Re: [R] evaluating expressions with sub expressions
Thanks so much everyone!
Bert's shorted example does what I need, but I'm filing away Gabor's
solution for when I inevitably need it some day. I've never found the
handling of variables in R to be very straightforward; sometimes I pine
for Maple to do my algebra for me...
If its good enough to have one level of substitution then esub in my
post (originally due to Tony Plate -- see reference in my post) is all
that is needed:
esub(mat[[2]], list(g1 = g1[[1]]))
but I think the real problem could require multiple levels of
substitution in which case repeated application of esub is needed as
you walk the expression tree which is what proc() in my post does.
For example, suppose mat[[2]] is a function of g1 which is a function
of Tm which is a function of z. Then continuing the example in the
original post this does the repeated substitution needed (which would
be followed by an eval, not shown here, as in my original post):
Tm - expression(z^2)
sapply(mat, proc)
[[1]]
[1] 0
[[2]]
f1 * s1 * (1/z^2)
To answer your question, quote() produces a call object but expression
produces a call wrapped in an expression which is why there is special
handling of expression objects in the proc() function in my post.
On Fri, Jan 29, 2010 at 4:38 PM, Bert Gunter gunter.ber...@gene.com
wrote:
Folks:
Stripped to its essentials, Jennifer's request seemed simple: substitute
a
subexpression as a named variable for a variable name in an expression,
also
expressed as a named variable. A simple example is:
e - expression(0,a*b)
z1 - quote(1/t) ## explained below
The task is to substitute the expression in z1, 1/t, for b in e,
yielding the substituted expression as the result.
Gabor provided a solution, but it seemed to me like trying to swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled around a bit
(guided by Gabor's approach and an old Programmer's Niche column of Bill
Venables) and came up with:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f
[[1]]
[1] 0
[[2]]
a * (1/t)
## f is a list. Turn it back into an expression
f - as.expression(f)
## check that this works as intended
f
expression(0, a * (1/t))
a - 2
t - 3
eval(f)
[1] 0.667
Now you'll note that to do this I explicitly used quote() to produce the
variable holding the subexpression to be substituted. You may ask, why
not
use expression() instead, as in
z2 - expression(1/t)
This doesn't work:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f
[[1]]
[1] 0
[[2]]
a * expression(1/t)
f - as.expression(f)
## Yielding ...
f
expression(0, a * expression(1/t)) Not what we want!
## And sure enough ...
eval(f)
Error in a * expression(1/t) : non-numeric argument to binary operator
I think I understand why the z - expression() approach does not work;
but I
do not understand why the z - quote() approach does! The mode of the
return
from both of these is call, but they are different (because
identical()
tells me so). Could someone perhaps elaborate on this a bit more? And is
there a yet simpler and more straightforward way to do the above than
what I
proposed?
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr,
sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an expression when one
of
the variables in the expression is defined separately as a sub
expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition of g1
vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for
variables
before adding this sub expression
Re: [R] evaluating expressions with sub expressions
On Sat, Jan 30, 2010 at 10:39 AM, Jennifer Young jennifer.yo...@math.mcmaster.ca wrote: handling of variables in R to be very straightforward; sometimes I pine for Maple to do my algebra for me... There are several interfaces to Computer Algebra Systems in R. Try this (but read instructions on home page http://ryacas.googlecode.com first particularly the part about ensuring you have the right version of XML installed). Below, Sym creates a symbolic variable: library(Ryacas) Loading required package: XML f1 - Sym(f1) s1 - Sym(s1) Tm - Sym(Tm) g1 - 1/Tm mat - f1 * s1 * g1 mat [1] Starting Yacas! expression(f1 * s1/Tm) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] evaluating expressions with sub expressions
Hallo I'm having trouble figuring out how to evaluate an expression when one of the variables in the expression is defined separately as a sub expression. Here's a simplified example mat - expression(0, f1*s1*g1) # vector of formulae g1 - expression(1/Tm) # expansion of the definition of g1 vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for variables before adding this sub expression I was using the following to evaluate mat sapply(mat, eval, vals) Obviously I could manually substitute in 1/Tm for each g1 in the definition of mat, but the actual expression vector is much longer, and the sub expression more complicated. Also, the subexpression is often adjusted for different scenarios. Is there a simple way of changing this or redefining mat so that I can define g1 like a macro to be used in the expression vector. Thanks! Jennifer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluating expressions with sub expressions
