Mathematica Asked on January 1, 2021
Consider the following code:
Options[pl] = {lim->{0,1}};
pl[n_, OptionsPattern[]] := Plot[x^n, Join[{x}, OptionValue[lim]] ]
This function works as expected on 12.0: running something like pl[2]
produces a plot over $[0,1]$, while pl[2, lim->{2,3}]
uses the given limits.
But if I instead use ParametricPlot
, as in
Options[pl] = {lim->{0,1}};
pl[n_, OptionsPattern[]] := ParametricPlot[{x,x^n}, Join[{x}, OptionValue[lim]] ]
then pl[2]
just gives
System`SampledPlotsDump`iParametricPlot[{x, x^2}, Join[{x}, OptionValue[pl, {}, lim]]]
with a similar error message when explicitly including the option. What could be going on here? Why does ParametricPlot
behave differently from Plot
?
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP