# How to evaluate function in the grid?

I have a function f[a_, b_]: = b^a + a b. a takes values {1,2,3,4}, and b - {10,11,12}.

How is it better to calculate all values? Grid-type output.

Also, for every b I need to calculate values of function from all a and then take an average. As a result of such computing I would like to have 2 numbers for every b:

{b1, 1/4 (f[a1,b1)+f[a2,b1)+f[a3,b1)+f[a4,b1)} (basically {b, average f for all a})

What is the shortest way to do that? Can I do this without loops?

• the first part of your question needs clarification. If you want all permutations of a+b from the two lists then Outer[Plus, {1, 2, 3, 4}, {10, 11, 12}] is somewhere to start but maybe best to show what output you want from first part of the question – Mike Honeychurch Nov 14 '15 at 4:52
• @MikeHoneychurch I have much more complicated function and a and b just parts. So, basically,it's not a sum of two values a and b. – Maria Nov 14 '15 at 4:55
• given the two example lists you have provided what output do you seek? You need to explain what you mean by "all values" – Mike Honeychurch Nov 14 '15 at 4:55
• Value of function in each point (for every a and b) with values of particular a and b. Grid-type output. – Maria Nov 14 '15 at 4:59
• in that case i think my answer addresses your question – Mike Honeychurch Nov 14 '15 at 5:01

a = {1, 2, 3, 4};
b = {10, 11, 12};


To get all the values:

vals = Outer[#2^#1 + #1*#2 &, a, b]
(* {{20, 22, 24}, {120, 143, 168}, {1030, 1364, 1764}, {10040, 14685, 20784}} *)


To get the averages:

vals = Outer[#2^#1 + #1*#2 &, a, b];
Transpose[{b, Mean[vals]}]
(* {{10, 5605/2}, {11, 8107/2}, {12, 5685}} *)


To see what is happening, evaluate

Clear[x, y, a, b, vals, f]
a = Array[x, 4]; b = Array[y, 3]
vals = Outer[f, a, b]
Transpose[{b, Mean[vals]}]

• Will it work if function is very complicated and after that I plug in a I do numeric inverse Laplace transform for every b? can I still use Outer function? – Maria Nov 14 '15 at 5:07
• @Maria. I don't really understand your question. The inverse Laplace transform of what? of a? In any case, Outer requires the list of a and b to be pre-determined. If you can define a function f[a_, b_] := ...  that takes an element of a and an element of b and does something to it, no matter how complicated, this should work. – march Nov 14 '15 at 5:14

For the second part:

{#, Mean@Thread[f[a, #]]} & /@ b

(* {{10, 5605/2}, {11, 8107/2}, {12, 5685}} *)


For the first part:

Outer[f, a, b]