2015.07.21 12:11 PM
q)((0 0;0 0);(0 0;0 5)) bin ((0 0;0 5); 2 1)
1 1
It seems that according to these results that:
2 1 > (0 0;0 5)
q)((0 0;0 0);(0 0;0 5)) bin ((0 0;0 5); 0 0 0; 0 0; 0; ((0 0;0 0);(0 0;0 0)))
1 1 1 -1 -1
And:
0 0 0 > (0 0;0 5)
0 0 > (0 0;0 5)
0 < (0 0;0 0)
((0 0;0 0);(0 0;0 0)) < (0 0;0 0)
Does not appear that when the shapes are the same, but the ranks are different that item-wise extension kicks in.
2015.07.21 02:01 PM
q)(::;((0 0;0 0);(0 0;0 0))) bin (::;0;0 0;(0 0;0 0);((0 0;0 0);(0 0;0 0)))
1 -1 1 1 1
2015.07.21 02:08 PM
q)5 2 3 1 4 bin 1 2 3 4 5
-1 1 3 4 4
2015.07.21 11:09 PM
If there is no match the result is the count of the left argument.
OK these two results I still don't understand--I would have thought they'd be -1q)(1 1;1 3;1 5)? 0 1
3
q)(1 1;1 3;1 5)? 0 0
3
2015.07.22 07:53 AM
2015.07.23 12:46 AM
2015.07.23 12:46 AM
2015.07.23 07:37 AM
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