Hi, Would this do as an alternative syntax? g1 - quote(1/Tm) mat - list(0, bquote(f1*s1*.(g1))) vals - data.frame(f1=1, s1=.5, Tm=2) sapply(mat, eval, vals) HTH, baptiste On 29 January 2010 17:51, Jennifer Young jennifer.yo...@math.mcmaster.ca wrote: Hallo I'm having trouble figuring out how to evaluate an expression when one of the variables in the expression is defined separately as a sub expression. Here's a simplified example mat - expression(0, f1*s1*g1) # vector of formulae g1 - expression(1/Tm) # expansion of the definition of g1 vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for variables before adding this sub expression I was using the following to evaluate mat sapply(mat, eval, vals) Obviously I could manually substitute in 1/Tm for each g1 in the definition of mat, but the actual expression vector is much longer, and the sub expression more complicated. Also, the subexpression is often adjusted for different scenarios. Is there a simple way of changing this or redefining mat so that I can define g1 like a macro to be used in the expression vector. Thanks! Jennifer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluating expressions with sub expressions
Hmm I *think* this will work, but may break in a further sub routine. It certainly works in this example, but my expression vector is used in many scenarios and it will take a while to check them all. For instance, I take the derivative of each element with respect to each variable using sapply(mat, deriv, names(vals)) This bit seems to still work, but I'd welcome a solution that doesn't change the structure of the expression vector to a list, just in case. Thanks for this solution. Hi, Would this do as an alternative syntax? g1 - quote(1/Tm) mat - list(0, bquote(f1*s1*.(g1))) vals - data.frame(f1=1, s1=.5, Tm=2) sapply(mat, eval, vals) HTH, baptiste On 29 January 2010 17:51, Jennifer Young jennifer.yo...@math.mcmaster.ca wrote: Hallo I'm having trouble figuring out how to evaluate an expression when one of the variables in the expression is defined separately as a sub expression. Here's a simplified example mat - expression(0, f1*s1*g1) # vector of formulae g1 - expression(1/Tm) # expansion of the definition of g1 vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for variables before adding this sub expression I was using the following to evaluate mat sapply(mat, eval, vals) Obviously I could manually substitute in 1/Tm for each g1 in the definition of mat, but the actual expression vector is much longer, and the sub expression more complicated. Also, the subexpression is often adjusted for different scenarios. Is there a simple way of changing this or redefining mat so that I can define g1 like a macro to be used in the expression vector. Thanks! Jennifer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition of g1
vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for
variables
before adding this sub expression I was using the following to evaluate mat
sapply(mat, eval, vals)
Obviously I could manually substitute in 1/Tm for each g1 in the
definition of mat, but the actual expression vector is much longer, and
the sub expression more complicated. Also, the subexpression is often
adjusted for different scenarios. Is there a simple way of changing this
or redefining mat so that I can define g1 like a macro to be used in
the expression vector.
Thanks!
Jennifer
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluating expressions with sub expressions
Folks:
Stripped to its essentials, Jennifer's request seemed simple: substitute a
subexpression as a named variable for a variable name in an expression, also
expressed as a named variable. A simple example is:
e - expression(0,a*b)
z1 - quote(1/t) ## explained below
The task is to substitute the expression in z1, 1/t, for b in e,
yielding the substituted expression as the result.
Gabor provided a solution, but it seemed to me like trying to swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled around a bit
(guided by Gabor's approach and an old Programmer's Niche column of Bill
Venables) and came up with:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f
[[1]]
[1] 0
[[2]]
a * (1/t)
## f is a list. Turn it back into an expression
f - as.expression(f)
## check that this works as intended
f
expression(0, a * (1/t))
a - 2
t - 3
eval(f)
[1] 0.667
Now you'll note that to do this I explicitly used quote() to produce the
variable holding the subexpression to be substituted. You may ask, why not
use expression() instead, as in
z2 - expression(1/t)
This doesn't work:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f
[[1]]
[1] 0
[[2]]
a * expression(1/t)
f - as.expression(f)
## Yielding ...
f
expression(0, a * expression(1/t)) Not what we want!
## And sure enough ...
eval(f)
Error in a * expression(1/t) : non-numeric argument to binary operator
I think I understand why the z - expression() approach does not work; but I
do not understand why the z - quote() approach does! The mode of the return
from both of these is call, but they are different (because identical()
tells me so). Could someone perhaps elaborate on this a bit more? And is
there a yet simpler and more straightforward way to do the above than what I
proposed?
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition of g1
vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for
variables
before adding this sub expression I was using the following to evaluate
mat
sapply(mat, eval, vals)
Obviously I could manually substitute in 1/Tm for each g1 in the
definition of mat, but the actual expression vector is much longer, and
the sub expression more complicated. Also, the subexpression is often
adjusted for different scenarios. Is there a simple way of changing this
or redefining mat so that I can define g1 like a macro to be used in
the expression vector.
Thanks!
Jennifer
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] evaluating expressions with sub expressions
You asked
And is there a yet simpler and more
straightforward way to do the
above than what I proposed?
