# [SciPy-user] Creating a Matrix from a Sum

Tom K. tpk@kraussfamily....
Fri Jul 3 20:18:40 CDT 2009

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David Warde-Farley-2 wrote:
>
>
> On 3-Jul-09, at 2:45 PM, David Warde-Farley wrote:
>
>> On 2-Jul-09, at 9:25 AM, Lorenzo Isella wrote:
>
> Then you want the max of i and j at each position in the matrix. For
> this, though there may be a more efficient way, you can use mgrid to
> get i and j indices at each position in the matrix and then use
> concatenate and max on them to concatenate along a third axis and then
> do a max along that same axis to recover a 2D array:
>
> 	idx = np.concatenate([M[:,:,np.newaxis] for M in mgrid[0:len(x),
> 0:len(x)]],axis=2).max(axis=2)
>

David this is interesting however I think the use of mgrid followed by
concatenation into a 3D array here is overkill.  The idx array is the same
as the max_ij array that I suggested - a one liner for it is
max_ij = np.maximum(np.arange(len(x))[:,np.newaxis], np.arange(len(x)))

This problem involves O(N) sums and O(N^2) assignments.   The vectorized
versions that David and I posted (my version 2) are doing just that much
work - however with overhead of several intermediate arrays.  A more
straight-forward approach in python is just a double for-loop, which may be
the most readable and perfectly sufficient unless your N is huge and/or this
function becomes the most significant item in a profile of your application.

sums = np.cumsum(x[::-1])[::-1]
P = np.zeros((len(x), len(x))
for i in range(len(x)):
for j in  range(len(x)):
P[i, j] = sums[np.max((i, j))]

#Here it is with list comprehensions:
P = np.array([[sums[np.max((i, j))] for j in range(len(x))] for i in
range(len(x))])

By the way I do NOT recommend using the first method I posted in my earlier
post - I think it is doing O(N^3) adds and O(N^3) assigns!
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