c0 and x are scalars, c - vector, and p - scalar. If c is [ ], then p = c0. If c is a scalar, then p = c0 + c*x . Else, p =

2 次查看（过去 30 天）
Subramanian Mahadevan
回答： Govind Mishra 2018-3-14
function [p] = poly_val(c0,c,x)
N = length(c);
n=1:1:N;
if (N<=1)
if(isempty(c))
p=c0;
else
p= c0+(c*x);
end
end
if(N>1)
p = c0+(sum(c(n).*(power(x,n))));
end
end

请先登录，再回答此问题。

采纳的回答

Thorsten
Thorsten 2016-11-14
function y = mypolyval(c0,c,x)
if isempty(c)
y = c0;
else
y = c0 + power(x, 1:numel(c))*c(:);
end
The case y = c0 + c*x is already covered by the else. And you can use matrix multiplication of a row and a column vector r*c instead of sum(r.*c').
  2 个评论
Raunil Raj
Raunil Raj 2018-3-6
really! I went through the same problem. I however don't understand as to how c(:) can convert any row vector or column vector into a column vector. Can someone please explain?

请先登录，再进行评论。

更多回答（6 个）

Jorge Briceño
Jorge Briceño 2018-1-29
Here is my solution:
function p = poly_val(c0,c,x)
format long
n=(1:1:length(c));
c=c(:)' & This part converts any array/matrix into a colunm vector and transpose...
% it afterwards, since you are working with row vector properties.
if isempty(c)
p=c0;
elseif isscalar(c)
p=c0+sum(c.*x);
else
p=c0+sum((c.*(x.^n)));
end
end
KSSV
KSSV 2016-11-10
编辑：KSSV 2016-11-10
function [p] = poly_val(c0,c,x)
N = length(c); % length of c
if N == 0 % if c is empty
p = c0 ;
elseif N == 1 % if c is a scalar
p = c0+c*x ;
else % if c is a vector
p = c0+(sum(c.*(power(x,N))));
end
  4 个评论
显示 2更早的评论

请先登录，再进行评论。

Gabir Yusuf
Gabir Yusuf 2017-8-8
if true
function p = poly_val(c0,c,x)
n=length(c);
if sum(size(c))==0
p = c0;
elseif isscalar(c)
p = c0 + c*x;
else
y=1:n;
z=x.^y;
if size(c)==[1 n]
p=sum(c.*z)+c0;
else
c=c';
p=sum(c.*z)+c0;
end
end
end
Anshuman Panda
Anshuman Panda 2017-8-19
function p=poly_val(c0,c,x) a=length(c); if a==0 p=c0; else if a==1 p=c0+c*x; else p=c0 + power(x , 1:a)*c(:); end end end
Darío Pascual
Darío Pascual 2018-3-12
function p=poly_val(c0,c,x)
N = length(c);
n=1:1:N;
d=size(c);
if(isempty(c))
p=c0;
end
if N==1
p= c0+(c*x);
end
if N>1
if d(1)==1
p = c0+(sum(c(n).*(power(x,n))));
else
c=c'
p = c0+(sum(c(n).*(power(x,n))));
end
end
Govind Mishra
Govind Mishra 2018-3-14
function [p] = poly_val(c0,c,x)
if(iscolumn(c)) c=transpose(c); end
N = length(c); n=1:1:N; if (N<=1) if(isempty(c)) p=c0; else p= c0+(c*x); end end if(N>1) p = c0+(sum(c(n).*(power(x,n)))); end end

类别

在 Help Center 和 File Exchange 中查找有关 Dimensionality Reduction and Feature Extraction 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by