Why does fplot think my function is not vectorized

11 次查看（过去 30 天）
Matt J
Matt J 2019-5-30
编辑： Matt J 2019-6-25
Executing the following as a script in R2018a
a=ones(1,10);
b=a;
b0=1;
fun=@(x) myFourier(x,a,b,b0);
fplot(fun);
function f=myFourier(x,a,b,b0)
n=numel(a);
arg=x(:).*(1:n);
f=b0+sin(arg)*a(:)+cos(arg)*b(:);
f=reshape(f,size(x));
end
results successfully in a plot, but throws warnings (EDIT: and results in non-vectorized execution!!!)
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your
function to return an output with the same size and shape as the input arguments.
> In matlab.graphics.function.FunctionLine>getFunction
In matlab.graphics.function.FunctionLine/updateFunction
In matlab.graphics.function.FunctionLine/set.Function_I
In matlab.graphics.function.FunctionLine/set.Function
In matlab.graphics.function.FunctionLine
In fplot>singleFplot (line 234)
In fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 193)
In fplot>vectorizeFplot (line 193)
In fplot (line 163)
In test (line 7)
But the input function is properly vectorized as is easily verified in simple tests,
>> fun(rand(1,5))
ans =
3.4371 2.5677 14.1698 2.4490 1.0828
>> fun(rand(5,1))
ans =
14.1745
8.3619
-0.2579
1.7024
1.5748
Why the warnings, then?

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2019-5-31
f1 = fun(1);
f2 = fun([1 2]);
f1 - f2(1)
ans =
4.44089209850063e-16
Your code produces different outputs for the same input, which violates the rule that the calculations must be independent.
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2019-6-1
fplot(@(x) x .* (1 + (length(x)>1)*eps))
generates the warning, so regardless of whether it "should", it does check for equality.
Matt J
Matt J 2019-6-5
编辑：Matt J 2019-6-5
Tech support got back to me and confirmed Walter's explanation of the problem, and said they will look into remedies for future releases. I passed on to them my suggestions for fixes as well.

请先登录，再进行评论。

更多回答（3 个）

dpb
dpb 2019-5-31
Because fplot is just looking at the source code, not the output and it doesn't recognize the reshape operation at the end--only that there are no "dot" operators in the evaluation from which it infers (usually correctly, but as you've shown not infallibly) that the function isn't "vectorized".
Whether there's any chance of being able to get this one right without way more parsing logic than would want to use for performance is, I'm guessing, pretty small. You can't just presume that if the user has a reshape call in the function that always works correctly, either.
  14 个评论
显示 12更早的评论
Matt J
Matt J 2019-6-2
编辑：Matt J 2019-6-2
Worse than that, because fplot incorrectly judges the function to be nonvectorized, it forms the plot by executing the function in non-vectorized fashion forcing the user to endure slow behavior. I think they should let fplot() operate like integral() already does. Let the user specify whether execution will be vectorized or not, and take responsibility for the result.
dpb
dpb 2019-6-2
This could well be the basis for an enhancement/quality of implemenation improvement request.

请先登录，再进行评论。

Catalytic
Catalytic 2019-6-7
Here's a workaround to trick fplot into doing the right thing.
Capture.PNG
Yair Altman
Yair Altman 2019-6-22
As far as I can tell (never mind exactly how), internally a check is made whether the output of fun(1:3) is exactly equaln to the output of [fun(1),fun(2),fun(3)]. Even a tiny FP eps difference will cause the warning to be evoked. If you want to disable the error in run-time, run the following command:
warning off MATLAB:fplot:NotVectorized
  1 个评论
Matt J
Matt J 2019-6-25
编辑：Matt J 2019-6-25
Thanks, Yair. But disabling the warning still leaves the bigger problem of execution efficiency. fplot will still go ahead and evaluate fun in non-vectorized (therefore slow) fashion based on the fun(1:3) test, even with the warning disabled.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Utilities for the Solver 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by