By the way— really kludgy, doesn’t exec() do what you want here: Note
The default *locals* act as described for function locals() <https://docs.python.org/3/library/functions.html#locals> below: modifications to the default *locals* dictionary should not be attempted. Pass an explicit *locals*dictionary if you need to see effects of the code on *locals* after function exec() <https://docs.python.org/3/library/functions.html#exec> returns. Though great still may not effect the current local namespace. But I’d be really surprised if there were no way to modify the local namespace — you can mess with pretty much anything else in Python. -CHB Sent from my iPhone On Aug 17, 2018, at 12:46 PM, Chris Barker <chris.bar...@noaa.gov> wrote: On Thu, Aug 16, 2018 at 12:37 PM, Chris Angelico <ros...@gmail.com> wrote: > > I've no idea what interpreter you're using, but it doesn't work for me. > That was in iPython, with python3.6.2 -- I wouldn't have expected it to be different in this case though -- really odd. OK -- tired again, and it indeed it failed -- so I'm really confused. You can see my console output -- it did indeed work at least once. > You've changed a cached dictionary but haven't actually created a local > which still makes me wonder WHY locals() returns a writable dict... -CHB -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov
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