On 08/10/2010 12:24 PM, Vitalie Spinu wrote:
On Fri, Oct 8, 2010 at 12:14 PM, Duncan Murdoch<murdoch.dun...@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.
Duncan Murdoch
Vitally.
>
> Duncan Murdoch
>
> Regards,
>> Vitally.
>>
>> [[alternative HTML version deleted]]
>>
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>
>
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