[R] lapply not reading arguments from the correct environment
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
I am building a list of dataframes, in each of which I want to keep
only column 2 (obviously I would not do it this way in real life but
that's just to demonstrate the bug).
If I execute the commands inline it works but if I clean my
environment, then define the function and then execute:
myfun()
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply. In
addition, if I define it outside the function and then execute the
function:
fooCollumn=1
myfun()
it works but uses the value defined in the general environment and
not the one defined in the function.
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
Thanks in advance.
JiHO
---
http://jo.irisson.free.fr/
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Re: [R] lapply not reading arguments from the correct environment
subset() was not defined inside myfun(); try this version instead:
myfun - function () {
foo - data.frame(1:10, 10:1)
foos - list(foo)
fooCollumn - 2
my.subset - function(...) subset(...)
cFoo - lapply(foos, my.subset, select = fooCollumn)
cFoo
}
myfun()
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: jiho [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 4:41 PM
Subject: [R] lapply not reading arguments from the correct environment
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
I am building a list of dataframes, in each of which I want to keep
only column 2 (obviously I would not do it this way in real life but
that's just to demonstrate the bug).
If I execute the commands inline it works but if I clean my
environment, then define the function and then execute:
myfun()
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply.
In
addition, if I define it outside the function and then execute the
function:
fooCollumn=1
myfun()
it works but uses the value defined in the general environment and
not the one defined in the function.
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
Thanks in advance.
JiHO
---
http://jo.irisson.free.fr/
__
R-help@stat.math.ethz.ch 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.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
On Fri, 18 May 2007, jiho wrote:
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
snip
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply.
snip
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
No, it isn't a bug (though it may be confusing).
The problem is that subset() evaluates its select argument in an unusual
way. Usually the argument would be evaluated inside myfun() and the value
passed to lapply(), and everything would work as you expect.
subset() bypasses the normal evaluation and explicitly evaluates the
select argument in the calling frame, ie, inside lapply(), where
fooCollumn is not visible.
You could do
lapply(foos, function(foo) subset(foo, select=fooCollum))
capturing fooCollum by lexical scope. In R this is often a better option
than passing extra arguments to lapply (or other functions that take
function arguments).
-thomas
__
R-help@stat.math.ethz.ch mailing list
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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] lapply not reading arguments from the correct environment
On 2007-May-18 , at 17:09 , Thomas Lumley wrote:
On Fri, 18 May 2007, jiho wrote:
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
snip
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply.
snip
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
The problem is that subset() evaluates its select argument in an
unusual way. Usually the argument would be evaluated inside myfun()
and the value passed to lapply(), and everything would work as you
expect.
subset() bypasses the normal evaluation and explicitly evaluates
the select argument in the calling frame, ie, inside lapply(),
where fooCollumn is not visible.
You could do
lapply(foos, function(foo) subset(foo, select=fooCollum))
capturing fooCollum by lexical scope. In R this is often a better
option than passing extra arguments to lapply (or other functions
that take function arguments).
Thank you very much, this works well indeed. I agree it is a bit
confusing, to say the least. The point is that supplying other
arguments in the ... of lapply worked for all other functions I tried
before (mean, sd, summary and even spline) so it is really a problem
with subset. Anyway, R is great even with such little flaws here and
there and as long as the community is there to support it, it will rule.
Cheers,
JiHO
---
http://jo.irisson.free.fr/
__
R-help@stat.math.ethz.ch 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] lapply not reading arguments from the correct environment
You need to study carefully what the semantics of 'subset' are. The
function body of myfun is not in the evaluation environment. (The issue
is 'subset', not 'lapply': select is an *expression* and not a value.)
Hint: using subset() programmatically is almost always a mistake. R's
subsetting function is '[': subset is a convenience wrapper.
On Fri, 18 May 2007, jiho wrote:
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
I am building a list of dataframes, in each of which I want to keep
only column 2 (obviously I would not do it this way in real life but
that's just to demonstrate the bug).
If I execute the commands inline it works but if I clean my
environment, then define the function and then execute:
myfun()
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply. In
addition, if I define it outside the function and then execute the
function:
fooCollumn=1
myfun()
it works but uses the value defined in the general environment and
not the one defined in the function.
