My question is: why doesn't the following work, and how do I fix it?
Plot[f[t], {t, 0, 2*Pi}] /. {{f -> Sin}, {f -> Cos}}
The result is two blank graphs.开发者_开发技巧 By comparison,
DummyFunction[f[t], {t, 0, 2*Pi}] /. {{f -> Sin}, {f -> Cos}}
gives
{DummyFunction[Sin[t], {t, 0, 2 *Pi}], DummyFunction[Cos[t], {t, 0, 2 * Pi}]}
as desired.
This is a simplified version of what I was actually doing. I was very annoyed that, even after figuring out the annoying "right way" of putting the curly brackets nothing works.
In the end, I did the following, which works:
p[f_] := Plot[f[t], {t, 0, 2*Pi}]
p[Sin]
p[Cos]
As an alternative to Peter's Hold
/ReleaseHold
strategy you could do
Plot[Evaluate[ f[t]/. {{f -> Sin}, {f -> Cos}} ], {t, 0, 2*Pi}]
which is a little cleaner to read. This ensures that f
is substituted before Plot
is evaluated.
This one is even shorter:
Plot[#[t], {t, 0, 2*Pi}] & /@ {Sin, Cos}
Mathematica is trying to evaluate Plot before the substitution. You can prevent that with the Hold and ReleaseHold functions:
ReleaseHold[Hold[Plot[f[t],{t,0,2*Pi}]] /. {{f -> Sin},{f -> Cos}}]
Hold[] will force the entire Plot subexpression to remain unsimplified while the substitution is performed, then ReleaseHold[] will let it proceed with the actual plotting.
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