[Tutor] getting diagonals from a matrix

[Glen Zangirolami]

I am not really sure of a better way but if your looking for a way to make
your code cleaner or more efficient
you can try Numpy - http://numpy.scipy.org/


On Tue, Mar 2, 2010 at 4:54 PM, David Eccles (gringer) <gmail at gringer.org>wrote:

> I've managed to drum up some code to obtain a list containing joined
> diagonal
> elements of a matrix (I'm making a word finder generator), but am wondering
> if
> there's any better way to do this:
>
> # setup so code snippet works properly
> sizeW = 4
> sizeH = 3
> puzzleLayout = ['spam'] * (sizeW * sizeH)
>
> # Starting with, say, an array with these indices:
> # 0 1 2 3
> # 4 5 6 7
> # 8 9 A B
>
> # I want the following items for a back diagonal (not in square brackets):
> # [-2],[3],8 (+5) | div 4 = -1, 1, 2
> # [-1],4,9 (+5)   | div 4 = -1, 1, 2
> # 0,5,A (+5)      | div 4 =  0, 1, 2
> # 1,6,B (+5)      | div 4 =  0, 1, 2
> # 2,7,[C] (+5)    | div 4 =  0, 1, 3
> # 3,[8],[D] (+5)  | div 4 =  0, 2, 3
>
> # in other words, increase sequence by sizeW + 1 each time (sizeW - 1
> # for forward diagonals), only selecting if the line you're on matches
> # the line you want to be on
>
> # back as in backslash-like diagonal
> puzzleDiagBack = [(''.join([puzzleLayout[pos*(sizeW+1) + i] \
> for pos in range(sizeH) if (pos*(sizeW+1) + i) / sizeW == pos])) \
> for i in range(-sizeH+1,sizeW)]
> puzzleDiagBackRev = [(''.join(reversed([puzzleLayout[pos*(sizeW+1) + i] \
> for pos in range(sizeH) if (pos*(sizeW+1) + i) / sizeW == pos]))) \
> for i in range(-sizeH+1,sizeW)]
> # fwd as in forwardslash-like diagonal
> puzzleDiagFwdRev = [(''.join([puzzleLayout[pos*(sizeW-1) + i] \
> for pos in range(sizeH) if (pos*(sizeW-1) + i) / sizeW == pos])) \
> for i in range(sizeW+sizeH-1)]
> puzzleDiagFwd = [(''.join(reversed([puzzleLayout[pos*(sizeW-1) + i] \
> for pos in range(sizeH) if (pos*(sizeW-1) + i) / sizeW == pos]))) \
> for i in range(sizeW+sizeH-1)]
>
> Cheers,
> David Eccles (gringer)
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