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I solved a set of equation and use reduce on the set of solutions. Changing the equal sign to arrows I get some set of rules but some of the entries are inequalities. E.g.

{\[ScriptA][2, 1] != 0, 
 0 -> \[ScriptA][2, 2] - Sqrt[\[ScriptA][2, 2]^2], \[ScriptA][4, 
   4] -> 1, \[ScriptA][4, 3] -> 
  0, \[ScriptA][3, 3] -> -1, \[ScriptA][1, 2] -> (
  1 - \[ScriptA][2, 2]^2)/\[ScriptA][2, 1], \[ScriptA][1, 
   1] -> -\[ScriptA][2, 2]}

How can I remove in a systematic way the first kind of entries ($ \cal a[2,1]\neq 0)$ for my rules? I tried Deletecase with several patterns but that does not seem to work.

Thanks!

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2
  • $\begingroup$ Select[{...}, #[[0]] == Rule &]? Cases[{...}, _Rule]? $\endgroup$ Aug 10, 2019 at 11:39
  • $\begingroup$ FullForm of a != b is Unequal[a,b], so you should use DeleteCases with this head: DeleteCases[{<your set of rules>},_Unequal]. $\endgroup$
    – Alx
    Aug 10, 2019 at 12:42

1 Answer 1

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DeleteCases[{\[ScriptA][2, 1] != 0, 
0 -> \[ScriptA][2, 2] - Sqrt[\[ScriptA][2, 2]^2], \[ScriptA][4,4] -> 1,
\[ScriptA][4,3] -> 0, \[ScriptA][3, 3] -> -1,
\[ScriptA][1, 2] -> (1 - \[ScriptA][2, 2]^2)/\[ScriptA][2, 1],
\[ScriptA][1, 1] -> -\[ScriptA][2, 2]},_Unequal]

gives

{0 -> \[ScriptA][2, 2] - Sqrt[\[ScriptA][2, 2]^2], \[ScriptA][4,4] -> 1,
\[ScriptA][4,3] -> 0, \[ScriptA][3, 3] -> -1,
\[ScriptA][1, 2] -> (1 - \[ScriptA][2, 2]^2)/\[ScriptA][2, 1],
\[ScriptA][1, 1] -> -\[ScriptA][2, 2]}
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