))
[1] TRUE
> is.integer(as.numeric(1:10))
[1] FALSE

If all else (except posting to R-help) fails, read the documentation which
says

     `is.integer' returns `TRUE' or `FALSE' depending on whether its
     argument is of integer type or not.

Note *NOT* `is an integer'.

To test for an integer *value* you could use x == round(x), assuming that
you want false for a value that is not integer by rounding error.
Otherwise use all.equal.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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Sven Garbade <garbade at psy.uni-muenchen.de> writes:
> Hi all,
> 
> I don't understand the behavior of is.integer():
> 
> > x <- integer()
> > is.integer(x)
> [1] TRUE
> > x <- 10
> > is.integer(x)
> [1] FALSE
> > x <- 1:10
> > is.integer(x)
> [1] TRUE
> 
> Why is.interger() returns FALSE if x has only one element? And how can
> someone check if x is an integer but contains only one value? (R 1.5.0
> on Linux i386)
This has nothing to do with is.integer or lengths. It is the ":"
operator that tries to make its result an integer vector whenever
possible. Constant are generally floating point numbers, even when
they happen to be integers.

Try is.integer(1:1) and is.integer(c(1,2,3)).

-- 
   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
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Hi all,

I don't understand the behavior of is.integer():
> x <- integer()
> is.integer(x)
[1] TRUE
> x <- 10
> is.integer(x)
[1] FALSE
> x <- 1:10
> is.integer(x)
[1] TRUE

Why is.interger() returns FALSE if x has only one element? And how can
someone check if x is an integer but contains only one value? (R 1.5.0
on Linux i386)

Thanks, Sven
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Hi,

On Wed, 29 May 2002, Sven Garbade wrote:

  |Hi all,
  |
  |I don't understand the behavior of is.integer():
  |
  |> x <- integer()
  |> is.integer(x)
  |[1] TRUE
  |> x <- 10
  |> is.integer(x)
  |[1] FALSE

This is OK -- x is now 10 (defaults to float), not any more integer().  Note
that assignment deleted the previous definition.

  |> x <- 1:10
  |> is.integer(x)
  |[1] TRUE

1:10 defaults to an integer vector, so your new vector is (a vector of)
integer(s).

  |Why is.interger() returns FALSE if x has only one element? And how can
  |someone check if x is an integer but contains only one value? (R 1.5.0
  |on Linux i386)

length(x) gives the number of values.  

is.integer() gives the type of argument (the way the argument is presented
in memory).  E.g:
> x <- 10
> is.integer(x)
[1] FALSE
> x == as.integer(x) 
  # use rather abs(x - as.integer(x)) < sqrt(Machine()$double.eps)
  # or something similar.
[1] TRUE
> length(x)
[1] 1

Perhaps it helps.

Ott

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Sven Garbade <garbade at psy.uni-muenchen.de>
writes:
>Hi all,
>
>I don't understand the behavior of is.integer():
>
>> x <- integer()
>> is.integer(x)
>[1] TRUE
>> x <- 10
>> is.integer(x)
>[1] FALSE
>> x <- 1:10
>> is.integer(x)
>[1] TRUE
>
>Why is.interger() returns FALSE if x has only one element? And how can
>someone check if x is an integer but contains only one value? (R 1.5.0
>on Linux i386)
That is down to the storage mode. Try:

        x <- 10
        typeof(x)

        y <- 1:10
        typeof(y)

You can test with:

        x == as.integer(x)

Mark

--
Mark Myatt


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