如何解决如何在MATLAB中的函数中具有顺序?
下面是代码,在没有功能/功能句柄的情况下可以正常工作。但是,我必须编写一个函数,以便可以在MATLAB中使用“ lsqcurvefit”的优化函数并优化参数constant(1)和constant(2)。
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,对于5星(*****)行,我收到错误消息“无法从function_handle转换为两倍”。
如果我删除了函数句柄,而是在我的代码末尾编写了一个函数,如下所示:
L=zeros(n,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,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)./hs)-1)^constant(2)); %*****
end
end
end
我收到错误消息
"Index exceeds matrix dimensions" for line with (*****).
矩阵i_dt的大小为(101,145),L为(101,145),height_meas(1,145)和height_pred(1,145)。
已知jj == 1的L(j,jj),但应针对L(:,2)至L(:,145)进行计算和优化。
***
感谢Praveen,“索引超过矩阵尺寸”的问题已解决,但现在我收到了错误
"Error using snls (line 183)
Finite difference Jacobian at initial point contains Inf or NaN values. lsqcurvefit cannot continue."
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',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
解决方法
在第一种情况下,您无法创建函数句柄数组。您可以使用单元阵列,但我不明白为什么需要它。
让我们来处理第二种情况。
%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,height_meas,lb,ub,options);
function L = Nelfun(constant,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
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。