On 2011-09-12 21:16, Artur Wroblewski wrote:
> On Mon, Sep 12, 2011 at 12:26 PM, Laurent Gautier<lgaut...@gmail.com> wrote:
>> Probably not.
>>
>> R is doing a lot of things behind the hood. Sometimes it is good, sometimes
>> it is bad.
>> The code snippet given to you has a quadratic time-complexity ( O(nm) ). It
>> can be make linearithmic ( O(n log(m) ) ) simply:
>>
>> from rpy2.robjects.vector import BoolVector
>> ref = set(differential)
>> select_b = BoolVector(tuple(x in ref for x in source.rx2('gene')))
>> mysubset = source.rx(select_b, True)
> If I reckon well BoolVector(...genexpr...) is not possible here due to
> R API limitation - we need length of iterable, isn't it?
You might have missed the call to tuple(). This will make it work
independently of having a generator with a length.
>>> tuple(x for x in [1,2,3])
(1, 2, 3)
> It seems like copying R on Python level is not always nice
> and can be quite inefficient.
>
> Above probably could be rewritten in more Pythonic way (it would
> be more efficient I believe, as well)
>
> mysubset = source.rx((x in ref for x in source.rx2('gene')), True)
>
> or
>
> mysubset = DataFrame(row for row in source if row['gene'] in ref)
>
> but of course is not supported by rpy2.
>
> Is there a chance to make rpy2 bit more Python integrated? :)
As Luca wrote it earlier, what is missing for it to work is that the
generator had a length.
This can probably be addressed by adding a custom iterator for R
vectors, and should I appear a little slow to have it implemented you
are welcome to submit a patch. ;-)
Best,
L.
> Best regards,
>
> w
>
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