It does work, thank you, but the literal 5<x now needs to be quoted by
expression():
> do.call(subset, list(dat, expression(5<x)))
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
This is ok, but the standard "subset(dat, 5<x)" looks more readable.
Anyway, thank you for your help, it's a nice paradigm.
Vadim
-----Original Message-----
From: bill.venab...@csiro.au [mailto:bill.venab...@csiro.au]
Sent: Saturday, June 26, 2010 1:08 AM
To: Vadim Ogranovich; r-help@r-project.org
Subject: RE: [R] subset arg in subset(). was: converting result of substitute
to 'ordidnary' expression
Here is another one that works:
> do.call(subset, list(dat, subsetexp))
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
>
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Vadim Ogranovich
Sent: Saturday, 26 June 2010 11:13 AM
To: 'r-help@r-project.org'
Subject: [R] subset arg in subset(). was: converting result of substitute to
'ordidnary' expression
Dear R users,
Please disregard my previous post "converting result of substitute to
'ordidnary' expression". The problem I have has nothing to do with substitute.
Consider:
> dat <- data.frame(x=1:10, y=1:10)
> subsetexp <- expression(5<x)
> ## this does work
> subset(dat, eval(subsetexp))
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
> ## and so does this
> subset(dat, 5<x)
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
> ## but this doesn't work
> subset(dat, subsetexp)
Error in subset.data.frame(dat, subsetexp) :
'subset' must evaluate to logical
Why did the last expression fail and why it worked with eval()?
Thank you very much for your help,
Vadim
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