For example
{a + b} == {c + d} // Thread
(*{a + b == c + d}*)
how to get {a + b} == {c + d} from {a + b == c + d} ?
This is my try
Equal @@ List /@ Join @@ {a + b == c + d}
Are there other ways to do this?
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asked Jul 12 '13 at 17:08
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One way would be
Distribute[{a + b == c + d}, Equal]
Same as
Thread[{a + b == c + d}, Equal]
More universal
{a + b == c + d}/. f_[g_[x_, y_]] -> g[f[x], f[y]]
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$\begingroup$ Thanks, I'm completely a mess on these. How do you change {a==b,c==d} to {a+c==b+d} again using Thread? $\endgroup$ – xslittlegrass Jul 12 '13 at 20:08
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$\begingroup$ @xslittlegrass, If you really need Thread for this, maybe:
{Thread[Total@{a == b, c == d}, Equal]}, another way would be:Inner[Plus, Sequence @@ {a == b, c == d}, Equal]$\endgroup$ – panda-34 Jul 12 '13 at 20:48 -
$\begingroup$ (+1) I like the
Threadway. It seems more natural in this case, where threading is basically transposition with different heads. In fact, it seems more universal than the replacement rule, certainly with respect to number of arguments, but also considering that one specifies the head forThreadin the first place, eitherListby default or explicitly by it optional second argument. $\endgroup$ – Michael E2 Jul 12 '13 at 21:17
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The Thread operation is not completely invertible in all cases, but if we assume that
- only level 1 is to be considered, and
- in the original
Thread, all arguments had the same head and length,
then the expression can be unthreaded. Perhaps others can extend this.
Here, in unthread[expr, h], if expr is of the form
f[h[a1, a2,...], h[b1, b2,...],...]
the the result will be
h[f[a1, b1,...], f[a1, b2,...],...]
The default for h is List. Thus unthread[f[h[a1, a2,...],..., h] undoes Thread[h[f[a1, b1,...],...], f].
Clear[unthread];
unthread::usage = "unthread[f[args], h] \"unthreads\" f over args each with head h";
unthread::tdlen = "Objects of unequal length in `` cannot be uncombined. :)";
unthread::head = "Objects in `` not all of type ``.";
unthread[expr_, h_: List] /;
(Length@DeleteDuplicates[Length /@ expr] == 1 ||
(Message[unthread::tdlen, expr]; False)) &&
(DeleteDuplicates[Head /@ List @@ expr] == {h} ||
(Message[unthread::head, expr, h]; False)) :=
With[{h2 = Head[expr]}, h @@ h2 @@@ Transpose[List @@ List @@@ expr]];
unthread[expr_, h_] := expr; (* return expressions that do not satisfy criteria *)
Examples
unthread[{a + b == c + d}, Equal]
(* {a + b} == {c + d} *)
unthread[g[f[a, aa], f[b, bb], f[c, cc]], f]
(* f[g[a, b, c], g[aa, bb, cc]] *)
unthread[g[f[a, aa], f[b, bb], f[c, cc, ccc]], f]
(* unthread::tdlen: Objects of unequal length in g[f[a,aa], f[b,bb], f[c,cc,ccc]]
cannot be uncombined. :) *)
(* g[f[a, aa], f[b, bb], f[c, cc, ccc]] *)
unthread[g[f[a, aa], f[b, bb], h[c, cc]], f]
(* unthread::head: Objects in g[f[a,aa], f[b,bb], h[c,cc]] not all of type f. *)
(* g[f[a, aa], f[b, bb], h[c, cc]] *)
An example in which two different Thread operations yield the same output, which shows Thread is not completely reversible.
Thread[f[g[a, b, c], g[aa, bb, cc], d], g]
(* g[f[a, aa, d], f[b, bb, d], f[c, cc, d]] *)
Thread[f[g[a, b, c], g[aa, bb, cc], g[d, d, d]], g]
(* g[f[a, aa, d], f[b, bb, d], f[c, cc, d]] *)
answered Jul 12 '13 at 20:04
Map[List, {a + b == c + d}, {2}]$\endgroup$ – Dr. belisarius Jul 12 '13 at 17:15{a + b == c + d} /. f_[x_, y_] :> f[{x}, {y}](only in special cases) $\endgroup$ – Kuba♦ Jul 12 '13 at 17:16Equal @@ Tuples@{a + b == c + d}=> {a + b} == {c + d} $\endgroup$ – user1066 Jul 12 '13 at 23:47