On 11/30/2019 8:51 PM, Andrew Barnert via Python-ideas wrote:
On Nov 30, 2019, at 16:36, Oscar Benjamin <oscar.j.benja...@gmail.com> wrote:
On Sat, 30 Nov 2019 at 22:24, Steven D'Aprano <st...@pearwood.info> wrote:
On Sat, Nov 30, 2019 at 06:16:49PM -0300, Soni L. wrote:
It'd be quite nice if dict.items() returned a namedtuple so all these
x[0], x[1], el[0], el[1], etc would instead be x.key, x.value, el.key,
el.value, etc. It would be more readable and more maintainable.
If you are doing
for item in somedict.items():
process(item[0])
process(item[1])
you could do this instead:
for key, value in somedict.items():
process(key)
process(value)
You can also make your own function to get the items as namedtuples.
That can work now with any class that defines items the current way.
from collections import namedtuple
Item = namedtuple('Item', ['key', 'value'])
def nameditems(d):
return (Item(*t) for t in d.items())
d = {'a': 1, 'b': 2}
for item in nameditems(d):
print(item.key, item.value)
Comparing that with Steve's example above though I don't see the
advantage of namedtuples here.
Presumably the main advantage is for cases where you can’t destructure the
tuple in-place:
sorted(d.items(), key=lambda it: it.value)
There’s no nice way to write that today. Maybe this makes it clear?
sorted(d.items(), key=(ValueGetter := operator.itemgetter(1)))
But normally you don’t bother; you just live with using [1] and assuming your
reader will know that [1] on a mapping item is the value. Which isn’t terrible,
because it almost always is obvious you’ve got a mapping item, and almost every
reader does know what [1] means there. But it’s not as nice as using .value
would be.
How I miss python 2's parameter unpacking:
>>> sorted({1:300, 2:4}.items(), key=lambda (key, value): value)
[(2, 4), (1, 300)]
Eric
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