% 读入数据并进行预处理 data = [0.1, 0.2, 0.3, 0.1; 0.3, 0.2, 0.1, 0.2; 0.5, 0.6, 0.5, 0.4]; logData = log(data); ageMean = mean(data, 2); timeMean = mean(data, 1); % 建立模型 k = 1.5;% 设定初值 gamma = logData - ageMean*ones(1,length(timeMean)) - k*(ones(length(ageMean),1)*timeMean); phi = mean(gamma,2); kappa = logData - ageMean*ones(1,length(timeMean)) - phi*ones(1,length(timeMean)) - k*(ones(length(ageMean),1)*timeMean); % 对kappa进行奇异值分解 [U,S,V] = svd(kappa); K = 1; % 选择前K个主成分 uk = U(:,1:K); sk = S(1:K,1:K); vk = V(:,1:K); kappaNew = uk * sk * vk'; phiNew = mean(kappaNew, 2); gammaNew = kappaNew - phiNew * ones(1, length(timeMean)); % 参数估计和预测 y = reshape(logData - phi*ones(1, length(timeMean)), 1, []); x(:,1) = reshape(gammaNew, 1, []); for i = 1:K x(:,i+1) = U(:,i)' * (logData - phi.*ones(size(logData,1),1)); end b = x\y; futureTime = [1:2]'+timeMean(end); % 对未来年份数据进行预测(2023, 2024年) ages = [1:size(data,1)]'; logDataPred = repmat(phiNew, [1, length(futureTime)]) + gammaNew*k + ageMean*ones(1,length(futureTime)) + [ones(size(ages)),U(:,1:K)]*b; dataPred = exp(logDataPred); % 输出结果 disp("数据:") disp(data); disp("拟合数据:") disp(dataPred); 代码中(:,i+1) = U(:,i)' * (logData - phi.*ones(size(logData,1),1));无法执行赋值,因为左侧的大小为 12×1,右侧的大小为 1×4。
时间: 2023-09-11 22:08:38 浏览: 144
这个错误是因为左侧的大小为12×1,而右侧的大小为1×4,无法进行赋值。可能是因为你在for循环中没有正确地将右侧的结果转置。你可以尝试修改代码为:
```
for i = 1:K
x(:,i+1) = (U(:,i)' * (logData - phi.*ones(size(logData,1),1)))';
end
```
将右侧的结果转置即可解决该问题。
相关问题
% 读入数据并进行预处理 data = [0.1, 0.2, 0.3, 0.1; 0.3, 0.2, 0.1, 0.2; 0.5, 0.6, 0.5, 0.4]; logData = log(data); ageMean = mean(data, 2); timeMean = mean(data, 1); % 建立模型 k = 1.5; % 设定初值 gamma = logData - ageMean*ones(1,length(timeMean)) - k*(ones(length(ageMean),1)*timeMean); phi = mean(gamma,2); kappa = logData - ageMean*ones(1,length(timeMean)) - phi*ones(1,length(timeMean)) - k*(ones(length(ageMean),1)*timeMean); % 对kappa进行奇异值分解 [U,S,V] = svd(kappa); K = 1; % 选择前K个主成分 uk = U(:,1:K); sk = S(1:K,1:K); vk = V(:,1:K); kappaNew = uk * sk * vk'; phiNew = mean(kappaNew, 2); gammaNew = kappaNew - phiNew * ones(1, length(timeMean)); % 参数估计和预测 y = reshape(logData - phi*ones(1, length(timeMean)), 1, []); x(:,1) = reshape(gammaNew, 1, []); disp("数据:") disp(U); disp("数据:") disp(logData - phi.*ones(size(logData,1),1)); for i = 1:K x(:, i+1) = (U(:, i)' * (logData - phi .* ones(1, length(timeMean))))'; end b = x\y; futureTime = [1:2]'+timeMean(end); % 对未来年份数据进行预测(2023, 2024年) ages = [1:size(data,1)]'; logDataPred = repmat(phiNew, [1, length(futureTime)]) + gammaNew*k + ageMean*ones(1,length(futureTime)) + [ones(size(ages)),U(:,1:K)]*b; dataPred = exp(logDataPred); % 输出结果 disp("数据:") disp(data); disp("拟合数据:") disp(dataPred);
这段代码主要是一个基于主成分分析(PCA)的线性回归模型,用于对给定的数据进行拟合和预测。在这段代码中,出现了 x(:, i+1) = (U(:, i)' * (logData - phi .* ones(1, length(timeMean))))'; 这行代码无法执行赋值的问题,你可以将其修改为以下代码:
```
for i = 1:K
x(:, i+1) = U(:, i) * (logData - phi .* ones(1, length(timeMean)));
end
```
这里将右侧的向量 (U(:, i)' * (logData - phi .* ones(1, length(timeMean))))' 修改为了 U(:, i) * (logData - phi .* ones(1, length(timeMean))),这样就可以保证左侧和右侧的向量大小一致,可以正确执行赋值操作。
请注意,修改后的代码可能会影响到后续的计算结果,你需要根据实际情况进行调整。如果还有其他问题或疑问,欢迎继续提问。
阅读全文