我正在嘗試將一個(gè)大維矩陣減少到僅 2D，我正在使用 2D 數(shù)組的示例，它有效，但我需要為更高維的散點(diǎn)做同樣的事情。我有兩個(gè)類，每個(gè)類都有 50x20 維特征空間的矩陣。對(duì)于我的例子，我有這些二維數(shù)組：rectangles = np.array([[1,1.5,1.7,1.45,1.1,1.6,1.8],[1.8,1.55,1.45,1.6,1.65,1.7,1.75]])triangles = np.array([[0.1,0.5,0.25,0.4,0.3,0.6,0.35,0.15,0.4,0.5,0.48],[1.1,1.5,1.3,1.2,1.15,1.0,1.4,1.2,1.3,1.5,1.0]])之后我找到了三角形和矩形類的均值# Calculate the mean vectors per classmean_rectangles = np.mean(rectangles,axis=1).reshape(2,1)    mean_triangles = np.mean(triangles,axis=1).reshape(2,1)通過(guò)矩形和三角形類給出的值，我用它們來(lái)計(jì)算散點(diǎn)：scatter_triangles = np.dot((triangles-mean_triangles),(triangles-mean_triangles).T)scatter_circles = np.dot((circles-mean_circles),(circles-mean_circles).T)# Calculate the SW by adding the scatters within classes SW = scatter_triangles+scatter_circles+scatter_rectanglesprint(SW)plt.show()我想知道如何找到類內(nèi)的散點(diǎn)并以完全相同的方式繪制它們，但對(duì)于更大的數(shù)據(jù)，恰好是 50x20 矩陣？