Re: comple list slices

<johnzenger <at> gmail.com> writes:

Although I don't know if this is faster or more efficient than your
current solution, it does look cooler:

def grouprows(inrows):
rows = []
rows[:] = inrows # makes a copy because we're going to be
deleting
while len(rows) > 0:
rowspan = rows[0]["rowspan"]
yield rows[0:rowspan] # "returns" this value, but control flow
unaffected
del rows[0:rowspan] # remove what we just returned from the
list, and loop

grouper = grouprows(copyrows)
print [x for x in grouper]

wow, i think this is much better then my solution. And you can easily call it
for subgroups:

grouper = grouprows(rows)
for x in grouper
grouperTwo = grouprows(x)
for y in grouperTwo

Do i need to copy the list in the iterator? (if I am not planning on using rows
again) The reference count for the list members will get bumped on yield right?


.

Relevant Pages