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Applying a constant function on a vector

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Jose
Jose el 22 de Nov. de 2012
Hi,
I am writing a program for school. I need the user to input a function as a string, and then I need to process it.
What I am doing:
f = vectorize(inline(user_input_string));
x = linspace(0, 20, 20);
y = f(x);
...
This works perfectly except when the user inputs a constant function (such as f = 1). In order for my project to work, y must be a vector. But if the user inputs a constant function, Matlab automatically sets y to a sclalar, instead of a vector.
What can I do?

Respuestas (5)

Star Strider
Star Strider el 22 de Nov. de 2012
Editada: Star Strider el 22 de Nov. de 2012
I'm not certain I completely understand your problem. However, you could test for a scalar and if something like f=1 was the input, perhaps set y to:
f = '1';
y = polyval(str2num(f), x);
If the test for a scalar was true, you could also consider something like:
user_input_string = sprintf('polyval(%s, x)', f);
Without knowing more, that's the best solution I can come up with.
  2 comentarios
Jose
Jose el 22 de Nov. de 2012
No, that won't solve it. I'll explain better:
- the user inputs a function
- i apply the function on a vector (linspace(0, 20, 100))
- then i want to get a vector i can use later on a internal product
- however, if the function e constant (let's say, f = 1), if i apply the function on the vector, i don't get another vector, i get a scalar.
Hope it's is clearer now.
Star Strider
Star Strider el 22 de Nov. de 2012
If the user inputs a constant, what output for user_input_string do you want?
The polyval function outputs a vector of constant values equal to the scalar input value with a length equal to the length of your x-vector. It's the only option I can think of that's compatible with your user_input_string variable.
The version I posted earlier assumes the scalar is read in as a string. If you read f in as a numeric value instead, replace the %s with %f:
user_input_string = sprintf('polyval(%f, x)', f);

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Walter Roberson
Walter Roberson el 22 de Nov. de 2012
if length(y) == 1; y = y(ones(size(x))); end

Matt J
Matt J el 22 de Nov. de 2012
Editada: Matt J el 22 de Nov. de 2012
f = inline(user_input_string);
fh=@(x) f(x);
x = linspace(0, 20, 20);
y = arrayfun(fh,x);

Jose
Jose el 23 de Nov. de 2012
Thanks guys!

Matt Fig
Matt Fig el 23 de Nov. de 2012
You can specify the variable in your call to INLINE. For example, this works even if the user enters 2:
f = vectorize(inline(input('Enter a func of x : ','s'),'x'));
  2 comentarios
Matt J
Matt J el 23 de Nov. de 2012
Editada: Matt J el 23 de Nov. de 2012
No, it doesn't work around the issue cited by the OP. The code still gives
>> f(1:10)
ans =
1
whereas the desired output is ones(1,10).
Matt Fig
Matt Fig el 23 de Nov. de 2012
Ah, good catch...
I guess we could get fancy.
S = input('Enter a func of x : ','s');
f = vectorize(inline([S,'+zeros(size(x),class(x))'],'x'));
But, yuck.

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