Re: [R] Why does debugging print() change output of function?
Another nice thing about using ~formula is that it stores the
environment in which the formula was made along with the formula.
Thus you know which envrionment should be used with evaluating it (and
don't have to guess that parent.frame() may be the right environmnet).
E.g.,
evalRHS - function(formula) {
RHS - formula[[length(formula)]]
eval(RHS, envir=environment(formula))
}
p1 - 1:3
p2 - 11:13
f1 - function(formula) {
p1 - 1000
evalRHS(formula)
}
f1(~p1+p2) # does not use the p1 defined in f1
# [1] 12 14 16
If you do want to override some of the variables in the formula you
can do that by replacing the environment of the formula with a child
of that environment. Do this in a function so it only replaces the
environment in a copy of the formula, not the original one.
f2 - function(formula) {
environment(formula) - list2env(list(p1=1000),
new.env(parent=environment(formula)))
evalRHS(formula)
}
f2(~p1+p2)
# [1] 1011 1012 1013
f1(~p1+p2)
# [1] 12 14 16
(These examples are not the best since the global environment is
searched in any case. Put the entire example inside a function to see
the strength of the formula approach.)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Sep 9, 2014 at 10:13 PM, David Winsemius dwinsem...@comcast.net wrote:
On Sep 9, 2014, at 4:07 PM, peter dalgaard wrote:
On 07 Sep 2014, at 00:31 , David Winsemius dwinsem...@comcast.net wrote:
The goal:
to create a function modeled after `subset` (notorious for its
non-standard evaluation) that will take a series of logical tests as
unquoted expressions to be evaluated in the framework of a dataframe
environment and return a dataframe of logicals:
...
A belated peep from the author of subset(): Don't!
I think we learned the hard way by now that it is much easier to pass
unevaluated expressions in the shape of formula objects or maybe expression
objects. Lots of pain can be avoided by slipping in a simple ~.
It's taken me several years to understand why you are probably correct in
this regard. I needed to learn that `~` is actually a function that creates
a language object.
is.function(`~`)
[1] TRUE
is.language( ~ x 5 x 10)
[1] TRUE
... and that it's rather easy to extract the object somewhat like but not
really an expression embedded in such an object:
is.expression( ~ x 5 x 10)
[1] FALSE
is.call( (~ x 5 x 10)[2] )
[1] TRUE
The task of learning the various types of language objects is not an easy
one.
--
David Winsemius, MD
Alameda, CA, USA
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Why does debugging print() change output of function?
On 07 Sep 2014, at 00:31 , David Winsemius dwinsem...@comcast.net wrote: The goal: to create a function modeled after `subset` (notorious for its non-standard evaluation) that will take a series of logical tests as unqiuoted expressions to be evaluated in the framework of a dataframe environment and return a dataframe of logicals: ... A belated peep from the author of subset(): Don't! I think we learned the hard way by now that it is much easier to pass unevaluated expressions in the shape of formula objects or maybe expression objects. Lots of pain can be avoided by slipping in a simple ~. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ 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] Why does debugging print() change output of function?
On Sep 9, 2014, at 4:07 PM, peter dalgaard wrote: On 07 Sep 2014, at 00:31 , David Winsemius dwinsem...@comcast.net wrote: The goal: to create a function modeled after `subset` (notorious for its non- standard evaluation) that will take a series of logical tests as unquoted expressions to be evaluated in the framework of a dataframe environment and return a dataframe of logicals: ... A belated peep from the author of subset(): Don't! I think we learned the hard way by now that it is much easier to pass unevaluated expressions in the shape of formula objects or maybe expression objects. Lots of pain can be avoided by slipping in a simple ~. It's taken me several years to understand why you are probably correct in this regard. I needed to learn that `~` is actually a function that creates a language object. is.function(`~`) [1] TRUE is.language( ~ x 5 x 10) [1] TRUE ... and that it's rather easy to extract the object somewhat like but not really an expression embedded in such an object: is.expression( ~ x 5 x 10) [1] FALSE is.call( (~ x 5 x 10)[2] ) [1] TRUE The task of learning the various types of language objects is not an easy one. -- David Winsemius, MD Alameda, CA, USA __ 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] Why does debugging print() change output of function?
