for (1):
longone=longone + char # where len(char)== 1
I known that string concatenation is time consuming, but a small test on timeit seems to show that packing and creating an array for those 2 elements is equally time consuming
- use cStringIO instead - or append all chars to a list and do "".join (listvar)
for (2): for k in hash.keys()[:]: # Note : Their may be a lot of keys here if len(hash[k])<2: del hash[k]
Here again, I think the hash.keys duplication can be time *and* memory consuming. But still better than (I suppose - no time it yet) hash=dict([(k,v) for (k,v) in hash if len(v)>1])
- Try if it isn't faster to iterate using items instead of iterating over keys
- use the dict.iter* methods to prevent building a list in memory. You shouldn't use these values directly to delete the entry as this could break the iterator:
for key in [k for (k, v) in hash.iteritems () if len (v) < 2]:
del hash (key)This of course builds a list of keys to delete, which could also be large.
- also: hash.keys()[:] is not necessary, hash.keys () is already a copy
Daniel -- http://mail.python.org/mailman/listinfo/python-list
