Reshaping subarray creation
One special case we still left out here. If the number of dimension specifier is less then the actual number of dimensions in the array, the subarray returned will reshape and extend the last dimension by trailing dimensions - joint together. In that special case the algorithm of subarray creation can be seen as follows:- reshape the original array by reducing its dimension to the number of given dimension specifications.
- In order to fit all elements inside the new shape, the last dimension will pick up all elements from trailing dimensions and therefore expand accordingly.
- indices into that last dimension will refer to the new dimension length.
Lets take an example again.
Example 7 (C#)
ILArray<double> A = ILMath.counter(4,3,2);
/* this recreates the 3-d array from the last example:
A
{<double> 44148697 [4x3x2] Phys.
(:,:,0)
1,00000 5,00000 9,00000
2,00000 6,00000 10,00000
3,00000 7,00000 11,00000
4,00000 8,00000 12,00000
(:,:,1)
13,00000 17,00000 21,00000
14,00000 18,00000 22,00000
15,00000 19,00000 23,00000
16,00000 20,00000 24,00000
}A now has 3 dimensions. We only specify 2 of them:
A[":;:"]
{<Double> 10972427 [4x6] Ref(3)
(:,:)
1,00000 5,00000 9,00000 13,00000 17,00000 21,00000
2,00000 6,00000 10,00000 14,00000 18,00000 22,00000
3,00000 7,00000 11,00000 15,00000 19,00000 23,00000
4,00000 8,00000 12,00000 16,00000 20,00000 24,00000
}
A["end;1:end"] -> evaluates to:
{<Double> 6130944 [1x5] Phys.
(:,:)
8,00000 12,00000 16,00000 20,00000 24,00000
}
A[1,"end:-5:0"]
{<Double> 23469 [5x1] Phys.
(:,:)
24,00000 19,00000 14,00000 9,00000 4,00000
}end
(23) stepping backwards with stepwidth 5 until 0 is exceeded. Those numbers are
than used to create the subarray as a column in the first dimension. That column
vector is than shifted by 1 dimension before returned. For a column vector such
a dimension shift turns the vector into a row vector.