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I am new to Mathematica. A quick question:

Given

a = {{1,0,0,2},{2,0,1,0}} 
aR = {{1,0,1,0},{0,1,0,1}} 

if element in a is 0, I would like to replace this element with the element from aR in the same position, so results should be `

{{1, 0, 1, 2}, {2, 1, 1, 1}}.

I tried this but does not work

(a[[#1, #2]] = If [a[[#1, #2]] == 0, aR[[#2, #1]], a[[#1, #2]]]) & @@@
   {Range[Part[Dimensions[a], 1]], Range[Part[Dimensions[a], 2]]};

Could you please help me?

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    $\begingroup$ a + (1 - Unitize@a)*aR... and perhaps more understandable ReplacePart[a, {x_, y_} /; a[[x, y]] == 0 :> aR[[x, y]]] or MapThread[If[# == 0, #2, #] &, {a, aR}, 2] $\endgroup$ – ciao Jun 10 '16 at 21:54
  • $\begingroup$ MapIndexed[If[#1 == 0, Extract[aR, #2], #1] &, a, {2}] $\endgroup$ – andre314 Jun 11 '16 at 8:25
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ciao doesn't like to write answers, so I will do it as a CW.

a + (1 - Unitize@a)*aR

Perhaps more understandable:

ReplacePart[a, {x_, y_} /; a[[x, y]] == 0 :> aR[[x, y]]] 

or

MapThread[If[# == 0, #2, #] &, {a, aR}, 2]

All give

{{1, 0, 1, 2}, {2, 1, 1, 1}}

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  • $\begingroup$ Just a note: the first method is about 50 times faster on bigger datasets $\endgroup$ – BlacKow Jun 10 '16 at 22:39
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    $\begingroup$ @BlacKow - a1 + BitXor[1, Unitize@a1]*a2 s/b a bit faster yet... $\endgroup$ – ciao Jun 10 '16 at 23:11
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    $\begingroup$ @ciao, well, when you put it that way: {{1, 0, 0, 2}, {2, 0, 1, 0}} ~BitOr~ {{1, 0, 1, 0}, {0, 1, 0, 1}}. (Of course, this assumes the entries in the two matrices are all nonnegative integers.) $\endgroup$ – J. M. is away Jun 10 '16 at 23:18
  • $\begingroup$ @J.M.--- might want to rethink that ;=} $\endgroup$ – ciao Jun 10 '16 at 23:51

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