The goal:
to create a function modeled after `subset` (notorious for its non-standard
evaluation) that will take a series of logical tests as unqiuoted expressions
to be evaluated in the framework of a dataframe environment and return a
dataframe of logicals:
mtest.data.frame -
function (x, ..., drop=FALSE)
{ tests - list(...); print(tests)
r - if (length(tests)==0)
stop(no 'tests')
else { cbind.data.frame(
lapply( tests, function(t){
e - substitute(t)
r - eval(e, x, parent.frame() )
if ( !is.logical(r) ) {
stop('tests' must be logical) }
r !is.na(r) } ) )
}
}
#--
testdata - structure(list(group1 = structure(1:7, .Label = c(Group A,
Group B, Group C, Group D, Group E, Group F, Group G
), class = factor), group2 = structure(c(3L, 3L, 2L, 1L, 1L,
2L, 3L), .Label = c(LS, SS, UNC), class = factor), valid1 =
structure(c(2L,
1L, NA, 1L, 2L, 2L, 1L), .Label = c(N, Y), class = factor),
valid2 = structure(c(1L, 1L, 2L, 1L, 1L, 2L, 1L), .Label = c(N,
Y), class = factor), valid3 = structure(c(4L, 3L, NA,
2L, 1L, NA, 5L), .Label = c(0.3, 0.7, 1.2, 1.4, 1.7
), class = factor), valid4 = structure(c(2L, 1L, 3L, 4L,
1L, 1L, 5L), .Label = c(0.3, 0.4, 0.53, 0.66, 0.71
), class = factor), valid5 = structure(c(4L, 1L, NA, NA,
3L, NA, 2L), .Label = c(11.2, 11.7, 8.3, 8.5), class = factor)),
.Names = c(group1,
group2, valid1, valid2, valid3, valid4, valid5), row.names = c(NA,
-7L), class = data.frame)
###
mtest.data.frame(testdata, valid2==N, valid3 1)
[[1]]
[1] tests are
[[2]]
[1] TRUE TRUE FALSE TRUE TRUE FALSE TRUE
[[3]]
[1] TRUE TRUENA FALSE FALSENA TRUE
This actually seemed to be somewhat successful, but when ...
Now if I take out the `print()` call for 'tests', I get an different answer:
mtest.data.frame -
+ function (x, ..., drop=FALSE)
+ { tests - list(...)
+ r - if (length(tests)==0)
+ stop(no 'tests')
+ else { cbind.data.frame(
+ lapply( tests, function(t){
+ e - substitute(t)
+ r - eval(e, x, parent.frame() )
+ if ( !is.logical(r) ) {
+ stop('tests' must be logical) }
+ r !is.na(r) } ) )
+ }
+ }
mtest.data.frame(testdata, valid2==N, valid3 1)
# i.e. no answer
--
David Winsemius
Alameda, CA, USA
__
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] Why does debugging print() change output of function?
On Sat, 6 Sep 2014 15:31:49 -0700 David Winsemius dwinsem...@comcast.net wrote: The only other difference I see is a missing semicolon in the second example, which, in the first precedes your print() instruction. JWDougherty __ 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] Why does debugging print() change output of function?
In your first example I get an error:
mtest.data.frame(testdata, valid2==N, valid3 1)
Error in mtest.data.frame(testdata, valid2 == N, valid3 1) :
object 'valid2' not found
I expect the error because list(...) ought to evaluate the ... arguments.
Use substitute() to get the unevaluated ... arguments up front and
don't use substitute() in the loop over the elements of test.
There are several ways to get the unevaluated ... arguments. E.g.,
f0 - function(x, ..., drop=FALSE) match.call(expand.dots=FALSE)$...
f1 - function(x, ..., drop=FALSE) substitute(...())
f2 - function(x, ..., drop=FALSE) as.list(substitute(list(...)))[-1]
Your function could be the following, where I also fixed a problem
with parent.frame() being
called in the wrong scope and improved, IMO, the names on the output data.frame.
