下面是一段代码，它在没有函数/函数句柄的情况下运行良好。然而，我必须编写一个函数，这样我才能使用"的优化函数；lsqcurvefit"；在MATLAB中对参数常数(1(和常数(2(进行优化。
L=zeros(n,length(S)); % Height of elements
L(:,1)=Linitial; % Height of elements for day zero
for jj=1:length(S)
if jj==1
continue
end
for j=1:n % n is 101 (previously defined...)
L(j,jj) = @(constant, i_dt) L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2)); %*****
end
end
height_pred = @(constant,i_dt) sum(L); %heigh_pred(1) is known and previously defined
options = optimoptions('lsqcurvefit','FinDiffType','central',...
'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);
当我运行上述代码时，我收到错误"；从function_handle转换为double是不可能的"；5颗星(*****(。
如果我去掉函数句柄，而是在代码末尾写一个函数，如下所示：
L=zeros(n,length(S)); % Height of elements
L(:,1)=Linitial; % Height of elements for day zero
for jj=1:length(S)
for j=1:n % n is 101 (previously defined...)
L(j,jj) = Nelfun(j,jj);
end
end
height_pred = @(constant,i_dt) sum(L); %heigh_pred(1) is known and previously defined
options = optimoptions('lsqcurvefit','FinDiffType','central',...
'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);
function L = Nelfun(constant,i_dt)
for jj=1:145
if jj==1
L(:,1)=0.0187600000000000;
continue
end
for j=1:101
L(j,jj-1);
L(j,jj) = L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2)); %*****
end
end
end
我收到错误
"Index exceeds matrix dimensions" for line with (*****).
矩阵i_dt的大小是(101145(，L是(101145%(，高度_meas(1145(和高度_pred(1145。
jj＝＝1的L(j，jj(是已知的，但是应该针对L(：，2(到L(：：，145(来计算和优化。
***
感谢Praveen；索引超过矩阵维度"；问题已经解决，但现在我收到错误
"Error using snls (line 183)
Finite difference Jacobian at initial point contains Inf or NaN values. lsqcurvefit cannot continue."
对于以下代码：
lb = [0.000000000001,0.05]; ub = [1,50]; constant0 = [1.06e-09,15.2];
height_pred = @(constant,i_dt) sum(Nelfun(constant,i_dt));
options = optimoptions('lsqcurvefit','FinDiffType','central',...
'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);
function L = Nelfun(constant,i_dt)
L=zeros(size(i_dt));
n=101; % number of nodes
ei=2.05613504572765;
e0=ei.*ones(n,1);
Hsolid=0.613847219422639;
hs = Hsolid./(n-1);
for JJ=1:size(i_dt,2)
if JJ==1
for j=1:n
Linitial0=hs.*(1+e0);
Linitial0(1)=0;
end
L(:,JJ) = Linitial0;
continue
end
for J=1:size(i_dt,1)
L(J,JJ) = L(J,JJ-1) - (i_dt(J,JJ).*constant(1).*((L(J,JJ-1)./hs)-1)^constant(2));
end
end
end
在第一种情况下，不能生成函数句柄数组。你可以使用细胞阵列，但我不明白你为什么需要它。
让我们来处理你的第二个案子。
%I don't know what is hs, I assume it a scalar
hs=myhs;
%call the Nelfun in your function handle
height_pred = @(constant,i_dt) Nelfun(constant,i_dt,hs);
options = optimoptions('lsqcurvefit','FinDiffType','central',...
'TolFun',1e-10);
% Constant0 should be 2 element array
constant0=[0,0];
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);
function L = Nelfun(constant,i_dt,hs)% why hs is not passed
L=zeros(size(i_dt)); % Always initialize before filling it up
for jj=1:size(i_dt,2) %use adaptive indexing
if jj==1
L(:,jj)=0.0187600000000000;
continue
end
for j=1:size(i_dt,1) %use adaptive indexing
L(j,jj) = L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2));
end
end