Zeroing out a factor of a term in the elements of a matrix

Question

Each element of my matrix is zero or has the form $k_1 a_{i,n} + k_2 b_{i,n}$. How can I replace all the $a_{i,n}$s by zero?

{{{0. - 0.000565035 Subscript[a, i, 1] + 0.00187682 Subscript[b, i, 1], 
   0. - 0.00176323 Subscript[a, i, 1] - 0.000565035 Subscript[b, i, 1]}, 
  {0., 0.}, 
  {0. - 0.00428076 Subscript[a, i, 3] - 0.00380882 Subscript[b, i, 3], 
   0. - 0.00421293 Subscript[a, i, 3] - 0.00428076 Subscript[b, i, 3]}}, 
 {{0., 0.}, 
  {0. + 0.00995524 Subscript[a, i, 2] + 0.00978341 Subscript[b, i, 2], 
   0. + 0.00932789 Subscript[a, i, 2] + 0.00995524 Subscript[b, i, 2]}, 
  {0., 0.}}, 
 {{-0.00428076 Subscript[a, i, 1] - 0.00380882 Subscript[b, i, 1], 
   -0.00421293 Subscript[a, i, 1] - 0.00428076 Subscript[b, i, 1]}, 
  {0., 0.}, 
  {0. - 0.00924663 Subscript[a, i, 3] - 0.0114643 Subscript[b, i, 3], 
   0. - 0.00745795 Subscript[a, i, 3] - 0.00924663 Subscript[b, i, 3]}}}

YourExpr/.Subscript[a,i,_]->0 does it. Note: you have a - followed by , on the last two lines, but I suspect that is just a scrape-n-paste typo. — Bill, Jun 10 '17 at 17:41
You might be interested in point 3 in https://mathematica.stackexchange.com/questions/18393/what-are-the-most-common-pitfalls-awaiting-new-users/18395#18395 — Michael E2, Jun 10 '17 at 17:50
If you meant to notify Bill, you should use @Bill to make sure your response is seen. (Author's of posts are always notified when someone makes a comment.) — Michael E2, Jun 10 '17 at 18:50

score 2 · Answer 1 · answered Jun 10 '17 at 22:20

Try this:

m = {{{0. - 0.000565035 Subscript[a, i, 1] + 
     0.00187682 Subscript[b, i, 1], 
    0. - 0.00176323 Subscript[a, i, 1] - 
     0.000565035 Subscript[b, i, 1]}, {0., 
    0.}, {0. - 0.00428076 Subscript[a, i, 3] - 
     0.00380882 Subscript[b, i, 3], 
    0. - 0.00421293 Subscript[a, i, 3] - 
     0.00428076 Subscript[b, i, 3]}}, {{0., 
    0.}, {0. + 0.00995524 Subscript[a, i, 2] + 
     0.00978341 Subscript[b, i, 2], 
    0. + 0.00932789 Subscript[a, i, 2] + 
     0.00995524 Subscript[b, i, 2]}, {0., 
    0.}}, {{-0.00428076 Subscript[a, i, 1] - 
     0.00380882 Subscript[b, i, 1], -0.00421293 Subscript[a, i, 1] - 
     0.00428076 Subscript[b, i, 1]}, {0., 
    0.}, {0. - 0.00924663 Subscript[a, i, 3] - 
     0.0114643 Subscript[b, i, 3], 0. - 0.00745795 Subscript[a, i, 3],
     0.00924663 Subscript[b, i, 3]}}};

Then

m1=m /. Subscript[a, i, n_] -> 0;

MatrixForm[m1]

Have fun!

@ Alexei Boulbitch, Thank you – Delaram Nematollahi Jun 10 '17 at 22:33 — Delaram Nematollahi, Jun 10 '17 at 22:33

score 0 · Answer 2 · answered Jun 11 '17 at 14:16

It can sometimes be more efficient to make a definition upon Subscript itself, within Block if needed, if your data is large.

(* m from Alexei's answer *)
big = RandomChoice[m, 50000];

r1 = big /. Subscript[a, i, n_] -> 0;                 // RepeatedTiming

r2 = Block[{Subscript}, Subscript[a, i, _] = 0; big]; // RepeatedTiming

r1 === r2

{0.470, Null}
{0.27, Null}
True

Zeroing out a factor of a term in the elements of a matrix

2 Answers2