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
It is a bug, in your function.
--
Brian D. Ripley, [EMAIL PROTECTED]
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, UKFax: +44 1865 272595
__
R-help@stat.math.ethz.ch 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] lapply not reading arguments from the correct environment
In particular, we can use [ directly instead of subset. This is the
same as your function except for the line marked ### :
myfun2 - function() {
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos, [, fooCollumn) ###
return(cFoo)
}
myfun2() # test
On 5/18/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
You need to study carefully what the semantics of 'subset' are. The
function body of myfun is not in the evaluation environment. (The issue
is 'subset', not 'lapply': select is an *expression* and not a value.)
Hint: using subset() programmatically is almost always a mistake. R's
subsetting function is '[': subset is a convenience wrapper.
On Fri, 18 May 2007, jiho wrote:
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun - function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
I am building a list of dataframes, in each of which I want to keep
only column 2 (obviously I would not do it this way in real life but
that's just to demonstrate the bug).
If I execute the commands inline it works but if I clean my
environment, then define the function and then execute:
myfun()
I get this error:
Error in eval(expr, envir, enclos) : object fooCollumn not found
while fooCollumn is defined, in the function, right before lapply. In
addition, if I define it outside the function and then execute the
function:
fooCollumn=1
myfun()
it works but uses the value defined in the general environment and
not the one defined in the function.
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
It is a bug, in your function.
--
Brian D. Ripley, [EMAIL PROTECTED]
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, UKFax: +44 1865 272595
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] lapply not reading arguments from the correct environment
On 2007-May-18 , at 18:21 , Gabor Grothendieck wrote:
In particular, we can use [ directly instead of subset. This is the
same as your function except for the line marked ### :
myfun2 - function() {
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos, [, fooCollumn) ###
return(cFoo)
}
myfun2() # test
On 5/18/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
You need to study carefully what the semantics of 'subset' are. The
function body of myfun is not in the evaluation environment. (The
issue
is 'subset', not 'lapply': select is an *expression* and not a
value.)
Hint: using subset() programmatically is almost always a mistake.
R's
subsetting function is '[': subset is a convenience wrapper.
Thank you very much. Indeed it is much better this way. I got used to
subset for data.frames because [ does not work with negative named
arguments while select does. E.g.:
x[,-c(name1,name2)]
does not work while
subset(x,select=-c(name1,name2))
works (it eliminates columns named name1 and name 2 from x). But I
guess in most cases an other syntax can achieve the same thing with
[, like:
x[,-which(names(x)%in%c(name1,name2))]
it's just a little less clear.
Thanks again.
JiHO
---
http://jo.irisson.free.fr/
__
R-help@stat.math.ethz.ch 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] lapply not reading arguments from the correct environment
On 5/18/07, jiho [EMAIL PROTECTED] wrote:
On 2007-May-18 , at 18:21 , Gabor Grothendieck wrote:
In particular, we can use [ directly instead of subset. This is the
same as your function except for the line marked ### :
myfun2 - function() {
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos, [, fooCollumn) ###
return(cFoo)
}
myfun2() # test
On 5/18/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
You need to study carefully what the semantics of 'subset' are. The
function body of myfun is not in the evaluation environment. (The
issue
is 'subset', not 'lapply': select is an *expression* and not a
value.)
Hint: using subset() programmatically is almost always a mistake.
R's
subsetting function is '[': subset is a convenience wrapper.
Thank you very much. Indeed it is much better this way. I got used to
subset for data.frames because [ does not work with negative named
arguments while select does. E.g.:
x[,-c(name1,name2)]
does not work while
subset(x,select=-c(name1,name2))
works (it eliminates columns named name1 and name 2 from x). But I
guess in most cases an other syntax can achieve the same thing with
[, like:
x[,-which(names(x)%in%c(name1,name2))]
it's just a little less clear.
which is not needed. Using builtin CO2:
CO2[ ! names(CO2) %in% c(Type, conc ]
or
CO2[ setdiff(names(CO2), c(Type, conc)) ]
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