Conditional function creating with @ handle

Question

Ashok Das 2019-2-15

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/445149-conditional-function-creating-with-handle

评论： Walter Roberson 2019-2-15

采纳的回答： Stephen23

If I want to make a function f(x,y) = x+y, we do

f= @(x,y) x+y

Now I want to create a function of the form, f(x,y) = x+y if x*y >1 ; x-y if x*y <=1.

How should I create it with a @ handle?

Thanks in advance

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Answer 1

Stephen23 2019-2-15

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https://ww2.mathworks.cn/matlabcentral/answers/445149-conditional-function-creating-with-handle#answer_361193

编辑：Stephen23 2019-2-15

Directly for this specific calculation:

>> f = @(x,y) x + y*(1-2*((x*y)<=1));
>> f(1,2)
ans =  3
>> f(2,1)
ans =  3

It works by simply defining the sign here:

(1-2*((x*y)<=1))

and using that sign to multiply with y. The inbuilt sign function cannot be used because it returns 0 when the input is 0.

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Ashok Das 2019-2-15

Thanks. :)

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Answer 2

Walter Roberson 2019-2-15

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https://ww2.mathworks.cn/matlabcentral/answers/445149-conditional-function-creating-with-handle#answer_361196

@(xx,yy) (xx+yy).*(xx*yy>0) + (xx-yy).*(xx*yy<0)

note that this will fail if xx or yy are infinite.

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Stephen23 2019-2-15

+1 general solution

Walter Roberson 2019-2-15

If you have the symbolic toolbox, it is often the case that such things are better rewritten in terms of piecewise()

If you happen to be using integration, then with complicated formula, numeric integration using multiplication by a logical condition can turn out to be much faster than symbolic integration. However, numeric integration is not always sufficiently sensitive to narrow regions, and narrow regions sometimes make a huge difference in integration. For example, numeric integration will almost always get the wrong results if there is a dirac delta in the calculation, unless you happen to drop in a waypoint right at the location of every delta (and integral2 does not permit waypoint specification at all.)

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