# [Rd] subset: evaluating expressions missing? (PR#6937)

Prof Brian Ripley ripley at stats.ox.ac.uk
Fri Jun 4 15:40:46 CEST 2004

```You should not use the system object `c' as an index! BTW, expression()
takes multiple arguments, so you can do

# OK
for (c2 in expression(x\$a==1, x\$b==1, x\$c==1)) print(x[eval(c2),])
for (c2 in expression(x\$a==1, x\$b==1, x\$c==1)) print(subset(x, eval(c2)))

# error, correctly
for (c2 in expression(x\$a==1, x\$b==1, x\$c==1)) print(subset(x,c2))
Error in r & !is.na(r) : operations are possible only for numeric or logical types
is.na() applied to non-(list or vector) in: is.na(r)

On Fri, 4 Jun 2004 martin.lenze at web.de wrote:

> Full_Name: Martin Lenze
> Version: Version 1.9.0  (2004-04-12), ISBN 3-900051-00-3
> OS: Microsoft Windows 2000 [Version 5.00.2195] SP4
> Submission from: (NULL) (82.83.167.79)

> now I switched to R 1.9.0 and did get a problem with subset, see sample:
>
> x <- data.frame(a=as.integer(round(runif(5),0)),
>   b=as.integer(round(runif(5),0)),c=as.integer(round(runif(5),0)))
> # correct results:
> for (c in c(expression(x\$a==1),expression(x\$b==1),
>   expression(x\$c==1))) print(x[eval(c),])
> # results I don't understand:
> for (c in c(expression(x\$a==1),expression(x\$b==1),
>   expression(x\$c==1))) print(subset(x,eval(c)))
> for (c in c(expression(x\$a==1),expression(x\$b==1),
>   expression(x\$c==1))) print(subset(x,c))
>
> Am I doing something wrong? Using subset this way with R 1.8.0 worked fine.

Not for me!  I get the same wrong results as in 1.9.0 with your code.

--
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 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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