[Rd] NULL assignment will change the expression's class into list

[Duncan Murdoch]

Vitalie Spinu wrote:
> On Fri, Oct 8, 2010 at 7:49 PM, Duncan Murdoch <murdoch.duncan at gmail.com>wrote:
>
>   
>>  On 08/10/2010 12:24 PM, Vitalie Spinu wrote:
>>
>>     
>>> On Fri, Oct 8, 2010 at 12:14 PM, Duncan Murdoch<murdoch.duncan at gmail.com
>>>       
>>>> wrote:
>>>>         
>>>>  Vitalie Spinu wrote:
>>>>
>>>>         
>>>>>  Hello Everyone!
>>>>>
>>>>>  NULL replacement will change expression object into list:
>>>>>
>>>>>
>>>>>
>>>>>           
>>>>>>  te<- expression(a=23*4, b=33-2)
>>>>>>  te
>>>>>>
>>>>>>
>>>>>>             
>>>>>  expression(a = 23 * 4, b = 33 - 2)
>>>>>
>>>>>
>>>>>
>>>>>           
>>>>>>  te[["a"]]<- quote(blabla) #ok
>>>>>>  te
>>>>>>
>>>>>>
>>>>>>             
>>>>>  expression(a = blabla, b = 33 - 2)
>>>>>
>>>>>
>>>>>
>>>>>           
>>>>>>  te[["a"]]<- NULL #change to list
>>>>>>  te
>>>>>>
>>>>>>
>>>>>>             
>>>>>  $b
>>>>>  33 - 2
>>>>>
>>>>>  I am on w32, version 2.11.1 (2010-05-31)
>>>>>
>>>>>
>>>>>           
>>>>  That's certainly an inconsistency, still present in a recent R-devel
>>>>         
>>> (but I
>>>       
>>>>  haven't checked the latest beta).  I don't know if it's a bug:  NULL
>>>>  assignments are handled specially in other situations (e.g. if te was a
>>>>         
>>> list
>>>       
>>>>  to start, the NULL assignment would remove the "a" entry).
>>>>
>>>>  A simple workaround is to use
>>>>
>>>>  te["a"]<- expression(NULL)
>>>>
>>>>  or te<- te[-1]
>>>>
>>>>  instead, depending on what you expected to happen.
>>>>
>>>>         
>>> As ussual with NULL assignment in recursive structures, I would expect to
>>> remove the elements altogether. And this is exactly what I need.
>>>
>>> I would say it's a bug, because NULL assignment in data.frames would not
>>> convert them to lists, for example.
>>>
>>>       
>> I think you're probably right.
>>
>>  Thanks for looking into it. It's quite inconvenient when you have to
>>     
>>> manipulate named expression. Have to use constructs like
>>> et<-et[!names(et)%in%"a"].
>>>
>>>       
>> Or simply follow te["a"] <- NULL
>>
>> with
>>
>> te <- as.expression(te)
>>
>> This is a pretty fast operation if te is an expression or a list formed by
>> mistaken conversion from one.
>>
>> D
>>     
>
> But is this a reliable way to do it? I have been struggling some time ago
> with this type of conversion.
> Can not thing of an example now. But as far as I could remember, the
> conversion to list of an expression is not always reversible with
> as.expression. Am I wrong?
>   

I don't know of any examples, but this is your construction.  I think 
it's very unlikely this will be fixed for 2.12.0, but it will probably 
be fixed for 2.12.1, if

1.  There is a 2.12.1

and

2.  It really isn't intentional behaviour.


Duncan Murdoch
>
>   
>> uncan Murdoch
>>
>>
>>  Vitally.
>>     
>>>>  Duncan Murdoch
>>>>
>>>>   Regards,
>>>>         
>>>>>  Vitally.
>>>>>
>>>>>         [[alternative HTML version deleted]]
>>>>>
>>>>>  ______________________________________________
>>>>>  R-devel at r-project.org mailing list
>>>>>  https://stat.ethz.ch/mailman/listinfo/r-devel
>>>>>
>>>>>
>>>>>           
>>>>         
>>>       
>
>