如何对31×6×2×3的4D数组按指定维度计算行列重复中位数？

Hey there! Let's solve this problem efficiently using MATLAB. I'll walk you through both a targeted approach for each shock and a more streamlined one that handles both shocks at once.

Step 1: Understand Your Array Dimensions #

Your 4D array df = randn(31,6,2,3) maps to:

• 31 rows (i)
• 6 variables/columns (j)
• 2 shock types (k)
• 3 repetitions (n)

Step 2: Streamlined Solution (Handle Both Shocks at Once) #

Instead of processing each shock separately, you can use MATLAB's median function with a dimension argument to compute medians across the repetition dimension (4th dimension) for all shocks in one go:

% Generate your test array (add rng for reproducibility)
rng(42);
df = randn(31,6,2,3);

% Compute median across the 4th dimension (repetitions)
median_results = median(df, 4);

The result median_results will be a 31×6×2 array:

• median_results(:,:,1): Medians for the first shock (31 rows × 6 columns, each value is the median of 3 repetitions for that row/column)
• median_results(:,:,2): Medians for the second shock (same structure)

Step 3: Separate Processing (If You Prefer) #

If you want to handle each shock individually (like your initial approach), here's how to do it:

% Process first shock
eg1 = squeeze(df(:,:,1,:)); % 31×6×3 array
median_eg1 = median(eg1, 3); % 31×6 array (median across repetitions, 3rd dimension)

% Process second shock
eg2 = squeeze(df(:,:,2,:)); % 31×6×3 array
median_eg2 = median(eg2, 3); % 31×6 array

% Optional: Combine results into a single 3D array
median_results = cat(3, median_eg1, median_eg2);

How It Works #

• The median(A, dim) function computes the median values along the specified dimension dim. For your case, we target the repetition dimension (4th for the full array, 3rd for the squeezed single-shock arrays).
• squeeze removes singleton dimensions (here, the shock dimension which is fixed to 1 or 2), simplifying the array to 3D before computing medians.

内容的提问来源于stack exchange，提问作者Rollo99