[Tutor] Making Doubly Linked List with Less Lines of Code.

[WolfRage]

OK thanks for the rapid response, I will start rewriting the functions 
in this way now, and will come back with what I wind up with.
Also Merry Christmas!

On 12/24/2014 04:56 PM, Steven D'Aprano wrote:
> 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.
>
>
>
>