ase')
> base.transform(DataFrame({'a': IntVector((1,2,3))}),
>a = parse('a+1'))
>
> Best,
>
>
> L.
>
>
>
Your snippet returns:
Out[10]:
[Vector]
a:
[Vector, Vector, Vector]
Guess each number in the original vector is jus
Hi all,
I would like to know if there is a way to pass an r expression to be
lazy evaluated by a function. I mean as in transform or within. For
example:
r.transform(r['data.frame'](a=r.c(1,2,3)), a=<>)
The part <> is the one I want to achieve somehow. I know I could just call:
r('transform(d
(s/mm/newcol/ in the error message above)
On Tue, Aug 21, 2012 at 2:25 PM, Carlos Pita wrote:
> Mh, this is interesting:
>
> dataf = r['[<-'](dataf, 'newcol', R.IntVector([1,2,3]))
>
> => Error in `[<-.data.frame`(list(a = 1:3, b = 4:6, mm = 1:3)
af = r['$<-'](dataf, 'newcol', R.IntVector([1,2,3]))
=> OK
I think what is needed is a rx, rx2 equivalent for $. Is there
something like that in rpy2?
Regards
--
Carlos
On Tue, Aug 21, 2012 at 1:56 PM, Carlos Pita wrote:
> Hi all,
>
>
>>>
Hi all,
>>empty.rx2["New"] = [str(item) for item in range(len(empty.rownames))]
>> RRuntimeError: Error in `[[<-.data.frame`(list(), "New", list("0", "1",
>> "2",
"3", "4", : argument "value" is missing, with no default
>> Anything I'm doing wrong?
> Not necessarily; I think
