Well, whether you factor out the loop-function is a separate issue. Lets
say we do:
smooth_signal = [average = compute_avg(average, x) for x in signal from
average=0]
Is just as readable and maintainable as your expanded version, but saves 4
lines of code. What's not to love?
On Thu, Apr 5, 2018 at 5:55 PM, Paul Moore <p.f.mo...@gmail.com> wrote:
> On 5 April 2018 at 22:26, Peter O'Connor <peter.ed.ocon...@gmail.com>
> wrote:
> > I find this a bit awkward, and maintain that it would be nice to have
> this
> > as a built-in language construct to do this natively. You have to admit:
> >
> > smooth_signal = [average = (1-decay)*average + decay*x for x in
> signal
> > from average=0.]
> >
> > Is a lot cleaner and more intuitive than:
> >
> > dev compute_avg(avg, x):
> > return (1 - decay)*avg + decay * x
> >
> > smooth_signal =
> > itertools.islice(itertools.accumulate(itertools.chain([initial_average],
> > signal), compute_avg), 1, None)
>
> Not really, I don't... In fact, factoring out compute_avg() is the
> first step I'd take in converting the proposed syntax into something
> I'd find readable and maintainable. (It's worth remembering that when
> you understand the subject of the code very well, it's a lot easier to
> follow complex constructs, than when you're less familiar with it -
> and the person who's unfamiliar with it could easily be you in a few
> months).
>
> The string of itertools functions are *not* readable, but I'd fix that
> by expanding them into an explicit loop:
>
> smooth_signal = []
> average = 0
> for x in signal:
> average = compute_avg(average, x)
> smooth_signal.append(average)
>
> If I have that wrong, it's because I misread *both* the itertools
> calls *and* the proposed syntax. But I doubt anyone would claim that
> it's possible to misunderstand the explicit loop.
>
> > Moreover, if added with the "last" builtin proposed in the link, it could
> > also kill the need for reduce, as you could instead use:
> >
> > last_smooth_signal = last(average = (1-decay)*average + decay*x for
> x in
> > signal from average=0.)
>
> last_smooth_signal = 0
> for x in signal:
> last_smooth_signal = compute_avg(last_smooth_signal, x)
>
> or functools.reduce(compute_avg, signal, 0), if you prefer reduce() -
> I'm not sure I do.
>
> Sorry, this example has pretty much confirmed for me that an explicit
> loop is *far* more readable.
>
> Paul.
>
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