12
$\begingroup$

Namely, while

Thread @ f[{a, b}, {x, y}]

gives

{f[a, x], f[b, y]}

the InverseThread (if it existed) should reverse the process that

InverseThread @ {f[a, x], f[b, y]}

gives

f[{a, b}, {x, y}] 
$\endgroup$
1
  • 17
    $\begingroup$ Thread[{f[a, x], f[b, y]}, f]. $\endgroup$
    – Szabolcs
    Nov 1, 2016 at 9:13

1 Answer 1

14
$\begingroup$

Let's make this a little more challenging:

invThread[body : _[h_[___] ..]] :=
  Replace[Thread[body, h], _[x_ ..] :> x, {1}]

f[{a, b}, {x, y}, z];
% // Thread
% // invThread
{f[a, x, z], f[b, y, z]}

f[{a, b}, {x, y}, z]
$\endgroup$

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