Creating Nested For Loops for modifying values to append to a List

Question

Basically, the title says it all. I am familiar with python, not so much with mathematica. For the program, I am starting out with three unique lists of values for which I am using a triple nested For loop to extract each term. Then, I am multiplying these extracted terms together and appending them to a new blank list. Since I feel I may not be explaining my intentions that well, my basic goal is to go from:

listA = {1, 2, 3}

listB = {4, 5, 6}

listC = {7, 8}

listFinal = {}

to then appending listFinal in a manner to obtain:

listFinal = {1*4*7, 1*4*8, 1*5*7, 1*5*8, 1*6*7, 1*6*8, 2*4*7, 2*4*8, etc...}

When I would create this loop in python, I would start with a blank list (listFinal), that I would then append these multiplied values to after indexing each value at the end of a triple nested For loop.

However, when I attempt to create this program in mathematica, I am left with a non-appended (blank) listFinal. Do you all know of a way to fix my nested For loop so it would function as intended, or use a different method of achieving my desired results? I've been searching through forums and have been met with the consensus that for loops are not the most efficient/best method of programming in mathematica. However, I am most familiar with using them, which is why my attempts so far have been with only for loops.

My original program is as follows:

m = List[.9, .8, .7, .8, .7, .6]
g = List[1, .95, .9, .95, .85, .75]
b = List[.85, .75, .65]

effectiveFlux = List[];

For[i = 1, i < 7, i++,
 For[j = 1, j < 7, j++,
  For[k = 1, k < 4, k++,
   var = m[[{i}]]*g[[{j}]]*b[[{k}]]
     Append[effectiveFlux, var]]]]

Thank you in advance!

Answer 1

It is understandable that you feel more comfortable using loops than list-processing functions at first. But have a little faith on what you've already read: loops have very restricted good applications in Mathematica.

m = List[.9, .8, .7, .8, .7, .6];
g = List[1, .95, .9, .95, .85, .75];
b = List[.85, .75, .65];

effectiveFlux = Flatten@Outer[Times, m, g, b]

Answer 2

ef2 = Times @@@ Tuples[{m, g, b}];
ef3 = Tuples[Hold[Times][m, g, b]] // ReleaseHold;
ef4 = Tuples[times[m, g, b]] /. times -> Times;
ef5 = Distribute[{m, g, b}, List, List, List, Times];

ef2

{0.765, 0.675, 0.585, 0.72675, 0.64125, 0.55575, 0.6885, 0.6075, 0.5265, 0.72675, 0.64125, 0.55575, 0.65025, 0.57375, 0.49725, 0.57375, 0.50625, 0.43875, 0.68, 0.6, 0.52, 0.646, 0.57, 0.494, 0.612, 0.54, 0.468, 0.646, 0.57, 0.494, 0.578, 0.51, 0.442, 0.51, 0.45, 0.39, 0.595, 0.525, 0.455, 0.56525, 0.49875, 0.43225, 0.5355, 0.4725, 0.4095, 0.56525, 0.49875, 0.43225, 0.50575, 0.44625, 0.38675, 0.44625, 0.39375, 0.34125, 0.68, 0.6, 0.52, 0.646, 0.57, 0.494, 0.612, 0.54, 0.468, 0.646, 0.57, 0.494, 0.578, 0.51, 0.442, 0.51, 0.45, 0.39, 0.595, 0.525, 0.455, 0.56525, 0.49875, 0.43225, 0.5355, 0.4725, 0.4095, 0.56525, 0.49875, 0.43225, 0.50575, 0.44625, 0.38675, 0.44625, 0.39375, 0.34125, 0.51, 0.45, 0.39, 0.4845, 0.4275, 0.3705, 0.459, 0.405, 0.351, 0.4845, 0.4275, 0.3705, 0.4335, 0.3825, 0.3315, 0.3825, 0.3375, 0.2925}

Equal @@ {ef2, ef3, ef4, ef5,effectiveFlux}

True

Answer 3

To make mental transition from For smoother, consider also Table way of looping:

tmp = Flatten@
  Table[m[[i]] g[[j]] b[[k]], {i, Length@m}, {j, Length@g}, {k, Length@b}]

Check

ef2 == effectiveFlux == tmp
(* True *)

Answer 4

efAlt = Distribute[{m , g , b}, List, List, List, Times]

and

efAlt == effectiveFlux == ef2

True

where

m = List[.9, .8, .7, .8, .7, .6];
g = List[1, .95, .9, .95, .85, .75];
b = List[.85, .75, .65];

In addition:

Distribute[{{1, 2, 3}, {4, 5, 6}, {7, 8}}, List]

{

{1, 4, 7}, {1, 4, 8}, {1, 5, 7}, {1, 5, 8}, {1, 6, 7}, {1, 6, 8},

{2, 4, 7}, {2, 4, 8}, {2, 5, 7}, {2, 5, 8}, {2, 6, 7}, {2, 6, 8},

{3, 4, 7}, {3, 4, 8}, {3, 5, 7}, {3, 5, 8}, {3, 6, 7}, {3, 6, 8}

}