m2 - function (x, ..., drop = FALSE, verbose = FALSE)
{
tests - substitute(...())
nms - names(tests) # fix up names, since data.frame makes ugly ones
if (is.null(nms)) {
names(tests) - paste0(T, seq_along(tests))
}
else if (any(nms == )) {
names(tests)[nms == ] - paste0(T, which(nms == ))
}
if (verbose) {
print(tests)
}
r - if (length(tests) == 0) {
stop(no 'tests')
}
else {
enclos - parent.frame() # evaluate parent.frame() outside of FUN()
data.frame(lapply(tests, FUN=function(e) {
r - eval(e, x, enclos)
if (!is.logical(r)) {
stop('tests' must be logical)
}
r !is.na(r)
}))
}
r
}
used as:
m2(testdata, group2==UNC, Eleven.Two=valid5==11.2)
T1 Eleven.Two
1 TRUE FALSE
2 TRUE TRUE
3 FALSE FALSE
4 FALSE FALSE
5 FALSE FALSE
6 FALSE FALSE
7 TRUE FALSE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Sep 6, 2014 at 3:31 PM, David Winsemius dwinsem...@comcast.net wrote:
The goal:
to create a function modeled after `subset` (notorious for its
non-standard evaluation) that will take a series of logical tests as
unqiuoted expressions to be evaluated in the framework of a dataframe
environment and return a dataframe of logicals:
mtest.data.frame -
function (x, ..., drop=FALSE)
{ tests - list(...); print(tests)
r - if (length(tests)==0)
stop(no 'tests')
else { cbind.data.frame(
lapply( tests, function(t){
e - substitute(t)
r - eval(e, x, parent.frame() )
if ( !is.logical(r) ) {
stop('tests' must be logical) }
r !is.na(r) } ) )
}
}
#--
testdata - structure(list(group1 = structure(1:7, .Label = c(Group A,
Group B, Group C, Group D, Group E, Group F, Group G
), class = factor), group2 = structure(c(3L, 3L, 2L, 1L, 1L,
2L, 3L), .Label = c(LS, SS, UNC), class = factor), valid1 =
structure(c(2L,
1L, NA, 1L, 2L, 2L, 1L), .Label = c(N, Y), class = factor),
valid2 = structure(c(1L, 1L, 2L, 1L, 1L, 2L, 1L), .Label = c(N,
Y), class = factor), valid3 = structure(c(4L, 3L, NA,
2L, 1L, NA, 5L), .Label = c(0.3, 0.7, 1.2, 1.4, 1.7
), class = factor), valid4 = structure(c(2L, 1L, 3L, 4L,
1L, 1L, 5L), .Label = c(0.3, 0.4, 0.53, 0.66, 0.71
), class = factor), valid5 = structure(c(4L, 1L, NA, NA,
3L, NA, 2L), .Label = c(11.2, 11.7, 8.3, 8.5), class =
factor)), .Names = c(group1,
group2, valid1, valid2, valid3, valid4, valid5), row.names = c(NA,
-7L), class = data.frame)
###
mtest.data.frame(testdata, valid2==N, valid3 1)
[[1]]
[1] tests are
[[2]]
[1] TRUE TRUE FALSE TRUE TRUE FALSE TRUE
[[3]]
[1] TRUE TRUENA FALSE FALSENA TRUE
This actually seemed to be somewhat successful, but when ...
Now if I take out the `print()` call for 'tests', I get an different answer:
mtest.data.frame -
+ function (x, ..., drop=FALSE)
+ { tests - list(...)
+ r - if (length(tests)==0)
+ stop(no 'tests')
+ else { cbind.data.frame(
+ lapply( tests, function(t){
+ e - substitute(t)
+ r - eval(e, x, parent.frame() )
+ if ( !is.logical(r) ) {
+ stop('tests' must be logical) }
+ r !is.na(r) } ) )
+ }
+ }
mtest.data.frame(testdata, valid2==N, valid3 1)
# i.e. no answer
--
David Winsemius
Alameda, CA, USA
__
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
Re: [R] Why does debugging print() change output of function?
On Sep 6, 2014, at 4:24 PM, William Dunlap wrote:
In your first example I get an error:
mtest.data.frame(testdata, valid2==N, valid3 1)
Error in mtest.data.frame(testdata, valid2 == N, valid3 1) :
object 'valid2' not found
I expect the error because list(...) ought to evaluate the ... arguments.
Thank you (and JWDougherty) for looking at this. I see that the difference lies
in the fact that I have vectors in my workspace that were used as
preliminaries in constructing my test case that are being accessed by my
logical expressions.
group1 - paste(Group, rep(LETTERS[1:7], sep=''))
group2 - c(UNC, UNC, SS, LS, LS, SS, UNC)
valid1 - c(Y, N, NA, N, Y, Y, N)
valid2 - c(N, N, Y, N, N, Y, N)
valid3 - c(1.4, 1.2, NA, 0.7, 0.3, NA, 1.7)
valid4 - c(0.4, 0.3, 0.53, 0.66, 0.3, 0.3, 0.71)
valid5 - c(8.5, 11.2,NA, NA, 8.3, NA, 11.7)
I should have executed rm(list=ls()) and repeated my testing before posting,
but you
Use substitute() to get the unevaluated ... arguments up front and
don't use substitute() in the loop over the elements of test.
