First clear solution for Moore Neighbourhood by skyJuice - python coding challenges

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First solution in Clear category for Moore Neighbourhood by skyJuice

def count_neighbours(grid, row, col):
    moor = [(-1,-1),(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1), (0,-1)]
    count = 0
    for i in moor:
        if(0<=row+i[0]
                        Sept. 30, 2014

                        
                            Comments:
                            
                                
                                    veky
 on Oct. 1, 2014, 4:42 p.m.

                                    <p>You can use builtin sum instead of coding it from scratch. And use unpacking in for.</p>

<pre class='brush: python'>return sum(grid[row+dr][col+dc] for dr, dc in moor
    if 0 &lt;= row+dr &lt; len(grid) if 0 &lt;= col+dc &lt; len(grid[row]))
</pre>

                                
                            
                                
                                    artemrudenko
 on Oct. 2, 2014, 5:43 a.m.

                                    <p>To get rid of i[0] and i[1] you can write:</p>

<pre class='brush: python'>for rshift, cshift in moor:
</pre>

<p>Aslo there is no any need to use redundant parentheses in the code below:</p>

<pre class='brush: python'>if(0&lt;=row+i[0]&lt;len(grid) and 0&lt;=col+i[1]&lt;len(grid[0])):
</pre>

<p>E.g. you can update it to(also don't forget about pep8 rules(<a href="http://legacy.python.org/dev/peps/pep-0008/">http://legacy.python.org/dev/peps/pep-0008/</a>#whitespace-in-expressions-and-statements):</p>

<pre class='brush: python'>if 0 &lt;= row+i[0] &lt; len(grid) and 0 &lt;= col+i[1] &lt; len(grid[0]):
</pre>

<p>Best Regards,
Artem</p>

                                
                            
                                
                                    skyJuice
 on Oct. 2, 2014, 7:18 a.m.

                                    <p>ok, thanks guys, i will think about this.</p>

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