Conditional function creating with @ handle
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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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採用された回答
Stephen23
2019 年 2 月 15 日
編集済み: Stephen23
2019 年 2 月 15 日
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.
その他の回答 (1 件)
Walter Roberson
2019 年 2 月 15 日
@(xx,yy) (xx+yy).*(xx*yy>0) + (xx-yy).*(xx*yy<0)
note that this will fail if xx or yy are infinite.
3 件のコメント
Walter Roberson
2019 年 2 月 15 日
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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