You could use S+, whose substitute() function
(a) descends into expressions and functions
(b) has an argument (evaluate=TRUE) so you
don't need to use do.call when the first
argument is not a literal
E.g.,
e - expression(a*b, function(x)x+b, log(b))
substitute(e, list(b=Quote(exp(1))), evaluate=TRUE)
expression(a * exp(1), function(x)
x + exp(1), log(exp(1)))
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
Sent: Friday, January 29, 2010 1:38 PM
To: 'Gabor Grothendieck'; 'Jennifer Young'
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
Folks:
Stripped to its essentials, Jennifer's request seemed simple:
substitute a
subexpression as a named variable for a variable name in an
expression, also
expressed as a named variable. A simple example is:
e - expression(0,a*b)
z1 - quote(1/t) ## explained below
The task is to substitute the expression in z1, 1/t, for b in e,
yielding the substituted expression as the result.
Gabor provided a solution, but it seemed to me like trying to
swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled
around a bit
(guided by Gabor's approach and an old Programmer's Niche
column of Bill
Venables) and came up with:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f
[[1]]
[1] 0
[[2]]
a * (1/t)
## f is a list. Turn it back into an expression
f - as.expression(f)
## check that this works as intended
f
expression(0, a * (1/t))
a - 2
t - 3
eval(f)
[1] 0.667
Now you'll note that to do this I explicitly used quote() to
produce the
variable holding the subexpression to be substituted. You may
ask, why not
use expression() instead, as in
z2 - expression(1/t)
This doesn't work:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f
[[1]]
[1] 0
[[2]]
a * expression(1/t)
f - as.expression(f)
## Yielding ...
f
expression(0, a * expression(1/t)) Not what we want!
## And sure enough ...
eval(f)
Error in a * expression(1/t) : non-numeric argument to binary operator
I think I understand why the z - expression() approach does
not work; but I
do not understand why the z - quote() approach does! The
mode of the return
from both of these is call, but they are different (because
identical()
tells me so). Could someone perhaps elaborate on this a bit
more? And is
there a yet simpler and more straightforward way to do the
above than what I
proposed?
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute,
list(expr, sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an
expression when one of
the variables in the expression is defined separately as a
sub expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition of g1
vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible
values for
variables
before adding this sub expression I was using the following
to evaluate
mat
sapply(mat, eval, vals)
Obviously I could manually substitute in 1/Tm for each g1 in the
definition of mat, but the actual expression vector is
much longer, and
the sub expression more complicated. Also, the
subexpression is often
adjusted for different scenarios. Is there a simple way of
changing
Re: [R] evaluating expressions with sub expressions
If its good enough to have one level of substitution then esub in my
post (originally due to Tony Plate -- see reference in my post) is all
that is needed:
esub(mat[[2]], list(g1 = g1[[1]]))
but I think the real problem could require multiple levels of
substitution in which case repeated application of esub is needed as
you walk the expression tree which is what proc() in my post does.
For example, suppose mat[[2]] is a function of g1 which is a function
of Tm which is a function of z. Then continuing the example in the
original post this does the repeated substitution needed (which would
be followed by an eval, not shown here, as in my original post):
Tm - expression(z^2)
sapply(mat, proc)
[[1]]
[1] 0
[[2]]
f1 * s1 * (1/z^2)
To answer your question, quote() produces a call object but expression
produces a call wrapped in an expression which is why there is special
handling of expression objects in the proc() function in my post.
On Fri, Jan 29, 2010 at 4:38 PM, Bert Gunter gunter.ber...@gene.com wrote:
Folks:
Stripped to its essentials, Jennifer's request seemed simple: substitute a
subexpression as a named variable for a variable name in an expression, also
expressed as a named variable. A simple example is:
e - expression(0,a*b)
z1 - quote(1/t) ## explained below
The task is to substitute the expression in z1, 1/t, for b in e,
yielding the substituted expression as the result.
Gabor provided a solution, but it seemed to me like trying to swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled around a bit
(guided by Gabor's approach and an old Programmer's Niche column of Bill
Venables) and came up with:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f
[[1]]
[1] 0
[[2]]
a * (1/t)
## f is a list. Turn it back into an expression
f - as.expression(f)
## check that this works as intended
f
expression(0, a * (1/t))
a - 2
t - 3
eval(f)
[1] 0.667
Now you'll note that to do this I explicitly used quote() to produce the
variable holding the subexpression to be substituted. You may ask, why not
use expression() instead, as in
z2 - expression(1/t)
This doesn't work:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f
[[1]]
[1] 0
[[2]]
a * expression(1/t)
f - as.expression(f)
## Yielding ...
f
expression(0, a * expression(1/t)) Not what we want!
