Steve Howell <showel...@yahoo.com> writes:
> On Jan 25, 1:32 pm, Arnaud Delobelle <arno...@googlemail.com> wrote:
>> Steve Howell <showel...@yahoo.com> writes:
>>
>> [...]
>>
>> > My algorithm does exactly N pops and roughly N list accesses, so I
>> > would be going from N*N + N to N + N log N if switched to blist.
>>
>> Can you post your algorithm? It would be interesting to have a concrete
>> use case to base this discussion on.
>>
>
> I just realized you meant the Python code itself. It is here:
>
> https://bitbucket.org/showell/shpaml_website/src/tip/shpaml.py
Hi - I didn't have the time to look at it before today. You scan
through a list called prefix_lines, and you use pop(0) as a means to
keep track of your position in this list. It seems to me that it would
be more effective to explicitely keep track of it - and it would remove
the need to the numerous copies of sublists of indent_lines that you
have to create.
I have rewritten part of the three relevant functions (html_block_tag,
get_indented_block and recurse, within indent_lines) to show you what I
mean (see below). I have tried to keep the changes to a minimum - you
can see that the code is no more complex like this. The advantage is
that only one list is created, prefix_lines, and there is no need to
mutate it or copy parts of it during the algorithm. I have not had the
time to test it though (only on one of the examples on the examples
webpage).
Code follows:
[...]
def html_block_tag(output, block, start, end, recurse):
append = output.append
prefix, tag = block[start]
if RAW_HTML.regex.match(tag):
append(prefix + tag)
recurse(block, start + 1, end)
else:
start_tag, end_tag = apply_jquery_sugar(tag)
append(prefix + start_tag)
recurse(block, start + 1, end)
append(prefix + end_tag)
[...]
def get_indented_block(prefix_lines, start, end):
prefix, line = prefix_lines[start]
len_prefix = len(prefix)
i = start + 1
while i < end:
new_prefix, line = prefix_lines[i]
if line and len(new_prefix) <= len_prefix:
break
i += 1
while i-1 > start and prefix_lines[i-1][1] == '':
i -= 1
return i
[...]
def indent_lines(lines,
output,
branch_method,
leaf_method,
pass_syntax,
flush_left_syntax,
flush_left_empty_line,
indentation_method,
get_block,
):
append = output.append
def recurse(prefix_lines, start, end):
while start < end:
prefix, line = prefix_lines[start]
if line == '':
start += 1
append('')
else:
block_end = get_block(prefix_lines, start, end)
if block_end == start + 1:
start += 1
if line == pass_syntax:
pass
elif line.startswith(flush_left_syntax):
append(line[len(flush_left_syntax):])
elif line.startswith(flush_left_empty_line):
append('')
else:
append(prefix + leaf_method(line))
else:
branch_method(output, prefix_lines, start, block_end,
recurse)
start = block_end
return
prefix_lines = map(indentation_method, lines)
recurse(prefix_lines, 0, len(prefix_lines))
--
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