遞歸計算數據的運行平均值

Question

我有兩個二維矩陣A和B ，其中行表示試驗，列表示試驗期間收集的樣本。

我處於A可用但B實時收集的場景中。 我想計算 { A的運行平均值和B } 的可用數據，因為B正在被采樣。 我想我可以通過計算A和B的加權平均值並在收集B試驗和樣本時更新權重來實現這一點。 具體來說，我認為我可以更新權重並遞歸使用我已經從上一次迭代中保存的值。 下面是我的代碼和輸出圖：

close all;
clear all;

%define the sizes of the matrices -- exact numbers aren't important for illustration
n1 = 5;
n2 = 10;
n3 = 12;

%define a matrix that will act as the history of data already collected
A = randi(10,[n2,n1]);
A_avg = mean(A,1); %averaged across n2 trials to get n1 values

%current acts as "incoming" data
B = randi(10,[n3,n1]); %n3 trials, n1 samples per trial

%preallocate matrices for final solutions
correct_means = zeros(n3,n1);
estimated_means = zeros(n3,n1);

for k1=1:size(B,1) %loop through trials
    %get running average in the case where we already have all samples
    correct_means(k1,:) = mean([A;B([1:k1],:)],1);
    for k2=1:size(B,2) %k2 should loop through samples
        %calculate averages as samples are incoming recursively (weighted averaging)
        if k1>1
            estimated_means(k1,k2) = (n2 / (n2+k1)) * A_avg(k2)...
                + ((k1-1)/(n2+k1)) * estimated_means(k1-1,k2) + (1/(n2+k1)) * B(k1,k2);
        elseif k1==1
            estimated_means(k1,k2) = (n2 / (n2+k1)) * A_avg(k2)...
                + ((k1-1)/(n2+k1)) * estimated_means(k1,k2) + (1/(n2+k1)) * B(k1,k2);       
        end
%         if k1==2, keyboard; end
    end
end

%plot the results
figure; hold on;
plot(nan, 'b', 'displayname', 'correct solution');
plot(nan, 'k--', 'displayname', 'my solution');
leg_tmp = legend('show');
set(leg_tmp,'Location','Best');

plot(correct_means, 'b', 'displayname', 'correct solution');
plot(estimated_means, 'k--', 'displayname', 'my solution');

ylabel('running averages');
xlabel('samples');

所附的情節概述了我嘗試的解決方案（黑色）以及我認為正確的答案（藍色）。 請注意，我只是在獲取所有試驗的所有樣本后繪制平均值，但我在收集數據時保存了運行平均值。 正如你所看到的 - 我的答案似乎有點偏離。

我的想法是A應該通過用於確定它是收集B時試驗總數的平均值的試驗的分數來更新。 同樣， B的當前樣本的權重只是 1 除以迭代中當前的試驗總數，並且B的先前樣本被遞歸調用並相應地加權。 這些權重加起來為 1，對我來說很有意義，所以我很難看出我在哪里搞砸了。

誰能看到我哪里搞砸了？

Answer 1

您應該考慮到代碼越長，它往往會積累越多的錯誤。

我提前道歉，對我來說，重寫業務邏輯比在代碼中找到錯誤更容易 - 所以雖然結果可能不是你想要的，但它確實提供了一個修復。 我希望你仍然會發現這很有用。

請看一個稍微簡化的版本，它似乎可以產生正確的結果：

function q60180320
rng(60180320); % For reproducibility

% define the sizes of the matrices
n1 = 5;
n2 = 10;
n3 = 12;

% define a matrix that will act as the history of data already collected
A = randi(10,[n2,n1]);
A_avg = mean(A,1); %averaged across n2 trials to get n1 values

% current acts as "incoming" data
B = randi(10,[n3,n1]); %n3 trials, n1 samples per trial
correct_means = cumsum([A;B],1)./(1:n2+n3).'; correct_means = correct_means(n2+1:end,:);

% preallocate matrices
estimated_means = zeros(n3+1,n1);
estimated_means(1,:) = A_avg; % trick to avoid an if-clause inside the loop

for k1 = 1:size(B,1) % Loop through trials
  %% Compute weights:
  totalRows = n2 + k1;
  W_old = (totalRows - 1)./totalRows;
  W_new = 1/totalRows;

  %% Incoming measurement (assuming an entire row of B becomes available at a time)
  newB = B(k1,:); 

  %% Compute running average
  estimated_means(k1+1,:) = W_old * estimated_means(k1,:) + W_new * newB;
end
estimated_means = estimated_means(2:end,:); % remove the first row;

% plot the results
figure; hold on;
plot(nan, 'k', 'displayname', 'correct solution');
plot(nan, 'w--', 'displayname', 'my solution');
leg_tmp = legend('show');
set(leg_tmp,'Location','EastOutside');

plot(correct_means, 'k', 'displayname', 'correct solution', 'LineWidth', 2);
plot(estimated_means, 'w--', 'displayname', 'my solution');

ylabel('running averages');
xlabel('samples');