New answers tagged pattern-matching
1
SeedRandom[1]
Mat = Table[vectelem @ RandomInteger[{1, 30}], {i, 10}, {j, 10}];
You can use Pattern + Condition (/;) + VectorQ + NumericQ to define your argument pattern:
ClearAll[pattern, Fmat1]
pattern = p : Pattern @@@ Array[{Symbol["q" <> ToString@#], _} &, 30] /;
VectorQ[p, NumericQ];
Fmat1[pattern] = Mat;
If a list of ...
8
A simple trick is to notice that in Mathematica, Exp[x] can be written as Exp@(x) where the @ symbol has very high precedence. A replacement of the string "exp" by the string "Exp@" therefore does the necessary conversion without needing any bracket-matching logic:
convert[s_String] := ToExpression[StringReplace[s, "exp" -> &...
7
Update: We can convert the strings to expressions without any string processing:
ClearAll[sR0]
sR0 = ToExpression[#, TraditionalForm] &;
sR0 /@ {str1, str2, str3, str4, str5} // Column
If you need to get strings as output wrap the expression with ToString[#, InputForm]& using the third argument of ToExpression:
ClearAll[sR01]
sR01 = ToExpression[#, ...
8
pattern = StringReplace["f_ath_r", "_" -> Blank[]]
"f" ~~ _ ~~ "ath" ~~ _ ~~ "r"
DictionaryLookup @ pattern
{"feather"}
Generalizing to allow __ and ___ in the input string:
toSPattern = StringReplace[p : Repeated["_", {1, 3}] :> ToExpression[p]];
DictionaryLookup[toSPattern@&...
0
You could post-process the definitions:
ClearAll[FixMinusPattern];
SetAttributes[FixMinusPattern, HoldAll];
FixMinusPattern[fn_] :=
DownValues[fn] =
Quiet@ReplaceAll[DownValues[fn],
HoldPattern[
Verbatim[HoldPattern][
fn[PatternSequence[otherArgs1___, Verbatim[Pattern][a_, type_],
otherArgs2___, -1*Verbatim[Pattern][a_, ...
0
A slightly pointless way of solving the problem with a single variable...
G[X : Repeated[_, {2}] /; {X} . {1, 1} == 0] := 0
G[4, -4]
(* 0 *)
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