R help - Nov 2010 - Calling substitute(expr, list(a=1)) when expr <- expression(a+b+c)

# The result I am after is the result after a substitution in an expression,
such as

substitute(expression(a+b+c), list(a=1))
expression(1 + b + c)
 # However, the way I want to do it is for a an expression "stored as a
variable" as

(expr <- expression(a+b+c))
expression(a + b + c)
 # a) The following does not work

(expr2 <- substitute(expr, list(a=1)))
expr
 # b) - whereas this does work:

ans <- eval(substitute(substitute(qqq, list(a=1)), list(qqq=expr[[1]])))
as.expression(ans)
expression(1 + b + c)
 # I have - at least - two problems:
# I am not sure I understand 1) why a) does not work and 2) why b) does work.
# Can anyone point me in the right direction?

# Thanks
# Søren

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On Fri, Nov 26, 2010 at 8:31 AM, S?ren H?jsgaard
<Soren.Hojsgaard at agrsci.dk> wrote:
> # The result I am after is the result after a substitution in an
expression, such as
>
> substitute(expression(a+b+c), list(a=1))
> expression(1 + b + c)
> ?# However, the way I want to do it is for a an expression "stored as
a variable" as
>
> (expr <- expression(a+b+c))
> expression(a + b + c)
> ?# a) The following does not work
>
> (expr2 <- substitute(expr, list(a=1)))
> expr
> ?# b) - whereas this does work:
>
> ans <- eval(substitute(substitute(qqq, list(a=1)), list(qqq=expr[[1]])))
> as.expression(ans)
> expression(1 + b + c)
> ?# I have - at least - two problems:
> # I am not sure I understand 1) why a) does not work and 2) why b) does
work.
> # Can anyone point me in the right direction?
>
It does not evaluate its argument so if expr is the first argument
about the only thing you can substitute is expr itself:
> substitute(expr, list(expr = 3))
[1] 3

Try this:
> expr <- expression(a+b+c)
> do.call("substitute", list(expr, list(a=1)))
expression(a + b + c)



-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

Thanks, but I think there is a small mistake in your code:
> expr <- expression(a+b+c)
> do.call("substitute", list(expr, list(a=1)))
expression(a + b + c)

I think it should be:
> as.expression(do.call("substitute", list(expr[[1]], list(a=1))))
expression(1 + b + c)
- or maybe it can be done in a simpler way?

Regards
S?ren


-----Oprindelig meddelelse-----
Fra: Gabor Grothendieck [mailto:ggrothendieck at gmail.com] 
Sendt: 26. november 2010 14:39
Til: S?ren H?jsgaard
Cc: r-help at stat.math.ethz.ch
Emne: Re: [R] Calling substitute(expr, list(a=1)) when expr <-
expression(a+b+c)

On Fri, Nov 26, 2010 at 8:31 AM, S?ren H?jsgaard
<Soren.Hojsgaard at agrsci.dk> wrote:
> # The result I am after is the result after a substitution in an
expression, such as
>
> substitute(expression(a+b+c), list(a=1))
> expression(1 + b + c)
> ?# However, the way I want to do it is for a an expression "stored as
a variable" as
>
> (expr <- expression(a+b+c))
> expression(a + b + c)
> ?# a) The following does not work
>
> (expr2 <- substitute(expr, list(a=1)))
> expr
> ?# b) - whereas this does work:
>
> ans <- eval(substitute(substitute(qqq, list(a=1)), list(qqq=expr[[1]])))
> as.expression(ans)
> expression(1 + b + c)
> ?# I have - at least - two problems:
> # I am not sure I understand 1) why a) does not work and 2) why b) does
work.
> # Can anyone point me in the right direction?
>
It does not evaluate its argument so if expr is the first argument
about the only thing you can substitute is expr itself:
> substitute(expr, list(expr = 3))
[1] 3

Try this:
> expr <- expression(a+b+c)
> do.call("substitute", list(expr, list(a=1)))
expression(a + b + c)



-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

On Fri, Nov 26, 2010 at 8:48 AM, S?ren H?jsgaard
<Soren.Hojsgaard at agrsci.dk> wrote:
> Thanks, but I think there is a small mistake in your code:
>
>> expr <- expression(a+b+c)
>> do.call("substitute", list(expr, list(a=1)))
> expression(a + b + c)
>
> I think it should be:
>> as.expression(do.call("substitute", list(expr[[1]],
list(a=1))))
Its already an expression so the extra as.expression would serve no purpose:
> identical(as.expression(do.call("substitute", list(expr,
list(a=1)))),
+ do.call("substitute", list(expr, list(a=1))))
[1] TRUE





-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com