Kirk Strauser wrote:
At 2008-12-15T19:06:16Z, Reckoner <recko...@gmail.com> writes:
The problem is that I don't know ahead of time how many lists there are or
how deep they go. In other words, you could have:
Recursion is your friend.
Write a function to unpack one "sublist" and call itself again with the new
list. For instance, something like:
def unpack(pattern):
# Find the first subpattern to replace
# [...]
results = []
for number in subpattern:
results.append(pattern.replace(subpattern, number))
return results
Calling unpack([1,2,3,[5,6],[7,8,9]]) would look cause it to call
unpack([1,2,3,5,[7,8,9]]) and unpack([1,2,3,6,[7,8,9]]), compile the
results, and return them.
Along these lines generators are the bees knees.
def expands(source):
'''Nested lists to list of flat lists'''
for n, val in enumerate(source):
if isinstance(val, list):
assert val, 'empty list @%s in %s undefined.' % (
n, len(source))
head = source[: n]
tail = source[n + 1 :]
for element in val:
for row in expands(head + [element] + tail):
yield row
break
else:
if source: # Just to make expands([]) return an empty list)
yield source
def answer(source):
'''Do the requested printing'''
for row in expands(source):
print '-'.join(str(x) for x in source)
--Scott David Daniels
scott.dani...@acm.org
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http://mail.python.org/mailman/listinfo/python-list
