- Try if it isn't faster to iterate using items instead of iterating over keys
items are huge lists of numbers. keys are simple small strings. And even if it is faster, how can I find the key back, in order to delete it ? for v in hashh.items(): if len(v)<2: del ???????
To elaborate on the memory requirements of .keys () vs. items ():
.keys () creates a new list of n objects. The objects are additional references to the existing keys.
.items () creates also a new list of n objects. These objects are tuples of references, one to the key and one to the value. Only references are used so it doesn't matter how large the value actually is. Whether the tuples are created for the items () call or already exist depends on the implementation of the dictionary. Trying to infer this by using sys.getrefcount got me inconclusive results.
I gonna try, but think that would be overkill: a whole list has to be computed ! Maybe whith genexps ...... for key in (k for (k,v) in hash.iteritems() if len(v)<2)
Using only iterators has problems:
for k,v in hash.iteritems ():
if len (v) < 2:
del hash [k]You are changing hash while you iterate over it, this very often breaks the iterator.
If you are memory bound, maybe a dabase like SQLite is really the way to go. Or you could write the keys to a remporary file in the loop and then write a second loop that reads the keys and deletes them from hash.
Daniel -- http://mail.python.org/mailman/listinfo/python-list
