On Wed, Dec 24, 2014 at 04:35:06PM -0500, WolfRage wrote:
> I wrote some code recently to make a linked list of Nodes for a 2d
> graph, so it consists of rows and columns. Now I wanted to make the code
> support being doubly linked, forwards and backwards. The difficult part
> of this is that the links are per row and per column. But the code I
> think is overly bloated. I am currently working on reducing the
> complexity of it. If any one has the time to look at it, if you have
> ideas for how I can re-write it to be much smaller I would appreciate
> the information. If you need more code let me know, but I tried to
> condense it since this singular function is around 325 lines of code.
Wow. It certainly is bloated.
I don't have time to look at it in any detail right now, as it is
Christmas Day here, but I'll give you a suggestion. Any time you find
yourself writing more than two numbered variables, like this:
> previous_col0_node = None
> previous_col1_node = None
> previous_col2_node = None
> previous_col3_node = None
> previous_col4_node = None
> previous_col5_node = None
> previous_col6_node = None
> previous_col7_node = None
you should instead think about writing a list:
previous_col_nodes = [None]*8
Then, instead of code like this:
> if col == 0:
> self.col0 = current_node
> previous_col0_node = current_node
> elif col == 1:
> self.col1 = current_node
> previous_col1_node = current_node
> elif col == 2:
> self.col2 = current_node
> previous_col2_node = current_node
etc.
you can just write:
for col in range(number_of_columns):
self.columns[col] = current_node
previous_col_nodes[col] = current_node
Look for the opportunity to write code like this instead of using range:
for col, the_column in enumerate(self.columns):
self.columns[col] = process(the_column)
Any time you write more than a trivial amount of code twice, you should
move it into a function. Then, instead of:
if row == 0:
if col == 0: a
elif col == 1: b
elif col == 2: c
elif col == 3: d
elif col == 4: e
elif row == 1:
if col == 0: a
elif col == 1: b
elif col == 2: c
elif col == 3: d
elif col == 4: e
elif row == 3:
# same again
you can write a function:
def process_cell(row, col):
if col == 0: a
elif col == 1: b
elif col == 2: c
elif col == 3: d
elif col == 4: e
# later on
for row in rows:
for col in cols:
process_cell(row, col)
Try those suggestions, and come back to us if you still need help.
--
Steven
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