Hi Barry,
this actually a good idea, to put them together! Probably even creating an
object containing both of them. Haven't thought about it before.
Best
Simon
On Jan 31, 2013, at 3:49 PM, Barry Rowlingson
wrote:
> On Thu, Jan 31, 2013 at 11:52 AM, Simon Zehnder wrote:
>> Dear R community,
Dear Barry,
thank you very much for this information. This looks pretty interesting!
Best
Simon
On Jan 31, 2013, at 4:09 PM, Barry Rowlingson
wrote:
>> it lets you do:
>>
>> (a~b~c) = foo()
>
> Mistook. should be:
>
> (a~b~c) %=% foo()
>
> because it defines the %=% operator.
>
> Barry
On Thu, Jan 31, 2013 at 6:52 AM, Simon Zehnder wrote:
> Dear R community,
>
> I do know, that an R function is constructing a copy of any object passed as
> argument into a function. I program on a larger S4 project for a package, and
> I arrived at a point where I have to think a little harder
On Thu, Jan 31, 2013 at 3:00 PM, Simon Zehnder wrote:
> Hi Barry,
>
> this actually a good idea, to put them together! Probably even creating an
> object containing both of them. Haven't thought about it before.
>
Hadley W asked for implementations of 'unstructuring assignments' the
other day,
> it lets you do:
>
> (a~b~c) = foo()
Mistook. should be:
(a~b~c) %=% foo()
because it defines the %=% operator.
Barry
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On Thu, Jan 31, 2013 at 11:52 AM, Simon Zehnder wrote:
> Dear R community,
>
> I do know, that an R function is constructing a copy of any object passed as
> argument into a function. I program on a larger S4 project for a package, and
> I arrived at a point where I have to think a little harder
Dear R community,
I do know, that an R function is constructing a copy of any object passed as
argument into a function. I program on a larger S4 project for a package, and I
arrived at a point where I have to think a little harder on implementation
style (especially to spare users complex obje
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