21.12.2018 23:13, John Snow wrote: > > > On 12/21/18 7:41 AM, Vladimir Sementsov-Ogievskiy wrote: >> Hmm. This made me check, is enumerate applicable to dicts, >> and, yes it is. >> > > enumerate on dicts gives you a numerical index paired with the key, > so... k is the numerical index and v is the key.
hmm, yes, I was wrong. Ok, than I'm fine with your code. Take my r-b. > >> so, it may be as easy as >> >> for k, v in enumerate(qmsg): >> if isinstance(v, list) or isinstance(v, dict): >> qmsg[k] = filter_qmp(v, filter_fn) > > ^ this will break if qmsg was a dict. > >> else: >> qmsg[k] = filter_fn(k, v) >> >> with this: >> Reviewed-by: Vladimir Sementsov-Ogievskiy <vsement...@virtuozzo.com> >> >> >> But does it what you want? >> So, for lists of strings, filter_fn will get pair of index in array >> and value? >> On the other hand, we can adjust the behavior as we want when we introduce >> such filters. > > I'm not sure what I want... > > Named index is the best, but isn't always available because like you > say, the object we are passed is not always a dictionary. We could > theoretically have a list of lists somewhere. What key would I pass > then? It's meaningless. > > I could change qmp_filter to pass the k,v to the filter when it finds > the "last dictionary," i.e. if it recursively finds no more dictionaries > underneath, it passes the value along regardless of whatever it finds. > This is a bit more complex because I think I'd need a co-mutual > recursive function to "travel" each value and report if there are > further dictionaries down below. > > Or, I could just leave the filter as-is where it only ever gets one, > non-dict/non-list item and receives a key. Sometimes the key will now be > a numerical index which is likely not useful... > > I might stage this as-is for now. Let's revisit it before 4.0 rc0. > > --js > -- Best regards, Vladimir
