Re: [R] How to pass selection criteria in a function
On Wed, 1 Dec 2010, cmccar...@bmcc.cuny.edu wrote:
Hi,
Suppose I have the following data
name score
Abel??? 88
Baker? 54
Charlie??? 77
stored a? table called myData.
I want to write a function that will create a table which is a subset of myData
containing those have a score 75.
I know I can do this with the following command:
subset(myData, score 75)
But I would like to do this via a function, something like:
newTable - function( data,? criteria){
??? subset( data, criteria) }
and then calling: newTable(myData, score 75)
But this doesn't work. I am sure there is a simple way to do this,
but I am stuck! Please help. Thanks!
Simple? Maybe not so much!
You are trying to pass objects without evaluating them. subset is rather
special in the way it works. Here is one way:
foo - function(x,...){
+ mc - match.call()
+ mc[[1]] - as.name(subset)
+ eval(mc)
+ }
foo(iris, Petal.Width2.4 )
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
101 6.3 3.3 6.0 2.5 virginica
110 7.2 3.6 6.1 2.5 virginica
145 6.7 3.3 5.7 2.5 virginica
Reading the code at the top of lm shows how this kind of strategy can be
used.
You might look at
?bquote
for another way to skin this cat.
HTH,
Chuck
Chris?McCarthy, PhD
Department?of?Mathematics?BMCC
Office:?N522???Extension:?5235
cmccar...@bmcc.cuny.edu
[[alternative HTML version deleted]]
Charles C. BerryDept of Family/Preventive Medicine
cbe...@tajo.ucsd.eduUC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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Re: [R] How to pass selection criteria in a function
Try this:
newTable - function(data, criteria) {
do.call(subset, list(data, substitute(criteria)))
}
On Wed, Dec 1, 2010 at 3:56 PM, cmccar...@bmcc.cuny.edu wrote:
Hi,
Suppose I have the following data
name score
Abel88
Baker 54
Charlie77
stored a table called myData.
I want to write a function that will create a table which is a subset of
myData containing those have a score 75.
I know I can do this with the following command:
subset(myData, score 75)
But I would like to do this via a function, something like:
newTable - function( data, criteria){
subset( data, criteria) }
and then calling: newTable(myData, score 75)
But this doesn't work. I am sure there is a simple way to do this,
but I am stuck! Please help. Thanks!
Chris McCarthy, PhD
Department of Mathematics BMCC
Office: N522 Extension: 5235
cmccar...@bmcc.cuny.edu
[[alternative HTML version deleted]]
__
R-help@r-project.org 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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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Re: [R] How to pass selection criteria in a function
Thank you all for your very fast replies. I tried Henrique's method (see one of the above posts) , and it works perfectly! Thanks again! -- View this message in context: http://r.789695.n4.nabble.com/How-to-pass-selection-criteria-in-a-function-tp3067765p3067829.html Sent from the R help mailing list archive at Nabble.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] How to pass selection criteria in a function
On Dec 1, 2010, at 1:12 PM, Charles C. Berry wrote:
On Wed, 1 Dec 2010, cmccar...@bmcc.cuny.edu wrote:
Hi,
Suppose I have the following data
name score
Abel88
Baker 54
Charlie77
stored a table called myData.
I want to write a function that will create a table which is a
subset of myData containing those have a score 75.
I know I can do this with the following command:
subset(myData, score 75)
But I would like to do this via a function, something like:
newTable - function( data, criteria){
subset( data, criteria) }
and then calling: newTable(myData, score 75)
But this doesn't work. I am sure there is a simple way to do this,
but I am stuck! Please help. Thanks!
Simple? Maybe not so much!
You are trying to pass objects without evaluating them. subset is
rather special in the way it works. Here is one way:
foo - function(x,...){
+ mc - match.call()
+ mc[[1]] - as.name(subset)
+ eval(mc)
+ }
foo(iris, Petal.Width2.4 )
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
101 6.3 3.3 6.0 2.5 virginica
110 7.2 3.6 6.1 2.5 virginica
145 6.7 3.3 5.7 2.5 virginica
I've never really understood how to use browser() to develop or
resolve problems with function definitions. Chuck's comments pointing
to the subset function offered another opportunity to teach myself
something new. Here's the console transcript:
newTable - function( data, criteria){ crit -
substitute(criteria);browser()}
# Obviously not a complete function
newT - newTable(myData, score 75)
Called from: newTable(myData, score 75) #so far so good
Browse[1] criteria # I realize now that this should
have been crit
Error: object 'score' not found # took another look at subset()
code
Browse[1] newTable - function( data, criteria){ crit -
substitute(criteria); logvec - eval(crit, data, parent.frame());
browser()}
Browse[1] criteria # still didn't catch on that crit
was the local object to examine
Error: object 'score' not found # just slow I guess.
In addition: Warning message:
restarting interrupted promise evaluation
Browse[1] c # was worried the the redefinition might not take
effect at the top-level
# Try # 2
newTable - function( data, criteria){ crit -
substitute(criteria); logvec - eval(crit, data, parent.frame());
browser()}
newT - newTable(myData, score 75)
Called from: newTable(myData, score 75)
Browse[1] logvec
[1] FALSE TRUE FALSE# now we're getting results
Browse[1] return(data[logvec, ])# see if the naive next step works
newT
name score
2 Baker54# very promising
newTable - function( data, criteria){ crit -
substitute(criteria); logvec - eval(crit, data, parent.frame());
return(data[logvec, ])}
newT - newTable(myData, score 75)
newT# SUCCESS
name score
2 Baker54
There is not much in the way of error checking, but it seems to be a
reasonable start (and looks to offer an example, albeit with a some
newbie errors, of an extremely useful R tool.)
Reading the code at the top of lm shows how this kind of strategy
can be used.
Charles C. BerryDept of Family/
Preventive Medicine
Thanks, Chuck.
--
David Winsemius, MD
West Hartford, CT
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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.
