Welcome on board. Here I would focus on the 'overall additions'.
Given:
a=:1 2 3 4 5
b=:1 2
From your description you want to find the 'cartesian sum' of a over
b (ie the outer product of all possible '+' combinations of a and b).
I would proceed as follows ... note also that NB. lines in the
following are comments):
NB. This is the 'cartesian sum' of all items of a added 'over' all
items of b, hence result is an array with shape <5 2> representing the
5x2 sums.
a+/b
2 3
3 4
4 5
5 6
6 7
NB. Now we can 'sum' up these items using +/ (note that this sums the
rows, or differently expressed, 'down the columns') ...
+/a+/b
20 25
NB. Now we can 'sum' up these (resulting) items using +/ again ...
+/+/a+/b
45
NB. You can also use ',' to 'list' the items after the cartesian sum,
meaning only one sum is then required...
+/,a+/b
45
By thinking in terms of 'array applications/results' the loops tend to
evaporate.
Hope this helps.../Rob Hodgkinson
PS: I saw Henry posted a response too just before I sent this, but
thought you may still find this helpful by way of explanation.
On 04/08/2008, at 9:43 AM, Ian Gorse wrote:
I want to be able to add each of the values in 'b' with each of the
values
in 'a', then sum them up
for example
(a[0]+b[0]) + (a[1]+b[0]) + (a[2]+b[0]) +(a[3]+b[0]) +
(a[4]+b[0]) ) = 2
3 4 5 6 = 20
which is (1+1) + (2+1) + (3+1) + (4+1) + (5+1) = 20
(a[0]+b[1]) + (a[1]+b[1]) + (a[2]+b[1]) +(a[3]+b[1]) +
(a[4]+b[1]) ) = 3
4 5 6 7 = 25
which is (1+2) + (2+2) + (3+2) + (4+2) + (5+2) = 25
total = 20 + 25 = 45
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