Update
In short, I have myCAans[[1 ;; 5]] like this:
{{0, 0, 0, True, 0, 0, 0, 0, True, 0, "T0", 39}, {0, 0, 0, True, 0, 0, 0, 1, True, 1, "B", 101500}, {0, 0, 0, True, 0, 0, 0, 2, True, 2, "B", 100}, {0, 0, 0, True, 0, 0, 0, 3, True, 3, "B", 1000}, {0, 0, 0, True, 0, 0, 0, 4, True, 4, "B", 1000}}
Dimensions[myCAans]
{3000, 12}
Each line of
Total[ Select[myCAans , #[[5]] == 0 && #[[10]] == 0 &][[;; , 12]] ]
Total[ Select[myCAans , #[[5]] == 0 && #[[10]] == 1 &][[;; , 12]] ]
Total[ Select[myCAans , #[[5]] == 0 && #[[10]] == 2 &][[;; , 12]] ]
Total[ Select[myCAans , #[[5]] == 1 && #[[10]] == 0 &][[;; , 12]] ]
Total[ Select[myCAans , #[[5]] == 5 && #[[10]] == 6 &][[;; , 12]] ]
works fine.
I would like to know how to efficiently find
Total[ Select[myCAans , #[[5]] == i && #[[10]] == j &][[;; , 12]] ]
for i from 0 to 3 and j from 0 to 3 for example. Something like
Total[ Select[myCAans , #[[5]] == i && #[[10]] == j &][[;; , 12]] ] -> XXXXXX <- Table[{i, j}, {i, 0, 3}, {j, 0, 3}]
OP
On a related question here, I am struggling to get my head around for this:
f@Table[{i, j}, {i, 0, 3}, {j, 0, 3}]
f @@ Table[{i, j}, {i, 0, 3}, {j, 0, 3}]
f @@@ Table[{i, j}, {i, 0, 3}, {j, 0, 3}]
f /@ Table[{i, j}, {i, 0, 3}, {j, 0, 3}]
Map[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 2}]
MapAt[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {1}] // TableForm
MapAt[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 1}] // TableForm
MapAt[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 2}] // TableForm
MapAt[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 3}] // TableForm
Map[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {1}] // TableForm
Map[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 1}] // TableForm
Map[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 2}] // TableForm
Map[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2, 3}] // TableForm
Even this
MatrixFunction[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}]]
The desired output is
Table[f[i, j], {i, 0, 3}, {j, 0, 3}] // TableForm
{{f[0, 0], f[0, 1], f[0, 2], f[0, 3]}, {f[1, 0], f[1, 1], f[1, 2], f[1, 3]}, {f[2, 0], f[2, 1], f[2, 2], f[2, 3]}, {f[3, 0], f[3, 1], f[3, 2], f[3, 3]}}
Final aim is to apply a pure function which takes two inputs over an array, like this
In163 is one of many trials without a success. I have tested
Total[Select[myCAans, (#[[5]] == 0 && #[[10]] == 0) &][[;; , 12]]]
Total[Select[myCAans, (#[[5]] == 0 && #[[10]] == 1) &][[;; , 12]]]
Total[Select[myCAans, (#[[5]] == 0 && #[[10]] == 2) &][[;; , 12]]]
Total[Select[myCAans, (#[[5]] == 0 && #[[10]] == 3) &][[;; , 12]]]
Total[Select[myCAans, (#[[5]] == 3 && #[[10]] == 3) &][[;; , 12]]]
which all works perfectly fine.
Any more examples where I can pick up mapping pure functions, using /@ @@ and &/@ much easier and faster?
Thanks.

Applyon deeper levels. In this case, you want to have level{2}. TryApply[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2}]. – Henrik Schumacher Jun 14 '18 at 22:36/@or@@or something similar? Do I have to use the wordsApplyI mean? – Chen Stats Yu Jun 14 '18 at 22:39f @@@ # & /@ array(wherearray = Table[{i, j}, {i, 0, 3}, {j, 0, 3}], but I would discourage that becaus it is less efficient. And for arrays of rank 3 or higher, you would finally hit the wall with such a syntax. – Henrik Schumacher Jun 14 '18 at 22:44Total[Select[myCAans, (#[[5]] == #1) && (#[[10]] == #2) &][[;; , 12]]]. I will look for further details. – Chen Stats Yu Jun 14 '18 at 22:55myCAansis? (A small sample would suffice.) – Henrik Schumacher Jun 14 '18 at 22:57f = {i, j} \[Function] Total[Select[myCAans, #[[5]] == i && #[[10]] == j &][[;; , 12]]]; Apply[f, Table[{i, j}, {i, 0, 3}, {j, 0, 3}], {2}]? Often, symbols instead of slots (#) are a bit better to track... – Henrik Schumacher Jun 14 '18 at 23:27