Getting multiple outputs from a function?
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Hi everyone. This is my code. It works well in this state but i want to change ''ord'' number from 1 to 4 and ''sfn'' number from 1 to ord+1. For example, when ord=3, sfn values should be 1, 2, 3 and 4, accordingly. How can i do this?
% shapefun( ord,sfn,x ) function gives the value of the 'sfn'th shape function of
% element with order 'order' at x coordinate.
% 'order' can be any integer above 0
% x € [-1,1]
% sfn is integers from 1 up to (order+1),
function result = shapefun(ord,sfn,x)
result=1;
if nargin==0
ord = 1;
sfn = 1;
x = linspace(-1,1,1000);
end
for i = 1:ord+1
if(i==sfn)
result = result.*1;
else
result = result.*(x-((i-1).*1./ord))./(((sfn-1).*1./ord)-((i-1).*1./ord));
end
end
end
1 Kommentar
Guillaume
am 31 Mär. 2019
Your code:
if(i==sfn)
result = result.*1;
Last I checked, multiplying by 1 doesn't do much. What is the point of that?
Antworten (1)
Anish Navalgund
am 31 Mär. 2019
Hi Isa,
You can use switch statements to control your order and for each order you can have for loops to calculate for each sfn.
The psuedo code is below.
order = input('order= ');
switch order
case 1
for sfn = 1:2
result = shapefun(ord,sfn,x);
end
case 2
for sfn = 1:3
result = shapefun(ord,sfn,x);
end
end
And so on...
Hope this helps.
6 Kommentare
Stephen23
am 31 Mär. 2019
Bearbeitet: Stephen23
am 31 Mär. 2019
"...my result matrix is being one row not two, i mean function take sfn as only 2 not both 1 and 2. Can you help about this case?"
Most likely you will need to use indexing on the output array, otherwise you simply discard the results of every iteration except for the last one. I say "most likely" as I do not understand your algorithm.
Anish Navalgund
am 31 Mär. 2019
@Isa, like @Stephen said you can use his method as the repetition of calling of the function is reduced. :)
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