rust-index: remove one collect when converting back
Turns out this is slightly faster. Sending the results back to Python is still
the most costly (like 75% of the time) of the whole method, but it's about
as fast as it can be now.
hg perf::phases on mozilla-try-2023-03-22
before: 0.267114
after: 0.247101
# Dummy extension to define a namespace containing revision names
from mercurial import namespaces
def reposetup(ui, repo):
names = {b'r%d' % rev: repo[rev].node() for rev in repo}
namemap = lambda r, name: names.get(name)
nodemap = lambda r, node: [b'r%d' % repo[node].rev()]
ns = namespaces.namespace(
b'revnames',
templatename=b'revname',
logname=b'revname',
listnames=lambda r: names.keys(),
namemap=namemap,
nodemap=nodemap,
)
repo.names.addnamespace(ns)