There are several ways to get the unevaluated ... arguments. E.g.,
f0 - function(x, ..., drop=FALSE) match.call(expand.dots=FALSE)$...
f1 - function(x, ..., drop=FALSE) substitute(...())
f2 - function(x, ..., drop=FALSE) as.list(substitute(list(...)))[-1]
These three version are somewhat confusing, the second one in particular makes
it appear that the ellipsis is a function, while the other ones make it appear
that they are an expression pointing to a list.
Your function could be the following, where I also fixed a problem
with parent.frame() being
called in the wrong scope and improved,
Yes, I was worried about that.
IMO, the names on the output data.frame.
m2 - function (x, ..., drop = FALSE, verbose = FALSE)
{
tests - substitute(...())
nms - names(tests) # fix up names, since data.frame makes ugly ones
if (is.null(nms)) {
names(tests) - paste0(T, seq_along(tests))
}
else if (any(nms == )) {
names(tests)[nms == ] - paste0(T, which(nms == ))
}
if (verbose) {
print(tests)
}
r - if (length(tests) == 0) {
stop(no 'tests')
}
else {
enclos - parent.frame() # evaluate parent.frame() outside of FUN()
data.frame(lapply(tests, FUN=function(e) {
r - eval(e, x, enclos)
if (!is.logical(r)) {
stop('tests' must be logical)
}
r !is.na(r)
}))
}
r
}
used as:
m2(testdata, group2==UNC, Eleven.Two=valid5==11.2)
T1 Eleven.Two
1 TRUE FALSE
2 TRUE TRUE
3 FALSE FALSE
4 FALSE FALSE
5 FALSE FALSE
6 FALSE FALSE
7 TRUE FALSE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
Thank you again, Bill.
--
David.
On Sat, Sep 6, 2014 at 3:31 PM, David Winsemius dwinsem...@comcast.net
wrote:
The goal:
to create a function modeled after `subset` (notorious for its
non-standard evaluation) that will take a series of logical tests as
unqiuoted expressions to be evaluated in the framework of a dataframe
environment and return a dataframe of logicals:
mtest.data.frame -
function (x, ..., drop=FALSE)
{ tests - list(...); print(tests)
r - if (length(tests)==0)
stop(no 'tests')
else { cbind.data.frame(
lapply( tests, function(t){
e - substitute(t)
r - eval(e, x, parent.frame() )
if ( !is.logical(r) ) {
stop('tests' must be logical) }
r !is.na(r) } ) )
}
}
#--
testdata - structure(list(group1 = structure(1:7, .Label = c(Group A,
Group B, Group C, Group D, Group E, Group F, Group G
), class = factor), group2 = structure(c(3L, 3L, 2L, 1L, 1L,
2L, 3L), .Label = c(LS, SS, UNC), class = factor), valid1 =
structure(c(2L,
1L, NA, 1L, 2L, 2L, 1L), .Label = c(N, Y), class = factor),
valid2 = structure(c(1L, 1L, 2L, 1L, 1L, 2L, 1L), .Label = c(N,
Y), class = factor), valid3 = structure(c(4L, 3L, NA,
2L, 1L, NA, 5L), .Label = c(0.3, 0.7, 1.2, 1.4, 1.7
), class = factor), valid4 = structure(c(2L, 1L, 3L, 4L,
1L, 1L, 5L), .Label = c(0.3, 0.4, 0.53, 0.66, 0.71
), class = factor), valid5 = structure(c(4L, 1L, NA, NA,
3L, NA, 2L), .Label = c(11.2, 11.7, 8.3, 8.5), class =
factor)), .Names = c(group1,
group2, valid1, valid2, valid3, valid4, valid5), row.names =
c(NA,
-7L), class = data.frame)
###
mtest.data.frame(testdata, valid2==N, valid3 1)
[[1]]
[1] tests are
[[2]]
[1] TRUE TRUE FALSE TRUE TRUE FALSE TRUE
[[3]]
[1] TRUE TRUENA FALSE FALSENA TRUE
This actually seemed to be somewhat successful, but when ...
Now if I take out the `print()` call for 'tests', I get an different answer:
mtest.data.frame -
+ function (x, ..., drop=FALSE)
+ { tests - list(...)
+ r - if
