Question

我对线性模型y = X * beta + eps进行模拟研究，其中尺寸（X）= [n d]。我考虑了基于两种方法的维数d的影响。我运行10个模拟数据并得到相应的beta估计值，然后我想计算10个模拟数据的beta平均值。

我的玩具matlab代码如下：

        nsim=10;   %iteration number
        dd=[4 6];  %two dimension cases,\beta=(\beta_1,\cdots,\beta_d)^T
        ddlen=length(dd);
        nmethod=2; %two methods
        seednum=0;

        BH  = cell(nsim,ddlen,nmethod); %estimation of beta w.r.t two dimension cases and two methods

        for di = 1:ddlen
            d = dd(di);
            for ni = 1:nsim
                seednum = seednum + di*ni;
                randn('seed', seednum);
                betahat=randn(d,1); 
                for method = 1:nmethod
                    if method==1
                        BH{ni,di,method} = betahat;
                    else
                        BH{ni,di,method} = 10*betahat;
                    end
                end
            end
        end

然后我们可以获得

BH(:,:,1) = 

    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]


BH(:,:,2) = 

    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]
    [4x1 double]    [6x1 double]

我希望10行（nsim = 10）的平均值得到类似

的东西

mean(BH(:,:,1))= 

    [4x1 double]    [6x1 double]

mean(BH(:,:,2)) = 

    [4x1 double]    [6x1 double]

有什么想法吗？谢谢！

Answer 1

如果我找到了你，你想要对矢量中相同位置的所有元素取平均值。因此，从BH(:,1,1)中的所有向量，我们得到一个4个均值的向量，每个向量用于向量中的一个位置。同样适用于BH(:,1,2)。对于BH(:,2,1)和BH(:,2,1)，我们会做同样的事情，但在向量中有6个元素。

您可以使用以下代码：

% split BH to 2 arrays:
bh4 = reshape(cell2mat(BH(:,1,:)),[],nsim,2); % all the 4 elements vectors
bh6 = reshape(cell2mat(BH(:,2,:)),[],nsim,2); % all the 6 elements vectors
meanBH4 = squeeze(mean(bh4,2)); % mean over all 4 element vectors
meanBH6 = squeeze(mean(bh6,2)); % mean over all 6 element vectors

但是，以正确的方式执行此操作的一步是定义两个数组，每个方法一个：

BH1  = zeros(nsim,ddlen,dd(1));
BH2  = zeros(nsim,ddlen,dd(2));

然后在你的循环中为它们分配值：

if method==1
    BH1(ni,di,:) = betahat;
else
    BH2(ni,di,:) = 10*betahat;
end

最后只取每个人的意思：

meanBH1 = mean(BH1,3)
meanBH2 = mean(BH1,3)

修改

要以更“'Matlabish'的方式写下所有这些，我建议如下：

nsim = 10; % iteration number dd = [4 6]; % two dimension cases,\beta=(\beta_1,\cdots,\beta_d)^T methods = 2; % two methods % preapering random seeds s = bsxfun(@times,1:numel(dd),(1:nsim).'); seednum = cumsum(s(:)); % initialize results array BH = nan(max(dd),nsim,numel(dd),methods); counter = 1; for k = 1:numel(dd) for n = 1:nsim % set a new random seed from the list: rng(seednum(counter)); % compute all betahats with this seed: betahat = randn(max(dd),2).*repmat([1 10],[max(dd) 1]); % assign the values to BH by dimension: for m = 1:methods BH(1:dd(k),n,k,m) = betahat(1:dd(k),m); end counter = counter+1; end end % compute the means over iterations: means = squeeze(mean(BH,2,'omitnan'))

所以你得到means作为结果。

P.S。我不知道为什么你在每次迭代besides that's not a recommended syntax上调用randn('seed', seednum)，但如果你可以删除它，那么你可以对大多数循环进行矢量化，你的代码会挤压到：

% compute all betahats: betahat = randn(nsim,max(dd),numel(dd),2); % apply dimensions: for k = dd betahat(:,k+1:end,1,:) = nan; end % apply methos 2: betahat(:,:,:,2) = betahat(:,:,:,2)*10; % compute the means over iterations: means = squeeze(mean(betahat,1,'omitnan'))

希望事情看起来更清楚......

Answer 2

我不知道这是否是最有效的方式，但你可以使用arrayfun：

% generate random array
BH = repmat({rand(4,1),rand(6,1)},[10 1 2]);
% generate indexes for the 2nd and 3rd dimensions
[n2,n1] = meshgrid(1:size(BH,2),1:size(BH,3));
% get mean across 1st (cell) dimension
[res] = arrayfun(@(n1,n2)mean([BH{:,n1,n2}],2),n1(:),n2(:),'UniformOutput',false);
% reshape to desired output
res = reshape(res,[1 size(BH,2) size(BH,3)]);

如果您想要推广到N维单元格数组：

% generate random array
BH = repmat({rand(4,1),rand(6,1)},[10,1,2,2,5]);
sz = size(BH);
% generate indexes for the 2nd and 3rd dimensions
n = cell(1,numel(sz) - 1);
[n{:}] = ndgrid(1:sz(2),1:sz(3),1:sz(4),1:sz(5));
n = cell2mat(cellfun(@(x) {x(:)},n));
idx = 1:size(n,1);
% get mean across 1st (cell) dimension
[res] = arrayfun(@(idx)mean([BH{:,n(idx,1),n(idx,2),n(idx,3),n(idx,4)}],2),...
    idx,'UniformOutput',false);
% reshape to desired output
res = reshape(res,[1 sz(2:end)]);

Answer 3

可替换地，

% split into seperate cell arrays
BH_1 = BH(:,:,1);
BH_2 = BH(:,:,2);

% create matrix of compatible vectors, and take mean and put result back into cell array
BH_1_mean = cat(2,{mean(cell2mat(BH_1(:,1)'),2)}, {mean(cell2mat(BH_1(:,2)'),2)});
BH_2_mean = cat(2,{mean(cell2mat(BH_2(:,1)'),2)}, {mean(cell2mat(BH_2(:,2)'),2)});

单元格数组在matlab中表示

3 个答案: