# Tag Info

5

You can use SlotSequence (##) and Apply (@@) Intersection[##, SameTest -> (Abs[#1] == Abs[#2] &)] & @@ {{1, 2, 3}, {1, 2, 3}, {-1, -2, -9}} It could be useful to inspect what following lines do: a = {{1, 2, 3}, {1, 2, -3}, {-1, -2, -9}}; Sequence @@ a ## & @@ a ## & @ a Sequence @ a # & @@ a # & @ a

4

A few points to make here: Always use Listable attributes of functions, that will speed things up. When unnecessary, do not use symbolic processing, use numeric processing instead. Thus, I'll first change the data to N form, then use Listable attributes of Mean and StandardDeviation to get the result in a shorter and faster code. imgd = N@imageData; ...

4

Two possible ways: listfun = {Sin, Cos, Tan}; Through[listfun[Pi]] (* {0, -1, 0} *) #[Pi] & /@ listfun (* {0, -1, 0} *)

1

For the better read,I post this solution as an answer Map[Greater[#,1]&,{{0.6,1.2},{5,0.1}},{-1}] {{False, True}, {True, False}}

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