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Lets define two identical functions $f$ and $g$ in a slightly different way

f = Interpolation[Table[{i, i^2}, {i, -5, 5}]];
g = Function[x, Interpolation[Table[{i, i^2}, {i, -5, 5}]][x]];
Timing[Do[g[1], {i, 1000}]]/Timing[Do[f[1], {i, 1000}]]
{12.0023, 1}

Does the second definition calculates the interpolating function every time $g$ is evaluated?

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  • $\begingroup$ I see now it is because of the HoldAll attribute of the function. A workaround: g = Function[x, Evaluate@Interpolation[Table[{i, i^2}, {i, -5, 5}]][x]] $\endgroup$ – user13427 Aug 8 '14 at 22:39
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Yes! You have passed the same function f into a pure function so it is evaluated every time you calculate g. Specially you define f by = so Mathematica calculates it once and then re-use it in Do loop. For the sake of demonstration if I define f by :=:

f1 := Interpolation[Table[{i, i^2}, {i, -5, 5}]];
f2 = Interpolation[Table[{i, i^2}, {i, -5, 5}]];
Timing[Do[f1[1], {i, 1000}]]/Timing[Do[f2[1], {i, 1000}]]

The result is

 {11.91, 1}
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    $\begingroup$ Thanks. I see , but this is not a SetDelayed issue, I think it is more related to the HoldAll attribute of Function, now I realize. $\endgroup$ – user13427 Aug 8 '14 at 22:38
  • $\begingroup$ You are right. As the Documentation says: "Function has attribute HoldAll. The function body is evaluated only after the formal parameters have been replaced by arguments." $\endgroup$ – Mahdi Aug 8 '14 at 22:41

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