Conditional function creating with @ handle

26 vues (au cours des 30 derniers jours)
Ashok Das
Ashok Das le 15 Fév 2019
Commenté : Walter Roberson le 15 Fév 2019
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

Réponse acceptée

Stephen23
Stephen23 le 15 Fév 2019
Modifié(e) : Stephen23 le 15 Fév 2019
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.

Plus de réponses (1)

Walter Roberson
Walter Roberson le 15 Fév 2019
@(xx,yy) (xx+yy).*(xx*yy>0) + (xx-yy).*(xx*yy<0)
note that this will fail if xx or yy are infinite.
  3 commentaires
Stephen23
Stephen23 le 15 Fév 2019
+1 general solution
Walter Roberson
Walter Roberson le 15 Fév 2019
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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