Hi Zach,
On 29.05.2017 22:57, Zach Morris wrote:
> Hi Marcus,
>
> Sorry about the trouble with Nabble, I see that my code got cut off. I
> will communicate via the mailing list from now on.
>
> If I understand your explanation correctly, the forecast function
> below tells the scheduler that this block needs at least noutput_items
> on each input port in order to produce noutput_items. In actuality, my
> arbitrary ratio block would produce anywhere between (0,
> noutput_items); would this be a valid implementation of forecast() for
> my arbitrary ratio block?
> def forecast(self, noutput_items, ninput_items_required):
> for i in range(len(ninput_items_required)):
> ninput_items_required[i] = noutput_items
>
yes, that would tell the scheduler that you'll need at least
noutput_items input items to produce noutput_items output items, indeed!
Now, the forecast really is just that: a forecast. If you can't produce
as many samples as promised, the scheduler won't be upset.
>
> This is my general_work() function right now: it just outputs the
> elements in the input vector that are above a threshold. My eventual
> goal is to receive a vector from an FFT and update statistics for each
> frequency 'bin' as long as its power is above a threshold, then output
> the statistics when its power drops below the threshold.
> def general_work(self, input_items, output_items):
> in0 = input_items[0][:len(output_items[0])]
This is **only** OK if you check that len(output_items[0]) <=
len(input_items[0]). You inherently do that with your forecast, but I'd
really recommend explicitly checking that condition.
Also, this the following code is really very redundant:
> out0 = output_items[0]
> ninput_items = len(in0)
well, you set in0 to be of len(output_items[0]).
>
> j = 0
> for i in range(0, ninput_items):
this is very un-pythonic.
> if (in0[i] > self.threshold).all():
> out0[j] = in0[i]
> j += 1
>
I'd just say:
def general_work(self, input_items, output_items):
super_threshold = filter(lambda x: x > self.threshold, in[0])
output_items[0][:] = super_threshold
self.consume_each(len(super_threshold))
return len(super_threshold)
Best regards,
Marcus
> self.consume(0, ninput_items)
> self.produce(0, j)
>
> return 0
>
> If I again understand your explanation, the 'self.consume(0,
> ninput_items)' is correct, because I 'use up' the elements of
> input_items on each loop. However, it sounds like I should either
> return WORK_CALLED_PRODUCE or return ninput_tems. Am I on the right
> track here? Is there a reason to prefer produce() over just returning
> the number of input items I consume?
>
> Thank you for your help.
>
> Zach
>
>
> On Sun, May 28, 2017 at 1:31 PM, Marcus Müller
> <marcus.muel...@ettus.com <mailto:marcus.muel...@ettus.com>> wrote:
>
> Hi Zach,
>
> sorry it took me so long to react:
> so, let's dissect a few things:
>
> First, the general purposes of these functions:
>
> * forecast() is called by the *scheduler* to ask your block "hey,
> if I'd
> need you to produce N items, how much would you need on your 0.
> (and 1.,
> and 2.,… if existing) input for that?". The scheduler usually
> repeatedly
> calls that with a decreasing N until the number of samples your block
> requires can be fulfilled by the number of input items that the
> scheduler has ready. If that never happens, usually, something is
> broken.
>
> * consume() is something that *you* call from within a
> (general_)work()
> to signal the scheduler how many items you've consumed from the
> input(s).
>
> * produce() is something that *you* call from within a
> (general_)work()
> to signal the scheduler how many items you've put into the output
> buffer. *If* you use produce, you should return the magic
> WORK_CALLED_PRODUCE values. You usually don't call produce() – you
> just
> return the number of consumed samples instead.
>
> * noutput_items is **not** a variable that you set – it's the info how
> many items you are asked to produce (parameter to forecast()) or how
> many items you're **allowed at most** to produce (when it's the
> parameter to a (general_)work())
>
> Again, the better parts of your mail have been broken by Nabble.
> Abandon
> Nabble. Subscribe directly:
>
> https://lists.gnu.org/mailman/listinfo/discuss-gnuradio
> <https://lists.gnu.org/mailman/listinfo/discuss-gnuradio>
>
> Hope that helps,
> Marcus
>
> On 05/26/2017 07:49 PM, Zach Morris wrote:
> > That worked to create a general block, but I'm still having some
> trouble with
> > implementing the forecast(), consume() and produce() functions.
> >
> > The goal of the block is to discard elements in an input vector
> that are
> > below a certain threshold, and eventually update some statistics
> based on
> > the remaining elements. For a vector of 256 elements, there could be
> > anywhere between 0 and 256 outputs (the arbitrary ratio you
> mentioned).
> >
> > The default forecast function has a 1:1 ratio:
> >
> > Should I replace "noutput_items" with my "vector_length" parameter,
> ensuring
> > that we always get (for example) 256 inputs?
> >
> > In the general_work function, I think we want to consume the
> entire input
> > vector once we figure out which elements are above the threshold:
> >
> > I'm not sure about the produce function; do I need to tell the system
> how
> > many items I produce, or is that covered at the end of general_work:
> >
> >
> > Thank you kindly,
> >
> > Zach
> >
> >
> >
> >
> > --
> > View this message in context:
> http://gnuradio.4.n7.nabble.com/Python-block-help-tp51706p64062.html
> <http://gnuradio.4.n7.nabble.com/Python-block-help-tp51706p64062.html>
> > Sent from the GnuRadio mailing list archive at Nabble.com.
> >
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