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Sjoerd Smit
Sjoerd Smit
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You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

This also avoids problems that can occur when you clone the function name naively that were observed in. See, e.g., this answer.

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

This also avoids problems that can occur when you clone the function name naively that were observed in this answer.

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

This also avoids problems that can occur when you clone the function name naively. See, e.g., this answer.

added 190 characters in body
Source Link
Sjoerd Smit
Sjoerd Smit
  • 24.8k
  • 48
  • 80

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

This also avoids problems that can occur when you clone the function name naively that were observed in this answer.

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}

This also avoids problems that can occur when you clone the function name naively that were observed in this answer.

Source Link
Sjoerd Smit
Sjoerd Smit
  • 24.8k
  • 48
  • 80

You can access the local version of the function directly with ResourceFunction["SymbolQ", "Function"] to avoid the evaluation semantics associated with ResourceFunction. The easiest way to do this, is to "clone" the function into a symbol convenient for you:

symbolQ = ResourceFunction["SymbolQ", "Function"];
x = 1;
ResourceFunction["SymbolQ"][x] // RepeatedTiming
symbolQ[x] // RepeatedTiming

{0.00076, True}

{5.*10^-7, True}