## And sure enough ...
eval(f)
Error in a * expression(1/t) : non-numeric argument to binary operator
I think I understand why the z - expression() approach does not work; but I
do not understand why the z - quote() approach does! The mode of the return
from both of these is call, but they are different (because identical()
tells me so). Could someone perhaps elaborate on this a bit more? And is
there a yet simpler and more straightforward way to do the above than what I
proposed?
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion of the definition of g1
vals - data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for
variables
before adding this sub expression I was using the following to evaluate
mat
sapply(mat, eval, vals)
Obviously I could manually substitute in 1/Tm for each g1 in the
definition of mat, but the actual expression vector is much longer, and
the sub expression more complicated. Also, the subexpression is often
adjusted for different
Re: [R] evaluating expressions with sub expressions
Inline below...
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Friday, January 29, 2010 2:12 PM
To: Bert Gunter
Cc: Jennifer Young; r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
If its good enough to have one level of substitution then esub in my
post (originally due to Tony Plate -- see reference in my post) is all
that is needed:
esub(mat[[2]], list(g1 = g1[[1]]))
--- Yes, this is essentially what I did using lapply
but I think the real problem could require multiple levels of
substitution in which case repeated application of esub is needed as
you walk the expression tree which is what proc() in my post does.
-- Indeed. But my point was to handle the simple case simply.
To answer your question, quote() produces a call object but expression
produces a call wrapped in an expression which is why there is special
handling of expression objects in the proc() function in my post.
-- Aha! And of course I should have realized that I could have easily
determined this myself with:
as.list(quote(a*b))
[[1]]
`*`
[[2]]
a
[[3]]
b
## But ...
e - as.list(expression(a*b))
e
[[1]]
a * b
mode(e[[1]])
[1] call
as.list(e[[1]])
[[1]]
`*`
[[2]]
a
[[3]]
b
So now all is clear to me.
Thanks to both Bill and Gabor for their informative replies.
-- Bert
Bert Gunter
Genentech Nonclinical Statistics
On Fri, Jan 29, 2010 at 4:38 PM, Bert Gunter gunter.ber...@gene.com wrote:
Folks:
Stripped to its essentials, Jennifer's request seemed simple: substitute a
subexpression as a named variable for a variable name in an expression,
also
expressed as a named variable. A simple example is:
e - expression(0,a*b)
z1 - quote(1/t) ## explained below
The task is to substitute the expression in z1, 1/t, for b in e,
yielding the substituted expression as the result.
Gabor provided a solution, but it seemed to me like trying to swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled around a bit
(guided by Gabor's approach and an old Programmer's Niche column of Bill
Venables) and came up with:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f
[[1]]
[1] 0
[[2]]
a * (1/t)
## f is a list. Turn it back into an expression
f - as.expression(f)
## check that this works as intended
f
expression(0, a * (1/t))
a - 2
t - 3
eval(f)
[1] 0.667
Now you'll note that to do this I explicitly used quote() to produce the
variable holding the subexpression to be substituted. You may ask, why not
use expression() instead, as in
z2 - expression(1/t)
This doesn't work:
f - lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f
[[1]]
[1] 0
[[2]]
a * expression(1/t)
f - as.expression(f)
## Yielding ...
f
expression(0, a * expression(1/t)) Not what we want!
## And sure enough ...
eval(f)
Error in a * expression(1/t) : non-numeric argument to binary operator
I think I understand why the z - expression() approach does not work; but
I
do not understand why the z - quote() approach does! The mode of the
return
from both of these is call, but they are different (because identical()
tells me so). Could someone perhaps elaborate on this a bit more? And is
there a yet simpler and more straightforward way to do the above than what
I
proposed?
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help@r-project.org
Subject: Re: [R] evaluating expressions with sub expressions
The following recursively walks the expression tree. The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html
esub - function(expr, sublist) do.call(substitute, list(expr, sublist))
proc - function(e, env = parent.frame()) {
for(nm in all.vars(e)) {
if (exists(nm, env) is.language(g - get(nm, env))) {
if (is.expression(g)) g - g[[1]]
g - Recall(g, env)
L - list(g)
names(L) - nm
e - esub(e, L)
}
}
e
}
mat - expression(0, f1*s1*g1)
g1 - expression(1/Tm)
vals - data.frame(f1=1, s1=.5, Tm=2)
e - sapply(mat, proc)
sapply(e, eval, vals)
The last line should give:
sapply(e, eval, vals)
[1] 0.00 0.25
On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
jennifer.yo...@math.mcmaster.ca wrote:
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub
expression.
Here's a simplified example
mat - expression(0, f1*s1*g1) # vector of formulae
g1 - expression(1/Tm) # expansion
