pop method question
Sun Mar 4 00:36:14 CET 2007
James Stroud <jstroud at mbi.ucla.edu> writes:
> for akey in dict1:
> if some_condition(akey):
> dict2[akey] = dict2.pop(akey)
> Which necessitates a key is a little cleaner than your latter example.
Yeah, I also think removing keys from a dict while iterating over it
(like in Steven's examples) looks a bit dangerous dangerous.
Assuming you meant "dict1.pop" instead ot dict2.pop above, your
example might be written
dict2 = dict((k, dict1.pop(k)) for k in dict1 if some_condition(k))
avoiding some namespace pollution etc.
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