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I have two list which are not equal dimension:

listA = {{{1, 2}, {2, 2}, {3, 4}}, {{1, 1}, {2, 2}, {3, 3}}};
listB = {{4, 4}, {5, 5}};

And I want to divide them or multiply the such as the results has the same dimensions, for the first dimension, it should look like:

(#*listB[[1]]) & /@ listA[[1]]

but I would like to do it all at once so that the result has the same dimension as listA. I bet it has probably been answered here but I cannot seem to find it. Therefore link to an existing answer is also great.

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    $\begingroup$ How about Table[(#*listB[[i]]) & /@ listA[[i]], {i, Length[listB]}] $\endgroup$
    – bill s
    Sep 23, 2016 at 21:18
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    $\begingroup$ MapThread[(Function[{x}, x #2] /@ #1) &, {listA, listB}]? $\endgroup$
    – march
    Sep 23, 2016 at 21:48
  • $\begingroup$ Related: (23395) $\endgroup$
    – Mr.Wizard
    Sep 24, 2016 at 6:30

2 Answers 2

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Some overhead but the most clean solution I could muster:

Diagonal@Outer[Times, listB, listA, 1, 2]

{{{4, 8}, {8, 8}, {12, 16}}, {{5, 5}, {10, 10}, {15, 15}}}

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Using Simon Woods's smartThread one may write:

smartThread[listA*listB, 1]

{{{4, 8}, {8, 8}, {12, 16}}, {{5, 5}, {10, 10}, {15, 15}}}

For fun here is a way using Inactive and Activate, assuming list elements are atomic:

Map[Inactive, Quiet[listA*listB], {-2}] // Activate
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